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I am trying to solve the following question.

Suppose you select a card at random from a standard deck, and then without putting it back, you select a second card at random from the remaining 51 cards. What is the probability that both cards have the same rank, or both have the same suit, or one is red and one is black (diamonds and hearts-red, clubs and spades-black?

For this problem, can I use the following formula for the union of these three events?

$$P(E\cup F\cup G)=P(E)+P(F)+P(G)-P(E\cap F)-P(E\cap G)-P(F\cap G)+P(E\cap F\cap G)$$

With probability that both cards have same rank being 13/52 * 12/52, probability that both have same suit being 26/52 * 25/52, and probability that one is red and one is black being 1/2 * 1/2? Am I on the right track?

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  • $\begingroup$ Your idea will work, but you have some of the details wrong. The probability that two cards have the same rank is $13/52 \cdot 12/\color{red}{51}$. (In any case, it looks like Rhys Hughes has a better way.) $\endgroup$ – awkward Mar 24 at 12:49
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Looks to me like you're overcomplicating this. Answering the following question should help you

  • Does the identity of the first card you pick change the number of cards which you can pick second that will result in a match (i.e. your criteria being met)?

The answer to this question is no. It doesn't matter, because every card has the same properties relative to the others (3 cards with same value, 12 with same suit, 25 with same colour, etc).

So assume your card is the Ace of Spades, for argument's sake.

Then a "winning" card is any of the following:

  1. Any red card (26)
  2. The Ace of Clubs (1) [note Aces of Hearts and Diamonds are red cards]
  3. Any other spade (12)

So there are $39$ winning cards out of $51$, giving a probability of $\frac{39}{51}$

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  • $\begingroup$ I think it does. For instance, you have a 13/52 chance in picking a particular suit on the first draw and 12/52 chance of the same card in the second draw since this is without replacement. Is that correct? $\endgroup$ – Robin Mar 24 at 5:22
  • $\begingroup$ Not quite. Let's say your first card is a spade. You're saying that picking a spade in the second card is less likely, which is true. However, if your first card is a diamond, the second card is less likely to be a diamond as well. $\endgroup$ – Rhys Hughes Mar 24 at 5:24
  • $\begingroup$ Ah so the "first" card doesn't matter, as only the second card has to the same suit. So after we draw the first card, the probability getting a card of the same suits is 12/51 and the overall probability of getting 2 cards of the same suit is 12/51? $\endgroup$ – Robin Mar 24 at 5:31
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    $\begingroup$ Exactly that. Then just add on the probabilities for the other conditions. $\endgroup$ – Rhys Hughes Mar 24 at 5:34
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A deck of card has 52 cards.

Total number of ways of selecting 2 cards from 52 cards=52C2=26*51

Total number of ways of selecting same rank cards=13*4C2=13*2*3

This is because number of cards of given rank(say Ace=4) and there are 13 unique ranks(Ace,King,...,2).

Total number of ways of selecting same suits=4*13C2=4*13*6

There are 13 cards in each of the 4 suits.

Total number of ways of selecting 1 red and 1 black card=26C1*26C1=26*26

Because 1 card is to be selected from 26 cards and there are two ways(red then black or black then red)

When two cards are of same rank and should be black and red, total numbers =13(ace,king, ...)*2(for red color ace)*2(for black color ace)

When two cards are of same suit and should be black and red, total numbers=0(If colours are different then suits will be different)

When two cards are of same rank and same suit, total numbers=0(Within a suit,no cards have same rank)

Also there can't be two cards such that all 3 conditions are satisfied.

Probability

$P=\frac{13*2*3+4*13*6+26*26-13*2*2}{26*51}$

$=\frac{13*2(3+12+26-2)}{26*51}$

$=\frac{39}{51}$

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