# Legacy Weapon – Debunking Force of Will

Force of Will is a more complicated spell than people give it credit for. Everything a person knows about the card, down to theory and deck construction, might be incorrect.

Believe it or not, I’ve also been wrong before. Want an example? The following is a simple thought experiment that I failed, as did my two doctorate-seeking, Magic-playing roommates. Apparently, many accomplished physicists also miss the correct answer. Can you do better?

## The Men Who Err at Goats

You’re on a game show, say Let’s Make a Deal, and there are three doors. Behind one of the doors there is a car, and behind the other two there are goats. After you choose a door, the show host opens one of the other two doors, revealing a goat and giving you the option to switch.

Now, is it mathematically correct to switch doors or stay?

I’ll give you a moment.

The answer is to switch. Your initial guess, with one third chance of being correct, is less than that of the other two doors combined, which is two thirds. Now that the show host has shown you one of the other two doors is a goat, with zero chance, the door you would switch to retains a two thirds chance of being correct.

Still don’t believe me?

 Your Pick Door B Door C Result if Staying Switching Car Goat Goat Win Fail Goat Car Goat Fail Win Goat Goat Car Fail Win

As you can see, not switching is only correct one third of the time (when you correctly guess the car) and switching is correct two thirds of the time. This assumes, of course, that the show host always opens a door, instead of timing it in accordance with some scheme.

The reason that I opened with the above thought problem is because I found it somewhat humbling. As intelligent types, we can have difficulty admitting our own ignorance, especially when the problems are seemingly simple. The sheer volumes of people that get the goat puzzle wrong hints that we shouldn’t trust our instincts regarding probabilities. Part of the recent drive to avoid results-oriented thinking relates to the dangers of trusting experience over something more substantial. We delude ourselves into seeing patterns when there is only randomness. You can’t intuit how often your two land Belcher deck whiffs by goldfishing, you can’t feel out your average net worth of Lead the Stampede, and, regardless of how many games you’ve played, you can’t know how consistent your blue count for Force of Will is without doing a bit of math.

Ah, Force of Will. Back in the day, people debated whether it was worth the card disadvantage. Fortunately, general Magic theory realized that the gain in tempo (trading zero mana for however much mana the opponent has spent) is a valuable tool for countering the opponent’s win or forcing your own win through disruption.

These days, many err in that they don’t understand the function that the Force plays in a deck. Without this crucial knowledge, they won’t know when to spend it or when to hold it, or even the correct matchup to be shuffling the card away or even boarding it out.

You didn’t read that wrong. The best players in the world, established pros or Legacy ringers, board Force of Will out regularly, but certainly not willy-nilly. If you’re just learning a matchup, a good rule of thumb is that if both players consistently end the games with few to no cards in hand, the Force might have been better off as a card that at least trades one for one. If both players typically end with gas in hand, the tempo from Force is more likely to be critical.

Here are the better decks that I’ve brewed and championed since my breakout performance in Columbus with their respective blue counts. They’ve all had some success, and I’ve had a blast cashing events and shifting metagames. Due to their exposure, they’ve all received some commentary from the peanut gallery, and I’ve heard arguments ranging from weak matchups against fringe decks to citations of the lowest recommended blue count for Force of Will at twenty two: a number that, if it were true, would have shut me down before I even started.

## UG Survival (17)

I liked the design behind this deck a lot, and so did a large segment of the Magic-playing community, and the deck was highly netdecked until it was eventually banned. Zack Strait ran my stock list to the top eight of multiple opens well after versions with Necrotic Ooze became popular, and Patrick Chapin claimed that, based on math taken from the open series, it was the best build for the Survival mirror match.

The popularity of the deck didn’t quiet the critics, however, and a few members of the insatiable Magic community condemned the low blue count, sometimes complaining about the power level of the blue cards in the next sentence. Apparently, my apple pie wasn’t using enough cinnamon, and what spice I did have sucked. The argument that seventeen blue cards were too few bewildered me at the time. Wasn’t fifteen the standardized minimum? Sure seventeen was less than Solidarity’s thirty eight or whatever, but it wasn’t unheard of.

Some cut the creature count to fit in Brainstorms, with the notion that the Brainstorm could simply find them a creature if they needed one. The argument reminds me of a Family Guy joke where Peter Griffin is given the choice between winning a boat or the contents of a mystery box.

“A boat’s a boat, but a mystery box could be anything. It could even be a boat!”

Or, as I now imagine it,

“A crit’s a crit, but a Brainstorm could draw into anything. It could even draw a crit!”

People also ridiculed the lone Gaea’s Cradle, comparing it to a Forest with fading when they should’ve been thinking of it as a cross between a green Dark Ritual and a City of Traitors. To be fair, the accomplished Brian Kibler was of that number, so maybe they were right.

## Painter (14)

This deck falls below my own recommended minimum of fifteen blue sources. Still, people haven’t questioned it much, perhaps because of the group of fellow ringers that began the season running the deck alongside me, or perhaps because Painter’s Servant allows for Lion’s Eye Diamond and City of Traitors to be pitched to Force of Will. That doesn’t help resolve Goblin Welder or the Painter itself, however, which comes up often.

The main reason this deck gets away with so few blue sources is because it is, primarily, a Sensei’s Divining Top deck. If it needs a blue card, it can typically find one, and it can always just let Force of Will chill on top of the deck until one shows up.

Why is it in this deck? To “force” the combo through disruption. Once, in testing, I had a friend move to Force of Will an Aether Vial on turn one against a Merfolk opponent, until I stopped him. Since the deck doesn’t typically need Force until turn three, if ever, and because it sees more cards with top, and because Painter’s Servant can turn blanks into Force fodder, the deck gets away with an extremely low blue count.

## Bant (16)

Like UG Survival, Bant is another Noble Hierarch deck, and can actually get away with hardcasting Force of Will on occasion. Unlike UG Survival, Bant isn’t trying to protect an engine, but it still needs the card for several matchups. Hitting a turn one Aether Vial or Goblin Lackey is important, especially on the draw. Also, it can be a stopper for otherwise unbeatable cards like Replenish, Pernicious Deed, or Humility.

Note that, as the Pithing Needles come in against Aether Vial decks, the Force of Wills lose importance, and often aren’t what you want to be doing. I leave them in against goblins, since countering Goblin Ringleader or Goblin Warchief is still a fine play, but I pull them against Merfolk, where the plan is to kill their key creatures, needle the right card, and crush them with equipment.

While a Bant deck could be made to run more blue spells, since there are plenty of blue playables like Stifle, Daze, and Preordain that I’m not utilizing, this isn’t necessarily the path to success. Running do-nothing or underpowered blue cards to “make Force good” is a poor mentality to take towards deckbuilding. You want good threats, ones that win if they resolve, to protect with Force, and you want to be able to win if you don’t draw Force too. Gerry Thompson recently made the finals of a Legacy event with Dazes in his deck, though he didn’t remember Dazing anything all day long. If Gerry Thompson doesn’t like a card, I probably don’t want it in my deck, blue count be damned.

This means that there are times when I have to hold the Brainstorm rather than cast it, or hardcast the Force on turn three or four as my last card in hand, or even bounce a Vendilion Clique with Karakas to pitch. Typically, however, I have a three or four drop to pitch on turn zero. The math works out that the card is consistent, the blue count adequate.

## The Math

On the left is the number of blue cards in a deck, including four Forces. The number on the right is approximately the chance to open with Force of Will and a blue card to pitch to it.

15: 31.4% (My recommended minimum)
16: 32.6%
17: 33.6%
18: 34.5%
19: 35.3%
20: 35.9%
21: 36.5%
22: 37.0% (A seemingly arbitrary minimum I heard once)

Since we start with a forty percent chance of opening with at least one Force, subtraction allows us to figure out the percentage of hands where we open with a Force but no other blue card.

Note the diminishing returns involved with adding more blue cards for the Force of Will math. Moving from fifteen to sixteen changes the percentages by over a full percent, while going from twenty one to twenty two increases our odds by a mere half percent. Clearly, the blue count matters more the lower it goes.

These numbers are useful if the Force of Will is necessary on turn zero, but often it isn’t. Painter, for example, is a deck that typically only needs protection on turn three, and most decks even later.

By multiplying the above numbers by .4, since forty percent is the likelihood of getting a Force in the opening hand, we get the odds of having the card and something to fuel it. The missing percentage (100-the number) is equal to how often we have the Force in our opener without being able to cast it.

15: 78.5%
16: 81.5%
17: 84%
18: 86.25%
19: 88.25%
20: 89.75%
21: 91.25%
22: 92.5%

This means that, with fifteen blue cards in the deck, twenty one and a half percent of the time we draw Force in our opening grip we won’t be able to cast it. Is this a relevant number? Yes. Is this the final number? No.

Of that twenty one and a half percent, now we need to subtract mulliganed hands, since a hand with a dead card is essentially a mulligan anyway, and if the rest of the cards are weak it’s likely to get shipped. Of the kept hands, the Force will only be necessary at all times in game ones against Belcher, which constitutes a small portion of the metagame. Against other, slower decks we have to approximate the turn we need a Force, and how often we’ll have the extra blue card by that turn. Now, of all the remaining games in which we never have the blue card for Force of Will, we can subtract the games in which the card isn’t necessary to win. This would include games in which we get an active Umezawa’s Jitte against a creature deck, Wasteland an opponent into oblivion, or simply have threats the opponent can’t handle.

What we’re left with is the number of times we actually needed more blue cards in our deck, which is a very small number indeed, and one that shrinks even further with tight play and proper mulliganing.

If you’re interesting in getting the numbers yourself, but you’re no math wiz, the deep analysis option on Magic Workstation has a probability function that I found quite useful, and can let you do anything you need to do (though it rounds to whole numbers.)

Excel is the more typical math-y program, and you can accomplish most of your probability goals with hypergeometric distribution. Simply open up excel and punch in =HYPGEOMDIST(A,B,C,D)

Now, simply replace A with the number of times you draw the card in question, B is the starting hand (7 as an opener, 6 after a mulligan, etc,) C is the number of the card that you can draw (1-4 unless a basic land,) and D is the number of cards in your library before you start drawing.

This means that =HYPGEOMDIST(0,7,4,60) should punch out 60%, or rather the chances of not drawing a four of in our opening grip.

Moving on.

## Forceful Play

Our play changes the way Force of Will functions in the different matchups.

Something that surprises a lot of newer Legacy players is that you can actually hardcast Force of Will if the situation calls for it. In some matchups the play might even be ideal. The first time I encountered the need was when I first started getting into Vintage back in 2004. I netdecked a storm combo deck that supported Force off of a meager fourteen blue cards, but the designer of the deck noted that Force could often be cast off of a Dark Ritual. While such a play is not ideal, I did learn that decks with lots of mana acceleration, such as City of Traitors, Moxen, Noble Hierarch, or Green Sun’s Zenith, are able to get away with running Force with fewer blue cards because they’ll be able to hardcast Force more often when it matters.

When a matchup consistently reaches the long game, two for oneing yourself to counter an opponent’s spell only makes sense if the card is otherwise dead, like a Sower of Temptation against a control deck. If you pitch relevant cards, like Brainstorm and Jace, you’ll probably regret it when the game reaches the point where your more powerful cards might’ve pulled you ahead. Now you’re stuck with a hand full of Sower of Temptations, and, in all that time, you could’ve found a spot to hardcast the Force anyway, netting an additional card towards the battle of attrition.

This is clearly different than where your opponent leads with a turn one Aether Vial, or goes for the Show and Tell. Here, simply staying alive takes priority, the games are more likely to be brutally short, and any old blue card will do. In matchups where Force of Will is most important, say against Belcher or other types of combo, casting Brainstorm when it’s your only blue card to pitch is a suicidal move.

Speaking of Brainstorm, it’s clear that a deck that stockpiles cards, like Landstill or High Tide, is going to have a much easier time paying for Force than a deck that dumps a hand full of cantrips, such as Tempo Thresh. Counterspell will chill in hand longer than Preordain, and you should plan your turns accordingly.

Finally, but not unimportantly, your sideboard should have a good mixture of blue spells for the matchups where Force of Wills are important. If you’re pulling a blue card against combo, say Sower of Temptation, you need a good blue replacement, such as Spell Pierce or Meddling Mage.

This is an advantage to running a deck with a higher blue count. A deck with twenty-two blue spells can lose seven in sideboarding and still have a consistent Force of Will, while a deck with only fifteen needs to manage its blue count more carefully. If sideboarding nulls the efficiency of Force of Will, perhaps the Forces were the cards you should’ve been pulling in the first place!

And at that, I’ll leave you. The Wescoe has yet another theory on how the game show host can scum people into choosing goats, and from his excitement it should be a good one.

Reach me in the comments or at [email protected]

-Caleb Durward

### 99 thoughts on “Legacy Weapon – Debunking Force of Will”

1. why are you giving us middle school math equations in your articles? :/ were not that dumb…

2. And comments about misapplication of probability and poor consideration of all pertinent factors…NOW!

3. I’m somewhat inclined to agree with the troll. The article seemed patronizing to me. I appreciate the sentiment though.

4. Actually, you’re only correct to switch if the host knows what’s behind the doors. If he just opened one of the doors without knowing and you saw a goat, it’d be 50/50. The way you state it, it’s ambiguous. Nice Monty Hall problem. Now onto the actual article, after this distraction.

5. ill just say it is not pure math. your odds are 50/50 after a goat is shown. Period. Swap or no

6. That is a terrible problem. Go do some research, 10,000 door openings should do it (5000 for not switching etc) – what do you get? Will switching give you a 2/3 chance? Math problems without reality are silly; in reality the opened door is no longer part of the equation.

7. Hey wrist band guy, you need a table entry for when the other goat is chosen. The goats are not the same goat and thus there should be two entries for when you choose the car.

8. You didn’t state the Monty Hall problem correctly. The whole point is that the host knows beforehand which prize is behind which door and intentionally chooses one with a goat. Otherwise, it’s implied that they just pick a random door and a goat happens to be behind it.

The whole problem always comes back to a stupid confusion about how the problem is set up.

9. everyone that is saying Adam’s math is incorrect is actually the incorrect ones. Adam is dead on. This is a math problem that has confused QUITE A FEW mathematicians. It’s not something that is easy to wrap your head around, but that is the correct math.

Thank you for the article Adam.

10. ERR Caleb! i was just reading both of them x.x

11. … How have none of you idiots ever seen the movie 24? Go… Watch… Learn. You are bad at math, and Caleb is good at math (or retaining information from movies).

12. @mikkyld
@ Sum Guy

The article’s author is actually quite right, and it’s not just ’empty math’ without real-world application. You’re making a reasoning error in assigning equal probability across all unknowns at all phases of a problem’s completion.

Seriously, read up on the Monty Hall Problem.

The problem doesn’t require the host to know which door has what prize, only that a situation has come up wherein the door that was opened has a goat, and one of the other possible ones doesn’t. The solution of this puzzle isn’t something like “well, he’s trying to trick you and so choose the other one,” which is what I feel that people are proposing. It’s actual hard statistical practice.

14. I like that the first reactions to this article are of willful ignorance and complaining about genuinely useful information. Peeve of mine, but hey, Magic players and their hive mind.

Good job, Caleb. This article contains a lot of things Magic players would benefit from through a furthered understanding.

15. Think of it this way: The first time you choose a door, you are given 2:1 odds. Then the host gives you a second chance with 1:1 odds. Take the better odds. Your second pick has a higher mathematical expectation.

16. And the odds the for the second guess are actually better than 1:1. The real question posed in your second chance is “were you right the first time?” The answer is “no” two times out of three, meaning the host is offering you 1:2 odds to change away from your original choice. He doubled your odds for free.

17. Actually, it’s even more likely. In Monty Hall, switching doors gives you a 2/3ds probability of winning the car.

Think of it this way. There are exactly two possibilities that could happen if you go into the problem knowing you’ll switch, and it all depends on your initial door choice.

You have a 1/3 probability of initially choosing the car door. The other two doors have goats, so it doesn’t matter which other door the host reveals – the other will also have a goat. 1/3 of the time, you lose no matter what happens.

You have a 2/3 probability of initially choosing a door with a goat. The other two doors have one goat and one car. The host always chooses to reveal the goat, meaning the other door will always have the car. Hence, if you initially choose a goat door, you will always win the car.

So your odds are even better. Switching gives you a 2/3 probability of winning the car.

18. @Eric – It’s still dependent on the host knowing what’s behind the doors. The proof involves a statement that the host cannot open the door with the car behind it. From the Wikipedia article on the Monty Hall problem: “If on the other hand the car is not behind Door 1, it is equally likely to be behind Door 2 or Door 3, and the host is forced to open the only door of this pair which has a goat behind it.”

Good article; I wish it sounded less arrogant.

19. The math was not incorrect, the question was incorrectly phrased. In the actual question, the host knows where the car is, and intentionally does not open ‘that’ door. This is actually a VITAL part of the problem.

20. Scientific American did an article some years ago that refutes the question as phrased. If the host doesn’t know what they’re choosing, switching is incorrect.

21. @Brady The problem was stated correctly – as mentioned, it doesn’t matter whether or not the host knows what was there, simply that the new door had a goat behind it. The problem is not a product of “tricky wording,” it’s simply a neat bit of probability made famous by it’s non-intuitive nature.

For those who think the math is wrong, see Caleb’s useful diagram that you apparently ignored. It’s probably the easiest way of seeing the solution to the problem.

As for the article, I found it quite interesting and well-written – thanks!

22. “The problem was stated correctly – as mentioned, it doesn’t matter whether or not the host knows what was there, simply that the new door had a goat behind it.”

I’m not sure how to argue against this, it’s simply wrong. It does matter whether or not the host knows what is behind the door he is picking.

Situation 1: Host knows.

1/3 of the time you will pick the car originally and the host can open either door remaining because they both have goats. Switching is bad in this case.

2/3 of the time you will pick a goat originally. The two doors remaining are a goat and a car. The host knows which is which, and opens the one with a goat. Switching is good in this case. This is the classic Monty Hall problem.

Situation 2: Host doesn’t know.

1/3 of the time you will pick the car originally and the host can open either door remaining because they both contain goats. Switching is bad in this case.

1/3 of the time you will pick a goat originally, and the host will randomly open a goat behind the other door. Switching is good in this case.

1/3 of the time you will pick a goat originally, and the host will randomly open a car behind the other door. Switching is meaningless in this case.

If the host has the knowledge of what is behind the doors, switching is good 2/3 of the time and bad 1/3 of the time.

If the host is just as clueless as you are, then switching is good 1/3 of the time, bad 1/3 of the time, and meaningless 1/3 of the time. We are told that a goat is revealed, so we can ignore the 1/3 where a car is revealed. This leaves us a scenario where 50% of the time it is a good idea to switch and 50% of the time it is a bad idea.

In short, the answer to the Monty Hall question is dependent on what information you believe the host has access to.

23. @lol Troll: Aside from reaching the coveted middle school demographic, many adults lack the basics, some of which are magic players. Typically, when I see a “Math” section of an article, I realize that it’s necessary for some but skip it myself. I suggest doing the same next time.

@The Goat Problem: My friend Wilson complains that, “That thought experiment presupposes that I value a car over a goat.” He’s right. Yet, you still knew what I meant, and had no difficulty in attempting the problem, or understanding how it fits within the context of the article.

I didn’t explain that the host knew he was revealing a goat because, like we can assume we’re valuing the car higher, we can also assume it’s illogical to open a door with the car behind it and ask the contestant if he wants to switch. It’s further implied at the end, where I state that the math changes if the host is acting on a scheme (which is impossible without knowing which door the car is behind.)

24. While I’m not sure what’s going on, I’m 100% sure statistics doesn’t work like that. If you pick a door you have a 1/3 chance of getting it right. One door is eliminated and now each door has a 50% chance of being the correct door. It doesn’t make a difference at all if you switch doors or not because they are equally likely to be the right door. That is just truth and this made me mad.

25. The real problem with the thought experiment is that it presupposes that I value a car over a goat.

26. If you doubt it just have someone do the experiment for you three napkins, one with a post-it hidden underneath and have someone play the game-host for you. Do like 30-40 of each and you’ll prolly see (though, more for accuracy).

If not it’s as simple as described. You chose right only 1/3rd of the time, but when the host reveals that one of the two other doors is a dud then it must be your door or the other that is the right one, but you only had 1/3rd chance of getting yours right which results in the other being right 2/3rds of the time… Its how math works, no reason to get mad

27. Luis Scott-Vargas

and the inevitable reaction to math in articles…

Good article, good example, and I like the Monty Hall problem for exactly this reason; it’s unintuitive until you break it down, as this thread demonstrates.

28. I think Force of will is the most overrated and misplayed card in magic. Its mostly only good at stopping combo. I lol so hard when some one exiles their Jace to counter my turn one vial (true story) Thats a 200\$ for 1\$ ! (on Modo) And of course I won that game any how.

29. I feel goated (sorry, just read LSV’s card evaluations for White) to respond to this, but I don’t really care about the first part, to be honest. I think that the best part of the article is the many caveats to the “blue card count issue”. Also, for what it’s worth, you made me reconsider my personal take on the minimum number, so thanks for that Caleb.
However, I do have two small issues with the article:
1) “A crit’s a crit, but a Brainstorm could draw into anything. It could even draw a crit!” <- BAD! according to you.
"A blue card’s a blue card, but a *Sensei's Divining Top* could draw into anything. It could even draw a blue card!" <- Fine, according to you.
It seems you are a being a bit inconsistent here..
I would argue that either is probably ok (Top for blue cards, Brainstorm for creatures), IF not taken too far. Too far can be a lot of things, of course, but the point is that if you can reasonably expect a Brainstorm/Top/whatever to be *either* a creature, a blue card, a land etc. as needed, then it is, unsurprisingly, pretty good.
2) "This deck falls below my own recommended minimum of fifteen blue sources"
You critize people for recommending an "arbitrary" minimum 17 blue cards, but go on to suggest 15 yourself. How is that not just as arbitrary? Sure, it is not completely unfeasible to have that blue card when you need it, as you yourself demonstrated, but why wouldn't 14, or 16 or indeed 17 be the correct minimum? I don't think "the correct blue card count " exists at all, except perhaps for a specific deck in a specific metagame. All you can say is that Force will be more or less consistent based on that number, but we all knew that anyways.

30. The monty hall description was fine. Perhaps only lacking in the fact that the host needs to know where the goats are but that is a minor variant.

The math on your force of will section is a bit off though. First of all it is worded quite weird, what exactly do you mean with the first set of percentages? The chance to have FoW AND an other blue card or just the chance to have an other blue card??

The only thing that is relevant in this entire discussion for numbers for FoW (on turn 0) is the chance to draw a blue card in 6 cards… You seemed to have used 7 but you need to account for the conditional probability, ie. you already have FoW in hand thus only 6 other cards.
Note by the way that the chance of having FoW at all is completely irrelevant… YOu are only looking at the chance of having a FoW miss, if you don’t have it in hand it can’t miss either…

Because of the poor math it is unfortunately a poor article.

LSV is dead on in his anticipation of whining about math but it’s just silly if the majority of the article is based on it and it is simply wrong.

31. Just watch the movie 21…. about ripping off casinos it has the same question

32. Since he states that you are on the gameshow “Let’s Make a Deal”, and on that gameshow the host knew what was behind the doors, it’s reasonable to assume in this case.

33. Just another probability question for all of the detractors. How likely is it that a game show goes on with a host who has a 1/3 chance of revealing the prize by accident? Just a rough stab at it, but I’m going to go with roughly 0%.

34. RainbowPenguin said:
However, I do have two small issues with the article:
1) “A crit’s a crit, but a Brainstorm could draw into anything. It could even draw a crit!” <- BAD! according to you.
"A blue card’s a blue card, but a *Sensei's Divining Top* could draw into anything. It could even draw a blue card!" <- Fine, according to you.
It seems you are a being a bit inconsistent here..

The difference is, Sensei's Diving Top is not a 1 shot trade: 1 card for 1 attempt to locate a creature/blue card like Brainstorm. Over a sequence of turns, SDT will let you dig deeper, and draw that card you need off the top at will if necessary. I know you probably know that, but I just think you can't compare the two as equal as the perform two different functions and provide a different resource for digging for cards.

35. Wouldn’t it have been easier to state the problem, a little more clearly maybe, say the result and link to proof, that way you wouldn’t have people arguing with you about your inexact proof via chart?

PS Haters, look up Monty Hall he is basically right, it is unintuitive, but so are lots of things.

36. As a few people said, the fact that the host knows where the prize is and, whatever your initial choice, will always open a door and reveal a goat is a crucial part of the problem.

If the host just randomly opens a door, and it happens to reveal a goat, then switch or stay still have 1/3 chance of success each.

37. Ah, Monty Hall problem
“Brady April 27, 2011 @ 12:50 am” explanation was the best one Imo. It’s actually pretty simple problem once you realize the ‘trick’, but yeah, it’s interesting how many people don’t get it initially (including mathematicians)

38. “The Math

On the left is the number of blue cards in a deck, including four Forces. The number on the right is approximately the chance to open with Force of Will and a blue card to pitch to it.

15: 31.4% (My recommended minimum)
16: 32.6%
17: 33.6%
18: 34.5%
19: 35.3%
20: 35.9%
21: 36.5%
22: 37.0% (A seemingly arbitrary minimum I heard once)

Since we start with a forty percent chance of opening with at least one Force, subtraction allows us to figure out the percentage of hands where we open with a Force but no other blue card.

Note the diminishing returns involved with adding more blue cards for the Force of Will math. Moving from fifteen to sixteen changes the percentages by over a full percent, while going from twenty one to twenty two increases our odds by a mere half percent. Clearly, the blue count matters more the lower it goes.

These numbers are useful if the Force of Will is necessary on turn zero, but often it isn’t. Painter, for example, is a deck that typically only needs protection on turn three, and most decks even later.

By multiplying the above numbers by .4, since forty percent is the likelihood of getting a Force in the opening hand, we get the odds of having the card and something to fuel it. The missing percentage (100-the number) is equal to how often we have the Force in our opener without being able to cast it.

15: 78.5%
16: 81.5%
17: 84%
18: 86.25%
19: 88.25%
20: 89.75%
21: 91.25%
22: 92.5%”

What exactly did you do to get these last set of numbers? According to you, you multiplied the first set each with 0.4. As far as I know, and according to my calculator, multipliying a number with anything <1 will leave you with a smaller number. So how do you turn 31.4% into 78.5%?

What am I missing here?

39. I’m a dork so here are the actual numbers. Basically Pr(F>=1, B>=1) + Pr(F>=2, B=0) where F is the number of FoW and B is the number of blue cards that aren’t FoW in your opening hand.

First column: number of blue cards INCLUDING FoW. Second column: probability(%) of being able to cast FoW with opening hand.

On the play:
4 6.32194161616783
5 9.92494386169984
6 13.2004004485471
7 16.1725740180937
8 18.8643538546642
9 21.2973087069491
10 23.4917385737159
11 25.4667254538060
12 27.2401830604175
13 28.8289054996736
14 30.2486149134770
15 31.5140080866495
16 32.6388020183585
17 33.6357784578277
18 34.5168274043355
19 35.2929895714971
20 35.9744978158341
21 36.5708175296289
22 37.0906859980655
23 37.5421507206551
24 37.9326066969489
25 38.2688326765352
26 38.5570263733234
27 38.8028386441134
28 39.0114066314503
29 39.1873858707659
30 39.3349813618047
31 39.4579776043371
32 39.5597675981570
33 39.6433808073662
34 39.7115100889440

On the draw:
4 8.22541450059983
5 12.7574927968665
6 16.7951625517223
7 20.3842023338163
8 23.5669357254846
9 26.3824306488834
10 28.8666908754118
11 31.0528398747568
12 32.9712971598963
13 34.6499472843934
14 36.1143016483163
15 37.3876532691189
16 38.4912246738145
17 39.4443090687788
18 40.2644049435156
19 40.9673442647185
20 41.5674144169649
21 42.0774740463744
22 42.5090629635670
23 42.8725062622555
24 43.1770128098054
25 43.4307682660969
26 43.6410227870242
27 43.8141735689643
28 43.9558423905516
29 44.0709483080913
30 44.1637756609460
31 44.2380375432297
32 44.2969348981443
33 44.3432113912915
34 44.3792042192949

And here are your chances of NOT being able to cast FoW IF you have one in your opening hand Pr(F=1,B=0)/Pr(F=1): (A quick aside, your chances of drawing exactly 1 card when there are 4 in your deck is 33.6% and 36.3% when on the play and draw respectively, not 40%)

On the play:
4 100
5 89.2857142857143
6 79.5454545454545
7 70.7070707070707
8 62.7024966647608
9 55.4675932034423
10 48.9419940030373
11 43.0689547226728
12 37.7952051647945
13 33.0708045191952
14 28.8489996869575
15 25.0860866843109
16 21.7412751264028
17 18.7765557909842
18 16.1565712620097
19 13.8484896531511
20 11.8218814112266
21 10.0485991995426
22 8.50266086115143
23 7.16013546202226
24 5.99903241412675
25 4.99919367843896
26 4.14218904784943
27 3.41121450999364
28 2.79099368999480
29 2.26768237312078
30 1.82877610735546
31 1.46302088588437
32 1.16032690949450
33 0.911685428888537
34 0.709088666913306

On the draw:
4 100
5 87.5000000000000
6 76.3636363636364
7 66.4646464646465
8 57.6862969315800
9 49.9208338830980
10 43.0689547226728
11 37.0393010614986
12 31.7479723384274
13 27.1180597057401
14 23.0791997495660
15 19.5671476137625
16 16.5233690960661
17 13.8946512853283
18 11.6327313086470
19 9.69394275720580
20 8.03887935963408
21 6.63207547169811
22 5.44170295113691
23 4.43928398645380
24 3.59941944847605
25 2.89953233349460
26 2.31962586679568
27 1.84205583539657
28 1.45131671879730
29 1.13384118656039
30 0.877812531530623
31 0.672989607506811
32 0.510543840177581
33 0.382907880133185
34 0.283635466765323

40. hamiltonianurst

The way I’ve heard the Monty Hall problem ‘explained’, for those who don’t believe it’s true even after seeing a probability table (this is ignoring the previously argued possibility that the host could just reveal the car, and that such a problem would be interesting):

Pick a card, any card, and try to pick the Ace of Spades from this 52-card deck. After you choose face-down, I will turn over 50 cards you did not choose, leaving only the card you have currently selected and one extra one you did not. You may choose to switch. I think it’s pretty trivial even for those who don’t understand math to see why you’d want to switch in this case, unless they think they’re really good at picking that Ace of Spades every time.

41. I think this whole Monty Door debate can be made easier for people that don’t understand it by the following example:

“It may be easier to appreciate the solution by considering the same problem with 1,000,000 doors instead of just three (vos Savant 1990). In this case there are 999,999 doors with goats behind them and one door with a prize. The player picks a door. The game host then opens 999,998 of the other doors revealing 999,998 goats—imagine the host starting with the first door and going down a line of 1,000,000 doors, opening each one, skipping over only the player’s door and one other door.

The host then offers the player the chance to switch to the only other unopened door. On average, in 999,999 out of 1,000,000 times the other door will contain the prize, as 999,999 out of 1,000,000 times the player first picked a door with a goat. A rational player should switch.
Intuitively speaking, the player should ask how likely is it, that given a million doors, he or she managed to pick the right one.”\

From a mathematical standpoint it does not matter how many doors there are, 3 or 1 million. The fact remains that probability dictates you have a greater chance of success if you change your pick.

There is a 2/3 chance you did not pick the right door (this is the key to understanding this problem). So by selecting from the pool that has a 2/3 chance of being correct you increase your chances of selecting correctly from 1/3 to 2/3 given that all the doors that are incorrect from the pool you did not choose from are revealed.

42. chromatone: force of will pitching jace to counter vial seems perfectly fine to me. The kind of deck that plays force and jace also plays several hard counterspells in it, all of which become terrible cards if you resolve a vial. Also, as good as Jace is in general, it is actually not that great against the types of decks that play vial.

I agree with Caleb that Force should often be sided out depending on the matchup, but another factor that is not often considered is starting out with Force in the board, depending on the expected metagame. I played Green Sun’s Zenith CounterTop Bant at an SCG Open recently with 4 Force in the board so that I could play more creature hate maindeck. In 9 rounds, I played 4 matches where I wanted Force and 5 where I didn’t so it ended up being the right call.

43. Good article, I always enjoy watching people argue about math when going off of nothing but their gut. (I’m guilty of it too, but it’s more fun when I do it to really smart Math majors). There are 1,000 doors. Behind one of them is a car, behind each other is nothing (or a goat, I’m going to go with nothing though). You pick one of the doors. The game show host then reveals 998 other doors, each with nothing behind them. Is it correct to switch? Obviously, because otherwise you’re assuming you got it right in your 1/1000 guess, but the reason the other door stayed is because it HAS to be the right one if your initial guess is wrong, so it has a 999/1000 probability of being right. Yes, in a vacuum, the probability of each door would be 50/50, but we’re not in a vacuum. We have a round of information and a structure built into the game that makes it different.

44. Wilson is perhaps right in regards to the goat. Percieved value, pshh. 😛

Just remember there are no goats allowed in the city limits so choose wisely based on where you live.

On topic, I remember back in the dark ages (circa 99-2000) when Magic theory was much less developed and people would look at you like you road in to town on the crazy bus for hard casting Force of Will. Then it started that sub-game in their head that if you hardcast it you MUST have another in hand so they would play around it all game long.

45. I think the reason the Monty Hall problem confuses so many people is that the game structure of “the host will always open a goat door if he can, and not the car door” is not stated well, or at all.

If you take the 1,000,000 doors example, and the host reveals 999,998 doors at random, the probability of your initial choice being correct is now 0.5 (although realistically in most situations he’d have revealed the car well before then). If the host has revealed 999,998 goat doors, and you know he will always reveal goats before cars, that’s when switching becomes overwhlemingly the right decision.

46. If the host always opens a door with a goat behind it, your chances are 50/50 regardles. At that point, as sum guy said, the opened door is no longer part of the equation. As said above, this is only if a door with a goat is ALWAYS the one opened.

47. sorry but it doesn’t matter if the host knows or doesn’t know. It is true that your first choice gave you 1/3 odds to win and that your second choice gives you 1 in 2 odds to win. The thing that is being missed is that keeping your original choice IS a choice – it is not necessary to change to gain the advantage of the new odds.

Only when looked at a priori are your odds changed as caleb stated – when viewing it from the position of the new choice, your odds are 50/50. Period.

There are psychological implications of the new offer in that most assuredly the host does know which door has which behind it and thus obviously chose a goat to reveal. That fact does not mathematically change your odds however, unless you can get a read on the host.

This reminds me of my algebra teacher showing me a trick of equations where he could set a=1 and b=2 in separate equations and then do everything as he is supposed to do in solving the equations to reach the conclusion that a=b.

It is not math to blame but faulty set up of a problem.

48. “If the host always opens a door with a goat behind it, your chances are 50/50 regardles. At that point, as sum guy said, the opened door is no longer part of the equation.”

The opened door is not part of the equation ONLY if you choose to ignore that information, revoke your initial pick, and randomly select between the two remaining doors. That would make your pick 50/50. The actions “staying” and “switching” are based off of the probability of your initial pick, which has a 1/3 chance of being right. Since you KNOW that the host can and will always reveal a door which has a goat and is not your choice, no matter what door you pick, that means your initial pick is still 1 in 3. With two doors left 1-1/3=2/3=the EV of the “switching” action.

49. “sorry but it doesn’t matter if the host knows or doesn’t know. It is true that your first choice gave you 1/3 odds to win and that your second choice gives you 1 in 2 odds to win.”

Again, 1 in 2 odds is correct if the second choice is independent of the first, i.e. if after the first round, you flip a coin to determine which of the remaining two doors to pick. However, in the second choice, the two doors now have information attached based on the first choice. One door is “the door that the host did not reveal even with the choice to”, and the other is “the door the host could not reveal”. This information gives “the door the host did not reveal even with the choice to” 2 in 3 odds of winning.

50. Build your deck assuming Force of Will can pitch any card to cast it. Then you go through your matchups and figure out what you need to counter and when to improve your matchup, then you scale your blue cards to reasonably achive that.
Example, if you can’t beat a turn 1 Lackey, you need to be able to Force on turn 1. If you can’t beat a turn 4 Moat, you need to be able to Force by turn 4.

51. As far as the math problem, the assumption that the host knows what is behind the doors is critical to the problem. In Deal or No Deal when you are down to two cases left, 1 dollar and 1 million dollars, the odds that your case or the case up on the state holds the million is 50%. This is different because no one has knowledge of which case has what.

52. I find it hilarious that even in the face of overwhelming evidence that people can’t understand that one instance in a collection of data is only independent if one chooses to ignore all the other data.

It would be like trying to figure out the chances that your opponent has a mana leak in hand by counting his graveyard and then ignoring the cards in his graveyard….

I flip a coin and get heads 6 times in a row. What is the chance that my next flip will yield a result of 7 heads in a row. If you say 50% you have no place in this conversation.

53. ***Counting his library then ignoring the cards in his graveyard….

54. @Quick Question: We don’t need the numbers for exactly one Force, but at least one Force. I appreciate the dorkiness.

@Fyren: My apologies! I had corrected this mistake, but I guess it didn’t save. You need to divide by .4, not multiply.

@RainbowPenguin: The time I cited a minimum as seemingly arbitrary was 22, not 17. 17 makes more sense, because it’s closer to the threshold of inconsistency. It goes back to the diminishing returns that I mentioned. The difference between 15 and 14 blue cards is relevant, consistency wise, but 22 and 21 not so much, making a minimum of 20+ seem arbitrary. I agree that there is no hard minimum, as you can see I break my own recommendation in the section you’re quoting!

55. @ Don
It is 50%, there is no debate.
Each coin flip is independent of each other, they are not related. When you are discussing the odds of 7 coins all flipping the same thing, you are talking about multiple SINGULAR events happening in sucession.
This is commonly called the Gambler’s Fallacy.

56. These are independent events however. Once one door is removed, there is now .50 chance of the car (or in your case head’s) coming up. The first 6 flips, has no bearing on the 7th, and the initial choice has no bearing on the outcome of switching or not. What does happen, is your first choice has now gone from 1/3 to 1/2 for the second. So technically you do have a higher chance of picking the right door, but there is no incentive to switch. You still have a .5 probability of picking the right door, no matter which door you stay or switch to….The probability table published here is a logical fallacy…. and flipping 6 heads in a row doesn’t mean the 7th will have a higher likelihood of being tails…

57. Upon further reading on conditional probability, as Brady said earlier, it depends on if the host has actually given you information… Coin flipping however, it’s still a “random” chance of .5 for each flip, and not dependent on previous information (which is a different scenario than is addressed in the Monty Hall problem)….. http://en.wikipedia.org/wiki/Monty_Hall_problem.

58. @ Deathcloud
The car is not the same as the coin.
In the car example, you have a 1 in 3 chance of picking the car, there is a 2 in 3 chance of not picking the car. The host reveals one of the goats behind the door you did not pick, since he knows what doors have what, the odds do not change. The remaining door that you did not pick and did not have the goat has a 2/3 chance of having a car, while the one you did pick only has a 1/3.

An easy way to think about this is to expand it dramaticly. There are now one million doors, you pick one, the host opens 999,998 doors that he knows are not the car, would you still argue that it is 50/50?

59. (I’m a math major and am taking probability but you dont need to be a math major or have taken probability to get this)

If you present this question in exactly the same way as it is written in the article to a mathematician or anyone else that is not retarded they will tell you you have an equal chance of getting a car if you switch or not.

It’s based on conditional probability. knowing that one of the door is a goat, then there is a 50% chance the car is in the door you have chosen and a 50% chance it is in the one you can switch to.

If you maybe explain in more detail how the host goes about revealing a door..etc.etc. then maybe you will have a different answer.

So yea…poorly wording a problem in an attempt to seem clever is actually retarded

60. There are now 100 doors and 99 goats and one car. You pick a door and have a 1% chance of getting a car. Because the host knows the locations of everything, he opens the door to 98 goats. He does will NOT open your door and will not open the door to the car. So now your door still has a 1% chance of having a goat behind it, since you’ve done nothing to your door. However, the host has left one door among the 98 goat he revealed and it has a 99% chance of being the car. So it’s pretty clear to switch here.

Basically, because the host knows where everything is, he will tell you 99/100 where the car is hidden by revealing all other doors. It’s confusing when it is not made clear that the host will only reveal goat doors (and cannot accidentally reveal the car) and will always reveal a door you have not picked. That’s the key.

61. the easiest way to understand the 3-doors quizmaster question, is to imagine there are 1000 doors to choose from, with 999 goats and 1 car. Then its easy to realise that your first choice is worse then any other choice after that.

62. Man, I’ve been waiting for someone to mention HYPGEOMDIST in Excel in one of these articles for a loooooooooong time.

63. scenario 1: host knows. let’s say you always pick door 1 for simplicity sake.

CGG – car is yours if you choose not to switch
GCG – car is yours if you do switch
GGC – car is yours if you do switch

switching is the right move, IF THE HOST KNOWS.

scenario 2: host doesn’t know, you pick door 1, he reveals door 3 always (for simplicity sake)

CGG – car is yours if you choose not to switch
GCG – car is yours if you do witch
GGC – host reveals car, and therefore non applicable

in this scenario, it’s irrelevant whether you switch or not.

also, people like Don who think the 7th coin flip is not 50/50 is pure lol to me. reading the ignorance on these things is half my pleasure.

64. @ CalebD

The calculations for having FoW miss are wrong and excessive you seem to have not understoond conditional chances correctly.

It is totally irrelevant what the chances of having FoW in the opener are (the 40%) , you are only interested in knowing the chance it misses. Therefore the ONLY thing you need to calculate is the chance of having a blue card in SIX cards. You are assuming afterall you already have FoW in hand…
Thus the proper method is:

15 cards chance for FoW to miss =>

45/59 * 44/58 * 43/57 * 42/56 * 41/55 * 40/54 = 18%

You calculate the chance for having 0 other blue cards then fow. You have 14 blue cards left and 59 cards left in deck after having 1 fow in hand thus its 45/59 * 44/58 etc.
Conclusion the FoW has fuel in 100-18 = 82% of the cases with 15 blue cards including FOW.

65. I find it very hard to believe that any real mathematicians or statisticians were confused by the Monty Hall problem or chose the wrong answer, even though I know many sources claim so. It is a very simple problem and any logical approach towards solving it will quickly come to the right conclusion. It is only the idiotic and uneducated masses making deductions faster than they can think who will get it wrong – and we have plenty of examples in the comments of this very article.

66. No chance at all for the car! While you were all wasting hours discussing this here, the second goat stole the car and ran away with it!

67. I think I can explain the discrepancy everyone is having.

Taking into account the host knowing what is behind each door and revealing only doors with goats. The only reason The Monty Hall problem works is that it ends in two choices.

Meaning the host has to reveal every door with a goat unless the door you chose has the car. If the problem ended in 3 choices the probability would not change because then the host would have a choice of what doors to reveal. Where as the only way the host has a choice is if you already picked the car.

The bottom line is the host has to reveal goats only and only if you picked the car can the host hide any goats. so the logical decision is to switch simply because 2/3 of the time it has to be the car, because 2/3 of the time the host can only hide the car.

68. Threads like this are exactly why I hate discussing math with my non-mathematical friends. Here’s a fun one to try if you have a few hours to spare arguing with someone: See if you can get an average person to understand that .999… = 1. Not as complex as Monty Hall but people have a serious tendency to believe that there can’t be 2 decimal representations of the same number.

69. 1/3 doors initially to choose a car
1/2 doors to choose a car after you see a goat
it takes a a huge amount of dumb to think one can accurately gauge probabilites of factors that have already been decided. I mean, you can do it, but that’s like asking the probability of the cubs winning last years world series. Wording things to make people look stupid makes you a douchebag.

70. the monty hall problem and .9 repeating = 1 in the same thread? we’re going infinite on comments

71. The problem is simple if you see it this way, i.e. the right way.

You are on the ‘switch’ plan from the beginning. You know the host will remove a wrong door. So if you choose to switch, it is like you ‘chose’ the 2 doors that you did not pick initially, as you know that you will always get to see those 2 doors (out of 3). So 2/3 odds to get the car if you choose to switch.

If you choose not to switch, whatever the host does is irrelevant and your odds are the same as offered initially, 1/3.

72. The chart is extremely flawed. There are only two options to consider, not three. You already know that you didn’t pick one of the goats so it is inaccurate to consider the statistics for both initially wrong choices. After the first goat is revealed there are two options: One is that your original choice is the car and staying is a win, the other is that your initial choice was one of the two goats (which one is completely irrelevant) and that staying is a loss. The host has eliminated the third option on the chart. The fact that we can’t define whether Wrong Choice A or Wrong Choice B was eliminated doesn’t mean we get to count both against the remaining odds.

The whole thing is just a mathmatical illusion, a sly manipulation of presentation of data. If anyone else ever gets a chance to test the theory in psych lab I would highly recommend it. Plus it’s funny to watch math majors get angry when you start posting your data.

73. @CalebD

Do you have a reason for why 15 U cards is the optimal number for FoW? I’ve most often heard 16 or 17, citing that the ROI for adding another U card falls to 1% would suggest it being worth the addition. Having written that out, it too seems arbitrary (also sensible), but as a fan of cutting corners whenever and wherever possible, 15 sounds intriguing.

74. CONSTANTS
Doors labeled #1, #2, and #3.
Contestants initial choice is always #1.
The Host cannot reveal the contestants door and must always reveal a goat.

NOT SWITCHING (the Host will always reveal either #2 or #3 and no switch is made)
Car Behind Door #1 – Win
Car Behind Door #2 – Loss
Car Behind Door #3 – Loss

SWITCHING
If the car is behind #1, #2 or #3 is revealed. The switch is made to the other, therefore:
Car Behind Door #1 – Loss

If the car is behind door #2, #3 must be revealed. The switch is to #2, therefore:
Car Behind Door #2 – Win

If the car is behind door #3, #2 must be revealed. The switch is made to #3, therefore:
Car Behind Door #3 – Win

Think of it this way…there are three choices, the host always eliminates one wrong choice and 2/3 of the time the contestant eliminates one wrong choice, allowing the host to narrow the remaining 50/50 to the correct choice. I spent 15 minutes on a whiteboard in my dorm room with this thing and it’s only just starting to make sense.

75. Melbourne_junkie

The game-show story would have been a nice way to lead into the math if it hadn’t been used recently in other MtG articles.

Good topic though, decent attempt at tackling it too.

To the FoW thing. I like this kind of article and I fully understand that when your deck contains senseiDT then you will need less blue cards, because is permanent brainstorm. But I must disagree with you in the term of Brainstorm could search into anything, even a crit. Brainstorm can search into something what is even better than the creature you put in instead of brainstorm (because you said someone cuts the number of creature to put in brainstorm). So for me, brainstorm is not a bad option instead of some creatures (of course, you can survival them and have like better brainstorm, but BS could find answers too at the lower cost and is also a card for Fow (if you put blue creatures instead of it then this argument is less valuable. ok)) But you didnt said that its a bad option, you said basically it sounds like nonsense to you so I would like to hear an explanation for this one point of your view.Hopefuly you get what I mean. thanks

77. Regarding the question of getting a seventh head, after getting six heads in a row – I’d be a little suspicious at that point! My Bayesian prediction is that there’s a better than 50% chance the coin comes up heads on the seventh flip. At the very least, I’d like to see the other side of the coin before making my wager.

78. I AM THE HEAVENS AND EARTH!!!!!!!!!!!!!!!!!!!!!!!!!

79. Chapin was right about the mirror, I literally lost 1 survival mirror out of 16 over the two opens.

80. Wow, let’s do some basic probability here. The first failing is that the chart assumes that the first door choice and second are linked in any way. They are not. The first, you have a 1/3 chance of selecting the correct door. Then one of the doors that was wrong is removed. So now we have a new incident in which you have a 1/2 chance of selecting the correct door. What you chose in the first pass DOES NOT MATTER IN THE SLIGHTEST. They are unrelated. In fact, assuming the game is fair, the first choice, made from the full 3 doors is just a waste of time, Like many gameshow tricks, in order to artificially heighten tension and make the show longer so they can milk out the prize for a bit more. I don’t care what the rest of your article has to day if you start with something you saw in 21, which was incorrect on just about every piece of “math” it presented. You would know this if you’d taken 11th grade pre-calculus, or any probability class ever.

81. RIon you are wrong… the problem has two probability things going on, not one. There is the probability of you picking the car, which is irrelevant and the probability of the host hiding the car. Because only one door has the car the host 2/3rds of the time has to be hiding the car. So you switch to what the host is hiding.

82. lol wasnt this the same problem in the beginning of the movie “21”

83. Thanks for the article, Caleb. Unlike your middle schoolers, who apparently revel in their excellence at knowing something about numbers, I studied Humanities and therefore appreciate your taking the time to explain these concepts. It’s sad, I know, but I did learn a thing about letters during my time not looking at statistical issues. For example, and for the edification of all the number crunchers who gave you a hard time, the subject to which your efforts refer is known as Mathematics, or “Maths” for short. I’m aware that in the US, “Math” is the common term; however, much like the practice of wearing trousers with your ass hanging out the back, it’s still wrong.
I look forward to your next article.

84. This is the best article I have read, not because of the content, but the argument that followed. Next time, you really should teach people about the The inspection paradox.

85. The article was very interesting. Reading so many people declare their intelligence and how wrong or right each of them were is priceless. I’m not a mathematician, math major nor have I ever been been better than above average in math. I do enjoy stats and have had to use certain principles for a genetics class for my minor in biology. Anyone who argued that the the choices are independent is incorrect. Say for instance a couple has 4 children and they were all girls. What is the probability that the 5th will be a boy? In short you would calculate all the previous instances of children to give you the percentages for a boy being born. However, you can do independent analysis of said situations but not when you have previous information like other doors or children. You can’t just calculate from two doors because staying with the choice you made gets added into the equation. Hence, you have 2/3 a chance to be right and not 50% to be right. This assumes that the host will reveal a goat and that most likely you have picked a goat door. Meaning you switch and most likely have a car.

86. On a side note pigeons who have been in this experiment have learned unlike most humans that switching wins you out most of the time. lol

87. Just 4 Fun:

the 0,999999999…. = 1 problem is pretty simple to solve with a variable:

0,9999999999… = x

9,9999999…= 10x

9 = 10x-x (cuz x =0,999999…)

9 = 9x

1= x = 0,99999999… true

88. The original problem hinged on the host knowing what was behind the doors. Regardless though, 2 out of 3 times you have will have picked a goat so it’s in your best interest to switch no matter if the host knows or not.

89. Pingback: MTGBattlefield