1
$\begingroup$

4 players participate in a game. 3 of them draw one card from a deck of 52 cards while the last player is allowed to draw two cards from the same deck (i.e. total of 5 cards drawn). What is the probability of that last player drawing at least one of the top 2 ranking cards among the 5 drawn cards.

Cards are not replaced after being drawn and suits are also ranked (i.e. no two cards are ranked the same).

Thanks

$\endgroup$
3
$\begingroup$

The fact that there are $52$ cards is irrelevant. Imagine that a referee picks the $5$ cards, and then chooses $2$ of these to give to the fourth player.

There are $\binom{5}{2}$ equally likely ways the referee can choose these two cards. And there are $\binom{3}{2}$ ways the referee can choose $2$ cards from the bottom $3$. So the probability the player gets neither of the two top cards drawn is $\frac{\binom{3}{2}}{\binom{5}{2}}$.

This is $\frac{3}{10}$. So the probability the fourth player gets at least one of the top two cards is $\frac{7}{10}$.

Another way: The referee has the $5$ cards, and to keep things exciting hands two of them to the fourth player, one at a time. The probability the first card given is a "low" one is $\frac{3}{5}$. Given that the first given card was low, the probability the second card given is low is $\frac{2}{4}$.

So the probability they are both low is $\frac{3}{5}\cdot\frac{2}{4}=\frac{3}{10}$. Thus the probability the fourth player gets at least one of the two high cards is $\frac{7}{10}$.

Remark: One important thing to remember is that all sequences of $5$ cards are equally likely. Thus whether players 1 and 2 get their cards first or last is irrelevant for the calculation of the probabilities.

An important example is the following problem. Players 1 and 2 draw a card one at a time, without replacement. (a) What is the probability Player 1 draws a Queen? Obviously $\frac{4}{52}$. (b) What is the probability Player 2 draws a Queen? It should be equally obvious that the answer is $\frac{4}{52}$. But it takes a while for this fact to become obvious.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.