An interesting gambling problem in probability

Question

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!

Answer 1

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, $$ \mathsf{P}(X_2<X_1\mid X_1=v)=\frac{4(v-1)}{51}, $$ and so, the probability that the first player wins is also $8/17$.

Answer 2

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)$=\left(\frac1 2 \cdot \frac{48}{51}\right) = \frac{24}{51} = \frac8 {17}$