Probability doubt for a card game

Question

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    AK   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?

Answer 1

You have 3 distinguishable cards and 1 partition (dividng the hands between the two players). Therefore, the number of cases should be $4! = (4)(3)(2)(1) = 24$. The permutation method is correct. In essence, the cards and the "partition" can be shuffled to any configuration, and as such, while P1 having AK7 isn't different from him having K7A in terms of the net result, these are two different states resulting from different shuffles. In your first calculation, you neglect the multiplicity of each configuration, and that's why you get an erroneous result.

Answer 2

As I mention in comments, in order for this question to be well-defined you need to make explicit what you mean by "random." Nonetheless, I think the first answer is the one that you want.

To make the question well-defined we will deal out cards by taking each card and assigning it to one of the two players by 50/50 chance.

Let's think about this in two ways. First we can say it doesn't matter which order you assign the cards in, because for each card there's still a 50/50 chance that it goes to each player. So there are 8 possible deals all of which are equally likely. As you checked 6 of those are survives, so 3/4 of the time you survive.

There's a second way you could think about it, which is to shuffle the cards and then assign each randomly starting at the first. Now there's not 8 possibilities, there's 6*8 = 48 possibilities. This is what you're trying to get at with your second method of calculating, but you're missing some of the options, because there are three different ways that say AK 7 could happen: the 7 could be the first card, second card, or third card.

So your second answer is wrong because the possibilities that you list aren't all equally likely. The AK 7 hand will happen 3 times as often as the AK7 hand.