ggarr wrote:

If I have 7 bets going on 7, 2 man, fights, what is the probability of me winning all 7 bets?

I'm thinking 1/2^7. Is it that simple?

This is a classic binomial probability problem.

It is exactly same as flipping a coin n times and betting on heads/tails everytime. For each time you flip, the probability of winning be p and probability of losing be q. Then, p + q = 1.

For you to win r times within those n attempts, the total probability is given by [nCr]*[p^(r)]*[q^(n-r)].

This is also applicable when n coins are flipped simultaneously at the same time. This explains that we are handling the flipping of a coin as mutually exclusive and independent event.

Substitute p = 1/2, q = 1/2, n = r = 7 for your problem, and the answer is evident. Though it looks simple, the final answer actually indicates that you have the remotest chances of winning all 7 bets. Any number of bets less than 7 have more probability of winning.

The same situation can be extrapolated even for problems with dice. When you roll just one dice, the probability of getting a chosen number p is 1/6 while probability of not getting a chosen number is 5/6. When you roll two dice, you'd need to sum up the individual numbers to find the probability of getting the desired result, and thereby, the probability of not getting the desired result and calculate your success probability for r successes in n attempts.

Hope that helps.

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