## Poll

3 votes (13.04%) | |||

4 votes (17.39%) | |||

7 votes (30.43%) | |||

2 votes (8.69%) | |||

2 votes (8.69%) | |||

5 votes (21.73%) | |||

1 vote (4.34%) | |||

4 votes (17.39%) | |||

7 votes (30.43%) | |||

6 votes (26.08%) |

**23 members have voted**

July 26th, 2016 at 4:54:31 PM
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I've been tasked with assessing the general value of Internet casino bonuses given certain parameters. I find it is necessary to have a general formula for the probability of ruin. I've seen such formulas before but they all seem to be based on infinite steps until a given goal is met or ruin. My dilemma is that I need a formula when there is a finite number of steps.

Let's have an example. You have $25. You play a game where you have a 49% chance to win $1 and a 51% chance to lose $1. What is the probability you can survive 100 bets without going broke?

I know it would be easy to simulate this but is there any approximation formula I can use based on:

Thank you.

Let's have an example. You have $25. You play a game where you have a 49% chance to win $1 and a 51% chance to lose $1. What is the probability you can survive 100 bets without going broke?

I know it would be easy to simulate this but is there any approximation formula I can use based on:

- Initial bankroll
- Probability of winning
- Number of bets required

Thank you.

It's not whether you win or lose; it's whether or not you had a good bet.

July 26th, 2016 at 5:39:16 PM
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Why not simulate X different scenarios, graph the results, then look at graph or whatever to approximate a formula with one variable (i.e.: starting BR). Then do the same for other variables like # of spins etc.

July 26th, 2016 at 6:02:19 PM
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Quote:RSWhy not simulate X different scenarios, graph the results, then look at graph or whatever to approximate a formula with one variable (i.e.: starting BR). Then do the same for other variables like # of spins etc.

That's easier said than done. I don't want to waste too much time on this, especially if some esoteric web site out there already has a formula.

It's not whether you win or lose; it's whether or not you had a good bet.

July 26th, 2016 at 6:36:24 PM
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Is this a serious question? Based on the poll options this appears to be a joke question I don't get. Can someone please explain.

July 26th, 2016 at 7:07:01 PM
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It's serious. The question is, if you start with bankroll B and make even-money bets that have probability P of winning, what is the overall probability that you will get through some number N bets without going broke?

There is a formula for figuring out the probability of reaching some target bankroll before going broke, but we're looking for one that has a limit on how many bets there are.

Looking at the Wizard's original example, but with win probability p (and loss probability q = 1-p):

Up through 25 bets, the probability of a bankroll change of (n-25) is the nth term (starting with 0) of the binomial expansion of (p+q)

In each subsequent bet, the probability of a bankroll change of n is p times the previous row's (n-1) + q times the previous row's (n-1), except that a term for -25 is ignored as once that is reached, the betting ends.

As usual with me, (n)C(k) is the number of combinations of n things taken c at a time; also known as combin(a,b).

The sum of the probabilities for the bankrupt point (bankroll = -25) is:

q

25 p q

((25)C(1) + (26)C(2)) p

(2 (25)C(1) + 2 (26)C(2) + (27)C(2)) p

(5 (25)C(1) + 5 (26)C(2) + 3 (27)C(2) + (28)C(3)) p

(14 (25)C(1) + 14 (26)C(2) + 9 (27)C(2) + 4 (28)C(3) + (29)C(4)) p

and so on, up to p

The terms correspond to something called Catalan's Triangle, but right-to-left

(i.e. normally in Catalan's triangle, row 5 is 1, 5, 9, 14, 14, but here is it 14, 14, 9, 5, 1).

However, I am not aware of a simple formula that would calculate the sum, even if you go by column (i.e. add up the (25)C(1) terms and multiply by 25, then add up the (25)C(2) terms and multiply by (25)C(2), and so on).

There is a formula for figuring out the probability of reaching some target bankroll before going broke, but we're looking for one that has a limit on how many bets there are.

Looking at the Wizard's original example, but with win probability p (and loss probability q = 1-p):

Up through 25 bets, the probability of a bankroll change of (n-25) is the nth term (starting with 0) of the binomial expansion of (p+q)

^{25}.

In each subsequent bet, the probability of a bankroll change of n is p times the previous row's (n-1) + q times the previous row's (n-1), except that a term for -25 is ignored as once that is reached, the betting ends.

As usual with me, (n)C(k) is the number of combinations of n things taken c at a time; also known as combin(a,b).

The sum of the probabilities for the bankrupt point (bankroll = -25) is:

q

^{25}

25 p q

^{26}

((25)C(1) + (26)C(2)) p

^{2}q

^{27}

(2 (25)C(1) + 2 (26)C(2) + (27)C(2)) p

^{3}q

^{28}

(5 (25)C(1) + 5 (26)C(2) + 3 (27)C(2) + (28)C(3)) p

^{4}q

^{29}

(14 (25)C(1) + 14 (26)C(2) + 9 (27)C(2) + 4 (28)C(3) + (29)C(4)) p

^{5}q

^{30}

and so on, up to p

^{37}q

^{62}, as anything larger would be more than 100 bets.

The terms correspond to something called Catalan's Triangle, but right-to-left

(i.e. normally in Catalan's triangle, row 5 is 1, 5, 9, 14, 14, but here is it 14, 14, 9, 5, 1).

However, I am not aware of a simple formula that would calculate the sum, even if you go by column (i.e. add up the (25)C(1) terms and multiply by 25, then add up the (25)C(2) terms and multiply by (25)C(2), and so on).

Last edited by: ThatDonGuy on Jul 26, 2016

July 26th, 2016 at 7:48:14 PM
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Wiz,

OK, you're serious. Got it. I can't wriggle out by saying, "You're not drunk as long as you can hold on to the floor."

But, seems to me, after 100 iterations, casino should have won 51 and player should have won 49, for a 2-unit loss for the player. Straight-line extension shows when player goes broke.

Of course, this is straight probability and a simulation could fluctuate significantly. But, OK. Let's look at some simulation results that are pretty close to your parameters. I've spent a LOT of time slicing and dicing your Baccarat simulation data (250,000 8-deck shoes). On the 20,219,242 hands you simulate, Banker wins 32,111 more hands than the player. Per 100 hands, that works out to the Banker winning 1.75 more units per 100 units bet (assuming 1 unit wagered per bet). Pretty close to my seat-of-the-pants suggestion of 2 units, right?

So, if you know the expected player loss per hand and the player bankroll, one can estimate the player's bankroll at any "finite step" one selects.

Seems so simple, I MUST be overlooking something. But, it's hard to "overlook" anything when you're holding on to the floor. However, I'm not as drunk as most tinkle peep I am, and I've got all day sober to Sunday up.

OK, you're serious. Got it. I can't wriggle out by saying, "You're not drunk as long as you can hold on to the floor."

But, seems to me, after 100 iterations, casino should have won 51 and player should have won 49, for a 2-unit loss for the player. Straight-line extension shows when player goes broke.

Of course, this is straight probability and a simulation could fluctuate significantly. But, OK. Let's look at some simulation results that are pretty close to your parameters. I've spent a LOT of time slicing and dicing your Baccarat simulation data (250,000 8-deck shoes). On the 20,219,242 hands you simulate, Banker wins 32,111 more hands than the player. Per 100 hands, that works out to the Banker winning 1.75 more units per 100 units bet (assuming 1 unit wagered per bet). Pretty close to my seat-of-the-pants suggestion of 2 units, right?

So, if you know the expected player loss per hand and the player bankroll, one can estimate the player's bankroll at any "finite step" one selects.

Seems so simple, I MUST be overlooking something. But, it's hard to "overlook" anything when you're holding on to the floor. However, I'm not as drunk as most tinkle peep I am, and I've got all day sober to Sunday up.

July 26th, 2016 at 8:04:21 PM
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Quote:ThatDonGuyNow with Math!

That is further than I got, thank you. However, for the purposes at hand, I need a simple formula that doesn't involve a summation symbol. Approximations are perfectly fine. For example there is a formula given here: Risk of Ruin and Drawdown Calculation Tool, however it is for the case of infinite bets and a player advantage. I can't find anything based on a finite number of bets.

It's not whether you win or lose; it's whether or not you had a good bet.

July 26th, 2016 at 9:18:21 PM
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Quote:mipletThere are a bunch listed here. Probability of reaching a goal before going bankrupt given a time constraint is probably close to what you want.

Those are some amazing calculators but they seem to be simulation based.

It's not whether you win or lose; it's whether or not you had a good bet.

July 26th, 2016 at 9:30:38 PM
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using a simple transition matrix in ExcelQuote:WizardYou have $25. You play a game where you have a 49% chance to win $1 and a 51% chance to lose $1. What is the probability you can survive 100 bets without going broke?

I gets 0.019520497 for ruin (in agreement with a sim I also ran)

so 1 - (I gets) = the survival probability

also shows the complete ending distribution too

starting with many different bankrolls

for $25

like ending at 23 = 0.079604491

ending at 21 = 0.078072581

and so on (kids love that feature these days. I am not a kid anymore))

the Don Schlesinger formula is a good one to use too. I have not looked at it in some time

there is a paper that shows that method

and the Binomial Ruin Distribution

named

"The Impact of a Finite Bankroll on an Even-Money Game"

Kelvin Morin – Manitoba Lotteries Corporation

even mentions the late Alan Krigman

should come up in Google

<<<<< >>>>>

I thinks BruceZ has tackled that B4 2 (maybe even with some R code)

maybe knots

BruceZ still rocks around Chicago I thinks

Sally pretty

added:

my online folder (link in my blog) has the Alan Krigman Excel sheet for

I) Survival criterion -- 100 rounds

Risk of ruin = 2.0124%

Prob of survival = 97.9876%

so a small error is there

the Excel sheet is RoR-AlanK.xls

free to download

have fun!

Last edited by: mustangsally on Jul 27, 2016

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