The Monty Hall Problem
A certain probability puzzle is well-known in math circles for its unusual ability to cause people to refuse to believe the answer when it is explained to them. It’s usually known as the “Monty Hall problem” (after the host of Let’s Make a Deal):
Monty Hall has given you a choice of three identical doors. Behind one is a car and behind the other two are goats. You choose a door, but before it’s opened to reveal your prize, Monty adds a twist. He opens one of the other doors to reveal a goat (he always does this to add to the suspense). He then offers you the choice of staying with the door you chose, or switching to the remaining unopened door. Should you stay, or should you switch, and what’s your chance of winning in either case?
The answer is surprising to most: switching doubles your odds of winning the car (2/3 chance of winning, as opposed to 1/3 if you stay). The key fact is that Monty’s knowledge of which of the other doors (if any) was a car causes him to always remove a goat from the prize pool. The chance that the initial door you chose contained a car was 1/3 to start, and it’s unchanged by Monty’s ritual. But if the car was behind either of the other two (2/3 probability in total), Monty will remove the losing door and leave the winning one, and switching will win.
(If you don’t buy that the probability is anything other than 50% when everything started out equal and there are now two doors remaining, there are myriad sources on the internet trying to explain in different ways).
The Jaina Proudmoore Problem
Let’s say it’s late in the Hearthstone game and you’re about to try playing your bomb legendary. You’re pretty sure you’ll win if it sticks for a turn, but if it gets answered immediately you might be hopelessly behind. You want to guess whether your opponent Jaina is holding the Polymorph you know she has in her deck (assume for the moment she only has 1 for simplicity; the same concept applies with 2 but the math is more intricate). Her hand has 4 cards, and you mouse over her deck and see she has 16 cards left. After her draw next turn, 5 of 20 unseen cards will be in her hand, so you’d think she had a 25% chance holding the card.
Monty Hall, however, tells you differently. Even though she only has 5 cards in her hand next turn, she’s been selecting out non-Polymorph cards and playing them all game. Just as Monty selected out non-winning doors and removed the pool, making the remaining doors of the ones he could have chosen more likely to be winners, Jaina has been casting non-Polymorph cards from her hand, making the ones she’s left in her hand more likely to be Polymorphs.
For now consider the simplest case, where nothing had been cast so far this game that Jaina would have been likely to Polymorph. Applying the logic from above, there’s a 50% chance that the Poly started the game in the bottom 15 cards in the deck, and that probability has not been changed by any subsequent events. There must, then, be a 50% chance that it’s among the 5 cards in her hand. Quite a significant difference from the 25% that seemed completely intuitive before considering this effect.
An important subtlety is that Jaina might not be as selective as Monty. Monty never prematurely revealed a car, but Jaina may have cast a Polymorph before the critical turn, if she had it and the game state called for it. The assumption above that the game state on prior turns never looked like one that would have drawn a Polymorph elides a deep point of probability, namely, the mysterious way that Jaina’s/Monty’s selectivity shifts probability between chosen and unchosen cards/doors. I think getting into the depths of how that happens might be beyond the scope of this post, but for now observe the following.
If you know Jaina is going to save her Poly for your bomb no matter what (and this might not be a bad assumption in a constructed card game where people are familiar with each other’s decks), then the situation is identical to Monty Hall: the probability in the above example would have been 50%. But if you made the opposite assumption, that Jaina is not smart and dispenses cards from her hand at random (ignore the vagaries of the mana curve for the moment), the probability would in fact be the 25% we naively estimated at the start. The best way to see the difference is to really understand why the answer to the Monty Hall problem is what it is. It might also be helpful to look at that bit of logic from the end of the previous section (about the 50% chance that Poly started in the bottom 15 cards of the deck) and try to see why it doesn’t apply in the case of the Jaina that plays randomly.
The other complexity is what I alluded to at the start: the arithmetic is more complicated with two Polymorphs in the deck (but as I said, the logic is unchanged). To work the same example with two Polymorphs:
- The naive estimate is that out of 20C2 (20 choose 2, referring to combinations) places the Polys could be, 15C2 choices have them in the bottom 15, so the chance of having one in hand is 1-(15C2 / 20C2), which evaluates to 17/38, or around 45%.
- The Monty-corrected estimate would be that out of 30C2 possible placements at the start of the game, 15C2 have them undrawn in the first 15 cards, so the chance of having one in hand is 1-(15C2 / 30C2), which evaluates to 22/29, or around 76%.
The result is the same as before, a much higher chance in the latter case.
Here is a full table of the probabilities for various deck and hand sizes, for reference:
The probability subtleties discussed at the beginning of the last section mean that guessing the probability that your opponent has a certain card is never an exact science. It depends on your judgment of how likely it would have been for them to use it at an earlier point in the game if they’d had it. The Monty Hall effect applies most strongly when it is highly unlikely or impossible (for example due to mana cost) that the card in question would have been played already. Where it applies, it causes the probability of the card being in your opponent’s hand to be substantially higher. So it’s both practical and perhaps fascinating to realize that you can’t rely on what you thought you knew about probability: all unseen cards are not equally likely to be the card you care about.