Another What are the Odds?

[astarck]

Given that two players each hold a pocket pair what are the odds that they split the pot in the way described below?

Example 1:
Player 1 - AA
Player 2 - QQ
Board - KKKKA

Example 2:
Player 1 - JJ
Player 2 - 44
Board - 7777Q

Example 3:
Player 1 - 88
Player 2 - 77
Board - TTTT9

As you can see there are MANY combinations for this to work. I'm not wondering the odds in the three specific examples above, they were mearly examples. I am wondering the odds that given two players each hold a different pocket pair they split the pot with 4-of-a-kind.

I think I have described what I was thinking well enough to get my question across.

[kitaristi0]

The odds are slim-to-none /images/graemlins/laugh.gif

[astarck]

[ QUOTE ]
The odds are slim-to-none /images/graemlins/laugh.gif

[/ QUOTE ]

Yeah, I was sort of figuring that much.

[Douche Bag]

[ QUOTE ]
The odds are slim-to-none /images/graemlins/laugh.gif

[/ QUOTE ]

What a douche bag.... /images/graemlins/mad.gif

[Poker60181]

well, youd obviously need 4 of a kind to flop. so. my guess is it would be 4/50 + 2/47 + 1/46... which would be ~ 4%.

4% sounds pretty accurate to me for a situation like this to come up. i've never seen it.

[MikeL05]

[ QUOTE ]
well, youd obviously need 4 of a kind to flop. so. my guess is it would be 4/50 + 2/47 + 1/46... which would be ~ 4%.

4% sounds pretty accurate to me for a situation like this to come up. i've never seen it.

[/ QUOTE ]

Seriously, you're not funny.

[BruceZ]

[ QUOTE ]
I am wondering the odds that given two players each hold a different pocket pair they split the pot with 4-of-a-kind.

[/ QUOTE ]

It depends on the rank of the higher pocket pair. The higher it is, the less likely the pot will be split, since splitting requires that the kicker be greater than or equal to the higher pocket pair.

If the higher pair is a pair of N's (J = 11, Q = 12, K = 13, A = 14) then the probability of a quad split is:

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

That is, for each of the N-3 possible quads less than N, there are (14-N)*4 kickers greater than N, plus 2 of rank N. Then for each of the 14-N quads greater than N, there are (13-N)*4 kickers greater than N, plus 2 of rank N.

<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 />

[astarck]

Finally an answer I can believe.

Thanks.

Edit - Are the numbers you gave already assuming Player 2 has a pocket pair as well?

[BruceZ]

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

[/ QUOTE ]

Yes.

[Bco1/75]

Thanks, this re-assures my assumption that this will never happen to me again....

Holding JJ....
Flop J-5-5...detecting weakness I check and Both villans check
Turn 5...Bet, Call, Call....
Flop 5 Check, bet, bet I fold they show weak 'A' and split the pot.

Thanks,
Bco

[Erik Blazynski]

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...

[astarck]

[ 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.

[AaronBrown]

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.

[BruceZ]

[ 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 />

[astarck]

Very interesting.