Probability of going broke

[gaming_mouse]

M,

This will have nothing to do with the difficulty of the problem. But pzhon has essentially posted the solution. He's just saying that the problem, as originally stated, is not completely well-defined. See his example of when your bankroll is less than the size of the minimum bet. Are you already bankrupted or are you allowed to make a partial bet (with the same odds) at that point?

gm

[jason1990]

I'm a little surprised by these asymptotics. Thinking in terms of the Central Limit Theorem, when n is large, the path of the bankroll looks more and more like a Brownian motion with drift. Using the hitting time probabilities for a Brownian motion with drift, the probability of going broke is approximately

e^{-2mb/s^2}
= e^{-100000n/1105000}
= (0.913476)^n.

This does not agree with your asymptotics. Have you computed c? Do you know it's not zero? Also, there seems to be a uniqueness issue in the difference equation. There doesn't seem to be enough boundary conditions to uniquely determine the solution. When dealing with the Brownian motion case, you do not actually compute the probability of going broke, but rather the probability of going broke before reaching some upper threshold. Then you let the threshold go to infinity. The analogous thing here would give you two-sided boundary conditions which would be enough to determine the solution. Anyway, the reason I'm so suspicious is that I find it hard to imagine that the Invariance Principle could not be used to make the Brownian motion approach rigorous.

But in any event, I still hold that asymptotics are "easy". What is most difficult is coming up with exact solutions for small n.

And, for what it's worth, I would say that if he has 500 and cannot accept a bet, then he is broke.

[jason1990]

[ QUOTE ]
But pzhon has essentially posted the solution. He's just saying that the problem, as originally stated, is not completely well-defined.

[/ QUOTE ]
I don't think this is quite what he's saying, nor is he claiming to have a solution. I think what he's saying is that when n is large (that is, when the banroll is much larger than the bets), then his solution is approximately correct. (Although I, personally, am not convinced of this, as I mentioned in another post.) His solution involves an unknown constant c, which depends on the interpretation of the problem, and which he says is a pain to compute. But this still leaves open the problem of finding exact solutions when the bankroll is small.

[gaming_mouse]

Jason,

Could you explain in more detail why the Gambler's Ruin solution cannot be used to solve this problem, by letting the upper bound grow larger.

Were you simply saying that because the win and loss sizes are not equal, the same method cannot be used?

Otherwise, it seems that by using an extremely large, but finite, upper bound, you get a very close approximation to having no upper bound (since at some point it becomes nearly impossible to go broke). Presumably, there would be limiting behaviour as n --> infinity.

Thank in advance,
gm

[MHoydilla]

If the amount of bankroll is less than the bet then it would be considered as being broke. So if 1000 bets and I have 800 I am broke.

[jason1990]

Yes, I was assuming that. In case it's unclear, when I talk about the case where the banroll is small compared to the size of the bet, I don't mean smaller. For example, if you have a bankroll of 1000000 and the bets are 100, then the bankroll is big. But if you have a bankroll of 1000000 and the bets are 100000, then the banroll is small.

[jason1990]

[ QUOTE ]
Were you simply saying that because the win and loss sizes are not equal, the same method cannot be used?

[/ QUOTE ]
Yes, that's all I was saying. There is limiting behavior as the upper bound goes to infinity and this is exactly what you do to compute these probabilities. But modifying the problem to one of equal wins and losses with unequal probabilities will only give you a good approximation when the bankroll is relatively large. And this is because they are both well approximated by the risk of ruin formula.

[gaming_mouse]

Jason,

The recursive equation which solved the ordinary Gambler's Ruin probler in the article I linked to above was:

p_h = q * p_{h-1} + (1-q) * p_{h+1} (1)

In the case we are interested in here, what is the reason we cannot simply rewrite:

p_h = .5 * p_{h-1100} + .5 * p_{h+1000}

and then solve in the same way? Is there not some extended version of theorem that specifies the form of the general solution?

gm

[jason1990]

To solve the first equation, you must find the (possibly complex) roots of a quadratic polynomial, specify 2 boundary conditions, and solve a 2x2 system of linear equations.

To solve the second equation, as pzhon showed, you must find the roots of a 21-degree polynomial, specify 21 different boundary conditions, and solve a 21x21 system of linear equations.

You can do that if you want. You might want to use a computer. But if you're going to break out the computer, you might as well just run a simulation.

[gaming_mouse]

Jason,

Got it. Although, solving the system of equations with a computer is cleaner than running a simulation. Because, with a simulation, at what point do you stop a given sample?

You're bankroll may have tripled, but there is still some (small) chance you break -- so do you let the simulation keep going? I suppose you could decide beforehand on some stopping point, after which there is only a tiny probability of breaking, and you could estimate the stopping point using the Brownian motion approximation you suggested.

Thanks for all your explanations,
gm