poker and probability
Tuesday, August 24th, 2010
Probability problem involving a standard set of poker cards?
Do you know how to set one of these two problems? I thank you. (Jack, Queen, King are the same Aces equal to 10 and 11) What is the probability of receiving two cards totaling 21? What is the probability that you are deealt 5 cards that are all the same color?
For the first: the case of two cards totaling 21 years is required: a.) an ace and a king b.) an ace and a queen c.) Ace one one Jack D.) an ace and 10 All these are mutually exclusive events, which will be useful later. The probability of an ace (a) = 4 / 52 probability of a king (B) = 4 / 51 The probability of a king and as = a * b = 4 / 52 * 4 / 51 = 0.00603 Do the same with the other scenarios, you get: The probability of a queen and an ace = 0.00603 The probability to have a cat and as = 0.00603 The probability of obtaining 10 and as = 0.00603 The probability of getting two 21 value cards = the probability that a of the above situations occurs when mutually exclusive, their probabilities are added. The probability of obtaining a total card value of 21 = 0.00603 + 0.00603 + 0.00603 + 0.00603 = 0.0181 for the second: There are 13 of each color on the cover. The probability of obtaining a special suit = 13/52 = 0.25 The probability of obtaining the same color = 12/51 = 0.235 The probability of obtaining the new suit = 11/50 = 0.22, the probability of making it the fourth time = 10/49 = 0.204, the probability of obtaining the fifth time = 9 / 48 = 0.1875 The total probability of getting five cards of the same color = 0.25 * 0.22 * 0.235 * 0.204 * 0.1875 = 0.000494
Re: Poker flush hand probability calculations Solution video
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