Hearthstone Probability Problem
03-22-2016
, 08:21 AM
For those of you unfamiliar with Hearthstone, here's what you need to know to take a crack at this problem:
- Each player constructs a deck with exactly 30 cards in it
- During game play, cards are either...
- When it is your turn, you have the option of playing anywhere from 0 to all of the cards in your hand.
- You cannot have more than 10 cards in hand.
- You can always see how many cards are in your opponents hand, and you can keep track of exactly what turn they drew each card in their hand.
- You can always see how many cards are left in your opponent's deck.
So here it is - Let's say that you know for certain that your opponent is running exactly one copy of, "Card A," and, "Card B," in their deck. For the sake of this problem, these cards cannot be played until your opponent has less than 15 cards remaining in their deck. How do you calculate the probability that your opponent has, "Card A," and, "Card B," in their hand once they have exactly 15 cards remaining in their deck?
Allow me to clarify and give some examples. The key here is going to be how many cards are in your opponent's hand, and what turn they drew those cards, once they have 15 cards remaining in their deck. Obviously the baseline odds of them drawing both, "Card A," and, "Card B," in the first half of their deck is going to be (1/2)(1/2) = 1/4.
But let's say they have 0 cards in their hand when they have exactly 15 cards remaining in their deck. Obviously the odds of them holding both, "Card A," and, "Card B," in their hand is 0.
Let's say that instead they hold 2 cards in their hand. If those 2 cards were the 2 very last cards that they drew, the odds of those 2 cards being, "Card A," and, "Card B," should be (2/17)(1/16) = 1/136. But if those 2 cards are cards that they've been holding onto since the very beginning of the game, the odds of them being, "Card A," and, "Card B," should be much higher, probably even higher than 1/4?
And things would get more complicated if they had more cards in their hand.
I feel like the answer to this problem is going to lie somewhere in between, "there's not enough information," and, "you can calculate a very accurate, but not exact, answer for every scenario." I'm just curious about your thoughts. How would you go about solving this? What extra information would you need to solve this? I appreciate all feedback
- Each player constructs a deck with exactly 30 cards in it
- During game play, cards are either...
- - Played (Both players see the card)
- In Hand (Only the player holding the hand sees the card)
- In Deck (Neither player sees the card)
- When it is your turn, you have the option of playing anywhere from 0 to all of the cards in your hand.
- You cannot have more than 10 cards in hand.
- You can always see how many cards are in your opponents hand, and you can keep track of exactly what turn they drew each card in their hand.
- You can always see how many cards are left in your opponent's deck.
So here it is - Let's say that you know for certain that your opponent is running exactly one copy of, "Card A," and, "Card B," in their deck. For the sake of this problem, these cards cannot be played until your opponent has less than 15 cards remaining in their deck. How do you calculate the probability that your opponent has, "Card A," and, "Card B," in their hand once they have exactly 15 cards remaining in their deck?
Allow me to clarify and give some examples. The key here is going to be how many cards are in your opponent's hand, and what turn they drew those cards, once they have 15 cards remaining in their deck. Obviously the baseline odds of them drawing both, "Card A," and, "Card B," in the first half of their deck is going to be (1/2)(1/2) = 1/4.
But let's say they have 0 cards in their hand when they have exactly 15 cards remaining in their deck. Obviously the odds of them holding both, "Card A," and, "Card B," in their hand is 0.
Let's say that instead they hold 2 cards in their hand. If those 2 cards were the 2 very last cards that they drew, the odds of those 2 cards being, "Card A," and, "Card B," should be (2/17)(1/16) = 1/136. But if those 2 cards are cards that they've been holding onto since the very beginning of the game, the odds of them being, "Card A," and, "Card B," should be much higher, probably even higher than 1/4?
And things would get more complicated if they had more cards in their hand.
I feel like the answer to this problem is going to lie somewhere in between, "there's not enough information," and, "you can calculate a very accurate, but not exact, answer for every scenario." I'm just curious about your thoughts. How would you go about solving this? What extra information would you need to solve this? I appreciate all feedback
