How would you play this hand? (continued)
Here's the information on how the expected value was computed for the How would you play this hand? posting.
It's a lot less complicated talking about keeping both the 7's and the 3's, so let's start there.
If you keep those 4 cards, then there are only 47 possible outcomes. You start out with a deck of 52 cards, but the 5 you see - two 7's, two 3's, and the king - are already used. That leaves 47 cards to fill in that spot of the card you didn't hold.
Out of those 47 cards, there are two 7's remaining, two 3's remaining, and four 2's remaining. That's 8 cards that would give you a full house.
The expected value is the odds of getting that full house (which is 8 divided by 47) times the payout (which is a variable in our case). If the payout is 3, then the expected value is 8*3/47 = 0.5106 and for a payout of 4 it is 8*4/47 = 0.6809
Keeping just the 7's is a lot more complicated. There are now 3 cards that will be filled in by using the 47 remaining cards. That is 16,215 possible ways to combine those cards. You might think it's 47*46*45 = 97,290 - but that's wrong. In that 97,290 there are repeats (a Jack of Spades, 4 of Hearts, 6 of Diamonds is the same as Jack of Spades, 6 of Diamonds, 4 of Hearts in video poker, but are counted separate in the 97,290). It's actually 97,290 divided by the 6 ways to mix and match any three cards and get the same result (JS/4H/6D, JS/6D/4H, 4H/JS/6D, 4H/6D/JS, 6D/4H/JS, and 6D/JS/4H). On your calculator, this is the "nCr" button. Put in 47 nCr 3 and you will get 16,215 - the number of ways to get a combination of 3 cards out of 47 card choices.
To get a five-of-a-kind, you can get that with the other two 7's and one of the four 2's (4 ways), or with one of the two 7's and two of the four 2's (2*6=12 ways), or with three 2's (4 ways). That's 20 possible combinations and each pays 15, so 20*15/16215 = 0.0185 for just the five-of-a-kind.
To get a four-of-a-kind, you can get that with the other two 7's and one other card (except that other card can't be a 2). There's 41 "other" cards (out of the 47 to choose from, the two 7's are gone and we can't pick from the four 2's). So there's 41 ways to get a four-of-a-kind with all 7's. You can also get a four-of-a-kind with one of the two 7's, one of the four 2's, and one "other" card. There's 2*4*41=328 ways to get that four-of-a-kind. You could also get a four-of-a-kind with two of the four 2's (6 ways = 4 nCr 2 on your calculator) and one "other" card. There's 6*41=246 ways. Adding that all up, there are 615 combinations and each pays 5, so 615*5/16215=0.1896 for just the four-of-a-kind.
To get a three-of-a-kind, you can get with one matching (either a 7 or a 2) and two "other" cards. Let's look at the two "other" cards first. I already said there were 41 other cards to worry about. There are 41 nCr 2=820 ways to combine those extra cards. To get the three-of-a-kind with all 7's, there are 2 ways to get that last 7, so 2*820=1640 ways total. To get the three-of-a-kind with a 2, there are 4 ways to get a 2, so 4*820=3280 ways total. Adding it all up, there are 4920 combinations and each pays 1, so 4920/16215=0.3034 for just the three-of-a-kind.
Full house is the most complicated. Remember that we are keeping the 7's only. There are only two 3's remaining in the deck (the other two were used already) and there are only three K's remaining in the deck. To get a full hour, you could get one of the two 7's along with two cards that match. But those two cards that match can't be 2's (or else that would be a five-of-a-kind). There are 9 faces that have four cards remaining in the deck (4,5,6,8,9,10,J,Q,A), and there are 6 ways to get two of those four cards (4 nCr 2), so that's 9*6=54 ways. Additionally, there is one way to get the remaining 3's from the deck, so we're up to 55 ways. And there are 3 ways to get two of the three remaining K's from the deck, so we're up to 58 ways. Multiply that by the 2 ways to get one of the two remaining 7's, and that is 116 ways to get a full house with 7's and another face.
But there are other ways to get a full house. You could get it with one of the four 2's and two cards that match. I already showed there are 58 ways to get two other cards that match, times 4 ways to get a 2, and that's 232 ways to get this full house.
The last way to get a full house is with three running cards (none of them being 7's or 2's). Since there aren't three 3's left in the deck, that can't happen. There's only one way to get the remaining three K's left. For the other 9 faces, each one has only 3 ways to get it (getting 3 of the 4 suits). That is 9*3=27+1=28 total ways.
Adding up all the full house options, there are 376 total combinations. If the payout is 3, then this is 376*3/16215=0.0696 and if the payout is 4, then this is 376*4/16215=0.0927
You add up all the expected values for each of the possible outcomes. When you keep both pair, there is only 1 option (well, 2 options - nothing or a full house). When you keep just the pair of 7's, you add up the expected value for each possible outcome. Adding that up, you see that when the full house payout is 4, it is better to keep both pair and when the full house payout is 3, it is better to keep just one pair.
One special note - if you are played deuces wild super bonus video poker, then the payouts for five-of-a-kind fluctuate based on the face (five 3's worth more/less than five A's, for example). In this case, the payout of 15 may go up or down depending on the cards making up the five-of-a-kind. So, in that case, it might be better to keep just a pair even with a full house payout of 4. But I don't know if I've ever seen a super bonus with a full house payout of 4, anyway.
It's a lot less complicated talking about keeping both the 7's and the 3's, so let's start there.
If you keep those 4 cards, then there are only 47 possible outcomes. You start out with a deck of 52 cards, but the 5 you see - two 7's, two 3's, and the king - are already used. That leaves 47 cards to fill in that spot of the card you didn't hold.
Out of those 47 cards, there are two 7's remaining, two 3's remaining, and four 2's remaining. That's 8 cards that would give you a full house.
The expected value is the odds of getting that full house (which is 8 divided by 47) times the payout (which is a variable in our case). If the payout is 3, then the expected value is 8*3/47 = 0.5106 and for a payout of 4 it is 8*4/47 = 0.6809
Keeping just the 7's is a lot more complicated. There are now 3 cards that will be filled in by using the 47 remaining cards. That is 16,215 possible ways to combine those cards. You might think it's 47*46*45 = 97,290 - but that's wrong. In that 97,290 there are repeats (a Jack of Spades, 4 of Hearts, 6 of Diamonds is the same as Jack of Spades, 6 of Diamonds, 4 of Hearts in video poker, but are counted separate in the 97,290). It's actually 97,290 divided by the 6 ways to mix and match any three cards and get the same result (JS/4H/6D, JS/6D/4H, 4H/JS/6D, 4H/6D/JS, 6D/4H/JS, and 6D/JS/4H). On your calculator, this is the "nCr" button. Put in 47 nCr 3 and you will get 16,215 - the number of ways to get a combination of 3 cards out of 47 card choices.
To get a five-of-a-kind, you can get that with the other two 7's and one of the four 2's (4 ways), or with one of the two 7's and two of the four 2's (2*6=12 ways), or with three 2's (4 ways). That's 20 possible combinations and each pays 15, so 20*15/16215 = 0.0185 for just the five-of-a-kind.
To get a four-of-a-kind, you can get that with the other two 7's and one other card (except that other card can't be a 2). There's 41 "other" cards (out of the 47 to choose from, the two 7's are gone and we can't pick from the four 2's). So there's 41 ways to get a four-of-a-kind with all 7's. You can also get a four-of-a-kind with one of the two 7's, one of the four 2's, and one "other" card. There's 2*4*41=328 ways to get that four-of-a-kind. You could also get a four-of-a-kind with two of the four 2's (6 ways = 4 nCr 2 on your calculator) and one "other" card. There's 6*41=246 ways. Adding that all up, there are 615 combinations and each pays 5, so 615*5/16215=0.1896 for just the four-of-a-kind.
To get a three-of-a-kind, you can get with one matching (either a 7 or a 2) and two "other" cards. Let's look at the two "other" cards first. I already said there were 41 other cards to worry about. There are 41 nCr 2=820 ways to combine those extra cards. To get the three-of-a-kind with all 7's, there are 2 ways to get that last 7, so 2*820=1640 ways total. To get the three-of-a-kind with a 2, there are 4 ways to get a 2, so 4*820=3280 ways total. Adding it all up, there are 4920 combinations and each pays 1, so 4920/16215=0.3034 for just the three-of-a-kind.
Full house is the most complicated. Remember that we are keeping the 7's only. There are only two 3's remaining in the deck (the other two were used already) and there are only three K's remaining in the deck. To get a full hour, you could get one of the two 7's along with two cards that match. But those two cards that match can't be 2's (or else that would be a five-of-a-kind). There are 9 faces that have four cards remaining in the deck (4,5,6,8,9,10,J,Q,A), and there are 6 ways to get two of those four cards (4 nCr 2), so that's 9*6=54 ways. Additionally, there is one way to get the remaining 3's from the deck, so we're up to 55 ways. And there are 3 ways to get two of the three remaining K's from the deck, so we're up to 58 ways. Multiply that by the 2 ways to get one of the two remaining 7's, and that is 116 ways to get a full house with 7's and another face.
But there are other ways to get a full house. You could get it with one of the four 2's and two cards that match. I already showed there are 58 ways to get two other cards that match, times 4 ways to get a 2, and that's 232 ways to get this full house.
The last way to get a full house is with three running cards (none of them being 7's or 2's). Since there aren't three 3's left in the deck, that can't happen. There's only one way to get the remaining three K's left. For the other 9 faces, each one has only 3 ways to get it (getting 3 of the 4 suits). That is 9*3=27+1=28 total ways.
Adding up all the full house options, there are 376 total combinations. If the payout is 3, then this is 376*3/16215=0.0696 and if the payout is 4, then this is 376*4/16215=0.0927
You add up all the expected values for each of the possible outcomes. When you keep both pair, there is only 1 option (well, 2 options - nothing or a full house). When you keep just the pair of 7's, you add up the expected value for each possible outcome. Adding that up, you see that when the full house payout is 4, it is better to keep both pair and when the full house payout is 3, it is better to keep just one pair.
One special note - if you are played deuces wild super bonus video poker, then the payouts for five-of-a-kind fluctuate based on the face (five 3's worth more/less than five A's, for example). In this case, the payout of 15 may go up or down depending on the cards making up the five-of-a-kind. So, in that case, it might be better to keep just a pair even with a full house payout of 4. But I don't know if I've ever seen a super bonus with a full house payout of 4, anyway.
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