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?
 A: 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.
A: 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
