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  #1  
Old 07-03-2007, 10:56 AM
Moseley Moseley is offline
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Default Odds of at least 1 player of 6 having pocket pair

I understand that 1-[(1326-78)/1326)^6 is a formula that give you a pretty accurate approximation of the chances of at least 1 of 6 players having a pocket pair.

What if I'm at a 6 handed game, UTG with AK, and wanted to know the chances of at least 1 of the 5 remaining players having a pocket pair.

Would I?:
reduce the number of possible hands to 50*49/2 = 1225?
reduce the number of pairs to 72, since I have an ace and king, reducing the possible ways to get a pair of those to 3 each?
And use 1-[(1225-72)/1225]^5 ?

1-[.9412]^5

.9412*.9412*.9412*.9412*.9412 = .7387

1-.7387 = .2617 or 26.17% chance at least one of my 5 opponents have a pocket pair?
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  #2  
Old 07-03-2007, 01:17 PM
Cobra Cobra is offline
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Default Re: Odds of at least 1 player of 6 having pocket pair

I did not check your math but the formula you are using is correct. As you stated this is a good approximation. By using inclusion-exclusion with five terms you can get an exact answer. This would be very labor intensive and not worth the time.

Cobra
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  #3  
Old 07-03-2007, 10:39 PM
EverettKings EverettKings is offline
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Default Re: Odds of at least 1 player of 6 having pocket pair

Do you really care for an exact number? Just assuming that everyone is 16:1 to get a pair and doing 1-(16/17)^5 should get you close enough for practical purposes.
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  #4  
Old 07-04-2007, 10:15 AM
Moseley Moseley is offline
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Default Re: Odds of at least 1 player of 6 having pocket pair

[ QUOTE ]
Do you really care for an exact number? Just assuming that everyone is 16:1 to get a pair and doing 1-(16/17)^5 should get you close enough for practical purposes.

[/ QUOTE ]

Close enough?! Man, your formula gives me an answer of 26.15% compared to my answer of 26.17% !!

Your formula doesn't doesn't take in account the fact that I have an ace and a king and comes within 2/100's of my answer.
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  #5  
Old 07-04-2007, 10:16 AM
Moseley Moseley is offline
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Default Re: Odds of at least 1 player of 6 having pocket pair

[ QUOTE ]
I did not check your math but the formula you are using is correct. As you stated this is a good approximation. By using inclusion-exclusion with five terms you can get an exact answer. This would be very labor intensive and not worth the time.

Cobra

[/ QUOTE ]

Yes! I want to avoid that formula, but I need one that is easy to use. I need to be able to show someone in black and white just how strong AK is v. 5 opponents.
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  #6  
Old 07-04-2007, 10:33 AM
Moseley Moseley is offline
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Default Re: Odds of at least 1 player of 6 having pocket pair

[ QUOTE ]
Do you really care for an exact number? Just assuming that everyone is 16:1 to get a pair and doing 1-(16/17)^5 should get you close enough for practical purposes.

[/ QUOTE ]

Now let's assume that I open/raise with 88 and the five remaining players will call with 99+,AQo+,AJs+.
That's 72 different hands.
(1326-72)/72 = 17.4167:1 odds that at least one of 5 players have one of the above hands.
Using your formula:

1-(17.4167/18.4167)^5 = 24.36% chance.

My formula:

1-[(1225-72)/1225]^5 = 26.13%

Now we have a bit of a difference. My formula is easier in this situation, however, I don't want to use it, if yours is more accurate.

Anyone know which one is a more accurate approximation?
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  #7  
Old 07-04-2007, 12:54 PM
BruceZ BruceZ is offline
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Default Re: Odds of at least 1 player of 6 having pocket pair

[ QUOTE ]

Now let's assume that I open/raise with 88 and the five remaining players will call with 99+,AQo+,AJs+.
That's 72 different hands.
(1326-72)/72 = 17.4167:1 odds that at least one of 5 players have one of the above hands.

[/ QUOTE ]

That should be 16.4167:1 odds, or 1 IN 17.4167

[ QUOTE ]

Using your formula:

1-(17.4167/18.4167)^5 = 24.36% chance.

[/ QUOTE ]

This becomes 1-(16.4167/17.4167)^5 = 25.59%, which is closer to yours, but you should use your method.


[ QUOTE ]
My formula:

1-[(1225-72)/1225]^5 = 26.13%[/

Now we have a bit of a difference. My formula is easier in this situation, however, I don't want to use it, if yours is more accurate.

Anyone know which one is a more accurate approximation?

[/ QUOTE ]

Your formula should be more accurate. Also, this type of approximation is extremely accurate, usually within a few tenths of a percent.
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  #8  
Old 07-05-2007, 04:01 PM
tvta tvta is offline
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Default Re: Odds of at least 1 player of 6 having pocket pair

here is a nice little formula i learned from phil gordons site. this is assuming you hold a pocket pair.

C = number players behind you
N = number of pairs higher than yours

(C*N)/2 = probability(%) a higher pair is behind you

so if you have 1 opponent remaning and you are dealt kk:

1/2 = 0.5%, close enough to 221-1

if you have 22 with 1 opponent behind:

12/2 = 6%, or odds to be dealt any pocket pair


similar to the ak situation; i think its pretty sweet
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