Hi everybody,

I played poker the other night and we used casino war in order to determine the dealer for the first hand. I tied with another player four times in a row and this seemed most unlikely. Let's quantify that!

For clearity: We used a shuffled poker deck (52 cards: 13 faces a 4 suits) and two players alternately picked cards from the pile (without putting the cards back afterwards). Cards with the same face, e.g. two Kings, tie regardless of the suit. What is the probability of four ties in a row?

I've come up with an answer but since it's combinatorics I am not sure if I'm correct. In my opinion, three cases have to be distinguished (The cards Jack, Queen, King, and Aces are abbreviated by their first letter.):

(a) Each two of the four pairs are different, e.g. JJQQKKAA: The probability of this happening is

$\displaystyle P_1 = \frac{4}{52} \frac{3}{51} \frac{4}{50} \frac{3}{49} \frac{4}{48} \frac{3}{47} \frac{4}{46} \frac{3}{45}$.

We are not restricted to these specific cards but can pick four of the 13 faces: So the binomial coefficient $\displaystyle \binom{13}{4}$ arises as a factor. Another factor of $\displaystyle \frac{4!}{1!1!1!1!} = 4!$ is in the game, since the sequence of pairs can be permutated (JJQQKKAA, JJQQAAKK, etc). So eventually, the probability of picking four different pairs is $\displaystyle 4! \binom{13}{4} P_1 $.

(b) There are only two different kinds of cards, e.g. JJJJAAAA: The probability of this happening is (analog to (a)) $\displaystyle P_2 = 4!\cdot 4!\cdot 44! / 52!$. Again, lifting the restriction of having Jacks and Aces gives rise to a factor of $\displaystyle \binom{13}{2}$. This time, $\displaystyle \frac{4!}{2!2!} = 6$ permutations(*) are possible, thus there is a factor of 6. This leaves us with the probability $\displaystyle 6 \binom{13}{2} P_2$.

(*) Those are: {JJJJAAAA, JJAAAAJJ, AAAAJJJJ, AAJJAAJJ, JJAAJJAA, AAJJJJAA}

(c) Exactly one pair is picked twice, e.g. JJJJQQKK: The probability for this is $\displaystyle P_3 = 4! \cdot 4\cdot3 \cdot 4\cdot3\cdot 44! / 52!$. Just as before, an extra factor of $\displaystyle \frac{4!}{2!1!1!} \cdot\binom{13}{3} $ is mandatory. This results in $\displaystyle 12\cdot \binom{13}{3}\cdot P_3$. The factor 12 is due to the possible permutations {JJJJQQKK, JJJJKKQQ, JJQQKKJJ, JJQQJJKK, JJKKQQJJ, JJKKJJQQ, QQJJJJKK, QQJJKKJJ, QQKKJJJJ, KKJJJJQQ, KKQQJJJJ, KKJJQQJJ}.

The sum of these probabilities gives the probability to tie four times in a row in a game of casino war, which is

$\displaystyle 4! \binom{13}{4} P_1 + 6 \binom{13}{2} P_2 + 12 \binom{13}{3} P_3$.

Am I correct or does anybody disagree? I tried different approaches and this one seems/for me/to be the least confusing one.

Thanks in advance for your help!

needsomemathhelp