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I was wondering about the probability of coin flips. How many times would i need to toss a coin to have at least 75% chance of getting a head.

I know the answer would be $1-\left(\left(\frac{1}{2}\right)^2\right)$ = 2 throws but how would you work it out without trial and error.

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  • $\begingroup$ One flip you gain .5, another flip you get half of that which makes .75 so 2 $\endgroup$ – suomynonA Dec 7 '16 at 3:01
  • $\begingroup$ Log(1-3/4)/log(1/2)? $\endgroup$ – Kitter Catter Dec 7 '16 at 3:06
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You can use complementary counting.

Having at least a $75%$ chance of getting a head is equivalent to a $(100-75)\%=25\%$ or lower chance of getting no heads.

Getting no heads means flipping all tails. The probability of flipping $k$ consecutive tails is $\frac{1}{2^k}$. We want this probability to be less than or equal to $25\%$ (or $\frac{1}{4}$).

$k=2$ is the smallest solution, so you would need to toss the coin twice.

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If you want to get a%, you need $log_{1\over2}1-{a\over100}$ times, then find the number that is closest to it that is greater than it.

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