4 card poker how to math (combinations and permutations)
January 19th, 2023 at 10:00:13 PM
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I'm trying to recreate some of Wayne Hong's math and can't seem to crack it. Can anyone spot the mistakes in my math? I'll be referencing wizardofodds com/games/4-card-split/ on the 4 card blind paytable. Basically poker with 4 dealt cards. going down the list here of what I get. + means I got the same answer, - means they are different
4 combinations. this ones easy
13 combinations. Another easy one
not too hard. 10 combinations of straights from A-J to 5-2. 1 for each suit gets 40
adding kickers finally messed me up. My initial thought was any of the 52 cards * 3 matchings left * 2 matches left * the 48 kickers / by the ways to permute it (3 identical objects out of 4 is 4!/3!(4-3)!) which is 4 yields 3,744. If I were to instead divide by 6 I would get wayne's answer of 2,496 but why would I do this? An alternate method would be to take the combinations of 3 of a Kind out of 3 cards and multiply it by the 48 kickers giving you 2,496. I was sort of able to get this one.
I might actually be right on this one. you take any 52 cards * 12 of that suit left * 11 * 10 / 24 permutations giving you 2,860 just 44 off of Wayne's 2,816. If it's true I'm right, this would add an extra 0.00114 units of return and take almost 11% of the house edge off the game. 60 does sound like 16 so I can see why.
I start by calculating one combination of a straight say A-K-Q-J. 16 cards for any of them * 12 for the ones remaining * 8 * 4 * 10 for the other straights / 24 permutations - 40 for straight flush gives 2,520. you subtract after because there are 40 combinations for SF and we still had permutations.Wayne gets 2,772 which I actually cannot understand this time
any of 52 cards * any nonmatching so 48 * any matching 6 * other matching 3 (44,928) / by the permutations of 2 out of 4 (4!/2!(4-2)! = 6) gets me 7,488, quite far from wayne's 2,808. If I divide 44,928 by 16 I get the right answer for some reason.
4 combinations. this ones easy
13 combinations. Another easy one
not too hard. 10 combinations of straights from A-J to 5-2. 1 for each suit gets 40
adding kickers finally messed me up. My initial thought was any of the 52 cards * 3 matchings left * 2 matches left * the 48 kickers / by the ways to permute it (3 identical objects out of 4 is 4!/3!(4-3)!) which is 4 yields 3,744. If I were to instead divide by 6 I would get wayne's answer of 2,496 but why would I do this? An alternate method would be to take the combinations of 3 of a Kind out of 3 cards and multiply it by the 48 kickers giving you 2,496. I was sort of able to get this one.
I might actually be right on this one. you take any 52 cards * 12 of that suit left * 11 * 10 / 24 permutations giving you 2,860 just 44 off of Wayne's 2,816. If it's true I'm right, this would add an extra 0.00114 units of return and take almost 11% of the house edge off the game. 60 does sound like 16 so I can see why.
I start by calculating one combination of a straight say A-K-Q-J. 16 cards for any of them * 12 for the ones remaining * 8 * 4 * 10 for the other straights / 24 permutations - 40 for straight flush gives 2,520. you subtract after because there are 40 combinations for SF and we still had permutations.Wayne gets 2,772 which I actually cannot understand this time
any of 52 cards * any nonmatching so 48 * any matching 6 * other matching 3 (44,928) / by the permutations of 2 out of 4 (4!/2!(4-2)! = 6) gets me 7,488, quite far from wayne's 2,808. If I divide 44,928 by 16 I get the right answer for some reason.
4 tens * 3 other tens * 48 kickers * 44 more kickers that don't match. divide by 6 permutations and multiply by the 4 other pairs (*5) for 21,120. Wayne got 31,680. Interestingly enough if I divided by 4 like you do for 3 of a kind, I would get 31,680
January 20th, 2023 at 2:12:03 AM
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Quote: richodudeI'm trying to recreate some of Wayne Hong's math and can't seem to crack it. Can anyone spot the mistakes in my math? I'll be referencing wizardofodds com/games/4-card-split/ on the 4 card blind paytable. Basically poker with 4 dealt cards. going down the list here of what I get. + means I got the same answer, - means they are different
4 combinations. this ones easy
13 combinations. Another easy one
not too hard. 10 combinations of straights from A-J to 5-2. 1 for each suit gets 40
adding kickers finally messed me up. My initial thought was any of the 52 cards * 3 matchings left * 2 matches left * the 48 kickers / by the ways to permute it (3 identical objects out of 4 is 4!/3!(4-3)!) which is 4 yields 3,744. If I were to instead divide by 6 I would get wayne's answer of 2,496 but why would I do this? An alternate method would be to take the combinations of 3 of a Kind out of 3 cards and multiply it by the 48 kickers giving you 2,496. I was sort of able to get this one.
I might actually be right on this one. you take any 52 cards * 12 of that suit left * 11 * 10 / 24 permutations giving you 2,860 just 44 off of Wayne's 2,816. If it's true I'm right, this would add an extra 0.00114 units of return and take almost 11% of the house edge off the game. 60 does sound like 16 so I can see why.
I start by calculating one combination of a straight say A-K-Q-J. 16 cards for any of them * 12 for the ones remaining * 8 * 4 * 10 for the other straights / 24 permutations - 40 for straight flush gives 2,520. you subtract after because there are 40 combinations for SF and we still had permutations.Wayne gets 2,772 which I actually cannot understand this time
any of 52 cards * any nonmatching so 48 * any matching 6 * other matching 3 (44,928) / by the permutations of 2 out of 4 (4!/2!(4-2)! = 6) gets me 7,488, quite far from wayne's 2,808. If I divide 44,928 by 16 I get the right answer for some reason.4 tens * 3 other tens * 48 kickers * 44 more kickers that don't match. divide by 6 permutations and multiply by the 4 other pairs (*5) for 21,120. Wayne got 31,680. Interestingly enough if I divided by 4 like you do for 3 of a kind, I would get 31,680
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Some of these are because you are mixing up combinations and permutations.
Here is how I would do these where c(x,y) is the combin(x,y) function in Excel:
Royal flush: c(4,1)
pick 1 of 4 suits
4 of a kind: c(13 ,1)
one for each rank
Straight flush c(4,1) * 10
pick 1 of 4 suits * 10 highest cards to start the straight
3 of a kind: c(13,1) * c(4,3) * c(12 ,1 ) * c(4,1)
pick 1 of 13 ranks and pick 3 of 4 suits in that rank. Pick 1 of the 12 kicker ranks and 1 of the 4 kicker suits.
Flush: c(4,1) * (c(13,4)-11)
Pick 1 of 4 suits * pick 4 of 13 ranks but subtract the 11 combinations that make a royal or straight flush
Straight: 11 * (4*4*4*4-4)
there are 11 different straights. Each rank can be 1 of 4 suits, but 4 of them will make royals or straight flushes.
2 pairs: c(13,2) * c(4,2) * c(4,2)
pick 2 of 13 ranks and pick 2 of 4 suits from each rank
10s or better: c(5,1) * c(4,2) * c(12,2) * c(4,1) *c(4,1)
pick 1 of 5 ranks 10 or better, pick 2 of 4 suits. Pick 2 of 12 ranks and 1 of 4 suits for each rank
Did this at 2am on my phone so hopefully no errors. Feel free to ask any questions.
“Man Babes” #AxelFabulous
January 21st, 2023 at 3:02:10 AM
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Quote: mipletQuote: richodudeI'm trying to recreate some of Wayne Hong's math and can't seem to crack it. Can anyone spot the mistakes in my math? I'll be referencing wizardofodds com/games/4-card-split/ on the 4 card blind paytable. Basically poker with 4 dealt cards. going down the list here of what I get. + means I got the same answer, - means they are different
4 combinations. this ones easy
13 combinations. Another easy one
not too hard. 10 combinations of straights from A-J to 5-2. 1 for each suit gets 40
adding kickers finally messed me up. My initial thought was any of the 52 cards * 3 matchings left * 2 matches left * the 48 kickers / by the ways to permute it (3 identical objects out of 4 is 4!/3!(4-3)!) which is 4 yields 3,744. If I were to instead divide by 6 I would get wayne's answer of 2,496 but why would I do this? An alternate method would be to take the combinations of 3 of a Kind out of 3 cards and multiply it by the 48 kickers giving you 2,496. I was sort of able to get this one.
I might actually be right on this one. you take any 52 cards * 12 of that suit left * 11 * 10 / 24 permutations giving you 2,860 just 44 off of Wayne's 2,816. If it's true I'm right, this would add an extra 0.00114 units of return and take almost 11% of the house edge off the game. 60 does sound like 16 so I can see why.
I start by calculating one combination of a straight say A-K-Q-J. 16 cards for any of them * 12 for the ones remaining * 8 * 4 * 10 for the other straights / 24 permutations - 40 for straight flush gives 2,520. you subtract after because there are 40 combinations for SF and we still had permutations.Wayne gets 2,772 which I actually cannot understand this time
any of 52 cards * any nonmatching so 48 * any matching 6 * other matching 3 (44,928) / by the permutations of 2 out of 4 (4!/2!(4-2)! = 6) gets me 7,488, quite far from wayne's 2,808. If I divide 44,928 by 16 I get the right answer for some reason.4 tens * 3 other tens * 48 kickers * 44 more kickers that don't match. divide by 6 permutations and multiply by the 4 other pairs (*5) for 21,120. Wayne got 31,680. Interestingly enough if I divided by 4 like you do for 3 of a kind, I would get 31,680
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Some of these are because you are mixing up combinations and permutations.
Here is how I would do these where c(x,y) is the combin(x,y) function in Excel:
Royal flush: c(4,1)
pick 1 of 4 suits
4 of a kind: c(13 ,1)
one for each rank
Straight flush c(4,1) * 10
pick 1 of 4 suits * 10 highest cards to start the straight
3 of a kind: c(13,1) * c(4,3) * c(12 ,1 ) * c(4,1)
pick 1 of 13 ranks and pick 3 of 4 suits in that rank. Pick 1 of the 12 kicker ranks and 1 of the 4 kicker suits.
Flush: c(4,1) * (c(13,4)-11)
Pick 1 of 4 suits * pick 4 of 13 ranks but subtract the 11 combinations that make a royal or straight flush
Straight: 11 * (4*4*4*4-4)
there are 11 different straights. Each rank can be 1 of 4 suits, but 4 of them will make royals or straight flushes.
2 pairs: c(13,2) * c(4,2) * c(4,2)
pick 2 of 13 ranks and pick 2 of 4 suits from each rank
10s or better: c(5,1) * c(4,2) * c(12,2) * c(4,1) *c(4,1)
pick 1 of 5 ranks 10 or better, pick 2 of 4 suits. Pick 2 of 12 ranks and 1 of 4 suits for each rank
Did this at 2am on my phone so hopefully no errors. Feel free to ask any questions.
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This is pretty helpful. I actually got straight and flush right but I was just dumb and forgot there were 11 straights not 10. Also I forgot to subtract the straight flushes which I thought I did. For 3 of a kind (and really all of them) you offered a better way of approaching the problem which I like.
I still have questions about 2 pairs and 10's though. your math checks out, but I don't understand why you would approach it the way you did. My way of going one card at a time (52*48*6*3 for 2-pair) and converting from permutations to combinations messed me up
