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poker hand is defined as drawing 5 cards at random without replacement from a deck of 52 playing cards. find the probability of each of the following poker hands: (a) four of a kind (four cards of equal face value and one card of a different value). (b) full house (one pair and one triple of cards with equal face value). (c) three of a kind (three equal face values plus two cards of different values). (d) two pairs (two pairs of equal face value plus one card of a different value). (e) one pair (one pair of equal face value plus three cards of different values)

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Final answer:

In poker, the probability of different hands can be determined using combinations and permutations. The probability of getting a four of a kind, full house, three of a kind, two pairs, or one pair can be calculated using specific formulas.

Step-by-step explanation:

To find the probability of different poker hands, we need to understand the concept of combinations and permutations.

(a) Four of a kind:

To get four of a kind, we have 13 choices for the value of the four cards, and for each value, we choose 4 cards from the 4 suits. The remaining card can be any of the remaining 48 cards. So the probability is (13 * 4 * 48)/(52 * 51 * 50 * 49).

(b) Full house:

For a full house, we have 13 choices for the triple and for each triple, we choose 3 cards from the 4 suits. We then have 12 choices for the pair and for each pair, we choose 2 cards from the 4 suits. So the probability is (13 * 4 * 12 * 6)/(52 * 51 * 50 * 49 * 48).

(c) Three of a kind:

The calculation is similar to (a), but with only 3 cards in the group. So the probability is (13 * 4 * 48 * 44)/(52 * 51 * 50 * 49).

(d) Two pairs:

To get two pairs, we have 13 choices for the first pair and for each pair, we choose 2 cards from the 4 suits. We then have 12 choices for the second pair and for each pair, we choose 2 cards from the remaining 3 suits. The remaining card can be any of the 44 remaining cards. So the probability is (13 * 4 * 12 * 4 * 44)/(52 * 51 * 50 * 49 * 48).

(e) One pair:

Similar to (d), we have 13 choices for the pair and for each pair, we choose 2 cards from the 4 suits. The remaining 3 cards can be any of the remaining 48 cards. So the probability is (13 * 4 * 6 * (48 * 47 * 46))/(52 * 51 * 50 * 49 * 48).

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