Probability of going broke

[MHoydilla]

I would like to know the chances of going broke if taking bets on an even money proposition but charging 10% juice of each wager lost if I had 1,000,000.
If the average bet placed againest me was 1000 meaning if I lost the bet I lose 1000 and if I win the bet I win 1100 what is the chance that I could lose my initail 1,000,000. (Please show how you did this)
What if the average bet was 5000 so -5000 or +5500.
What if was 10000.
All outcomes of bets would be 50/50.

[MHoydilla]

Bump, I need some help on this ASAP.

[jimdmcevoy]

Ok, suppose you got a 1,000,000 and if some one bets they have a 50/50 at winning 1000 or losing 1100.

Let's call the number of these bets you take n.

As in if you took 15 bets, n=15.

Now suppose you win m out of these n bets.

You will have a .5^n*n!/[(n-m)!*m!] chance winning m bets out to the total n bets you took.

m can equal 0,1,2,3.... all the way up to n

So, you want to work out the chance that you will end up with 0 or less?

that means you want to know the chance that:

1,000,000 + 1000*(-n+1.1*m)<=0

With some algebra this is the same as saying what is that chance that:

m>=(n-1000)/1.1

So once you know n, you can work out what m (the total number of bets you won) has to be in order to remain positive. Suppose you need m>5 to be positive. Then you work out the probability that m=0, add that to the probability that m=1, and again for m=2, and m=3, and m=4, and that is the probability you will go negative.

But remember this is not really 'going broke', you are going negative. The only way to avoid risking going negative is if you take bets that you will be able to pay back even if every bet you take you lose.

[gaming_mouse]

jim,

the analysis you give is only for n bets. the real question is: what is the chance that you EVER go broke? that is, let n approach infinity. what is the chance that your bankroll is 0 or less at least once? that's a much harder question to answer. unfortunately, i don't remember the solution either.

gm

[BaronVonCP]

I don't see how its possible to not be broke. If each bet has a negative -EV

[jimdmcevoy]

I assumed he meant taking n bets at a time.

But after a little thought I agree with you, I made a bad assumption, finding the answer to your(his?) question would be more usefull. I'll give it a try tommorow.

[gaming_mouse]

[ QUOTE ]
I don't see how its possible to not be broke. If each bet has a negative -EV

[/ QUOTE ]

Baron,

read the question again. he wins 1100 but loses only 1000 on a 50/50 bet. so with a very large bankroll (relative to 1000) he would be unlikely ever to go broke. but the exact calculation is not so easy.

gm

[Precision1C]

For a reasonably large number(say 10+) of samples your result will look like normal binomial distribution*$1000 shifted positive $50*N where N is the number bets. To determine risk of ruin you use standard deviation to determine on how widely you can tolerate the short term swings, 1 std dev is around 68%, 2 std dev is 90+%, and 3 std dev is 99.9%. By my calculations with a $1,000,000 bankroll your odds of ruin are miniscule, way beyond 5 std deviations.

F(N)=$50*N - 3(sigmaN)*1000

The above equation represents the maximum loss for N bets if you are one in a thousand kind of unlucky.

sigmaN is the standard deviation for a binomial distribution for coin flips

sigmaN=SQRT(N*.5*.5)=.5SQRT(N)

F(N)=$50*N- 1500SQRT(N)

To find the minimum value of F(N) take the derivative with respect to N and set to 0

F'(N)=50-750/SQRT(N) skipping some basic algebra steps

when F'(N)=0 N=[750/50]^2=225

plugging in N=225 for F(N)

F(225)= $50*225 - $1500*SQRT(225)=$-11,250

Notice that this sum is vastly smaller than $1,000,000 and even if you use 5 standard deviations instead of 3 for the equations. If you somehow lose $1,000,000 watch out for asteroids and falling airplane debris because they are bigger dangers. This is the reason that your competent sports book isn't going broke. However this test is affected by assumptions of event independence, for example if you take in 100*$1,000 on the same bet that isn't the same as setting N=100 with a $1,000 bet but is equivalent to setting N=1 with a $100,000 bet.

[gaming_mouse]

This is not correct, although it may provide a reasonable approximation.

The problem is, we are not interested in the simple sum of the trials, because it is possible to have bad runs (which bring the bankroll negative) followed by good runs, which bring the overall sum over n back to something we would expect.

That is, if X_i are the independent Bernouilli trials, the distribution of the sum of the X_i (call it S_n, which follow a Binomial distribution) is not sufficient to answer our question. We need the distribution of the minimum of the following sequence:

X_1, X_1 + X_2, ... , S_n.

Quite a different animal indeed. As I said though, I've forgotten how you solve for this distribution.

gm

[jason1990]

If the bet size is 1000, then your winrate is

m = .5(1100) + .5(-1000) = 50

and your variance is

s^2 = .5(1100^2) + .5(1000^2) = 1,105,000.

You can use the risk of ruin formula

-(s^2/(2m))log(p) = b

to find that for a banroll of b = 1,000,000, your risk of ruin is

p = e^{-(2bm)/s^2}
= e^{-100,000,000/1,105,000}
= 5 * 10^{-40},

which is astronomically small.

You can do something similar for other bet amounts, but there is a danger. The risk of ruin formula is derived by assuming that your bankroll follows a Brownian motion with drift. Since a random walk converges to a Brownian motion as the step size goes to zero, this approximation will be good when the bet size is much smaller than the bankroll. But if the bet size is comparable to the bankroll, this is no longer a good approximation. Let me just warn you: getting an exact answer in the discrete setting would be an extremely daunting task. Good luck to all of you who would like to try it.