# A two-outcome game on a limited budget: the probability of being busted over n tries

#### smart_mask

##### New Member
Hello,

My question does not regard a piece of homework, but I have become stuck trying to solve a statistical problem and I would really appreciate the help of this forum.

The problem is as follows:

A gambler has a budget of 100$. She plays a game consisting of consecutive rounds in each of which she has a 49% probability of winning. In each round she bets 1$. If she wins a round she gains an extra 1$, if she loses she forfeits her initial bet. She plays in this manner either for 1000 rounds, or until she loses all her money. I would like to know what is the probability that the gambler will lose all her money under such a scenario. Optionally, I would also like to know how to calculate the player's expected loss. This is what I have been able to figure out until now: If the gambler were to play on an unlimited budget she would always complete the 1000 rounds and would be expected to lose on average 20$ in the process (she would win 490 rounds and lose 510). Her outcomes would be best described by the binomial distribution. Moreover, because of the large number of tries (1000) they would be very well approximated by the normal distribution. In order to calculate the probability that the gambler would have a negative budget at the end of 1000 turns one would need to simply calculate the standard deviation using the sample size, and to use the standard normal distribution table for calculating the area in the left tail corresponding to a last balance of 0 or less. This gives us about P=0,00573 or 0,573%.

However this number is less than the true probability of the gambler going bust, because it is based on the final outcomes after the end of 1000 turns, and these outcomes include cases in which the gambler's money went into negative territory at some point before the end of 1000 turns and subsequently recovered to a positive number. According to the conditions of the problem such outcomes should be included in the probability of the player going bust.

The only technically possible way to solve the problem that I could think of is to add up all the intermediate probabilities of the player going bust.

The earliest chance of the player going bust is in the 100th round if she had the bad luck of losing 100 times in a row (P1). She will also go bust if she wins just 1 round out of 102 (P2), 2 rounds out of 104 (P3), 3 out of 106 (P4) ... 450 out of 1000 (P450). The probability of the player going bust will be equal to the sum of all these probabilities (P1+P2+P3+...P450). P1, P2, P3 etc. can all be calculated using the binomial distribution, and as the probabilities decrease to very insignificant numbers as we go from P450 towards P1 we would not need to calculate and sum up all the 450 factors, but only 10-20 or so.

This solution is, however, highly impractical because it requires the calculation of very large factorials and very large powers of numbers under 1 (numbers out of range for my spreadsheet software). For example the probability that the gambler would lose 440 out of 980 rounds and run out of money on the 980th turn is calculated as P(440)=980!/(440!*540!)*0,51^540*0,49^440. This is a very very large factorial.

Also, I could not find an acceptable way to simplify the sum P1+P2+...P450.

I was wondering whether I've missed something and if anybody knows how to compute such a probability in a more elegant fashion. I would appreciate any assistance. Thank you if you have borne through this interminable post.

Dan

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#### smart_mask

##### New Member
Re: A two-outcome game on a limited budget: the probability of being busted over n tr

A solution using spreadsheet software is in fact possible and I had half found it. Instead of using the binomial distribution to calculate intermediate sums one can approximate them with the standard normal distribution, just as I have done for the outcomes after 1000 tries. In Excel one can then easily generate a table of the intermediate probabilities and sum the factors. The solution is a surprisingly high 40,24% probability of going bust for the gambler. Many factors are significant, so summing up 10 or 20 as I had thought feasible is not appropriate (the estimate gets within 1% of the real figure around the 180th factor). The problem of calculating the expected loss of the player under these conditions remains

#### BGM

##### TS Contributor
Re: A two-outcome game on a limited budget: the probability of being busted over n tr

Usually this type of problem is related to the finite time ruin probability, and not sure if there is an elegant solution or not.

First of all the problem can be modeled as a Markov chain, with states $$\{0, 1, 2, ...\}$$ where the states represent the current wealth of the gambler. The transition probabilities are $$p_{00} = 1, p_{i,i+1} = p_{i, i-1} = \frac {1} {2}, i = 1, 2, 3, ...$$ such that the state 0 is absorbing.

Now the interested quantities is $$p_{100,0}^{(1000)}$$, which is the $$(100, 0)$$ entry of the transition matrix to the power 1000. You can arbitrarily truncated the infinitely large transition matrix to keep the states up to 1100 only because it is the max possible wealth that the gambler could have after 1000 round. So it is computable in this sense.

If you decompose the probability of ruin as the sum of the probability of ruin at 100, 102, 104, ..., 1000 round (the time of first reaching 0), I think you are unable to use the Binomial.