Probability question

[Rutager]

While watching a friend of mine losing his money at a lousy casino game I'd never play because it was negative expectation, my mind started to wander and I came up with a simple probability question that I can't quite figure out. Maybe some of the math experts on this board can answer this question for me.

Let's say you are playing a fair game with a casino (ie- If you bet a dollar, you have an even chance of winning a dollar or losing a dollar.) Say your bankroll is only 1 dollar (1 bet) to start out. Clearly, if you played this game forever, you would eventually go broke. My question is what is the expected number of rounds you'd have to play before you go broke? I suspect it is infinite, but I cannot quite prove why. Any ideas? Just curious.

-Rutager

 

[Rob McGarvey]

"my mind started to wander" That was your first mistake! grin I'm sure one of the math geniuses here can answer your question with a formula. Why don't you take a crack at solving your own problem with the math you already know??

50% on the first play
25% if there is a second play
Now we either have 3 units or 1
If 3 then it is 12.5%
If 1 it's 50%

We should be using the probability to help us decide what to do. The answer to that you already have.......don't play with your own money.

Now, if a tree falls in a forest........

 

[Imaya]

Well as I see, you fear is to be broke right ? I will explain quickly the way I learned. By the way, my teacher is Mr. Brown so if anybody, knows him, let me know, I would like to get news from him. Ok so the minimum requirement is to get 500 times the minimum bet. Where I play it's 5$ so i'm going in with 2500$. I NEVER put more than 5% of my bankroll on one hand. Basicly, I went 37 times (8hours shift) in casinos and the results: down:0 times even:2 times up:35times. I was about 25K over last year...this is not bad for a rookie. I know the probabilities tell me that I SHOULD went down few times but for now, I cross my fingers

 

[The Mayor]

Although I don't recall the answer, this is known as the Gambler's ruin paradox. If your bankroll is finite and the casino's is infinite, and you flat bet, then even in a fair game (100% return), you will go broke. The over/under for this is something that is solved using a type of mathematics known as a Markov Process, which is a type of discrete differential equation.

I wish I could do better. For now, let's just say, it is 100% certain you will go broke. Hence the "Gambler's Ruin" paradox.

--Mayor

 

[Adam N. Subtractum]

"Gambler's Ruin" formulas...

E(n) = expected number of plays before ruin or win limit
n = number of plays
z = initial capital
a = desired final capital (a > z, z-a is the desired win)
p = probability of a win
q = probability of a loss
r = probability of a tie

For an even game (p = q) use:

z(a-z)
-------- = E(n)
1 - r

For an infinite win limit (a), where p < q, use:

z
------------ = E(n)
(q-p)(1-r)

For an infinite win limit (a), where p = q:

E(n) = infinity

ANS

 

[Rob McGarvey]

Theoretically Speaking

Is your formula saying he will play for an infinity with one unit? I think not. How many units (standard deviations) will it take to protect the player from RoR and allow him to play forever, since we know that with 1 unit he is faced with imminent destruction?

cough, cough, MathProf, cough ;>

 

[Rutager]

I am somewhat familiar with Markov Chains. However, I have no idea how to handle markov chains with infinite states. But I'm sure you know far more than I.

And to clarify, I am not saying I want to play an even money game. I'm just curious because my question 'what is the expected number of rounds before going broke' seems to be simple, but I cannot solve it.

Rutager

 

[Rutager]

Re: "Gambler's Ruin" formulas...

Thanks for you help.

From the equation you gave:

z(a-z)
-------- = E(n)
1 - r

It is clear that as a-->infinity, E(n) --> infinity. Which seems to support my 'guess' that the expected number of steps before going broke is infinite. This seems very counterintuitive(insert card counter/counterintuitive joke here).

But how did you come up with that formula? Also, how did you you come up with the formula for p < q ?

I wish I still had all my college math textbooks right about now.

Rutager

 

[The Mayor]

Re: "Gambler's Ruin" formulas...

And in general the question is one of the expected return time for a random walk. As they say "you can never go home again."

Thanks for posting the details.

--Mayor

 

[Adam N. Subtractum]

Re: "Gambler's Ruin" formulas...

"It is clear that as a-->infinity, E(n) --> infinity. Which seems to support my 'guess' that the expected number of steps before going broke is infinite. This seems very counterintuitive..."

You're right, this may seem counterintuitive to some. What needs to be realized is that the formula provides the *average* expected time to ruin or win limit...an average that obviously has variance. Once this is understood the results become quite intuitive.

"But how did you come up with that formula? Also, how did you you come up with the formula for p < q ?"

The formulas are found in Epstein's "Theory of Gambling & Statistical Logic", which is highly recommended for any gaming theorist.

"I wish I still had all my college math textbooks right about now."

Check out Epstein's book, Griffin's "Theory of Blackjack", and the archives at BJMath.com, I've found just about everything I've ever needed in these sources.

ANS

 

[Rob McGarvey]

The longer you play the more likely your outcome represent the actual EV of the game you are playing, with or without your fingers crossed. A 1% edge.

 

[Rob McGarvey]

Theory of Relativity

Is a wonderful theory, but too few people will ever need to understand it. If you actually think that you can start off with one unit and play a 50/50 game forever, test the theory, and report your results. I'm sure you will report back that the first time I did it it took x tries to lose my one unit,

the second time it took me x tries
the third time it took me x tries
the fourth time it took me x tries
the fifth time it took me x tries

and you will get an infinite # of failures of the theory before the theoretically possible infinity takes place.