Comments on: How Many Different 5 Card Poker Hands Would Contain Only Cards Of A Single Suit?

By: kro

kro — Sat, 21 Nov 2009 02:02:10 +0000

5,108 it’s called a flush.
36 straight flush
4 royal flush

By: Bob K

Bob K — Fri, 20 Nov 2009 22:14:47 +0000

First, you choose which suit you want to have 5 cards of, so that would be 4C1, next, there are 13 cards in a suit and you want to choose 5 of those, or 13C5. And then multiply the two choices as they are independent
4C1*13C5
4!/3!1! * 13!/5!8!
4 * 1287
5148

By: icemetal

icemetal — Fri, 20 Nov 2009 20:55:08 +0000

Think of it this way: There are 13 cards of a single suit. So find the number of combinations of 13 cards, no repetition. Then realize there are 4 suits, so multiply that number by 4.
The formula for combinations is n!/(r!(n-r)!) where n=number of possible cards (13) and r=number of chosen cards (5). Substituting gives us:
13!/(5!(13-5)!) = 13!/(120*8!)
If we work this out, we get 1287. Now we remember there are 4 suits, and multiply this by 4 to get:
5148 combinations.
-IMP

By: a²+b²=c²

a²+b²=c² — Fri, 20 Nov 2009 20:43:58 +0000

There are 4 suits, and for each suit, 13 cards. You’re choosing 5 of them, so 4(13C5) = 5148 hands.
You can also solve this by looking up the frequency of the types of hands in poker. What you want is the number of flushes plus straight flushes plus royal flushes.