October 14th, 2019 at 7:37:31 PM
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What are the odds of being dealt two pair, with aces? I believe the frequency of any two pair is 1/21 on the deal...so is 1/13 of those hands aces and another pair, making the frequency 1/273?

October 14th, 2019 at 9:35:44 PM
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1/21 is a close estimate but not exact. You are off by a factor of 2. 2/13 of the two pair combinations contain a pair of aces. If your first pair is any one of the 13 ranks, the second can be any of the other 12, making 156 in all if the ordering is important. Then 12 of these 156 will have the first pair being aces and 12 will have the 2nd pair being aces. 24/156 reduces to 2/13.Quote:KatchWhat are the odds of being dealt two pair, with aces? I believe the frequency of any two pair is 1/21 on the deal...so is 1/13 of those hands aces and another pair, making the frequency 1/273?

If you always name your two pair by the highest ranking pair, then you have a different way of counting:

12 will be aces up;

11 will be kings up; etc. until

1 will be 3s and 2s.

Counting these ordered sets up will give you 78 with 12 of them being aces up, with 12/78 also reducing to 2/13.

October 15th, 2019 at 6:24:42 AM
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Sorry, I'm a little lost in following...are you saying the odds of being dealt aces up is 2/13*21?

October 15th, 2019 at 6:51:38 AM
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2 pair odds is different for every game. What game are you playing?

2 pair in 3 card poker is 0% odds. VP it occurs in hands of 5-10 cards available, each having its own odds. Most other poker games are somewhere in between.

So, if you already know in YOUR particular game some kind of 2 pair occur 1 in 21 hands, he's saying 1 in 13 times the first of those pairs will be Aces. And 1 in 13 times, the 2nd of those pairs will be Aces. So probably the formula is 1/21(1/13 + 1/13) overall, and the answer within the stipulation that you HAVE 2 pair, 2/13 chance one of those pairs is Aces. That includes BOTH pairs being Aces, btw, so you probably need to subtract that occurrence out.

I think it may be more complicated than that, though, because the chance of the second pair also being Aces is NOT equivalent to the chance of it being any other different pair. But now we're back to having to know what game it is and how many total cards are available, to figure out how to subtract that second pair of Aces.

If you eliminate the second pair Aces (which are NOT a 1/13 chance of receiving, and also a higher- ranked hand at 4OAK), you're left with 1/13 x 1/12 x 2. 1/13 for the first pair, 1/12 for the 2nd pair, X2 because the Aces can occur in either order. I think this is the most correct way to calculate it in isolation. This number would be taken x the rate of expected occurrence in your particular poker game for a total expectation.

I could be wrong. I'll let the math guys sweep up my mess as necessary.

2 pair in 3 card poker is 0% odds. VP it occurs in hands of 5-10 cards available, each having its own odds. Most other poker games are somewhere in between.

So, if you already know in YOUR particular game some kind of 2 pair occur 1 in 21 hands, he's saying 1 in 13 times the first of those pairs will be Aces. And 1 in 13 times, the 2nd of those pairs will be Aces. So probably the formula is 1/21(1/13 + 1/13) overall, and the answer within the stipulation that you HAVE 2 pair, 2/13 chance one of those pairs is Aces. That includes BOTH pairs being Aces, btw, so you probably need to subtract that occurrence out.

I think it may be more complicated than that, though, because the chance of the second pair also being Aces is NOT equivalent to the chance of it being any other different pair. But now we're back to having to know what game it is and how many total cards are available, to figure out how to subtract that second pair of Aces.

If you eliminate the second pair Aces (which are NOT a 1/13 chance of receiving, and also a higher- ranked hand at 4OAK), you're left with 1/13 x 1/12 x 2. 1/13 for the first pair, 1/12 for the 2nd pair, X2 because the Aces can occur in either order. I think this is the most correct way to calculate it in isolation. This number would be taken x the rate of expected occurrence in your particular poker game for a total expectation.

I could be wrong. I'll let the math guys sweep up my mess as necessary.

If the House lost every hand, they wouldn't deal the game.

October 15th, 2019 at 7:28:36 AM
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There are 12 ranks for the other pair and 11 for the singleton. There are 6 ways to choose two suits out of 4. There are obviously 4 suits for the singleton.

Thus, the number of combinations is 12*11*combin(4,2)*combin(4,2)*4 = 19,008

There are combin(52,5) = 2,598,960 total combinations.

Thus, the probability is 19008/2598960 = 0.0073 = 1 in 136.73

Thus, the number of combinations is 12*11*combin(4,2)*combin(4,2)*4 = 19,008

There are combin(52,5) = 2,598,960 total combinations.

Thus, the probability is 19008/2598960 = 0.0073 = 1 in 136.73

It's not whether you win or lose; it's whether or not you had a good bet.

October 15th, 2019 at 11:19:41 AM
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Thanks, that's what I was looking for. In answer to the clarification questions, was specific to VP, trying to get inputs to calculate singular strategy change effects on EV since none of the still comercially available software I have access to has this capability.