# *Rate of Ruin & Optimal N

#### coolcat

##### New Member
For a fair coin toss game, both prob(ruin) & prob(consecutive losses) increases with no. of trials.

Is the Rate of ruin .i.e.where prob(ruin) changes fastest, affected most by the number of consecutive losses?
If so, is there any condition or optimal N, maximum no. of consecutive losses to indicate a good time to exit the game?

In a bias coin toss game where prob(head)=0.7, what is the probability of 30 tails clustering (not necessary consecutively) in the first 50 tosses out of 100 trials?

Appreciate any suggestions.

#### BGM

##### TS Contributor
Firstly, define what is "ruin" here - do you mean the number of tail (loss) observed is exceeding a certain number (the initial wealth)?

Secondly, do you know the probability of success/failure in each trial before you playing the game?

Thirdly, you need to define an objective function to maximize before finding the corresponding optimal stopping rule.

#### coolcat

##### New Member
Firstly, define what is "ruin" here - do you mean the number of tail (loss) observed is exceeding a certain number (the initial wealth)?

Secondly, do you know the probability of success/failure in each trial before you playing the game?

Thirdly, you need to define an objective function to maximize before finding the corresponding optimal stopping rule.

Last edited:

#### BGM

##### TS Contributor
By assuming the independence of the trials, the ruin probability is independent of the past history.

Again, you want the optimal condition to quit - but optimal in what sense?

#### coolcat

##### New Member
By assuming the independence of the trials, the ruin probability is independent of the past history.
Again, you want the optimal condition to quit - but optimal in what sense?
When the no. of consecutive losses, N, has the greatest effect on the chances of current balance going to zero.Find a mathematical expression for this N in terms of Cnow & bet size, b (as % of Cnow).

Last edited:

#### coolcat

##### New Member
To re-phrase,
When is the prob(streak of losses just enough to ruin its bankroll B) > 50%?
i.e. Is it possible to express this prob in terms of current bankroll B, b, N, p, q, etc?

Also, how do you get prob. of at least X tails in M tosses with prob. of tail = q, out of K trials

#### BGM

##### TS Contributor
Sorry I do not understand how the ruin probability is related to the number of consecutive losses observed right now, assumed that each trial is independent. Or you mean anything else?

This just recall my first post in TS - a person also asking the question if there is a 10 consecutive losses, will the next trial has a higher probability to win...

Maybe I state one example:

You have $10 now and bet$1 each time. Two possible sample paths are

1. WWWWWLLLLL
2. WLWLWLWLWL

and both of them end up with \$10 after 10 trials. And their ruin probability is the same after 10 trials.

#### coolcat

##### New Member
Also, how do you get prob. of at least X tails in M tosses with prob. of tail = q, out of K trials
Its ok, thanks. How about the above question?
For e.g,
In a bias coin toss game where prob(head)=0.7, what is the probability of 30 tails clustering (not necessary consecutively) in the first 50 tosses out of 100 trials?

Last edited:

#### BGM

##### TS Contributor
Not sure what do you mean by "clustering";

If you want to find what is the probability of 30 tails in the first 50 trials, then you can use the Binomial distribution to help; it is independent of the next 50 trials.

#### coolcat

##### New Member
Not sure what do you mean by "clustering";
If you want to find what is the probability of 30 tails in the first 50 trials, then you can use the Binomial distribution to help; it is independent of the next 50 trials.
How to use the Binomial distribution to help?

#### BGM

##### TS Contributor
Assume each coin toss is independent and have the identical probability $$p$$ to have a head. Then the number of heads in $$n$$ trials $$\sim \text{Binomial}(n, p)$$.

Assume each coin toss is independent and have the identical probability $$p$$ to have a head. Then the number of heads in $$n$$ trials $$\sim \text{Binomial}(n, p)$$.