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CASHIZ
11-16-2004, 11:14 PM
WHAT ARE THE ODDS THAT IF I HOLD A 10 IN THE CUTOFF, THAT A PLAYER IN ETHIER THE BLINDS OR THE BUTTON WILL HOLD AN ACE WITH A BETTER KICKER(J,Q,K). THANKS

gaming_mouse
11-17-2004, 05:19 AM
Assuming only that you hold a T, there are 51 cards left in the deck. Thus your oppo has nCr(51,2) = 1275 possible holdings.

Of those holdings, how many contain an AJ AQ AK or AA?

AA - 6 hands
AK - 16 hands
AQ - 16 hands
AJ - 16 hands
--------------
total: 54 hands

This assumes you do not hold an A,K,Q, or J yourself. If you do, the calculation will be slightly different.

So the chance that Button has one of these hands is 54/1275.

We now calculate the chance that none of the players has one of these hands, and subtract it from 1 to get the final answer:

1 - (1221/1275)*(1220/1274)*(1219/1273)=.1221

So about a 12% chance that at least one of them has AJ, AQ, AK or AA.

gm

gaming_mouse
11-17-2004, 05:29 AM
I just realized you meant that you had AT, not you had "a 10" -- caps lock is confusing.

To adjust for that:

Oppo now had nCr(50,2) = 1225 possible holdings

Oppo cards you're worried about (no longer including AA):

AK - 12 cards
AQ - 12 cards
AJ - 12 cards
-------------
total: 36 cards

1 - (1189/1225)*(1188/1224)*(1187/1223)=.0859

So final answer is closer to 8%

ALSO NOTE: This is a VERY CLOSE approximation (within a percent for sure), but it is NOT EXACT. The reason it's not exact is that I'm assuming that the three players not having the cards in question are independent events, when in fact they are LOOSELY dependent.

gm

BruceZ
11-17-2004, 07:59 AM
[ QUOTE ]
I just realized you meant that you had AT, not you had "a 10" -- caps lock is confusing.

To adjust for that:

Oppo now had nCr(50,2) = 1225 possible holdings

Oppo cards you're worried about (no longer including AA):

AK - 12 cards
AQ - 12 cards
AJ - 12 cards
-------------
total: 36 cards

1 - (1189/1225)*(1188/1224)*(1187/1223)=.0859

So final answer is closer to 8%

ALSO NOTE: This is a VERY CLOSE approximation (within a percent for sure), but it is NOT EXACT. The reason it's not exact is that I'm assuming that the three players not having the cards in question are independent events, when in fact they are LOOSELY dependent.

gm

[/ QUOTE ]

Even assuming independence, the denominator would not just change by 1 for each player. The first one would be C(50,2) = 1225, but the second should be C(48,2) = 1128, and the third should be C(46,2) = 1035. The numerator would change by the same amount, but only if we assume that an A,K,Q, or J was not dealt. It is not possible to get an exact answer by this method easily, but we can get an exact answer by the inclusion-exclusion principle (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&amp;Board=probability&amp;Number=417 383) as:

3*36/1225 -
C(3,2)*36*30 / C(50,2) / C(48,2) +
C(3,3)*36*(6*25+24*24) / C(50,2) / C(48,2) / C(46,2)

=~ 8.58%.

BruceZ
11-17-2004, 08:37 AM
[ QUOTE ]
3*36/1225 -
C(3,2)*36*30 / C(50,2) / C(48,2) +
C(3,3)*36*(6*25+24*24) / C(50,2) / C(48,2) / C(46,2)

=~ 8.58%.

[/ QUOTE ]

Slight correction to 3rd line:

3*36/1225 -
C(3,2)*36*30 / C(50,2) / C(48,2) +
C(3,3)*36*(6*26+24*24) / C(50,2) / C(48,2) / C(46,2)

=~ 8.58%.

gaming_mouse
11-17-2004, 10:28 AM
Even assuming independence, the denominator would not just change by 1 for each player. The first one would be C(50,2) = 1225, but the second should be C(48,2) = 1128,

Right you are. Silly mistake, though it didn't seem to affect the answer much. Also, nice use of the OR forumula to get the exact answer.

gm

CASHIZ
11-17-2004, 01:16 PM
so what u are saing is that there's about a 8%-9% chance that ethier the button, small blind or big blind have my AT beat? How about mid postion just behind the cutoff? How much do the percentages change with each additional player?
And thanks, u guys are a big help.

gaming_mouse
11-17-2004, 02:10 PM
[ QUOTE ]
so what u are saing is that there's about a 8%-9% chance that ethier the button, small blind or big blind have my AT beat? How about mid postion just behind the cutoff? How much do the percentages change with each additional player?
And thanks, u guys are a big help.

[/ QUOTE ]

Yes, that is correct.

Also, not the chance that you're beat, it's the chance that you're dominated. You still have three outs.

To get an exact answer (using the method Bruce describes) gets more difficult with each player you add. However, you can get a good approximation (within a few percent I think) using an even cruder version of the formula I suggest:

1 - (1221/1275)^n

where n is the number of opponents you interested in. This will approximate the chance that at least one has you dominated. The "^" means "to the power of".

HTH,
gm