I have a questions about combinatorics.
my question : You are playing a version of poker where you can see three of your cards and three of your opponents. She has 6♣, 8♣, 10♣ and you have A♦, A♣, A♠. What is the probability that you will win? (both have 5 cards)
I'm not familiar with poker. I found the rules but i'm confused since i have no experience.
Could you help me??
The ranking of the possible hands
Straight flush-your opponent draws 7C, 9C
Four of a kind-you draw the fourth ace
Full house-you draw a pair
Flush-your opponent draws two clubs that are not 7,9
Straight-your opponent draws a 7 and a 9, not both clubs
Three of a kind-neither of you improve
For the complete list, which you do not need here, see Wikipedia
As for calculating the respective probabilities... this becomes quite tedious.
The probability that your opponent completes her straight flush and draws both $7\clubsuit$ and $9\clubsuit$ will be $\dfrac{1}{\binom{46}{2}}$
The probability that your opponent did not complete her straight flush and that you did complete your four-of-a-kind will be $\dfrac{2}{46}\cdot\dfrac{\binom{45}{2}-1}{\binom{45}{2}}$
The probability that your opponent did not complete her straight flush and that you did complete your full house will be $\dfrac{2\cdot 3}{\binom{46}{2}}+\left(\dfrac{5\cdot 3 + 7\cdot \binom{4}{2}}{\binom{46}{2}}\right)\cdot\dfrac{\binom{44}{2}-1}{\binom{44}{2}}$
The probability that your opponent did complete a straight but not a straight flush while you did not complete your four-of-a-kind or full house will be $\dfrac{4^2-1}{\binom{46}{2}}\cdot\dfrac{\binom{7}{2}\cdot 4^2+7\cdot 5\cdot 4\cdot 3 + \binom{5}{2}\cdot 3^2}{\binom{44}{2}}$
The probability that your opponent did complete a flush but not a straight flush while you did not complete your four-of-a-kind or full house will be $\dfrac{\binom{9}{2}-1}{\binom{46}{2}}\cdot\dfrac{\binom{7}{2}\cdot 4^2+7\cdot 5\cdot 4\cdot 3 + \binom{5}{2}\cdot 3^2}{\binom{44}{2}}$
Finally, all remaining probability will correspond to you winning with a three-of-a-kind of Aces, the probability of this occurring will be $1$ minus all of the previously mentioned probabilities.
The probability of you winning will then be the sum of the second, third, and fifth of the above probabilities while the probability of your opponent winning will be the sum of the remaining probabilities.