Double or nothing
October 29th, 2017 at 3:19:19 PM
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This poker/math question has two parts.
I have been playing a game of online Double or Nothing hold'em tournaments that cost $1.50 to play. If I win, I get my money back plus $1.20. I have been winning 75% of the games. How much money should I have to only run one chance out of 20 of losing all of my money?
Part 2 is this.
In addition to that amount of money, how much do I need to play the next most expensive game?
This game cost $3.50 to play and you win your money back plus $2.90 if you win. Assuming I win 75% of those games and I'm willing to take a risk of one chance in six of losing this additional money and going back to the $1.50 games.
30 years ago when playing blackjack there was a formula for figuring this out involving standard deviation, but I have forgotten it.
Thank you, Lee Hansen.
I have been playing a game of online Double or Nothing hold'em tournaments that cost $1.50 to play. If I win, I get my money back plus $1.20. I have been winning 75% of the games. How much money should I have to only run one chance out of 20 of losing all of my money?
Part 2 is this.
In addition to that amount of money, how much do I need to play the next most expensive game?
This game cost $3.50 to play and you win your money back plus $2.90 if you win. Assuming I win 75% of those games and I'm willing to take a risk of one chance in six of losing this additional money and going back to the $1.50 games.
30 years ago when playing blackjack there was a formula for figuring this out involving standard deviation, but I have forgotten it.
Thank you, Lee Hansen.
October 29th, 2017 at 4:26:09 PM
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Is there a limit - either total number of games, or some "stop when I reach this amount of money" level - after which you stop?
October 29th, 2017 at 6:16:18 PM
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No limits here. Just want to build bankroll as quickly as possible while limiting my chances of losing to the above risk.
Later there are $15 and $30 games.
My SWAG(Scientific Wild Ass Guess) is a bankroll of$35 will be required to stay within those risks but I could be WAY off.
Later there are $15 and $30 games.
My SWAG(Scientific Wild Ass Guess) is a bankroll of$35 will be required to stay within those risks but I could be WAY off.
October 29th, 2017 at 7:19:56 PM
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Quote: LeftyBlue30 years ago when playing blackjack there was a formula for figuring this out involving standard deviation, but I have forgotten it.
I think you are referring to the Risk of Ruin formula:
The probability P of going bankrupt given a starting bankroll B and a bet that has a mean result of M and a standard deviation of S is:
P = (2 / (1 + M/R) - 1)B/R
where R = sqrt(M2 + S2)
Usually, you use it to determine P based on an initial bankroll B, but in this case, you want to determine the initial bankroll B based on a value P (in this case, 1/20), so the formula is rewritten as:
B = R ln P / ln (2 / (1 + M/R) - 1)
In your initial game, your value is +1.2 75% of the time, and -1.5 25% of the time, so M = (3/4 x 6/5 - 1/4 x 3/2) = 21/40, S = 27 sqrt(3) / 40, and R = 3 sqrt(73) / 20.
I get the initial bankroll to be 4.4112.
However, your actual stop point is when you drop below the bet amount, so add 1.5 to it; the initial bankroll really should be 5.9112.
October 30th, 2017 at 7:33:45 AM
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THANK YOU!!!
The fact I only need $6 is surprising. I'll figure $10 to avoid the emotional elements (I just, for the first time, lost three in a row).
Now for the second part of my problem I just plug in the numbers for the $3.50 games and add that to $10. When I reach that bankroll I can play the next level game.
ETCETERA!
You have been MOST helpful.
Lee Hansen aka Lefty Blue
The fact I only need $6 is surprising. I'll figure $10 to avoid the emotional elements (I just, for the first time, lost three in a row).
Now for the second part of my problem I just plug in the numbers for the $3.50 games and add that to $10. When I reach that bankroll I can play the next level game.
ETCETERA!
You have been MOST helpful.
Lee Hansen aka Lefty Blue
October 30th, 2017 at 8:46:50 AM
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Quote: LeftyBlueThe fact I only need $6 is surprising.
Not really - with $4.50 or more, you need to lose the first three in a row to go bust without winning, and that has a probability of 1/64, which is well below your threshhold of 1/20.
October 31st, 2017 at 10:34:52 AM
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So, I went to punch in numbers for a different win rate and realized I was a little lost.
If you get the time, could you walk me through it with a little more detail. It has been 50 years since high school algebra. Thank you again Lee
If you get the time, could you walk me through it with a little more detail. It has been 50 years since high school algebra. Thank you again Lee
October 31st, 2017 at 11:11:21 AM
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Quote: LeftyBlueSo, I went to punch in numbers for a different win rate and realized I was a little lost.
If you get the time, could you walk me through it with a little more detail. It has been 50 years since high school algebra. Thank you again Lee
Here are the numbers for your second game (75% chance of winning 2.9; 25% chance of losing 3.5; bankroll needed for a 1/6 chance of busting out)
Mean = (75% x (+2.9) + 25% x (-3.5)) = 1.3
StdDev = sqrt(75% x (2.9 - 1.3)2 + 25% x ((-3.5) - 1.3)2)
= sqrt(7.68) = 2.7712813
R = sqrt(Mean2 + SD2) = sqrt(1.69 + 7.68) = 3.0610456
Given P = 1/6:
B = R ln P / ln (2 / (1 + Mean / R) - 1)
= 3.0610456 x ln (1/6) / ln (2 / (1 + (1.3 / 3.0610456)) - 1)
= 6.05
Again, you need to add one loss to that, as this includes situations where you can bet with less than 3.5 left in your bankroll:
Initial bankroll = 6.05 + 3.5 = 9.55
Here is a table for various probabilities of busting out on the 3.50/2.90 game:
| Prob of busting | Bankroll |
|---|---|
| 1/2 | 5.84 |
| 1/3 | 7.21 |
| 1/4 | 8.18 |
| 1/5 | 8.93 |
| 1/6 | 9.55 |
| 1/7 | 10.07 |
| 1/8 | 10.52 |
| 1/9 | 10.92 |
| 1/10 | 11.27 |
| 1/11 | 11.59 |
| 1/12 | 11.89 |
| 1/13 | 12.16 |
| 1/14 | 12.41 |
| 1/15 | 12.64 |
| 1/16 | 12.86 |
| 1/17 | 13.06 |
| 1/18 | 13.26 |
| 1/19 | 13.44 |
| 1/20 | 13.61 |
October 31st, 2017 at 2:06:29 PM
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Thanks!
November 1st, 2017 at 10:38:00 AM
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Do you have a risk of ruin formula for a negative expectation game and a fixed number of bets - like Blackjack for instance?
For example, I am going to play for 4 hours tonight, estimate 200 bets total. How much bankroll do I need for a 90 % chance of survival?
Lets call blackjack a bet with .43 chance of winning, winning payout 1.31 to 1, house edge 0.67%, SD 1.14.
For example, I am going to play for 4 hours tonight, estimate 200 bets total. How much bankroll do I need for a 90 % chance of survival?
Lets call blackjack a bet with .43 chance of winning, winning payout 1.31 to 1, house edge 0.67%, SD 1.14.
It’s all about making that GTA
November 1st, 2017 at 11:13:07 AM
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I don't think there is a generic formula for a situation like this.
There probably is for "flat betting," but with blackjack, you take things like splitting and doubling - which not only affect the potential win, but the potential loss - into account.
There probably is for "flat betting," but with blackjack, you take things like splitting and doubling - which not only affect the potential win, but the potential loss - into account.
November 1st, 2017 at 11:30:50 AM
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That is why I equated it to a binomial.Quote: ThatDonGuy
There probably is for "flat betting," but with blackjack, you take things like splitting and doubling - which not only affect the potential win, but the potential loss - into account.
It’s all about making that GTA
November 1st, 2017 at 4:05:44 PM
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Still, I don't think there's a universal formula that takes the boundary condition (200 games) into account.
I think I can do it through simulation, although I am not entirely sure of the accuracy of my normal random variable generator.
Assuming a HE of 0.67% means a mean of -0.0067 (using a base bet of 1), I get 89% chance of survival with an initial bankroll of 24, and 90.2% with 25.
Remember, that's 24x and 25x the base bet - on a $10 table, that's $240 and $250, respectively.
I think I can do it through simulation, although I am not entirely sure of the accuracy of my normal random variable generator.
Assuming a HE of 0.67% means a mean of -0.0067 (using a base bet of 1), I get 89% chance of survival with an initial bankroll of 24, and 90.2% with 25.
Remember, that's 24x and 25x the base bet - on a $10 table, that's $240 and $250, respectively.
November 1st, 2017 at 9:20:02 PM
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Well Don, I guess you'll just have to make a spreadsheet. Hell, I'd pony up a dollar for this.
The result would be bankroll required.
User would input odds and return on winning bets and on losing bets.
Minimum and maximum bet (this would apply to blackjack) and anything else you can think of.
You game?
Thanks from all the "low rollers".
The result would be bankroll required.
User would input odds and return on winning bets and on losing bets.
Minimum and maximum bet (this would apply to blackjack) and anything else you can think of.
You game?
Thanks from all the "low rollers".
November 2nd, 2017 at 6:07:06 AM
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The main problem with this is, the formula only works if the expected return > 0 - otherwise, eventually you will lose everything.
For games with expected return < 0 and a cap on how many times you play, this has to be worked out "one step at a time" (if you have a specified set of results and probabilities) or through simulation (if you have a mean and standard deviation).
There is a formula for "even money, win or lose" games with expected return < 0 if there is a stopping point based on how large your bankroll gets.
For games with expected return < 0 and a cap on how many times you play, this has to be worked out "one step at a time" (if you have a specified set of results and probabilities) or through simulation (if you have a mean and standard deviation).
There is a formula for "even money, win or lose" games with expected return < 0 if there is a stopping point based on how large your bankroll gets.
November 2nd, 2017 at 12:42:23 PM
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For 10% risk of ruin I get 28 units via simulation.
I don't know of a formula for this situation either, but I have found a pretty accurate way to estimate it. When the house edge is low, say 1.5% or less (the only games I would play), the risk of ruin is very close to double the probability of finishing the session at zero or less (have to add to initial bankroll to finish session).
In this case, after 200 bets the expected result is 28 - 200 * .0067 = 26.66 +/- the standard deviation of 16.17 ((.43*.57*200)^.5 * 2.31).
So zero is 26.66 / 16.17 = 1.65 deviations south of the expectation. There's a 5% chance of finishing the session 1.65 or more deviations left of the expectation, half of the 10% risk of ruin.
I don't know of a formula for this situation either, but I have found a pretty accurate way to estimate it. When the house edge is low, say 1.5% or less (the only games I would play), the risk of ruin is very close to double the probability of finishing the session at zero or less (have to add to initial bankroll to finish session).
In this case, after 200 bets the expected result is 28 - 200 * .0067 = 26.66 +/- the standard deviation of 16.17 ((.43*.57*200)^.5 * 2.31).
So zero is 26.66 / 16.17 = 1.65 deviations south of the expectation. There's a 5% chance of finishing the session 1.65 or more deviations left of the expectation, half of the 10% risk of ruin.
Last edited by: Ace2 on Nov 2, 2017
It’s all about making that GTA
November 3rd, 2017 at 8:57:11 AM
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So Don,
Any chance of a spreadsheet as described above?
I have set aside a $ to buy.
In all seriousness it could actually be a somewhat profitable thing.
But if not, thanks again for all your help.
Lee aka LeftyBlue
Any chance of a spreadsheet as described above?
I have set aside a $ to buy.
In all seriousness it could actually be a somewhat profitable thing.
But if not, thanks again for all your help.
Lee aka LeftyBlue
