# Probability that at least one of two drawn cards has greater face value than at least one of your two, if both your two cards are the same?

I have just been messing around with some card odds and have been having a little difficulty. The situation is that both you and your single opponent are dealt two cards. You look at your own cards and both have the same number on their face (ie 3 and 3), but you cannot see your opponents. The question is: what is the probability that your opponent has at least one card with a face number that is higher than both of your cards.

I have taken a few shots at this question, and each time I've gotten muddled and have had to restart and try a new approach. Here is my primary attempt, and most of my retries have been variations upon what follows.

Call your opponent's face card values $O_1$ and $O_2$, and your own cards $H_1$ and $H_2$. We have three possible orderings of our opponents cards, with the following probabilities:

P($O_1$=$O_2$)=1/17 ,P($O_1$>$O_2$)=8/17 and likewise P($O_2$>$O_1$)=8/17

Now Im guessing that because we know our own cards, we can say $H_1$$\ge$$H_2$.

From here I said if $O_1$>$O_2$, then:

P($O_1$>$H_1$) = 4(13- $H_1$)/52 = 1 - $H_1$/13 and P($O_1$=$H_1$) = 3/52 This is as far as I've gotten before I start to lose track of what I'm doing and why.

Please post any mistakes found, the answer you found and an explanation as to how you attained it. If you have any questions about my working I shall attempt to explain my thought process, but I hope you can follow it for the most part (be it right or wrong).

• You should include your attempt to solve the question and indicate where you are stuck so that you receive responses that address the specific difficulties you are encountering. – N. F. Taussig Jan 10 '16 at 13:38

Cards numbered $1$-$13$, if you have two cards numbered $k$ then there are $4k-2$ cards that do not exceed yours. The probability that both of the cards drawn by your opponent do not exceed yours is $$\frac{\binom{4k-2}{2}}{\binom{50}{2}}=\frac{(4k-2)(4k-3)}{2450}$$ Therefore, the probability that they have at least one that exceeds yours is $$1-\frac{(4k-2)(4k-3)}{2450}$$
If you don't know your card ahead of time, each of the $13$ values is equally likely. Therefore, the probability would be $$1-\frac1{13}\sum_{k=1}^{13}\frac{(4k-2)(4k-3)}{2450}=\frac{788}{1225}$$