what's the expectation of this prop bet?
Standard poker deck, all cards are worth the same as in blackjack. Aces are always 11. If the three flop cards equal 21 it's push. How big of an edge does the person have who is betting the flop will be 22 or greater?
I'm going to assume there is "no information", ie: The first 3 cards can be any 3 cards of a normal 52-card deck, without the information of what my 2 card hand is.
edit: actually nvm i don't want to figure this out. but it is an interesting proposition. without doing any math, i'd say the person betting 3 cards is greater than 21 has a very strong edge.
| Value | Flops | Wins | Losses |
|---|---|---|---|
| 6 | 4 | 22096 | 0 |
| 7 | 24 | 22072 | 4 |
| 8 | 48 | 22024 | 28 |
| 9 | 92 | 21932 | 76 |
| 10 | 136 | 21796 | 168 |
| 11 | 200 | 21596 | 304 |
| 12 | 268 | 21328 | 504 |
| 13 | 352 | 20976 | 772 |
| 14 | 512 | 20464 | 1124 |
| 15 | 740 | 19724 | 1636 |
| 16 | 896 | 18828 | 2376 |
| 17 | 1080 | 17748 | 3272 |
| 18 | 1196 | 16552 | 4352 |
| 19 | 1336 | 15216 | 5548 |
| 20 | 1408 | 13808 | 6884 |
| 21 | 1508 | 12300 | 8292 |
| 22 | 1800 | 10500 | 9800 |
| 23 | 1856 | 8644 | 11600 |
| 24 | 1652 | 6992 | 13456 |
| 25 | 1472 | 5520 | 15108 |
| 26 | 1264 | 4256 | 16580 |
| 27 | 1108 | 3148 | 17844 |
| 28 | 920 | 2228 | 18952 |
| 29 | 784 | 1444 | 19872 |
| 30 | 840 | 604 | 20656 |
| 31 | 504 | 100 | 21496 |
| 32 | 96 | 4 | 22000 |
| 33 | 4 | 0 | 22096 |
If the push number is 22 or less, then it is a good bet; if it is 23 or more , then it is a bad bet.
This assumes all aces are 11, as opposed to "the first ace is 11, but the second and third aces are 1".
Quote: ThatDonGuyHere's what I get:
Value Flops Wins Losses 6 4 22096 0 7 24 22072 4 8 48 22024 28 9 92 21932 76 10 136 21796 168 11 200 21596 304 12 268 21328 504 13 352 20976 772 14 512 20464 1124 15 740 19724 1636 16 896 18828 2376 17 1080 17748 3272 18 1196 16552 4352 19 1336 15216 5548 20 1408 13808 6884 21 1508 12300 8292 22 1800 10500 9800 23 1856 8644 11600 24 1652 6992 13456 25 1472 5520 15108 26 1264 4256 16580 27 1108 3148 17844 28 920 2228 18952 29 784 1444 19872 30 840 604 20656 31 504 100 21496 32 96 4 22000 33 4 0 22096
If the push number is 22 or less, then it is a good bet; if it is 23 or more , then it is a bad bet.
This assumes all aces are 11, as opposed to "the first ace is 11, but the second and third aces are 1".
What about 21 is a push, > 21 wins, and < 21 loses?
How did you come up with your answer, specifically?
Quote: RSWhat about 21 is a push, > 21 wins, and < 21 loses?
How did you come up with your answer, specifically?
Of the 22,100 possible three-card flops, 12,300 are higher than 21, 8292 are lower than 21, and 1508 are exactly 21.
I got my numbers through a count of all 22,100 possible three-card flops.
Quote: ThatDonGuyOf the 22,100 possible three-card flops, 12,300 are higher than 21, 8292 are lower than 21, and 1508 are exactly 21.
I got my numbers through a count of all 22,100 possible three-card flops.
Specifically, though? How'd you figure 22,100 different 3-card flops combinations? How'd you come up with 12,3000 are higher than 21? Surely you didn't write down each combination "2, 2, 2 is 6. 2, 2, 3 is 7. 2, 2, 4 is 8...." and go from there.
I'm not trying to say you're wrong (or right). But I'm genuinely interested in how you came up with those figures.
Quote: RSSpecifically, though? How'd you figure 22,100 different 3-card flops combinations? How'd you come up with 12,3000 are higher than 21? Surely you didn't write down each combination "2, 2, 2 is 6. 2, 2, 3 is 7. 2, 2, 4 is 8...." and go from there.
I'm not trying to say you're wrong (or right). But I'm genuinely interested in how you came up with those figures.
Yea, that's what computers are for. Probably the easiest way to solve it.
But you could estimate the answer.
First, total combinations = (52*51*50) / (3*2*1) = 22100
Now let X be the value of any single card.
Then, E[X] = 380/52 = 7.308
And, var[X] = 8.52
So, E[X1+X2+X3] = 7.308*3 = 21.92
cov[Xi,Xj] = -var[X]/(N-1) = -8.52/51 = -0.167
And, var[X1+X2+X3] = var[X1]+var[X2]+var[X3]+2cov[X1,X2]+2cov[X1,X3]+2cov[X2,X3] = (8.52*3) - (6*0.167) = 24.56
Therefore, the standard deviation is...
sd = 24.561/2 = 4.96
Then, the Z-value = (21.5 - 21.92)/4.96 = -0.0847
And P(Z>-0.0847) = 0.534
53.4% of 22100 is 11801
So about 11801 combinations are greater than 21.
Similar process get 8553 combinations less than 21 and 1746 combinations that are 21.
Again, this is just an estimation. ThatDonGuy has the exact solution. Also, I'm a tad rusty on this stuff...I'm not 100% sure if everything I did was legal.
Quote: RSSpecifically, though? How'd you figure 22,100 different 3-card flops combinations? How'd you come up with 12,3000 are higher than 21? Surely you didn't write down each combination "2, 2, 2 is 6. 2, 2, 3 is 7. 2, 2, 4 is 8...." and go from there.
I'm not trying to say you're wrong (or right). But I'm genuinely interested in how you came up with those figures.
There are 52 x 51 x 50 / 6 = 22,100 different flops - 52 possibie cards for the first one; for each of these, 51 possible remaining cards for the second one; for each of these, 50 possible remaining cards for the third one, so that's 52 x 51 x 50, but in a flop, the order doesn't matter (As 2h 3c is the same flop as 2h 3c As), so divide by the six different orders in which three cards can appear.
Here's some "pseudocode" for counting how many of each value there are:
The cards are numbered 0-51; the rank = (the number divided by 4 and rounded down) + 1 (so 1 = Ace, 2 = 2, ..., 13 = King)
FlopCounts[34] is an array where all of the values are initially 0
for (Card1 = 0 to 49):
{
. . . . Rank1 = floor(Card1 / 4) + 1
. . . . if (Rank1 > 9) then Rank1 = 10
. . . . if (Rank1 = 1) then Rank1 = 11
. . . . for (Card2 = Card1 + 1 to 50)
. . . . {
. . . . . . . . Rank2 = floor(Card2 / 4) + 1
. . . . . . . . if (Rank2 > 9) then Rank2 = 10
. . . . . . . . if (Rank2 = 1) then Rank2 = 11
. . . . . . . . for (Card3 = Card2 + 1 to 51)
. . . . . . . . {
. . . . . . . . . . . . Rank3 = floor(Card3 / 4) + 1
. . . . . . . . . . . . if (Rank3 > 9) then Rank3 = 10
. . . . . . . . . . . . if (Rank3 = 1) then Rank3 = 11
. . . . . . . . . . . . FlopValue = Rank1+ Rank2 + Rank3
. . . . . . . . . . . . FlopCounts[FlopValue] = FlopCounts[FlopValue] + 1
. . . . . . . . }
. . . . }
}
When this is done, FlopCounts[N] contains the number of flops where the total is N
It is then a simple matter, for each N, of adding up FlopCounts[3] through FlopCounts[N-1] to determine the number < N, and adding up FlopCounts[N+1] through FlopCounts[33] to determine the number > N.
(Actually, in this case, you can start with FlopCounts[6], as that is the smallest possible flop total if all Aces are 11.)
Quote: scottuga44This is a bet I saw two guys make at a poker game. I was trying to figure out how big an edge the one guy had but was unsure how to set it up.
Standard poker deck, all cards are worth the same as in blackjack. Aces are always 11. If the three flop cards equal 21 it's push. How big of an edge does the person have who is betting the flop will be 22 or greater?
The person betting 22 or more has a 18.13574661% edge.
Quote: RSSpecifically, though? How'd you figure 22,100 different 3-card flops combinations? How'd you come up with 12,3000 are higher than 21? Surely you didn't write down each combination "2, 2, 2 is 6. 2, 2, 3 is 7. 2, 2, 4 is 8...." and go from there.
I'm not trying to say you're wrong (or right). But I'm genuinely interested in how you came up with those figures.
I used a spread sheet.
Quote: scottuga44This is a bet I saw two guys make at a poker game. I was trying to figure out how big an edge the one guy had but was unsure how to set it up.
Standard poker deck, all cards are worth the same as in blackjack. Aces are always 11. If the three flop cards equal 21 it's push. How big of an edge does the person have who is betting the flop will be 22 or greater?
I get 18.13574661%, too.
Flop Pays Permutations
6 -1 24
7 -1 144
8 -1 288
9 -1 552
10 -1 816
11 -1 1200
12 -1 1608
13 -1 2112
14 -1 3072
15 -1 4440
16 -1 5376
17 -1 6480
18 -1 7176
19 -1 8016
20 -1 8448
21 0 9048
22 1 10800
23 1 11136
24 1 9912
25 1 8832
26 1 7584
27 1 6648
28 1 5520
29 1 4704
30 1 5040
31 1 3024
32 1 576
33 1 24
Total 132600
Quote: mickeycrimmThere is also the hustle of betting a guy that a 6, 7, or 8 will come on flop which I put at about 60%.
I think you were about 4.71%
high Mickey, or 3 sheets to the wind ?
Quote: mickeycrimmThere is also the hustle of betting a guy that a 6, 7, or 8 will come on flop which I put at about 60%.
Second question from Ask the Wiz 196
Quote: questionPlaying last night, one of the players, an old crafty scruffy aggressive player, was challenging the table to make even money side bets on the flop. This old curmudgeon was betting that one of the three cards on the flop would be either an ace, deuce, or jack (sometimes he would change the 3 identifiable cards). What are the odds of this bet? Your sage wisdom would be greatly appreciated.
Quote: answerBefore any cards are seen, the probability of any three ranks not appearing on the flop are combin(40,3)/combin(52,3) = 9880/22100 = 44.71%. So this guy had a 10.59% advantage.
