# Odds of having at least one pair in a 4-card hand

I'm trying the calculate the odds of having at least one pair(or "better" such as three or four of same cards) in a 4-card hand. I made the calculations in 2 different ways but the math doesn't add up.

odds of having at least 1 pair in a starting hand = 32.50%

I group the cards of same value, so I have 13 4-card groups. I pick one out of these groups and pick 2 cards from that group. I pick 2 cards from the remaining 48 cards.

(13*COMBIN(4,2)*COMBIN(48,2))/COMBIN(52,4) = 32.50%


odss of having exactly one pair in a hand = 30.42%

Select one group out of 13 4-card groups and select 2 card from that group. Then one card from two of remaining 12-card groups:

(13*COMBIN(4,2)*COMBIN(12,2)*4*4/COMBIN(52,4)) = 30.42%


odds of having two pairs = 1.04%

Select 2 groups out of 13 4-card groups and select 2 cards from each selected group

COMBIN(13,2)*COMBIN(4,2)*COMBIN(4,2)/COMBIN(52,4) = 1.04%


Now I'm thinking

1-pair hand odds + 2-pair hand odds = at least 1-pair hand odds


but it doesn't add up. Where is my mistake?

• You're not counting 3-of-a-kinds and 4-of-a-kinds. Some people would count triples as 3 pairs and fourples as 6 pairs. – B. Goddard Feb 16 at 14:55
• There are indeed issues with your calculation, but the root cause is that as of now the setting of the question is ambiguous. You might want to specify how triples and quads should be treated. – Lee David Chung Lin Feb 16 at 14:56
• Do you want to account for flush and straight (i.e. poker hands)? – Hagen von Eitzen Feb 16 at 14:56

I found the error. When calculating odds of having at least 1 pair in a starting hand I count 2-pairs twice. For example AA22 and 22AA are counted twice. As calculated, odds of having two pairs is 1.04%. Subtracting that gives 31.46% which verifies the other calculations (30.42% + 1.04% = 31.46%)