0
$\begingroup$

I want to solve this problem.

Consider a deck of 52 cards, ordered such that Ace > King > Queen > ... > 2. I pick a card, then you pick one. What is the probability that my card is larger than yours?

I thought of doing it like this, but it is wrong and I don't see why.

Denote the cards by values 1 to 13, so the card "2" is denoted by 1 and the card "Ace" is denoted by 13. This is just a transformation.

Let $p$ be the desired probability. Then $$p = \sum_{i=1} P(\text{my card} > \text{your card } | \text{ I pick card of value i})P(\text{I pick card of value i})$$ by law of total probability.

We have $$P(\text{I pick card of value i}) = \frac{4}{52} = \frac{1}{13}$$ since there are 4 cards of value $i$ in a deck (eg. there are 4 aces in a deck).

We also have $$P(\text{my card} > \text{your card} | \text{I pick card of value i}) = \frac{i-1}{12} + \frac{i-1}{13}\times 3$$ because if I pick a card of value $i$ then the cards $1, ..., i-1$ are smaller than my card, and there are 12 cards from the suite my card is from and 13 cards in the 3 remaining suites.

So we get

$$p= \frac{1}{13}\times (\frac{1}{12} + \frac{3}{13})\times \sum_{i=1}^{13} i-1 = \frac{1}{13}\times (\frac{1}{12} + \frac{3}{13})\times 66$$ and this is bigger than $1$. What's the error in my method and how do I fix it?

$\endgroup$
14
  • 2
    $\begingroup$ Hard to follow. Note that either you get a tie or the two players have equal chances of winning. $\endgroup$
    – lulu
    Commented Mar 23, 2022 at 13:38
  • 1
    $\begingroup$ What did you do to get $\frac{i-1}{12}+\frac{i-1}{13}\times 3$? What makes you think those numbers specifically? Recall, there are $51$ cards left in the deck after the first player gets their card... so where does $12$ come from? Why not $51$? $\endgroup$
    – JMoravitz
    Commented Mar 23, 2022 at 13:41
  • 1
    $\begingroup$ No, because neither of you draws from different groups of the deck. Your opponent draws 1 card from the 51 left, and $4\times (i-1)$ are good for you. $\endgroup$
    – Ingix
    Commented Mar 23, 2022 at 13:45
  • 1
    $\begingroup$ "I thought I was splitting it using $P(A\text{ or }B)=P(A)+P(B)$." your interpretation of your $A$'s and $B$'s are off. $A$ the event "second card is smaller and comes from same suit", $B$ the event "second card is smaller and comes from different suit." Your value of $\frac{i-1}{12}$ is the conditional probability of the second card being smaller given that it comes from the same suit. As for $\frac{i-1}{13}\times 3$, you mean to have $\frac{(i-1)\times 3}{13\times 3}$ or more simply $\frac{i-1}{13}$ is the conditional probability of the second smaller given from different suit $\endgroup$
    – JMoravitz
    Commented Mar 23, 2022 at 13:52
  • 3
    $\begingroup$ The probability "second card smaller given that it comes from the same suit" is not the same thing as the probability "second card is smaller and it comes from the same suit." $\Pr(X\mid Y)\neq \Pr(X\cap Y)$. If you wanted to correct this line of thinking, you would need to condition those probabilities on their conditionals... multiply by the probability of that conditional being the case. $\frac{i-1}{12}\times\frac{12}{51} + \frac{i-1}{13}\times\frac{39}{51}$, which of course equals as alluded to before $\frac{(i-1)\times 4}{51}$ $\endgroup$
    – JMoravitz
    Commented Mar 23, 2022 at 13:54

1 Answer 1

Reset to default
1
$\begingroup$

The correct (and simpler) expression is $$P(\text{my card} > \text{your card} | \text{I pick card of value i}) = \frac{(i-1)\times 4}{51}$$

Simply: the numerator is the possible cards that are lower than $i$, and $51$ are the available cards.

AS noted in the comments, this can be expressed as

$$ \dfrac{(i-1) + (i-1)\times 3}{12+13\times 3}$$

where the first terms in each sum correspond to the selected rank. But this does not really help. And of course, the mistake amounts to assuming $\frac{a}{b}+\frac{c}{d}\neq \frac{a+c}{b+d}$ This is already noted by
JMoravitz' comment.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .