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If you have a $1$ in $36$ chance of winning an event.

What is the probability that you will win at least once in $36$ tries?

Any help would be much appreciated

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3  
Can you please explain what you've tried, or what lines you're thinking along? –  Tom Oldfield Jan 2 '13 at 4:44
1  
Hint: what's the probability that you won't? –  anon Jan 2 '13 at 4:45
    
That the odds are 1/36. It seems like you would win at least once. (1/36) * 36 is 1. But then I don't think it is in your favor to bet 36 times –  Joe Jan 2 '13 at 4:47
    
In general this probability is $1 - (1 - 1/n)^{n}$, which approaches $1 - 1/e = 0.63212...$ for large $n$. Here, with $n=36$, it's $0.63729...$. –  mjqxxxx Jan 2 '13 at 4:56

3 Answers 3

up vote 29 down vote accepted

It is $1$ minus the probability that you will win $0$ times.

On any trial, the probability that you lose is $\dfrac{35}{36}$.

The probability you win $0$ times, that is, lose $36$ times in a row, is $\left(\dfrac{35}{36}\right)^{36}$.

Remark: We assumed independence. If there is dependence between the various trials, there is little we can say. For example, if a lottery has $36$ tickets, and we buy $1$, the probability we win is $\dfrac{1}{36}$. But if we buy all $36$, the probability we win is $1$.

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It's correct! Thank you very much! –  Joe Jan 2 '13 at 4:51
5  
@Neal you should mark the answer as correct by ticking the green arrow. –  Nick Jan 2 '13 at 10:37

More generally, if you have a $1$ in $n$ chance of winning, where $n$ is a positive integer, you have an $(n-1)/n$ chance of losing.

So your chance of losing $k$ times in a row is $({n-1 \over n})^k$ since you have to lose all $k$ times.

The chance of losing $n$ times in a row is $({n-1 \over n})^n = (1-1/n)^n$, so the chance of not losing $n$ times in a row is $1$ minus this or $1-(1-1/n)^n$.

For $n = 36$, $(1-1/n)^n = 0.3627...$ so the chance is $1-0.3627... = 0.6372...$.

Note that, for moderately large $n$, $(1-1/n)^n \approx 1/e = 0.3678...$. You can derive or find more accurate estimates.

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To elaborate on Andre's answer, you first need to recognize this as an example of the discrete probability distribution. You also need to know that in such distribution, the cumulative distribution function (total probability) for all $36$ tries is $1$. In that case we have

$$1 = pr(0) + pr(1) + pr(2) + pr(3) + ... + pr(36)$$

So if we're looking for the probability of winning or losing at least once, we would have

$$pr(1) + pr(2) + pr(3) + ... + pr(36) = 1 - pr(0)$$

Hence to answer the question, we have to find $1$ - pr(Winning $0$ times). Assuming this is a binomial distribution (which I strongly believe it is), you can then use the ff formula to derive $pr(0)$

$$pr(k) = Cr(N, k)P^{(k)}Q^{(1 - k)}$$

where $N$ = no tries = $36$, $P$ = prob winning an event/a try = $\frac{1}{36}$, Q = prob losing an event = $(1- P)$ = $\frac{35}{36}$ and $k =$ the no of times we want to win = $0$. Therefore, $pr(0)$ = $\frac{36!}{(36!)(0!)}$$(1/36)^{0}(35/36)^{36}$ which gives us $\frac{35}{36}^{36}$.

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