Posted in Poker February 23rd, 2010
How many different 5 card poker hands would contain only cards of a single suit?
I know there are 2,598,960 different 5 card poker hands but I don’t know how many would only contain a single suit.
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4 Comments on this post
Posted by a²+b²=c² February 23rd, 2010 at 9:53 pm
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.
Posted by kro February 23rd, 2010 at 10:06 pm
5,108 it’s called a flush.
36 straight flush
4 royal flush
Posted by Bob K February 23rd, 2010 at 10:31 pm
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
Posted by icemetalpunk February 23rd, 2010 at 11:04 pm
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
