Odds of flopping a pair.

[CT11]

I've been going through some odds calculations because I wanted to make a chart like the one in "Pot Odds Made Easy" (http://www.cardplayer.com/poker_magazine/archives/?a_id=13913) but with a few more pieces of information and some more useful information. I fired up excel and started at it. Most of my numbers agree with Krieger's table but a few don’t. Primarily my issue is with his percentage calculation for AK flops at least a pair of 32.4%. When calculating two cards of different rank gets a pair or better I get 21.9%.

My calculation is this:

P[of one card on flop]=(N/50)*((1-(N-1))/49)*(1-(N-1)/48)+(1-N/50)*(N/49)*(1-(N-1)/48)+(1-N/50)*(1-N/49)*(N/48)

P[of two cards on flop]= (N/50)*((N-1)/49)*(1-(N-2)/48)+(N/50)*(1-(N-1)/49)*((N-1)/48)+(1-N/50)*(N/49)*((N-1)/48)

P[of all three on flop]= (N/50)*((N-1)/49)*((N-2)/48)

The sum of these should give me the odds of getting at least a pair on the flop.

Because there are 3 more A and 3 more K, N=3+3=6

Any one see a problem with this?

Thanks

~CT11

[BruceZ]

[ QUOTE ]
Primarily my issue is with his percentage calculation for AK flops at least a pair of 32.4%.

[/ QUOTE ]

He's correct. 1 - (44/50 * 43/49 * 42/48) = 32.4%.

[ QUOTE ]
When calculating two cards of different rank gets a pair or better I get 21.9%.

My calculation is this:

P[of one card on flop]=(N/50)*<font color="red">((1-(N-1))/49)</font>*(1-(N-1)/48)+(1-N/50)*(N/49)*(1-(N-1)/48)+(1-N/50)*(1-N/49)*(N/48)

P[of two cards on flop]= (N/50)*((N-1)/49)*(1-(N-2)/48)+(N/50)*(1-(N-1)/49)*((N-1)/48)+(1-N/50)*(N/49)*((N-1)/48)

P[of all three on flop]= (N/50)*((N-1)/49)*((N-2)/48)

The sum of these should give me the odds of getting at least a pair on the flop.

Because there are 3 more A and 3 more K, N=3+3=6

Any one see a problem with this?

[/ QUOTE ]

Yes, the term in red should be (1-(N-1)/49). That accounts for the difference.

[CT11]

Thakns so much. It's been driving me crazy. I updated my spreadsheet and I got what was expected.

~CT11