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MHoydilla
11-19-2004, 05:35 AM
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
11-19-2004, 06:51 AM
Bump, I need some help on this ASAP.

jimdmcevoy
11-19-2004, 08:56 AM
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)&lt;=0

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

m&gt;=(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&gt;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
11-19-2004, 10:13 AM
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
11-19-2004, 10:43 AM
I don't see how its possible to not be broke. If each bet has a negative -EV

jimdmcevoy
11-19-2004, 10:51 AM
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
11-19-2004, 11:07 AM
[ 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
11-19-2004, 04:23 PM
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
11-19-2004, 06:10 PM
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
11-19-2004, 06:41 PM
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.

Vintage
11-19-2004, 07:41 PM
eventually you will go on a bad run, just as if person types random letters for infinite years, he WILL write a novel.

jimdmcevoy
11-19-2004, 08:18 PM
Yah but before he writes a novel he will right a whole lot of garbage (you will have a whole lot of good runs).

MHoydilla
11-19-2004, 09:08 PM
Ok I sort of understand the answers, anyway I appreciate the responses. If anyone else can come up with exact answers to the 1000, 5000, and 10000 please do so.

jimdmcevoy
11-19-2004, 09:59 PM
Well, the best I can do is (sort of) solve a similar problem. Let's say there is a c chance that you will win one dollar and a (1-c) chance you will lose one dollar.

And let's define P(n) as the probability you will go broke if you play this game an infinite number of times.

So we can say straight away:

P(0) = 0
P(1) = c*P(2)

And you can also say:

P(n) = (1-c)*P(n-1) + c*P(n+1) for n&gt;0

but since we have P(1) in terms of P(2), with this general formula we can then get P(2) in terms of P(3), P(3) in terms of P(4) and so on.

In the end we will get a 'recurrence relation' :

P(n) = X(n)*P(n+1)

With some algebra using this equation and the other general equation you can show:

X(n+1) = c/[1-X(n)*(1-c)]

and we already know the starting values:

X(0)=0 and X(1)=c

Can anyone solve X(n) just in terms of n?

So anyway you can then show that:

P(n) = Product_{k=n}^{infinity} X(k)

This can be interpreted as P(n) equals the probability that you will get to n+1 before you go broke (which is X(n)) times the probability that you will get to n+2 from n+1 before you go broke (which is X(n=1)) times the.... and so on until you get to inifinity.

So if some one out there can solve the reccurence relation:

X(n+1) = c/[1-X(n)*(1-c)]
X(0) = 0

In terms of n, then maybe the infinite product:

P(n) = Product_{k=n}^{infinity} X(k)

can be put explicitly in terms of n, and then we have out answer.

Now as far as I can see the variance for each bet will be 1, so you can just vary c (the chance you will win the bet) to get the same expectation to variance ratio as the situation mentioned in the original post. I then reckon that the risk of ruin will be very very similar.

gaming_mouse
11-19-2004, 11:52 PM
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.

Yes, I can see that quite clearly. I was hoping there might be a clever solution involving recursion, but if there is I can't see it. Interesting problem...

gm

Precision1C
11-20-2004, 12:04 AM
An infinite number of trials is kind of irrelevant since you promised a great return on each bet so the majority of the danger is for cases of N of around 225. With only a \$10,000 bankroll this question would perhaps be interesting but with a \$1,000,000 bankroll the odds of going bankrupt at \$1,000 a bet at one time is going to be really small. Say you accept a bet once a second, every second, but get paid off before the next bet. Assuming you only pile the money up and never touch the account the sun will become a red giant then a white dwarf long before you go busted. Look at the odds of losing 1000 bets with a 50% chance of success, (.5)^1000 an incredibly small number. However lets say the bet was exactly fair, that you have an expected value of zero then eventually with a huge N your standard deviation would become large enough to swamp any fixed bankroll.
As an aside, the question you really likely are asking is that if you have a bankroll of \$1,000,000 and take in coin flip bets of \$1,000 with an EV of \$50 a bet and sweep out any money above \$1,000,000 how long can you continue to do this and how profitable is this? Well, in a perfect world where suckers just line up without advertising, no competition, and no overhead a setup like this is nearly infinitely profitable with nearly no risk. As I said the risk is on the order of Armageddon happening tomorrow so perhaps not taking this kind of deal is a parlay, betting that the end of the world will happen or the scheme will fail.

Note: If you get offered a deal like this I have some words of wisdom. "Son, if a man offers to bet you that the Jack of Spades will pop out of the deck and squirt cider in your ear don't take the bet unless you like cider in your ear"

gaming_mouse
11-20-2004, 12:14 AM
Jason,

Here is a discussion of the exact solution when you have a finite upper bound (a stop-win point), known as the gambler's ruin problem:

http://www.mathpages.com/home/kmath084.htm

Perhaps you could take the limiting value as n --&gt; infinity using that solution. I'm not sure whether or not that's feasible.

gm

jason1990
11-20-2004, 02:03 AM
I just quickly glanced at it, but if I'm not mistaken, this is a different problem. The OP is dealing with a random walk with drift. The Gambler's Ruin deals with an asymmetric random walk. In the discrete setting, these are two very different things. The Gambler's Ruin is "easy". The random walk with drift is not.

One of the chief difficulties in the OP's problem is that the bets are in units of 1000 (for example) but his bankroll can conceivably take on any value that is divisible by 100. If you modify the problem so that he jumps by +1000 with a probability greater than 1/2 and jumps by -1000 otherwise, then you avoid this difficulty and the problem becomes much simpler.

pzhon
11-20-2004, 04:39 AM
The asymptotic behavior is the same as for a biased even-money wager.

In both cases, there is a recursion for the probability of going bankrupt with solutions of the form x^n, and all solutions are linear combinations of these. The question is how to start the recursion. It's not clear what to do if your bankroll reaches \$500 and you can't accept a \$1000 wager. Is this counted as bankrupt?

If you let p[n] be the probability of going bankrupt with 100n dollars, then

p[n]=1/2 p[n+11] + 1/2 p[n-10]

For there to be a solution of the form p[n]=x^n, x must satisfy

x^n = 1/2 x^(n+11) + 1/2 x^(n-11)
x^21 - 2x^10 + 1 = 0

I believe only the roots of magnitude less than 1 matter, and the most important is the one closest to 1, 0.990957... Asymptotically, the probability of going bankrupt with 100n dollars is about c (0.990957)^n, where c depends on the choices made about dealing with a bankroll under \$1000.

It does seem to be a pain to compute c.

MHoydilla
11-20-2004, 10:47 AM
Well you guys are truly great for putting in the time and effort in this problem. I cant get into details of why just yet, but this info is of the outmost importance to me. Would the problem become easier with smaller numbers, such as instead of 1,000,000 you use 1,000 and instead of 10000 you use 10 dividing both by 1000? So what would be the chance of ruin if I have 1000 and take bets on a coin flip if I lose I pay 10 but I win I get paid 11? Say I was taking exactly, 50,000 bets, 10,000 bets, 5000 bets, or 1000 bets.

gaming_mouse
11-20-2004, 11:07 AM
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
11-20-2004, 11:13 AM
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
11-20-2004, 11:20 AM
[ 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
11-20-2004, 11:41 AM
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 --&gt; infinity.

Thank in advance,
gm

MHoydilla
11-20-2004, 12:04 PM
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
11-20-2004, 12:34 PM
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
11-20-2004, 12:40 PM
[ 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
11-20-2004, 01:49 PM
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
11-20-2004, 02:08 PM
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
11-20-2004, 02:20 PM
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

jason1990
11-20-2004, 02:52 PM
Well, it depends on what you're used to as to which numerical method is easier. I would be more comfortable with the simulation, because I would be less worried about round off errors. And you're right about needing to be careful about how long to run the simulation, but also about how many times. I would be more comfortable dealing with these technicalities than dealing with the technicalities of round off errors. I suppose it's just a matter of what your background/taste is.

pzhon
11-20-2004, 05:44 PM
[ QUOTE ]
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.

[/ QUOTE ]

The disagreement is smaller than it looks. We are using different definitions of n. I was talking about a bankroll of 100n. I believe you are talking about a bankroll of 1000n. To convert between the asymptotics, 0.9909571^10 = 0.913163.

[ QUOTE ]
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.

[/ QUOTE ]

I haven't computed c in this example. I will, later. I think there should be some principle that allows you to bound c over all similar problems, which I think is more interesting than the particular case.

I think there are enough boundary conditions. Ten roots of the characteristic polynomial have magnitude greater than 1, and the coefficients of the exponential terms corresponding to these roots must be 0. (They would correspond to sequences satisfying the recurrence bounded in the opposite direction.) I'm a bit unsatisfied by this, since I don't know why the number of roots with magnitudes greater than 1 is right. I think there must be a general result about the roots of this type of polynomial.

jason1990
11-20-2004, 08:58 PM
[ QUOTE ]
The disagreement is smaller than it looks. We are using different definitions of n. I was talking about a bankroll of 100n. I believe you are talking about a bankroll of 1000n. To convert between the asymptotics, 0.9909571^10 = 0.913163.

[/ QUOTE ]
Aha! Okay, then. My guess is that the remaining discrepancy is due to round off errors. So c=1, right? No need to compute?

As for what you said about boundary conditions, it reminded me of the heat equation. If you solve the heat equation on the entire real line (no boundary conditions) and all you specify is initial conditions, then you don't get a unique solution. But if you require the solution to have some bounded growth rate, then you do. (I'm a little fuzzy on the details, but I think this is generally correct.) It seems like something similar is happening here. I don't know if this is conceptually relevant, but I thought it was interesting.

pzhon
11-20-2004, 09:35 PM
[ QUOTE ]
[ QUOTE ]
The disagreement is smaller than it looks...
0.9909571^10 = 0.913163. [vs. 0.913476]

[/ QUOTE ]
Aha! Okay, then. My guess is that the remaining discrepancy is due to round off errors. So c=1, right? No need to compute?

[/ QUOTE ]
I don't see how to get c=1. I still haven't computed c, though.

I think the remaining difference is not just a round-off error. The bases have different algebraic degrees. I'm not sure what the source of the difference is. Perhaps it is that the Central Limit Theorem does not work rapidly enough, particularly at the tails, and that is where we would like to use it for this problem.

[ QUOTE ]

As for what you said about boundary conditions, it reminded me of the heat equation. If you solve the heat equation on the entire real line (no boundary conditions) and all you specify is initial conditions, then you don't get a unique solution. But if you require the solution to have some bounded growth rate, then you do. (I'm a little fuzzy on the details, but I think this is generally correct.) It seems like something similar is happening here. I don't know if this is conceptually relevant, but I thought it was interesting.

[/ QUOTE ]
Yes, this is close to what I was thinking. In both cases, there is some sort of spectrum, and for the answer to be "physical" there must be no energy (zero coefficients) in half of the spectrum, and with this restriction the solution may be unique.

If you vary the polynomial, the roots may change magnitude. Can a root have magnitude 1, and if so, what happens?

jason1990
11-21-2004, 11:51 AM
[ QUOTE ]
I don't see how to get c=1. I still haven't computed c, though.

I think the remaining difference is not just a round-off error. The bases have different algebraic degrees. I'm not sure what the source of the difference is. Perhaps it is that the Central Limit Theorem does not work rapidly enough, particularly at the tails, and that is where we would like to use it for this problem.

[/ QUOTE ]
Well, I was naively thinking our asymptotics must agree and this trivially gives c=1. But of course you're right about algebraic degree. Even in the simple case where the bets are +2 and -1 with equal probability, the true asymptotics are given by the difference equation and are

[(-1+sqrt{5})/2]^n

and the (incorrect) asymptotics given by the Brownian approximation are

[e^{-.4}]^n.

One problem which I originally thought would "disappear" in the limit is this: the player's wealth consists of two components, a simple random walk (SRW) which moves +1.5 and -1.5 with equal probability and a drift which moves at constant speed of +.5. "Scaling" the SRW gives Brownian motion. But this scaling means scaling both time and space *and* scaling them in different ways. However, in order to achieve a scaling limit for the drift, we must scale time and space in the *same* way.

So saying that the player's wealth looks a lot like a Brownian motion with drift when his bankroll is large is true to some extent, but it does not literally mean that his wealth converges to a Brownian motion with drift under some suitable scaling. This is, of course, because there is no way to consistently scale both components.

Obviously, this problem doesn't "disappear" in the limit. I believe it does "disappear" in many interesting applications, but certainly not here. I wonder also how they deal with this in mathematical finance. There they often model stocks and other risky commodities as diffusions with drift. But, in certain circumstances, just like here, these diffusions will have different asymptotic behaviors than their discrete counterparts. This is definitely something I'll be thinking more about.

Also, thinking more about boundary conditions, I think we simply need to add the one additional "boundary" condition that p(n)--&gt;0 as n--&gt;infty. I suppose this was obvious and probably what you were thinking all along, but I thought I'd say it explicitly.