Need help on a problem i can't solve

A.B.

New member
Joined
Nov 23, 2021
Messages
3
Hello Everyone 😊

I'm stuck at something, and I'd really appreciate if someone could help me out of this.

So, it's a betting problem. If win rate is 90% - from an online calculator available I calculated odds of 6 losses happening in a row in 1,000 occurrences is 0.090%. (The online calculator might be unreliable, so please feel free to correct me if the math is wrong)

But the calculator says nothing about wins & losses showing up randomly one after another. Say there're 5 losses in a row, then a win, then another 4 losses, followed by 2 wins, then another 5 losses, and so on.

I want to know the probabilities of such different instances showing up with both wins & losses mixed up. An idea about the grey area of betting.

So that I'll have a better idea of suited starting bankroll. I'm much more concerned about losing streaks showing up suddenly & frequently in between the wins - so in a sense, it's a bit like preparation for the worst case scenarios.

And if there's some excel spreadsheet or calculator anyone knows of where I can calculate about such different scenarios, and point me toward it - that'd be really helpful.

Thank you for taking the time to read through my lengthy post, and also thank you in advance for any kind of suggestion/solution/tip you can give me on this.

Have a lovely day!
 

JeffM

Elite Member
Joined
Sep 14, 2012
Messages
6,963
The probabilty of six losses in a row when the probability of winning is 90% is

[math]\dbinom{6}{6} * 0.9^0 * 0.1^6 = 0.000001[/math], which is once in 100,000 times.

The general formula for k losses out of n attempts when the probability of winning is p is

[math]\dbinom{n}{k} * p^{n-k}(1 - p)^k, \text { where } \dbinom{n}{k} = \dfrac{n!}{k! * (n - k)!}[/math].

This reduces to the probability of n losses in a row being

[imath](1 - p)^n[/imath], where p is the probability of winning.
 

A.B.

New member
Joined
Nov 23, 2021
Messages
3
The probabilty of six losses in a row when the probability of winning is 90% is

[math]dbinom{6}{6} * 0.9^0 * 0.1^6 = 0.000001[/math], which is once in 100,000 times.

The general formula for k losses out of n attempts when the probability of winning is p is

[math]\dbinom{n}{k} * p^{n-k}(1 - p)^k, \text { where } \dbinom{n}{k} = \dfrac{n!}{k! * (n - k)!}[/math].

This reduces to the probability of n losses in a row being

[imath](1 - p)^n[/imath], where p is the probability of winning.

Thank you so much!

I'm glad that I didn't fully rely on that online calculator - I really appreciate you giving me the correct probabilty.

I'm somewhat concerned about if say there are 5 losses in a row, then a win, then another 4 losses, followed by 2 wins, then another 5 losses. That kind of uncertainty that can suddenly show up bugs me.

Can you give me any suggestion on how might I calculate such scenarios? Or at least get a rough idea of them before I start the process?

Have a great evening!
 

JeffM

Elite Member
Joined
Sep 14, 2012
Messages
6,963
You seem to be trying to determine “risk of ruin“ without specifying your strategy or the number of “plays.”


This is a field I know little about, but I suspect specification of a betting strategy, payoffs, and an outside limit on the number of plays available are necessary to get ANYWHERE.

You used a probability of winning of 90%. That will be true for few games of chance that involve money (though it may be true of some computer games.)

In any case, I have given the formula for computing the a priori probability of n losses. You can use that to compute the probability that you will go bust.

If the probability of a win is 0.6, and you have enough money to play 3 times, the probability that you will go bust is

[math](1 - 0.6)^3 = 0.4^3 = 6.4\%.[/math]
Now you can recompute as play goes on. If you lose on the first play, a common strategy is to play again with a doubled bet. If you do that, your probability of going bust is now 40%.

That is, you do not have to compute everything in advance. As your bank roll changes and the size of your bet changes, you can calculate anew in light of the changed facts.

I do not know whether this is helpful. I hope it is but apologize if it is not.
 

A.B.

New member
Joined
Nov 23, 2021
Messages
3
You seem to be trying to determine “risk of ruin“ without specifying your strategy or the number of “plays.”


This is a field I know little about, but I suspect specification of a betting strategy, payoffs, and an outside limit on the number of plays available are necessary to get ANYWHERE.

You used a probability of winning of 90%. That will be true for few games of chance that involve money (though it may be true of some computer games.)

In any case, I have given the formula for computing the a priori probability of n losses. You can use that to compute the probability that you will go bust.

If the probability of a win is 0.6, and you have enough money to play 3 times, the probability that you will go bust is

[math](1 - 0.6)^3 = 0.4^3 = 6.4\%.[/math]
Now you can recompute as play goes on. If you lose on the first play, a common strategy is to play again with a doubled bet. If you do that, your probability of going bust is now 40%.

That is, you do not have to compute everything in advance. As your bank roll changes and the size of your bet changes, you can calculate anew in light of the changed facts.

I do not know whether this is helpful. I hope it is but apologize if it is not.

It was very helpful 😊 Thank you so much!
 
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