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

Question

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?

Answer 1

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%)