This is a homework question. Not something to hand in, but rather a recommended practice question.

A Card Game
Three students are playing a card game. They decide to choose the first person to play by each selecting a card from the 52-card deck and looking for the highest card in value and suit. They rank the suits from lowest to highest: clubs, diamonds, hearts, and spades.

a. If the card is replaced in the deck after each student chooses, how many possible configurations of the three choices are possible?

b. How many configurations are there in which each student picks a different card?

c. What is the probability that all three students pick exactly the same card?

d. What is the probability that all three students pick different cards?

I figured out the the answer to part A is $52^3 = 140608$, since each student chooses 1 card from a deck of 52.

I also know that the answer to part B is $52\times 51\times 50 =132600$, or $\frac {52!}{(52-3)!}$.

However, I am not sure how to find the probability that they all pick the same card. I initially thought it would be $\left( \frac 1{52}\right)^3$, but that does not match the answer in the back of the book, which is $0.00037$.

Sorry for the formatting (or lack thereof), I tried my best to make it fairly easy to read, but I am unsure of how to make fancy fractions and such.

  • 3
    $\begingroup$ For $c$: the first person takes whatever. The probability that the second matches the first is $\frac 1{52}$. The probability that the third matches the first is $\frac 1{52}$ independent of what the second person drew. So... $\endgroup$ – lulu Jun 11 '17 at 20:54
  • 1
    $\begingroup$ An answer of $(1/52)\cdot (1/52)\cdot (1/52)=(1/52)^3$ is the probability of all three specifically selecting the ace of spades, but remember there are many ways other than just everyone getting the ace of spades for them to all have the same card. As for part (d), if you were to read and understand the birthday problem you'll find the mathematics behind it is very similar. Alternatively, look more closely at the phrasing of parts (b) and (a) and how they relate to question (d). $\endgroup$ – JMoravitz Jun 11 '17 at 21:00
  • $\begingroup$ So (1/52)^2. Thank you! Are you going to post it as an answer so that I can accept it? $\endgroup$ – Bunyip Jun 11 '17 at 21:02
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    $\begingroup$ As a complete aside, you say "I am unsure of how to make fancy fractions and such." This page will give you the information you need to do so (at least on this site using MathJax). $\endgroup$ – JMoravitz Jun 11 '17 at 21:03

For part C:

Try thinking about it like this, player A is the one that "chooses" which card B and C must choose in order for all of them to pick the same card, so the probability that all three students pick exactly the same card is:

$\ \frac{1}{52}\frac{1}{52} = (\frac{1}{52})^2$

For part D:

Start as in C, player A chooses the card and B has a $\ \frac{51}{52}$ chance of its card being different from A's card, similarly C has a $\ \frac{50}{52}$ of its card being different from A and B, so the probability that all cards differ is:

$\ \frac{51}{52}\frac{50}{52}$



Part c only makes sense if the cards are drawn with replacement. You already have the total number of configurations. There are 52 configurations where all the cards are the same. With these two facts, the required probability is readily calculated.

For part d, which again only makes sense if cards are drawn with replacement, you can get the number of ways in which all three cards are different from the answer to part b. This will lead to the required probability.


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