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  #11  
Old 07-11-2005, 11:39 AM
Erik Blazynski Erik Blazynski is offline
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Default Re: Another What are the Odds?

4%??? think about that, 4 out of every hundred hands? Don't you think that you would have seen this before? unless you've only played like 25 hands in your life...
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  #12  
Old 07-11-2005, 01:49 PM
astarck astarck is offline
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Default Re: Another What are the Odds?

[ QUOTE ]
[ QUOTE ]
Are the numbers you gave already assuming Player 2 has a pocket pair as well?

[/ QUOTE ]

Yes.

[/ QUOTE ]

After thinking about this further, does it really matter that each player is holding a pocket pair? Sure it changes the odds a very small amount (since certain 4-of-a-kinds are not possible for the board) but I think the probabilities would be almost the same since both players are playing the board no matter what they hold.
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  #13  
Old 07-11-2005, 02:46 PM
AaronBrown AaronBrown is offline
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Default Re: Another What are the Odds?

On the other hand. . .

Once the flop was dealt, there were 903 possible combinations of turn and river cards. Only 5-5 and A-A lost for you, and each could only happen one way. So you had 2/903 or about 0.2% chance of losing. In both cases your opponents tie.

This will happen to you again. It's not likely to be quads with two opponents splitting the pot, but you'll see people pull out the only possible pair of cards that can beat you many times.
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  #14  
Old 07-11-2005, 04:37 PM
BruceZ BruceZ is offline
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Default Re: Another What are the Odds?

[ QUOTE ]
After thinking about this further, does it really matter that each player is holding a pocket pair? Sure it changes the odds a very small amount (since certain 4-of-a-kinds are not possible for the board) but I think the probabilities would be almost the same since both players are playing the board no matter what they hold.

[/ QUOTE ]

I ran the numbers again for both players holding no-pair hands, with no ranks in common. It now depends on the highest hole card, rather than the highest pair. The difference is that there are now 9 ranks of quads instead of 11, and each quad gets 1 more possible kicker, since there are 3 of the highest hole card rank.

New formula:

{ [(14-N)*4 + 3]*(N-5) + [(13-N)*4 + 3]*(14-N) } / C(48,5)

That is, for each of the N-5 possible quads less than N, there are (14-N)*4 kickers greater than N, plus 3 of rank N. Then for each of the 14-N quads greater than N, there are (13-N)*4 kickers greater than N, plus 3 of rank N.

<font class="small">Code:</font><hr /><pre>
high hole P(split) odds-to-1

5 0.018% 5435
6 0.017% 6050
7 0.015% 6821
8 0.013% 7818
9 0.011% 9156
10 0.0091% 11046
11 0.0072% 13920
12 0.0053% 18816
13 0.0034% 29021
14 0.0016% 63418
</pre><hr />

Here are the pocket pair data again for comparison.

Original formula:

{ [(14-N)*4 + 2]*(N-3) + [(13-N)*4 + 2]*(14-N) } / C(48,5)

<font class="small">Code:</font><hr /><pre>
high pair P(split) odds-to-1

33 0.027% 3705
44 0.025% 4057
55 0.022% 4481
66 0.020% 5006
77 0.018% 5669
88 0.015% 6535
99 0.013% 7712
TT 0.011% 9407
JJ 0.0083% 12057
QQ 0.0060% 16786
KK 0.0036% 27617
AA 0.0013% 77831
</pre><hr />
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  #15  
Old 07-11-2005, 06:00 PM
astarck astarck is offline
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Default Re: Another What are the Odds?

Very interesting.
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