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).



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.


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