Here's the solution for 78s making exactly 2 pair by the river. I wrote this fairly quickly so there may be an error somewhere but the answer seems very close to what I would expect....
Let's say you have 78s... you can end up with exactly 2 pair with a board in the following configurations
(ignoring suits (for now))
i) 78abc
ii) 78aab
iii) xaabc
iv) xaabb
v) aabbc
where a, b, c are different ranks and x is a 7 or 8
Case i)
There are 3 * 3 * 11c3 * 4 * 4 * 4 ways to make this board = 95040
... however some of these will also be straights or flushes, so we must subtract them.
Straight Flush... ways to make a SF = (3 * 3 * 4) + (3 * 4) = 48
Flushes... ways to make a F = (3 * 3 * 11c3) + (3 * 11c3) = 1980 ... this number will include all the SF so we must subtract 48 = 1932
Straights... ways to make a S = 3 * 3 * 4 * 4 * 4 * 4 = 2304 ... this number also will include the SF so we must also subtract 48 here too = 2256
so there are exactly 95040 - 48 - 1932 - 2256 = 90804 to make exactly 2 pair on this board
Case ii)
There are 3 * 3 * 11 * 6 * 10 * 4 ways to make this board = 23760
Straights and flushes are not possible on this board so we are done with case ii
Case iii)
There are 6 * 11 * 6 * 10c2 * 4 * 4 ways to make this board = 285120
... and again, some of these will also be straights or flushes, so we must subtract them.
Straight Flush... ways to make a SF = 6 * 3 * 4 = 72
Flushes... ways to make a F = 6 * 11 * 3 * 10c2 = 8910 ... this number will include all the SF so we must subtract 72 = 8838
Straights... ways to make a S = 6 * 4 * 6 * 4 * 4 = 2304 ... this number also will include the SF so we must also subtract 72 here too = 2232
so there are exactly 285120 - 72 - 8838 - 2232 = 273978 to make exactly 2 pair on this board
Case iv)
There are 6 * 11c2 * 6 * 6 ways to make this board = 11880
Straights and flushes are not possible on this board so we are done with case iv
Case v)
There are 11c2 * 6 * 6 * 9 * 4 ways to make this board = 71280
... and again, some of these will also be straights or flushes, so we must subtract them.
Straight Flush... ways to make a SF = 3 * 3 * 4 = 36
Flushes... ways to make a F = 11c2 * 3 * 3 * 9 = 4455 ... this number will include all the SF so we must subtract 36 = 4419
Straights... ways to make a S = 4 * 6 * 6 * 4 = 576 ... this number also will include the SF so we must also subtract 36 here too = 540
so there are exactly 71280 - 36 - 4419 - 540 = 66285 ways to make exactly 2 pair on this board
So adding all these cases together = 90804 + 23760 + 273978 + 11880 + 66285 = 466707
Total number of boards = 50c5 = 2118760
So the odds of ending up with exactly 2 pair when starting with 78s = 466707 / 2118760 = ~ .22027
What I mean is pairing both of your hole cards on an unpaired board -- ie j8 on a jxyz8 board.
So you having AK on a Awxyz board -- what are the odds of a single opponent having 2 pair (I know this is a bit different from the J8 example as the A is in your hand).
ok i've lost the ability to edit my posts. even the ones in which i'm wrong. oops!
sorry mickey. you are correct. the number seemed huge but i guess not. i'll try to post the numbers my software came up with but i'm having problems here with 2 separate computers so i don't think it is my problem...
MH, do you mind running the math for flopping trips with a starting pair. Thanks.
There are 48c3 = 17296 flops that do have none of the remaining cards of your rank. There are 46 flops that contain both your cards to give you quads... so there must be 2258 flops that contain exactly 1 of that cards needed to make trips... There are 50c3 = 19600 flops so 2258/19600 = .1152 or about ~1 in 8.7
So you having AK on a Awxyz board -- what are the odds of a single opponent having 2 pair (I know this is a bit different from the J8 example as the A is in your hand).
Answers?
Thanks
There are (45 * 44 / 2) or 890 different hands that your opponent can have.
There are 4 different hands with an Ace that have hit 2 pair (Aw, Ax, Ay, Az)
There are 6 different ways that he can have each of these hands.
There are 6 different hands without and Ace that have hit 2 pair (wx, wy, wz, xy, xz, yz)
There are 9 different ways that he can have each of these hands.
There are (45 * 44 / 2) or 890 different hands that your opponent can have.
There are 4 different hands with an Ace that have hit 2 pair (Aw, Ax, Ay, Az)
There are 6 different ways that he can have each of these hands.
There are 6 different hands without and Ace that have hit 2 pair (wx, wy, wz, xy, xz, yz)
There are 9 different ways that he can have each of these hands.
So, the answer is (4 * 6 + 6 * 9) / 890 or 7.88%
TY, I will assume the math is right pending any rebuttals:)
Ehrm...so the odds of J8 pairing both hole cards on an unpaired board is ~22%??
J8 on Jxyz8 board?
Not sure if the 22% is including all the paired boards, etc.
Thanks
No... ~22% is for you to make exactly 2 pair anyway. 2 pair using both hole cards would be covered by onlycase i in the original solution I gave for the example 78s. It would be closer to 90804 / 2118760 = 4.29%
There are (45 * 44 / 2) or 890 different hands that your opponent can have.
There are 4 different hands with an Ace that have hit 2 pair (Aw, Ax, Ay, Az)
There are 6 different ways that he can have each of these hands.
There are 6 different hands without and Ace that have hit 2 pair (wx, wy, wz, xy, xz, yz)
There are 9 different ways that he can have each of these hands.
So, the answer is (4 * 6 + 6 * 9) / 890 or 7.88%
45*44/2 = 990 different hands.
although 7.88% is correct for (4 * 6 + 6 * 9) / 990
This value would be an upper bound assuming your opponent would equally play all possible variations of wxyz... not very likely on any board, and very, very unlikely on a board of AJ742... unless he was the BB and you limped AK.
There are 48c3 = 17296 flops that do have none of the remaining cards of your rank. There are 46 flops that contain both your cards to give you quads... so there must be 2258 flops that contain exactly 1 of that cards needed to make trips... There are 50c3 = 19600 flops so 2258/19600 = .1152 or about ~1 in 8.7
Comments
Next, the answer is different for 56s compared to J8o, so you have to be more specific.
Finally, do you only want answers that include the cards in your hand... Ex. Holding 67o board AJJ99
Edit: Will post answer later this week if no one else anwers this... Its a very non-trivial answer.
Let's say you have 78s... you can end up with exactly 2 pair with a board in the following configurations
(ignoring suits (for now))
i) 78abc
ii) 78aab
iii) xaabc
iv) xaabb
v) aabbc
where a, b, c are different ranks and x is a 7 or 8
Case i)
There are 3 * 3 * 11c3 * 4 * 4 * 4 ways to make this board = 95040
... however some of these will also be straights or flushes, so we must subtract them.
Straight Flush... ways to make a SF = (3 * 3 * 4) + (3 * 4) = 48
Flushes... ways to make a F = (3 * 3 * 11c3) + (3 * 11c3) = 1980 ... this number will include all the SF so we must subtract 48 = 1932
Straights... ways to make a S = 3 * 3 * 4 * 4 * 4 * 4 = 2304 ... this number also will include the SF so we must also subtract 48 here too = 2256
so there are exactly 95040 - 48 - 1932 - 2256 = 90804 to make exactly 2 pair on this board
Case ii)
There are 3 * 3 * 11 * 6 * 10 * 4 ways to make this board = 23760
Straights and flushes are not possible on this board so we are done with case ii
Case iii)
There are 6 * 11 * 6 * 10c2 * 4 * 4 ways to make this board = 285120
... and again, some of these will also be straights or flushes, so we must subtract them.
Straight Flush... ways to make a SF = 6 * 3 * 4 = 72
Flushes... ways to make a F = 6 * 11 * 3 * 10c2 = 8910 ... this number will include all the SF so we must subtract 72 = 8838
Straights... ways to make a S = 6 * 4 * 6 * 4 * 4 = 2304 ... this number also will include the SF so we must also subtract 72 here too = 2232
so there are exactly 285120 - 72 - 8838 - 2232 = 273978 to make exactly 2 pair on this board
Case iv)
There are 6 * 11c2 * 6 * 6 ways to make this board = 11880
Straights and flushes are not possible on this board so we are done with case iv
Case v)
There are 11c2 * 6 * 6 * 9 * 4 ways to make this board = 71280
... and again, some of these will also be straights or flushes, so we must subtract them.
Straight Flush... ways to make a SF = 3 * 3 * 4 = 36
Flushes... ways to make a F = 11c2 * 3 * 3 * 9 = 4455 ... this number will include all the SF so we must subtract 36 = 4419
Straights... ways to make a S = 4 * 6 * 6 * 4 = 576 ... this number also will include the SF so we must also subtract 36 here too = 540
so there are exactly 71280 - 36 - 4419 - 540 = 66285 ways to make exactly 2 pair on this board
So adding all these cases together = 90804 + 23760 + 273978 + 11880 + 66285 = 466707
Total number of boards = 50c5 = 2118760
So the odds of ending up with exactly 2 pair when starting with 78s = 466707 / 2118760 = ~ .22027
What I mean is pairing both of your hole cards on an unpaired board -- ie j8 on a jxyz8 board.
So you having AK on a Awxyz board -- what are the odds of a single opponent having 2 pair (I know this is a bit different from the J8 example as the A is in your hand).
Answers?
Thanks
Sorry, clarified above.
I eagerly await a rebuttal proof!
sorry mickey. you are correct. the number seemed huge but i guess not. i'll try to post the numbers my software came up with but i'm having problems here with 2 separate computers so i don't think it is my problem...
1.7m hands ak vs j8
so ya, around 20 percent. good job mickey
There are 48c3 = 17296 flops that do have none of the remaining cards of your rank. There are 46 flops that contain both your cards to give you quads... so there must be 2258 flops that contain exactly 1 of that cards needed to make trips... There are 50c3 = 19600 flops so 2258/19600 = .1152 or about ~1 in 8.7
Pkrfce... no prob man.
J8 on Jxyz8 board?
Not sure if the 22% is including all the paired boards, etc.
Thanks
Getting back to the original question....
To calculate what percentage of time j8 will hit a jxyz8 board....
(Assuming xyz are not a J or an 8)
= 3 * 3 * 44 * 43 * 42 * 5! / 3! * (46! / 50!)
= 5.63%
There are (45 * 44 / 2) or 890 different hands that your opponent can have.
There are 4 different hands with an Ace that have hit 2 pair (Aw, Ax, Ay, Az)
There are 6 different ways that he can have each of these hands.
There are 6 different hands without and Ace that have hit 2 pair (wx, wy, wz, xy, xz, yz)
There are 9 different ways that he can have each of these hands.
So, the answer is (4 * 6 + 6 * 9) / 890 or 7.88%
TY, I will assume the math is right pending any rebuttals:)
No... ~22% is for you to make exactly 2 pair anyway. 2 pair using both hole cards would be covered by only case i in the original solution I gave for the example 78s. It would be closer to 90804 / 2118760 = 4.29%
This answer includes boards like J8AAA and J8KK3 and 789TJ and flush boards and even straight flush boards.
45*44/2 = 990 different hands.
although 7.88% is correct for (4 * 6 + 6 * 9) / 990
This value would be an upper bound assuming your opponent would equally play all possible variations of wxyz... not very likely on any board, and very, very unlikely on a board of AJ742... unless he was the BB and you limped AK.
this is sick; thanks for the effort