in a standard deck of 52 cards and you draw 5 cards, what is the probability of getting a pair of jacks or better? any help greatly appreciated!
4/52 i think
I forgot to elaborate - you are drawing 5 cards
By that do you mean, you are dealt 5 cards before you can turn in and draw some newer cards? Anyway, good old Wikipedia states that there are a total of 2,598,960 different hands that can be made from any 5 cards chosen from 52. The number of ways that the 5 cards will consist of a "no pair" hand (every hand is worse than 2 deuces) is1,302,540. I guess the next step is to determine how many ways there are to make one pair hands that are pair of tens or lower. Well, ALL ways that a one pair hand could be made are 1,098,240. Multiplying this by 9/13 will generate how many hands would be a pair of tens or lower and that's 760,320. Now we add 1,098,240 760,320 to get 1,858,560 ways to draw 5 cards form 52 so that each hand is a pair of 10's or worse. SO to get your answer for the probability of being dealt a five card hand which is a pair of jacks or better is 1,858,560 / 2,598,960 or p = 0.715116815957152 Thank you for your attention.
EDITED TO ADD: The p = 0.715116815957152 is the probability of being dealt a pair of tens or worse. So, the probability of being dealt a pair of Jacks or better is 1 - 0.715116815957152 or 0.284883184042848
Thank you very much, wolf! If you don't mind, I would like to know why we would multiply by 9/13?
There are 13 cards that rank from deuce to Ace 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King, Ace of these thirteen, 9 are beneath the rank of Jack. So, out of ALL the ways to make a pair, 9/13 of those are worse than a pair of Jacks.
Ohh fantastic. What a tremendous help, thank you very much!
Glad to help out. (Do I get a medal?) I've never asked for a medal before, but that answer has to be the LONGEST answer I've typed for anyone. :-)
Sorry for failing to respond! I am new at this so I was unaware of this medal system - of course you get a medal! =)
Thanks for the medal (it was a LONG answer - but I was glad to help out) :-)
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