If it is assumed that all (52 2) poker hands are equally likely, what is the pro

Question

If it is assumed that all (52 2) poker hands are equally likely, what is the probability of being dealt: (a) a flush? (A hand is said to be a flush if all 5 cards are of the same suit.) (b) one pair? (This occurs when the cards have denominations a, n, b, c, d where a,b,c,d are all distinct.) (c) two pairs? (This occurs when the cards have denominations a, b, b, b, c where a, b, c are all distinct.) (d) three of a kind? (This occurs when the cards have denominations a. a, a, b. c where a. b, c are all distinct .) (e) four of a kind? (This occurs when the cards have denominations a, a, u, a, b.)

Explanation / Answer

Solution:-

(a) a flush?

A flush consists of cards of all the same suit. First, of the 4 suits in a standard deck, we choose 1 of them for use in the flush. In each suit, there 13 cards. Of these 13 cards, we must choose 5 to make a full poker hand.

Therefore, the probability of being dealt a flush is

P(flush) = 4C1 * 13C5 / 52C5 = 0.002

(b) one pair?

First of all, from the 13 denominations in a standard deck of playing cards, we must select 1 denomination for the pair. Once this denomination is selected, we have 4 suits of this denomination to choose from.

A pair consists of 2 cards, so from these 4 suits, we must choose two of them to be in the pair.

So our answer will have the following in the answer 13C1 * 4C2

Now consider the other 3 cards in the hand. From the 12 remaining denominations, we must choose 3 denominations for the other cards. Then, for each card separately, we have four possible suits to choose from for each of the three denominations and we must choose one for each card.

12C3 * 4C1 * 4C1 * 4C1

So the probability of being dealt a pair is

13C1 * 4C2 * 12C3 * 4C1 * 4C1 * 4C1 * 52C5 0.4226

(c) Two pairs?

First, from the 13 denominations, we must choose 2 denominations, one for each pair. Once of the denomination is chosen, there are 4 possible suits to choose from for this denomination, and we must choose 2 of them to form a pair. Again, for the second pair, there are 4 possible suits to choose from for its denomination, and we must choose 2 of them to form the second pair.

So the following will be in our answer. 13C2 * 4C2 * 4C2

Now we must draw the fifth card of a different denomination. Of the 11 remaining denominations, we must choose 1. Of the 4 suits available for that denomination, we choose 1 for the final card. Another way we can look at this is to consider that we have already drawn 2 denominations, both of which cannot be drawn again. That is, of the 52 cards originally in the deck, 8 of them are off limits since their denominations have already been dealt into a pair. Of the 44 remaining cards, we must draw 1.

11C1 * 4C1 = 44C1

So the probability of being deal two pairs is 13C2 * 4C2 * 4C2 * 11C1 * 4C1 * 52C5 0.0475

(d) three of a kind?

A three of a kind is when we have three of one denomination and two of some other two denominations. Of the 13 denominations, we choose 1 to use in the three of a kind.

There are 4 suits of a denomination, and we must choose 3 of these suits to define a three of a kind.

13C1 * 4C3

Of the 12 remaining denominations, we must choose 2 of them for the other 2 cards. Of the 4 suits of each denomination, we must choose one for each individual card.

So the following will be in the answer 12C2 * 4C1 * 4C1

Then the probability of a three of a kind is 13C1 * 4C3 * 12C2 * 4C1 * 4C1 * 52C5 = 0.0211

(e) a four of a kind?

In a four of a kind, we have all four suits present of a particular denomination with one card of a different denomination. First, we draw one denomination of the 13 possible.

Then, we choose all 4 of the possible suits for the denomination: 13C1 * 4C4

For the final card, we choose a denomination from the remaining 12 denominations, and one suit out of the possible 4 for that denomination. 12C1 * 4C1

Thus, the probability of drawing a four of a kind is given by

13C1 * 4C4 * 12C1 * 4C1 * 52C5 = 0.0002

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