probability homework question

[Gene Fish]

anyone wanna help with some homework?? [img]/images/graemlins/smile.gif[/img]

What is the probability of getting 5 cards dealt (normal deck) and three of the cards are the same denomination (trips), and the other two cards are one above and one below...example: 8 8 8 9 7

thanks in advance!!

[BruceZ]

[ QUOTE ]
anyone wanna help with some homework?? [img]/images/graemlins/smile.gif[/img]

What is the probability of getting 5 cards dealt (normal deck) and three of the cards are the same denomination (trips), and the other two cards are one above and one below...example: 8 8 8 9 7

thanks in advance!!

[/ QUOTE ]

Count the number of ways to make such a hand, without regard to the order of the cards, and divide by the total number of possible hands C(52,5) = 2,598,960. Consider the number of possible denominations for the set, the number of ways to make a set for each denomination, and the number of ways to get the other 2 cards. You need to know if the ace can be either a high or low card, and whether it can be both in the same hand. If it can be either high or low, but not both in the same hand, then AAA2K is not allowed, but 222A3 and KKKQA are both allowed, and the final answer comes to about 0.03%.

[pococurante]

There are 2,598,960 possible 5-card combinations. 59,280 of those include 3 of a kind. We can toss out the 4560 combinations of trip aces, and the 4560 trip deuces, leaving you with 50160 live hands so far.

Given trips, there is a 8/49 chance that your next card will either be 1 below or one above. Then your next card has a 4/48 chance of completing the hand that you want.

Results: There are 682.5 combinations out of 2,598,960 that will win. Since there are no half-cards in poker, the answer is wrong, probably because the card combination numbers I looked up were off... this number should be very close though.

[pococurante]

I tried counting the possible hands, but it's not working.

Suppose you start with 3 of a kind in your hand. For each rank, there are 4 possibilities (trip aces without Ah, trip aces without As, etc.). So there are 52 ways to hold trips in your hand.

Now add a fourth card to your hand. There are 49 possibilities, so there are now 2548 ways to have them.

Now add a fifth card. 48 cards are there to choose from, meaning that you have 122,304 ways to have 3 of a kind in your hand. (You might also have quads or a full house right now... anything that includes 3 of a kind is possible.)

Can someone tell me why poker odds sites say there are 54912 trips, 624 quads, and 3744 full houses... for a total of 59280 ways to have 3 of a kind in a 5 card hand?

No point in doing the math if your starting numbers are all wrong.

[Gene Fish]

http://www.math.sfu.ca/~alspach/comp18/

this site has the same numbers as you do as well as explainations.

[Gene Fish]

Bruce, approx how did you come up with that 0.03%?

[pococurante]

A lot of sites have those same numbers. So why am I getting 122,000 when I do it myself?

[pococurante]

And getting back to the original question. I assume that 59280 is the correct number.

So let's say I give you a random 5 card hand that is one of those 59280 possible combinations. I then ask you to set the three of a kind off to the side. You now have 2 random cards out of the 49 possible cards you could have had.

Flip over random card #1. What are the odds that it is one rank above or one rank below your 3 of a kind... it should be 8/49, correct?

But 8/49 * 59280 equals 966.5. How is it possible for there to be 966.5 poker hands that qualify? There is no such thing as half of a hand. This makes no sense.

[Gene Fish]

oh wait...but if your trips are AAA then you can't have one above...maybe that's the half????

[pococurante]

I finally figured it out... my method is flawed. The way I was doing it, KKK 7c 8c counts as a different hand from KKK 8c 7c, which is obviously not true.