Bet size, bankroll and variance

[schroedy]

I am certain that your average gambler (investors and poker players included) is consistently placing bets that are too large for his bankroll.

I have taken this supposition to the point of speculating that your average Las Vegas tourist would lose his roll even if he were playing games with the odds switched (i.e., the house took the worst of it and gave the tourist the house edge). And the reason I believe this is that I believe the games have sufficient variance so that many tourists would "bust out" before they could start to realize their edge.

I am not advanced enough, however, to quantify the variance that exists in popular casino games like blackjack or dice.

But here is my question:

Take 1,000,000 tourists each with a $1,000 bankroll. Have them deal blackjack to a casino with a $1,000,000,000 (1 Billion dollar) bankroll. The stakes are $100/hand.

How many tourists go home broke by going down 10 bets before their edge can assert itself?

Change the stakes to $50 per hand, bust out level down 20 bets.

And $10 per hand, bust out level down 100 bets.

I don't know the numbers on variance for games like blackjack, dice or, for that matter, poker and I am not sure that I could do the math even if I did.

Thanks in advance.

[TomCollins]

Do they keep playing forever? Do they stop at a certain point? Do you want a confidence interval? If someone keeps playing forever, eventually they will hit a bad enough run that they will lose their roll. This may take trillions of hands, but in the long run...

[TomCollins]

For the sake of simulation, once a bankroll was 1000 bets, I assumed it was "safe" and you would not go broke. I also had players keep playing until they had 0 or 1000 bets.

For a player advantage of 50.5%, I ran the simulation with varying starting bet sizes, and here are the results for 10000 simulations:

For starting with 10 bets, 83% of the time, players went broke. 17% of the time, the players reached 1000 bets.

Starting with 20 bets, 67% of the time, players went broke. 33% of the time, the players reached 1000 bets.

Starting with 50 bets, 37% of the time, players went broke. 63% make it to 1000 bets.

Starting with 100 bets, 14% of the time, players went broke. 86% made it to 1000 bets.

With 1000 bets, over 99% of all players made it to 10000 bets, so its safe to say 1000 is a sufficient bankroll.

[schroedy]

Thank you for running that simulation.

<<Running off to show some of my trader friends.>>

[BugsBunny]

Just as a comparison, using BruceZ's bankroll formula.
Assume an earn of 1 bet/100
Assume a SD of 10/100

<font class="small">Code:</font><hr /><pre>
BR size - Risk of Ruin - Simulation Result
10 81.87% 83%
20 67.03% 67%
50 36.79% 37%
100 13.53% 14%
200 1.83% -
250 0.67% -
300 0.25% -
400 0.03% -
500 0.0045% -
1000 0.0000002% less than 1%
</pre><hr />

I'd say that's a pretty close correlation /images/graemlins/smile.gif

[pzhon]

[ QUOTE ]
If someone keeps playing forever, eventually they will hit a bad enough run that they will lose their roll. This may take trillions of hands, but in the long run...

[/ QUOTE ]

From your description of your simulation, I presume you don't believe this, but I'll respond anyway.

You have a positive probability of never going broke in a game like this with an advantage. You will get bad streaks that will cost you any fixed amount of money, but by the time you lose fifty hands in a row, your bankroll will probably be over a billion units, so you won't even notice such a bad streak. The danger is that you hit a shorter bad streak earlier.

If you gain one unit with probability p, risking one unit, then your chance of going broke decreases exponentially with your bankroll: ((1-p)/p)^bankroll. For the median result to be to win without ever going bankrupt, you need a bankroll of at least log(2)/(log p -log(1-p)), 35 when p=50.5%. A bankroll twice that size means you go bankrupt only 1/4 of the time; 10 times that size means you go bankrupt less than 1 time in a thousand.