#1
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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? |
#2
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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 |
#3
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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
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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. |
#5
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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. |
#6
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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? |
#7
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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. |
#8
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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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