Bankroll and Risk of Ruin
I mainly play BJ ($5-10), Craps $5-10 PL, w/2x odds, or maybe some 3-Card poker $5-$10. What kind of bankroll should I be taking with me on my trips?
Quote: gamblerDefine a "trip" or a "session". For me, when I travel to Las Vegas, I spend 4 or 5 days there and play 8 to 10 hours per day. Your bankroll requirements would be smaller if a session is only an hour or two. Also how many sessions would you play in a day?
Lets say a "session" would be 6 hours of actual playing time.
I'll assume 1% house advantage, approximately $10 standard deviation/hand, and 80 hands/hr.
In 6 hours, your average loss will be $10 x 0.01 x 80 x 6 = $48.
Over the same period, your standard deviation is $10 x sqrt(80 x 6) = $220.
So your typical outcome over this time frame will be between a loss of $270 and a win of $170.
The worst case loss you would expect would be three standard deviations from the mean, or about $700.
So your bankroll for this session should be between $300 and $700, depending on how aggressive you are. I'd split the difference, and go in with $500. This will give you about a 97% chance of surviving 6 hours without going broke.
Quote: PapaChubbyHere's the math for $10 BJ.
I'll assume 1% house advantage, approximately $10 standard deviation/hand, and 80 hands/hr.
In 6 hours, your average loss will be $10 x 0.01 x 80 x 6 = $48.
Over the same period, your standard deviation is $10 x sqrt(80 x 6) = $220.
So your typical outcome over this time frame will be between a loss of $270 and a win of $170.
The worst case loss you would expect would be three standard deviations from the mean, or about $700.
So your bankroll for this session should be between $300 and $700, depending on how aggressive you are. I'd split the difference, and go in with $500. This will give you about a 97% chance of surviving 6 hours without going broke.
Great thanks. I'd like to do the same calculation for craps, but I have no idea what a good average number of comeout rolls to use. Is it correct that I could just plug in the 0.6% HA for PL w/2x odds, along with the number of comeouts per hour to determine?
Expected loss at 1% = $25 x 80 x 8 x .01 = $160.
Typical outcome would be loss of $800 and a win of $470.
Three standard deviations = 1,900.
Bankroll should be about $2,060 per session to avoid ruin with a 3% uncertainty.
Expected loss at 1% = $25 x 80 x 8 x .01 = $160.
Typical outcome would be loss of $800 and a win of $470.
Three standard deviations = 1,900.
Bankroll should be about $2,060 per session to avoid ruin with a 3% uncertainty.
Quote: ruascottGreat thanks. I'd like to do the same calculation for craps, but I have no idea what a good average number of comeout rolls to use. Is it correct that I could just plug in the 0.6% HA for PL w/2x odds, along with the number of comeouts per hour to determine?
The method is certainly the same. But its difficult to calculate standard deviation and rate of play because of all the various possibilities.
I'd use .6% HA, and estimate 40 come outs/hr with a SD of $30. These are just some ballpark guesses on my part.
Average loss is $30 x 0.006 x 40 x 6 = $43.
SD = $30 x sqrt(40 x 6) = $465.
So you'd probably want about $1000 bankroll for 6 hours of $10 PL craps with 2x odds.
Quote: boymimboAnd for Maloru, if he's playing $25 blackjack for 8 hours, the SD is $25 x sqrt(80 x 8) = $632.
Expected loss at 1% = $25 x 80 x 8 x .01 = $160.
Typical outcome would be loss of $800 and a win of $470.
Three standard deviations = 1,900.
Bankroll should be about $2,060 per session to avoid ruin with a 3% uncertainty.
IMO, a 3-sigma bankroll is pretty conservative, and I think corresponds to a risk-of-ruin uncertainty more like 0.5%. I know a lot more about normal distributions than risk-of-ruin, but I think the Wizard's ROR table confirms ROR for the scenario you describe as ~ 0.5%. I'd use a 2-sigma number for more of a 3% ROR estimate, so I think a bankroll of around $1500 is reasonable.
I typically plan my gambling sessions to last something like an hour, and my goal is to either lose my stake or double it in that period of time. I aim for a risk-of-ruin around 30% with an associated possibility-of-success around 30% for each session. I'll typically buy in with about 10 bets ($100 buy-in for $10 BJ).
Of course, if you play my strategy 6 times per day, you're getting into the same ballpark as the original poster's approach. There are probably some subtle differences, like me leaving the table early if I bust out or succeed during the session.
I guess this strategy was more for taking trips to Vegas or something similar where the expectation is going to be to play more.
Quote: PapaChubbyThe method is certainly the same. But its difficult to calculate standard deviation and rate of play because of all the various possibilities.
Actually, you can calculate the SD quite easily. For $10 pass, double odds, the average bet is $23.34 and the SD is $28.57. So, for 240 bets, ev is -$33.94 and SD is $442.67.
Quote: PapaChubbyI'd use .6% HA, and estimate 40 come outs/hr with a SD of $30. These are just some ballpark guesses on my part.
Average loss is $30 x 0.006 x 40 x 6 = $43.
SD = $30 x sqrt(40 x 6) = $465.
So you'd probably want about $1000 bankroll for 6 hours of $10 PL craps with 2x odds.
Your estimates are very good!
I ran a sim of 20,000 sessions of 240 bets, and the bust rate was 2.78%; the bankroll was doubled 2.12% of the sessions.
There were 9317 winning sessions, 43 broke even and 10640 lost.
For the same number of bets and bankroll, I tried $10 pass with 5X odds. The bust rate was 27.56% and the bankroll was doubled 25.77%. There were 9722 winning sessions, 15 breakeven and 10263 losing sessions. Of course, more variance increases the chance of winning, given the same ev.
Cheers,
Alan Shank
Quote: PapaChubbyBTW, I should state for the record that I don't condone this sort of gambling strategy. The longer you play, the more hands you play and the more the house advantage accumulates. Also, if you devise your bankroll to minimize risk of ruin, you are also minimizing your chances of winning any significant amount of money.
I typically plan my gambling sessions to last something like an hour, and my goal is to either lose my stake or double it in that period of time. I aim for a risk-of-ruin around 30% with an associated possibility-of-success around 30% for each session. I'll typically buy in with about 10 bets ($100 buy-in for $10 BJ).
You cannot have an equal probability of winning a given amount as losing it, except in a zero-HA game. In craps, the probability of winning a given amount in a certain number of bets is always lower than the probability of losing that amount. The standard deviations are on either side of the expected loss. I'm pretty sure you understand that.
Cheers,
Alan Shank
Quote: goatcabinYou cannot have an equal probability of winning a given amount as losing it, except in a zero-HA game. In craps, the probability of winning a given amount in a certain number of bets is always lower than the probability of losing that amount. The standard deviations are on either side of the expected loss. I'm pretty sure you understand that.
Cheers,
Alan Shank
Thanks, I do understand this. Note that I said "around" 30% ;-) . This is exactly why I place largish bets and keep my number of hands played to a minimum. This keeps the variance large relative to the house edge, and allows me to keep the probabilities as close to even as possible. Sure, my chance of doubling up may be 25% vs. 35% risk of ruin, but its not 10% vs. 70% like so many players that I see.
Quote: goatcabinActually, you can calculate the SD quite easily. For $10 pass, double odds, the average bet is $23.34 and the SD is $28.57. So, for 240 bets, ev is -$33.94 and SD is $442.67.
Your estimates are very good!
I ran a sim of 20,000 sessions of 240 bets, and the bust rate was 2.78%; the bankroll was doubled 2.12% of the sessions.
There were 9317 winning sessions, 43 broke even and 10640 lost.
For the same number of bets and bankroll, I tried $10 pass with 5X odds. The bust rate was 27.56% and the bankroll was doubled 25.77%. There were 9722 winning sessions, 15 breakeven and 10263 losing sessions. Of course, more variance increases the chance of winning, given the same ev.
Cheers,
Alan Shank
Thanks for the simulation support! I'm a bit surprised to see the likelihood of doubling up so close to the risk of ruin. But I guess the mean is still less than 1/10th of a SD, so it shouldn't surprise me too much.
Extreme geekiness to follow...
I've been doing some thinking about the relationship between the normal distribution and risk of ruin. It seems to me that there is probably a pretty good correlation as long as the bankroll is something like 2 or 3 standard deviations.
Theoretically, the normal distribution tells you what the likelihood is that you are at a certain win/loss level at the end of the session. Risk of ruin adds the possibility that you exceeded this level at some time during the session, even though you might have won some back if you'd had the bankroll to continue.
For a large bankroll like 2 or 3 standard deviations, the likelihood of exhausting the bankroll at the end of the session is quite low. So the likelihood of exhausting it earlier in the session is even lower, and does not significantly affect the accuracy of the estimate.
This does not hold true for smaller bankrolls. Consider playing $10 blackjack with a $10 bankroll. Using a normal distribution to estimate risk of ruin will result in something like 50% for most typical session lengths. At the end of 10 or 100 hands, you'll be up about half the time and down about half the time. However, it seems obvious to me that the real risk of ruin in this situation is much higher. On many of those session which ended with a net win, the player was probably down more than $10 at some point during the session.
Quote: PapaChubbyThanks for the simulation support! I'm a bit surprised to see the likelihood of doubling up so close to the risk of ruin. But I guess the mean is still less than 1/10th of a SD, so it shouldn't surprise me too much.
Extreme geekiness to follow...
I've been doing some thinking about the relationship between the normal distribution and risk of ruin. It seems to me that there is probably a pretty good correlation as long as the bankroll is something like 2 or 3 standard deviations.
Theoretically, the normal distribution tells you what the likelihood is that you are at a certain win/loss level at the end of the session. Risk of ruin adds the possibility that you exceeded this level at some time during the session, even though you might have won some back if you'd had the bankroll to continue.
For a large bankroll like 2 or 3 standard deviations, the likelihood of exhausting the bankroll at the end of the session is quite low. So the likelihood of exhausting it earlier in the session is even lower, and does not significantly affect the accuracy of the estimate.
This does not hold true for smaller bankrolls. Consider playing $10 blackjack with a $10 bankroll. Using a normal distribution to estimate risk of ruin will result in something like 50% for most typical session lengths. At the end of 10 or 100 hands, you'll be up about half the time and down about half the time. However, it seems obvious to me that the real risk of ruin in this situation is much higher. On many of those session which ended with a net win, the player was probably down more than $10 at some point during the session.
There's a formula I found for the probability of getting behind and never catching up:
p = 1 - (2 * P)
where p is the probability of getting behind and never catching up and P is the probability of winning a single bet.
Having a relatively low "stop-loss" increases the number of losing sessions, partly because you give up your chance of combing back and ending up ahead.
I am preparing a massive blog post based on another computer program I wrote. It compares results of 40,000 sessions each, broken down this way:
flat bet w/stop loss w/win goal +.5 on win +.5 on loss
just pass
double odds
3,4,5X odds
The idea here is to isolate the effects of each factor, the odds, having a stop loss, having a win goal, increasing bets on a win and increasing them on a loss.
Coming soon!
Cheers,
Alan Shank
Quote: goatcabinThere's a formula I found for the probability of getting behind and never catching up:
p = 1 - (2 * P)
Alan: I have generally found that a simpler formula works almost as reliably in a commercial casino:
p = 1
;-)
Quote: PapaChubbyExtreme geekiness to follow...
I've been doing some thinking about the relationship between the normal distribution and risk of ruin. It seems to me that there is probably a pretty good correlation as long as the bankroll is something like 2 or 3 standard deviations.
Theoretically, the normal distribution tells you what the likelihood is that you are at a certain win/loss level at the end of the session. Risk of ruin adds the possibility that you exceeded this level at some time during the session, even though you might have won some back if you'd had the bankroll to continue.
The normal distribution approximates the binomial distribution, which gives the exact distribution of results of a series of win/loss events. So for a large # of trials that you can use the normal distribution. The extreme left of the normal curve gives you a good estimate of your RoR.
Quote: goatcabinActually, you can calculate the SD quite easily. For $10 pass, double odds, the average bet is $23.34 and the SD is $28.57. So, for 240 bets, ev is -$33.94 and SD is $442.67.
So based upon my rudimentary statistics knowledge, would there not be a 95% chance that my 6 hour sessions would fall between -$919.28 and +$851.40? (+/- 2 SD Assuming we use the normal distribution as an estimate)....and therefore only about a 2.5% chance that I would exceed a loss of $919.28?
Its been a handful of years since I've done any stats, but its SLOWLY coming back to me....
Quote: ruascottGreat thread guys, and thanks for your input.
So based upon my rudimentary statistics knowledge, would there not be a 95% chance that my 6 hour sessions would fall between -$919.28 and +$851.40? (+/- 2 SD Assuming we use the normal distribution as an estimate)....and therefore only about a 2.5% chance that I would exceed a loss of $919.28?
Its been a handful of years since I've done any stats, but its SLOWLY coming back to me....
Yes, that is exactly the way I am evaluating the problem.
Quote: ruascottGreat thread guys, and thanks for your input.
So based upon my rudimentary statistics knowledge, would there not be a 95% chance that my 6 hour sessions would fall between -$919.28 and +$851.40? (+/- 2 SD Assuming we use the normal distribution as an estimate)....and therefore only about a 2.5% chance that I would exceed a loss of $919.28?
Its been a handful of years since I've done any stats, but its SLOWLY coming back to me....
Check!
Cheers,
Alan Shank
Just curious, what would the EV and SD be of the same bet level, but betting on the 'wrong' side? ($10 DP w/2x odds).
Have you run simulations for the DP? I know the HA is slightly lower than PL, just curios on the variance.
Quote: ruascottAlan,
Just curious, what would the EV and SD be of the same bet level, but betting on the 'wrong' side? ($10 DP w/2x odds).
Have you run simulations for the DP? I know the HA is slightly lower than PL, just curios on the variance.
Very, very little difference. For 240 bets, the ev is -$33.66, SD $448.37. In 20,000 sessions, the bust rate was 3.09%, just a little bit higher than with pass. I think the reason for this is that you have to lay the long end of the odds, so if you have bad luck initially, the loss builds up faster. Keep in mind that the ev for 1980 DP bets is -27 units; for 1980 passline bets, it's -28. That's damn little difference.
Cheers,
Alan Shank
Quote: goatcabinThere's a formula I found for the probability of getting behind and never catching up:
p = 1 - (2 * P)
where p is the probability of getting behind and never catching up and P is the probability of winning a single bet.
O.K., Alan, my first reply on your formula was admittedly a bit smart-ass, but it was just something that immediately jumped to mind when I read it.
On second reading, I have a more serious comment/question: doesn't there have to be something much more to this? Doesn't it have to include something about the payout rate on individual winning wagers? For an unrealistic example, suppose there were a wager that had only 1/3 probability of winning but paid out 4 to 1. I think you would very likely win significantly almost every session with only a modest starting bankroll, but your formula suggest that you would have a 1/3 probability of getting behind and never catching up. I know I will never find such a wager opportunity in the real world, but I think this illustrates that something major is missing from the formula. Conversely, a wager with an attractive probability of winning but a very poor payout could frequently lead to ruin, I think. Comment?
Quote: Doc
On second reading, I have a more serious comment/question: doesn't there have to be something much more to this? Doesn't it have to include something about the payout rate on individual winning wagers? For an unrealistic example, suppose there were a wager that had only 1/3 probability of winning but paid out 4 to 1. I think you would very likely win significantly almost every session with only a modest starting bankroll, but your formula suggest that you would have a 1/3 probability of getting behind and never catching up. I know I will never find such a wager opportunity in the real world, but I think this illustrates that something major is missing from the formula. Conversely, a wager with an attractive probability of winning but a very poor payout could frequently lead to ruin, I think. Comment?
this pretty much describes poker.
Quote: DeMangoAlan are these programs written on the Win Craps program? If not what language are you using? I wonder if there is other programming help for WinCraps, the notes are just not enough to get through my thick skull!!
I use WinCraps for some of my simulations, but I also spent 15 years as a "C" programmer and I have a suite of programs to simulate various strategies, calculate ev, SD, skew, etc. and run binomial expansions.
As far as WinCraps auto-bet files are concerned, you can download a bunch of them free from www.cloudcitysoftware.com. The example files give you a pretty good starting point. I can also send you some of mine.
Cheers,
Alan Shank
Quote: DocO.K., Alan, my first reply on your formula was admittedly a bit smart-ass, but it was just something that immediately jumped to mind when I read it.
On second reading, I have a more serious comment/question: doesn't there have to be something much more to this? Doesn't it have to include something about the payout rate on individual winning wagers? For an unrealistic example, suppose there were a wager that had only 1/3 probability of winning but paid out 4 to 1. I think you would very likely win significantly almost every session with only a modest starting bankroll, but your formula suggest that you would have a 1/3 probability of getting behind and never catching up. I know I will never find such a wager opportunity in the real world, but I think this illustrates that something major is missing from the formula. Conversely, a wager with an attractive probability of winning but a very poor payout could frequently lead to ruin, I think. Comment?
Bzzzzzzzzzzzzzzzzzzt!!! Sorry. I forgot to mention that this works only for even-money bets, so it's applicable to the line bets in craps, without odds.
Good catch!
Cheers,
Alan Shank
silly
Sally
Just from practical experience: 40 avg. bets in BJ will get you through most but not all short sessions. These 40 should be the money you've written off already - think of it as betting it all the moment you buy in, and the rest being a game about how much of that or more you get to keep.
If your losses get over 40 bets, it's just not your day (this sounds like superstition, but it isn't baseless - you'll have bad beats getting to -40, and will likely go on tilt, spiking the losses). 40 bets is half an hour of non-stop losing, even I couldn't keep my cool through that. Unless I'm playing drunk, in which case see above.
If you want a comfortable buffer not to leave broke, get through multiple games, pay off the EV and so on, about 100 units.
Most use the normal distribution to calculate the standard deviation for a wager and number of trials.Quote: ruascottWhat is an appropriate bankroll for different games/betting styles that would make risk of ruin at least somewhat improbable in a given session?
EV and SD
"at least somewhat improbable"
This is classic!
This has NO meaning and is exactly why Papa and Sally shows the math, and why one should not go by experience and guessing about XX avg bets.
Learn to do the math. It ain't hard.
Is busting out 1 in 10 sessions right for you
or is 1 in 20 better?
What bankroll is needed to withstand the downswings??
would you rather just bust out 1 out of 100 sessions on average.
Only the math or simulations can give the correct answer.
40 average bets returns what kind of RoR exactly?
I guess so.
I gots a feeling more is coming
Enjoy!
40 average bets gives you a risk of ruin of about .0001 in an hour, or about .1 in a reasonably long session (6 hr with breaks).
It's also a psychological threshold, at standard bet of $25, 40*$25=$1,000, and unless you really don't care, you're more likely to start losing your game as you cross it.
The above implies that 40 bets is your soft stop-loss, which you can cross in splits, doubles and finishing hot decks, not the last red cent.
Yes I agree with you, average bets can be different for different styles of play in BJ. even in other games for the average gambler. And average bet is easy to use in most math Risk of Ruin formulas.Quote: P90Just to clarify, "average bet" implies you're either strictly flat-betting or using a weak count at break-even spread (e.g A-5 1-4) and varying your bet based on the count and nothing else ever. If you increase it to "get even", then obv you're going to bust much larger bankrolls with ease.
40 average bets gives you a risk of ruin of about .0001 in an hour, or about .1 in a reasonably long session (6 hr with breaks).
It's also a psychological threshold, at standard bet of $25, 40*$25=$1,000, and unless you really don't care, you're more likely to start losing your game as you cross it.
I bet 1 dollar Sally will be showing up later to show the math on your 40 bet average and the RoR numbers for a session of play, either 1 hour or 6 hours.
We will see who is closer to actual results, including the 2 posters on page one.
rules of thumbs are great for many and there are many rules of thumbs.
silly
Quote: guido111I bet 1 dollar Sally will be showing up later to show the math on your 40 bet average and the RoR numbers for a session of play, either 1 hour or 6 hours.
Oh, it is 0.0001/0.1. It's just that 40 is a good round number that gives you $1,000 bankroll, 1/10,000 to lose it in an hour, 1/10 to lose it in a session.
If you look at this table - https://wizardofodds.com/games/blackjack/appendix/12/ - you'll see 45 for 100 bets (an hour is a little less, so 40), or 39 for 400 (which is a 6 hr session with breaks).
