# What's the proper way to calculate probability for a card game?

I'm creating AI for a card game, and I run into problem calculating the probability of passing/failing the hand when AI needs to start the hand. Cards are A, K, Q, J, 10, 9, 8, 7 (with A being the strongest) and AI needs to play to not take the hand.

Assuming there are 4 cards of the suit left in the game and one is in AI's hand, I need to calculate probability that one of the other players would take the hand. Here's an example:

AI player has: J Other 2 players have: A, K, 7

If a single opponent has AK7 then AI would lose. However, if one of the players has A or K without 7, AI would survive. Now, looking at possible distribution, I have:

P1   P2   AI
---  ---  ---
AK7       loses
AK   7    survives
A7   K    survives
K7   A    survives
A    7K   survives
K    7A   survives
7    KA   survives
AK7  loses


Looking at this, it seems that there is 75% chance of survival.

However, I skipped the permutations that mirror the ones from above. It should be the same, but somehow when I write them all down, it seems that chance is only 50%:

P1   P2   AI
---  ---  ---
AK7       loses
A7K       loses
K7A       loses
KA7       loses
7AK       loses
7KA       loses
AK   7    survives
A7   K    survives
K7   A    survives
KA   7    survives
7A   K    survives
7K   A    survives
A    K7   survives
A    7K   survives
K    7A   survives
K    A7   survives
7    AK   survives
7    KA   survives
AK7  loses
A7K  loses
K7A  loses
KA7  loses
7AK  loses
7KA  loses


12 loses, 12 survivals = 50% chance. Obviously, it should be the same (shouldn't it?) and I'm missing something in one of the ways to calculate.

Which one is correct?

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## migrated from gamedev.stackexchange.comNov 12 '12 at 22:09

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If the cards are dealt, the results are independent of the card order. Unless a rule tells the player they must play a certain card, it doesn't matter what what order the player is holding their cards, what matters is simply what they play. – Benjamin Danger Johnson Nov 2 '12 at 17:50
The cards are dealt from a shuffled 24 card deck, with 3 suits. Each player has exactly 8 cards, but in various suit. The problem I gave above only considers a single suit. I want to calculate probability for lowest card in each suit and then play the suit where the player has the most chance of surviving. This means that other two players might have 0-3 cards each. AI only knows that they have 3 cards total, because from other 5 cards, 4 have already been played and the remaining one is in AI's hand. – Milan Babuškov Nov 2 '12 at 18:05
What is the point of ranks? Everyone must play the same? It's a bit tricky to come up with an exact formula without all the rules for play. – Benjamin Danger Johnson Nov 2 '12 at 18:13
Everyone must play the same suit and highest card loses. – Milan Babuškov Nov 2 '12 at 18:24

## 2 Answers

It depends on how the cards are drawn, which you haven't described.

For example, if each card is dealt to a random player, one at a time, then the first calculation is correct. On the other hand, if the cards are first shuffled, the deck is then split at a random position, and one player gets the bottom half while the other gets the top half, then the second calculation is correct.

In particular, using the first method of dealing, the probability of player 1 getting no cards at all is (1/2)3 = 0.125, while using the second method, it is 1/(3+1) = 0.25.

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The cards are first shuffled and then split. Although, I fail to see how that changes the outcome? – Milan Babuškov Nov 2 '12 at 18:24
I don't understand that too, how shuffling a random deck is going to change the probability? – Markus von Broady Nov 2 '12 at 19:21
It's not the shuffling it's how they're dealt. If you are handing out the cards one at a time then each card has a 50/50 chance of going to each player; but if you have three cards and you pick (with equal probability) either to split before the first card (all three to P2), before the second (one to P1, two to P2) before the third (two to P1, one to P2) or after the third (all to P1), then there are 4 possible outcomes with 6 combinations each that each have a 25% chance of happening. Essentially you're cutting out all the possibilities of the single card being dealt between the other two. – Lunin Nov 2 '12 at 20:33

Assuming you're handing out each of those remaining cards with a 50% chance between the two players, the chance of a specific having all three is (1/2) * (1/2) * (1/2), or 0.125. Since if either player has all three the AI will lose we have two of this situation that will end in AI failure, 0.125 * 2 = 0.25 = 25% chance of failure.

The reason you don't see this in your second chart is because you're missing permutations. In the all to one player examples you're showing every set of orders, however in the examples where one player gets two and the other gets one you are only looking at combinations where each player's individual order matters.

If player 1 gets AK and player 2 gets 7, there aren't two ways this can happen as in your chart, there are six (3 * 2 * 1)

• Player 1 gets A, Player 1 gets K, Player 2 gets 7
• Player 1 gets A, Player 2 gets 7, Player 1 gets K
• Player 1 gets K, Player 2 gets 7, Player 1 gets A
• Player 1 gets K, Player 1 gets A, Player 2 gets 7
• Player 2 gets 7, Player 1 gets A, Player 1 gets K
• Player 2 gets 7, Player 1 gets K, Player 1 gets A

If that particular order looks familiar it's because it's the same ordered combinations of cards from the 1 player gets all example in your second chart. The truth is for the situation you're describing, order doesn't matter so your first inclination was correct. Each of those possibilities has 6 ways to be dealt out, so in the end you're going to end up with the same probabilities

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