Poker Hand OddsDate: 08/07/97 at 14:31:42 From: Anonymous Subject: Poker Hand Odds I am trying to calculate the odds of getting a hand of exactly one pair (any pair) in five card stud poker, using one 52-card deck. I have calculated the total number of possible hands as 2,598,960 (52 C 5), but I get stuck trying to figure out the possibility of exact hands. Help! Date: 08/07/97 at 15:53:29 From: Doctor Anthony Subject: Re: Poker Hand Odds One pair [two of one face value, and 3 cards of different face values, not matching the pair] 13 x 4C2 x 12C3 x 4^3 Prob(1 pair) = ---------------------- = 0.422569 52C5 The denominator is the number of ways of selecting 5 cards from 52. We will explain the numerator, term by term, starting at the lefthand end. 13 is the number of ways of choosing the face value of the pair, e.g. two 10's or two 4's or whatever. There are 13 face values 2, 3, ..., Q, K, A. 4C2 is the number of ways of choosing the two 10's (say) from the four 10's in the pack. 12C3 is the number of ways of choosing 3 different face values from the 12 still available - we have used up one face value for the pair. For argument's sake, suppose the 3 remaining face values are 3, 7, K The 4^3 term arises in the following way. Suppose the face values are to be 3, 7, K as suggested above. We have 4 ways of choosing the 3, since there are four 3's in the pack. Similarly, there are 4 ways of choosing the 7 and 4 ways of choosing the K, giving 4 x 4 x 4 = 4^3 ways of selecting the other 3 cards with a given set of face values. Multiplying all these factors together gives the total number of ways that we could choose exactly one pair. And this divided by 52C5 gives the required probability. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 09/03/97 at 15:47:50 From: Doctor Naomi Subject: Re: Poker Hand Odds Hi there, Before talking about the cards, I would like to make a clarification. You asked for "odds" in your question, but do you mean "probability?" The probability of an event occurring is expressed as a fraction, e.g. 2/3, which means that on average an event will happen two times out of every three trials (so there is a 67 percent (rounded) chance of this event happening in any one trial). Odds tell you how many times an event will happen and how many times it won't, e.g. odds of 2 to 3 mean that on average, an event will happen 2 times for every 3 times that it doesn't happen (so there is a 40 percent chance of this event happening in any one trial because 2/(2+3) = 0.4). I assume that you are asking for the probability of getting exactly one pair in a hand of five cards randomly dealt from a 52-card deck. The probability of getting exactly one pair will be the number of hands with exactly one pair divided by the total number of 5-card hands. You have already found the denominator, (52 C 5). Now we must find the numerator, which, as you said, is a lot more tricky. The total number of 5-card hands with exactly one pair can be thought of as the following: (number of different face values that the pair might have) x (number of ways to pick two cards of the same face value from the four available in the deck) x (number of ways three more cards can be chosen such that none of them have the same face value and none share the same face value as the one chosen for the pair). Now we have to calculate these quantities. First you must find the number of different ways of choosing one face value for the pair (e.g., there could be two 10's or two queens, etc.). There are 13 face values (2,3,...,Q,K,A) to choose from, so we have (13 C 1) different face values that the pair might have. Now that we have chosen a particular face value (let's say we chose Jacks) we need to calculate the number of ways to choose two Jacks from the four Jacks available in the deck, yielding (4 C 2). Now comes the hard part: calculating the number of ways three more cards can be chosen such that none of them share the same face value as the one chosen for the pair. One way to get this number is by first choosing 3 different face values (one for each card), none of them matching the face value (Jacks) already picked for the pair. With the Jacks out of the running, there are 12 choices of face values left from which 3 can be chosen, giving us (12 C 3). Let's say that we chose Ace, Queen and King as the other three face values. We now need to multiply (12 C 3) by (the number of ways to choose 1 Ace from the 4 Aces in the deck) x (the number of ways to choose 1 Queen from the 4 Queens in the deck) x (the number of ways to choose 1 King from the 4 Kings in the deck), yielding (12 C 3)(4 C1) (4 C 1)(4 C 1). Putting this all back together, we have no. of 5-card hands with exactly one pair Prob(1 pair) = ----------------------------------------- total no. of 5-card hands (13 C 1)x(4 C 2)x(12 C 3)x(4 C 1)^3 = --------------------------------------- (52 C 5) I hope this helps you out! -Doctors Bill, Anthony, and Naomi, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/