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We have a standard deck of 52 cards and we draw four cards. What's the probability that among them will be two of the same rank, where one is black and the second is red.

My idea:

1) Four cards of the same rank

If we have four cards of the same rank, we have two black cards and two red cards. Therefore,we can choose rank in 13 ways.

2)Three cards of the same rank

If we have three cards of the same rank, we have two black cards (two red cards) and one red card (one black card). $ 2 \cdot { 4 \choose 3} \cdot 13 \cdot 48$

($48$ -number of ways to choose the fourth card)

3) Two cards of the same rank $ {4 \choose 2} \cdot 13 \cdot 2 \cdot 1 \cdot 48 \cdot 44$

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  • $\begingroup$ Your approach seems flawed. In case $1a$ you speak of three red and one black card of a single rank, which is not possible. Also you appear to omit the case of two cards of one rank (and opposite colors) AND two cards of a different rank (also of opposite colors). $\endgroup$
    – lulu
    Commented Nov 7, 2019 at 18:15
  • $\begingroup$ Also: there are clearly only $13$ ways to get all four of a single rank (one hand for each rank). Not sure what you are computing. $\endgroup$
    – lulu
    Commented Nov 7, 2019 at 18:20
  • $\begingroup$ @lulu You are absolutely right. I wrote nonsense. $\endgroup$
    – pawelK
    Commented Nov 7, 2019 at 18:35
  • $\begingroup$ But your overall method, going case by case, seems sensible. So...amend your post to correct the bad algebra. $\endgroup$
    – lulu
    Commented Nov 7, 2019 at 18:47
  • $\begingroup$ Worth remarking: for the case when you have exactly three of one rank, note that you MUST have two of opposite colors so you can ignore that constraint (in this particular case). $\endgroup$
    – lulu
    Commented Nov 7, 2019 at 18:48

1 Answer 1

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I think it’s easier to count the number of ways you can avoid having two cards of the same rank and different color. The tree diagram below shows the ways with example choices. I'll describe (less clearly) in words below the picture, too.

enter image description here

The first card can be any card (52 choices). The second card can either be a) the same rank as the first card (1 choice to avoid other color) or b) a different rank (48 choices). These are counted separately because the number of choices of third card differ. After a), the third card must be of a different rank than the one seen so far (48 choices), and the fourth card can then be either a new rank (44) or the same-color card of the rank that is unique in the first three (1), so (45 choices). After b), the third card can be either c) a third rank (44 choices, followed by a fourth card that is either a fourth rank or one of the three color matches of ranks already seen, so 43 choices) or d) one of the color-matches (2 choices, followed by a fourth card that is either a third rank or the other color match, so 45 choices).

So there are $52\cdot1\cdot48\cdot45+52\cdot48\cdot2\cdot45+52\cdot48\cdot44\cdot43$ ways to avoid the configuration you are trying to find the probability of. This should get you through to the end.

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