# Probability of deck of card

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 the lowest to the highest: clubs, diamonds, hearts and spades.

1. If the card is replaced in the deck after each student chooses, how many possible configurations of the three choices are possible?
2. How many configurations are there in which each student picks a different card?
3. What is the probability that all three students pick exactly the same card?
4. What is the probability that all three students pick different cards?
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Sounds like a nice homework problem, and if it is so, please add the homework tag. People here will be glad to help you but please tell us what you tried in your attempts to solve the problem and where you are stuck. –  Dilip Sarwate Aug 1 '12 at 11:07

By a configuration we will understand an ordered triple $(a,b,c)$, where $a$ is the first card chosen, $b$ the second card chosen, and $c$ the third card chosen.

$1.$ The first card chosen has any of $52$ values. For each of these values, the second card chosen has any of $52$ values. And for each such value the third card has any of $52$ values, for a total of $(52)(52)(52)$.

$2.$ The first card chosen has any of $52$ values. For each of these values, the second card chosen has any of $51$ values. And for each such value the third card has any of $50$ values, for a total of $(52)(51)(50)$.

$3.$ Call a configuration weird if all three cards are identical. We count the number of weird configurations. The first card can have any of $52$ values. For each such value, the second card can be chosen in only $1$ way, as can the third card. So the number of weird configurations is $(52)(1)(1)$.

Note that the configurations counted in $(1)$ are all equally likely, at least if there is very good shuffling between successive picks. So the probability of getting a weird configuration is $$\frac{(52)(1)(1)}{(52)(52)(52)}.$$

There are other ways to attack the problem. For example, whatever card is first chosen, if there is good shuffling between picks, the probability that the next card matches the first is $\frac{1}{52}$. And the probability that the third matches the first is also $\frac{1}{52}$. So the probability all three cards match is $\left(\frac{1}{52}\right)^2$.

$4.$ Call a configuration good if all three cards are different. The number of good configurations was counted in $(2)$. The total number of configurations was counted in $(1)$. Divide.

Or else the probability that the second card is different from the first is $\frac{51}{52}$. And given that the second was different from the first, the probability that the third card is different from both the first and second is $\frac{50}{52}$. So the probability that the cards are all different is $\frac{51}{52}\cdot\frac{50}{52}$.

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