# An interesting gambling problem in probability

Back in high school we used to play a simple gambling game with cards. Two people would randomly choose a card from a deck of $$52$$ cards. The person with a higher value card (Ace being the strongest and two the weakest) would then be the winner.

In this seemingly simple scenario is it possible to calculate the probability of a win? That is the first person choosing a higher value card. (I believe a similar line of thought would hold for the second person winning?)

Although appearing quite straightforward at first glance I couldn't figure out how to even get started.

Any help or ideas would be appreciated. Thank you!

• Intuitively, what do you think the answer is? – Andrew Chin Mar 1 at 8:56
• With information you give, the probability must be $1/2$. :) – NN2 Mar 1 at 8:56
• Also what about a draw? – Math Lover Mar 1 at 8:57
• Intuitively I was thinking that it should be close to 1/2, however as Math Lover pointed out there is also the possibility of a draw. Plus wouldn't the first draw and the second would have a different set of advantages? – ritvik1512 Mar 1 at 9:02
• The probability of a draw is straightforward. It is $\frac{3}{51} = \frac{1}{17}$. Probability of any of them winning is $\frac{8}{17}$. – Math Lover Mar 1 at 9:07

Let $$X_{i}$$ denote the value of the $$i$$-th draw (from 1 to 13). Conditionally on $$X_1$$, $$\mathsf{P}(X_2>X_1\mid X_1=v)=\frac{4(13-v)}{51}.$$ Thus, the probability that the second player wins is $$\sum_{v=1}^{13}\mathsf{P}(X_2>X_1\mid X_1=v)\mathsf{P}(X_1=v)=\sum_{v=1}^{13} \frac{4(13-v)}{51}\times \frac{4}{52}=\frac{8}{17}.$$ Similarly, $$\mathsf{P}(X_2 and so, the probability that the first player wins is also $$8/17$$.

• I guess it turned out to be quite simple after all. Thank you! – ritvik1512 Mar 1 at 9:32

You can make it even simpler by looking at it in the following way:

For any card drawn by the first person, there is a Pr of $$\frac 3 {51}$$ for a draw, hence a Pr of $$\frac{48}{51}$$ that it results in a win for one or the other.

In two randomly drawn cards, each will have an equal probability of being of higher value,

thus P(first person wins)= P(second person wins)$$=\left(\frac1 2 \cdot \frac{48}{51}\right) = \frac{24}{51} = \frac8 {17}$$

• The underlying ideas are good and the answer is right, but as written this answer isn't a rigorous way of translating the ideas to the answer. – Greg Martin Mar 1 at 17:28
• @Greg Martin: I have changed the presentation to make it more rigorous – true blue anil Mar 1 at 19:25