- Thread starter idontknow
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I think the easiest way to do this is using conditional probability.

$P[\text{A tosses more heads than B}] = P[\text{A tosses more heads than B | B tosses $k$ heads}]P[\text{B tosses $k$ heads}] = \\~\\

\sum \limits_{k=0}^n \left(\sum \limits_{i = k+1}^{n+1} \dbinom{n+1}{i}\left(\dfrac 1 2\right)^{n+1} \right) \dbinom{n}{k}\left(\dfrac 1 2\right)^n$

Mathematica gives a nutty result for $P[n]$ which evaluates to $P[n]=\dfrac 1 2,~\forall n > 0$

There's probably a simple way to explain this.

and when multiplied by $\left(\dfrac 1 2\right)^{2n+1}$ results in $\dfrac 1 2$ as seen

We can look at this a little differently and see if it's revealing at all.

$P[\text{A tosses more heads than B in $n+1$ tosses}]=\\

P[\text{A tosses more heads than B in $n$ tosses}] + \dfrac 1 2 P[\text{A has tossed the same number of heads as B in $n$ tosses}]$

$P[\text{A has tossed the same number of heads as B in $n$ tosses} = \dbinom{2n}{n} \left(\dfrac 1 2\right)^{2n}$

$P[\text{A tosses more heads than B in $n$ tosses}] = \dfrac{2^{2n}-\dbinom{2n}{n}}{2} \left(\dfrac 1 2 \right)^{2n}$

$P[\text{A tosses more heads than B in $n+1$ tosses}]=\\

\dfrac{2^{2n}-\dbinom{2n}{n}}{2} \left(\dfrac 1 2 \right)^{2n} + \dfrac 1 2 \cdot \dbinom{2n}{n} \left(\dfrac 1 2\right)^{2n} =\\~\\

\left(2^{2n}-\dbinom{2n}{n}\right) \left(\dfrac 1 2\right)^{2n+1} + \dbinom{2n}{n} \left(\dfrac 1 2 \right)^{2n+1} = \\~\\

2^{2n} \cdot \left(\dfrac 1 2\right)^{2n+1} = \dfrac 1 2$

That answer is not dependent on the amount of coins. I want to know whether the probability of having more heads from two more tosses depends on the amount of tosses. It does. The sequence is:

P (A gets more heads from two tosses than B from zero) = 3/4

P (A gets more heads from three tosses than B from one) = 3/4 - 1/16 = 11/16

P (A gets more heads from four tosses than B from two) = 3/4 - 1/16 - 1/32 = 21/32

Each term is the previous term minus half of the last part of the previous part. Can somebody write that in Sigma notation?