Simple Mathematical Bankroll Formula?

Is there a simple mathematical formula a gambler can use to calculate what their bankroll should be for a trip with a very tiny chance of going broke before the trip is over.

For example:

Average Bet * House Advantage/Edge on Game * Hours Played = Bankroll for trip?

$100 average bet * 2% house edge * 16 hours = ??? etc.

Thanks math people.

Quote: gambler

Is there a simple mathematical formula a gambler can use to calculate what their bankroll should be for a trip with a very tiny chance of going broke before the trip is over.

For example:

Average Bet * House Advantage/Edge on Game * Hours Played = Bankroll for trip?

$100 average bet * 2% house edge * 16 hours = ??? etc.

Thanks math people.

Sure. You're missing two elements: the number of bets per hour and the type of bet. For the type of bet, I'll assume you're playing something that is roughly a 1:1 proposition: blackjack, craps pass line, red/black, etc. For these games, the standard deviation of a single bet is roughly equal to the size of the bet. If you're playing jackpot games, individual roulette numbers, etc. the results will be different.

Your average loss for the trip will be:

Average Loss = Average Bet x house advantage x bets/hour x hours played.

The standard deviation for your trip will be:

Standard Deviation = Average Bet x sqrt(bets/hour x hours played)

The probability of going broke as a function of bankroll size is:

Average Loss + 1 x Standard Deviation: 16%
Average Loss + 2 x Standard Deviations: 2.5%
Average Loss + 3 x Standard Deviations: 0.15%

I'd use 2 standard deviations as a guideline.

So for blackjack at 70 hands/hour, $100 average bet, 1% house advantage for 16 hours:

Average loss = $1120
Standard deviation = $3347
Bankroll = $7813

You can also refer to the wizard's game comparison - https://wizardofodds.com/houseedge

He lists the standard deviations for most common games - but the previous post is a good approximation too.

The standard rule is that for 4 hours of play, 40x your minimum bet is needed to ride out the variance on a table game.

-B

Quote: toastcmu

The standard rule is that for 4 hours of play, 40x your minimum bet is needed to ride out the variance on a table game.

At the risk of stating the obvious ultimately you can't beat the EV. If you set your bankroll big enough that you can ride out the variance so that you come back to even by the end of the play, sooner or later you are going to lose the big bankroll. If you bring $5K to cover your $100 bets, then one of your trips you are going to be very upset.

So if you are trying to meet a 4 hour requirement to cover your comps, it will work until it doesn't work. The other factor is that people who are not on a time budget will often times gamble longer than if they lost their smaller bankroll (as opposed to going to the ATM or taking out a marker). The longer you gamble the more the casino grinds.

Quote: pacomartin

At the risk of stating the obvious ultimately you can't beat the EV. If you set your bankroll big enough that you can ride out the variance so that you come back to even by the end of the play, sooner or later you are going to lose the big bankroll. If you bring $5K to cover your $100 bets, then one of your trips you are going to be very upset.

The longer you gamble the more the casino grinds.

Can't disagree with you - the grind is what pays for everything in the casino. :)

It all depends on how "degenerate" you want to be and whether you consider comps part of you EV. If you play big money, unless you're an AP, you have to expect big losses to go with the big wins. Money management is key for me if I'm playing for extended periods. I know it doesn't affect the long run EV, but it makes me feel better in. The short term.

-B

Quote: PapaChubby

Sure. You're missing two elements: the number of bets per hour and the type of bet. For the type of bet, I'll assume you're playing something that is roughly a 1:1 proposition: blackjack, craps pass line, red/black, etc. For these games, the standard deviation of a single bet is roughly equal to the size of the bet. If you're playing jackpot games, individual roulette numbers, etc. the results will be different.

Your average loss for the trip will be:

Average Loss = Average Bet x house advantage x bets/hour x hours played.

The standard deviation for your trip will be:

Standard Deviation = Average Bet x sqrt(bets/hour x hours played)

The probability of going broke as a function of bankroll size is:

Average Loss + 1 x Standard Deviation: 16%
Average Loss + 2 x Standard Deviations: 2.5%
Average Loss + 3 x Standard Deviations: 0.15%

I'd use 2 standard deviations as a guideline.

So for blackjack at 70 hands/hour, $100 average bet, 1% house advantage for 16 hours:

Average loss = $1120
Standard deviation = $3347
Bankroll = $7813

Excellent info, Papa!
I cranked some numbers for $5 pass, 3, 4, 5X odds for 480 bets, which is about 16 hours at 100 rolls/hr. Because you're only subjecting $2400 to a 1.4% edge, the ev is just -$33.94, call it -$35. However, you have a pretty big standard deviation, $538. So, two SD worse than ev is just $1111.

Pacomartin says, "ultimately, you can't beat the ev". This is not true! I keep preaching about variance, which gives you the chance to beat that HA. Note that the SD works on both sides of the ev, so you have the same 2.5% chance of winning $1041 as of losing $1111. With an ev/SD of .0653, your chances of breaking even or better are about 47.5%.

For $10 pass, 3, 4, 5X odds, the ev is about -$70, SD $1076, +/- 2 SD are -$2222 and +$2082.
Cheers,
Alan Shank

I ran a simulation of a couple of different odds multiples with a $5 pass bet, 480 bets max, 10,000 sessions each. The results:

	3,4,5 X odds	10 X odds
bankroll	$1100	$2400
busts	470	389
win goal	$550	$1200
reached WG	2770	2918
winning sessions	4915	5033
losing sessions	5051	4949

There were a few breakeven sessions. So, these bankrolls kept busts down under 5%, and the variance allowed the player to win half his/her bankroll over 25% of the time. The way my program works is, having reached a win goal, continue play with 20% of the win goal amount, setting a new win goal of 50% more. At least one session for each method reached five successive win goals.

Cheers,
Alan Shank

Quote:

Pacomartin says, "ultimately, you can't beat the ev". This is not true! I keep preaching about variance, which gives you the chance to beat that HA.

Alan -

You're talking in the short term. You can certainly walk away in the short term up from a table due to the variance, but what we're talking about is that if you keep track over months, years, decades, the house edge will catch up with you - you'll have that losing session(s) where you may take your stop-loss and walk away, or lose the entire bankroll for that session. This loss is taken away from your 'gain' from the previous session and the more times you play, the closer you get to the EV over the long run.

-B

Quote: toastcmu

You're talking in the short term. You can certainly walk away in the short term up from a table due to the variance, but what we're talking about is that if you keep track over months, years, decades, the house edge will catch up with you - you'll have that losing session(s) where you may take your stop-loss and walk away, or lose the entire bankroll for that session. This loss is taken away from your 'gain' from the previous session and the more times you play, the closer you get to the EV over the long run.

-B

About the only thing you can say for sure is that, the longer you play, the more bets you make, the luckier you have to be to still be ahead, overall. Even after thousands of bets, assuming you play low-edge, relatively high variance, the extreme right tail will still be ahead of the game. Who are those people? >:-)

What is your definition of "the long term"? Some examples:

$5 pass, 3, 4, 5X odds, 5000 bets
ev -$354
SD: $1738
+SD required to be even or ahead: .2
probability: .4207

$5 pass, 3, 4, 5X odds, 10000 bets
ev -$707
SD: $2458
+SD required to be even or ahead: .29
probability: .3859

$5 pass, 3, 4, 5X odds, 15000 bets
ev -$1061
SD: $3010
+SD required to be even or ahead: .35
probability: .3632

You can see the trend, but also that, even after 15,000 bets, over one-third of the players would expected to be ahead. Or, to put it a bit differently, we would expect over one-third of the players to be ahead. No individual player would "expect" to be ahead after 15,000 bets.

Since the standard deviation increases with the square root of the number of bets, and the ev with the number of bets, there is a point where the expected loss equals the standard deviation. For this strategy, it is 120,871 bets. At that point, a player would have to be one standard deviation to the right of the ev to break even, a probability of about .16, as PapaChubby indicates in his post. Even after 483,487 bets, you only need to be two SD ahead of the ev, so it still means about 1 in 40 players would be ahead.

Of course, with all that variance, there are also going to be some poor people in the left tail. Who are those guys? >:-)

Now, without those odds bets, the picture is quite different.

$5 pass, no odds, 15000 bets
ev -$1061
SD: $612
+SD required to be even or ahead: 1.73
probability: .0418

The probability of breaking even or better is a function of the ev divided by the standard deviation; it tells you how lucky you have to be to overcome the HA for any given number of bets.

When you put the casinos on one side and all the players on the other, you know the players, taken as an aggregate, will lose. But variance ensures us that some players, even for a lifetime of play, can hope to be ahead.


Cheers,
Alan Shank