You flip a weighted coin repeatedly. Assume that, each time the coin is flipped, it lands on heads (H) with probability p > 1/2.

(a) What is the probability that you obtain more heads (H) than tails (T) during the first n flips.

wouldn't the probability just be p>.5 since that is given in the question. I don't know any other way to do this.

  • $\begingroup$ If you flip one coin then the probability of more heads than tails is $p$. If you flip, lest's say $100$ coins then the probability is almost $1$ (if $p-\frac12$ is not too small). Can you imagine why? $\endgroup$ – drhab Feb 6 '15 at 10:00

If n is odd

$$P(\text{more heads than tails}) = \sum_{i = \frac{n+1}{2}}^n {n\choose i} p^i(1-p)^{n-i}$$

If n is even

$$P(\text{more heads than tails}) = \sum_{i = \frac{n+2}{2}}^n {n\choose i} p^i(1-p)^{n-i}$$


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