Just One Extra Coin...

Either you throw more heads than your prof, or you throw more tails than your prof, since you have only one extra coin, but not both!

These two mutually exclusive possibilities occur with equal probability.

The probability that you obtains more heads than your professor is 1/2.

The probability is independent of the number of coins held by the players.

Take m as the number of coins you have. And n as the number of coins the professor has. For a better generalization, n>m.

1/2^n) / (1/2^m)

=(1/2^101)/(1/2^100)

=(1/2^101)*(2^100)

=1/2 paragraph 4

For sake of variety, here's another approach....

Let $A$ be the probability that both you and your professor get the same number of heads after each tossing $100$ coins. By symmetry, the probability that you have tossed more heads at this point in the proceedings is then $\dfrac{1 - A}{2}$ , and this status won't change regardless what the result is of your $101$ st toss.

If you and your professor are tied after $100$ tosses, however, then there is a probability of $\frac{1}{2}$ that your $101$ st toss will leave you with more heads. Thus the probability that you end up with more heads than your professor after tossing all your coins is

$\dfrac{1 - A}{2} + A*(\dfrac{1}{2}) = \dfrac{1}{2} = \boxed{0.5}$ .

This gets more complicated if you have $100 + n$ coins for $n \gt 1$ . For example, with $n = 2$ , if you are tied after than $100$ tosses then there is a $\frac{3}{4}$ probability that at least one of your extra two tosses will yield a head. That's easy enough to deal with, but we would also have to deal with the possibility that you have one less head after $100$ tosses, in which case there is a $\frac{1}{4}$ probability that your extra two tosses yield heads and hence leave you with the greater number of heads. As noted in my comment to Satyen's solution, I found that that the probability $P(n)$ of having more heads when given $n$ extra coins to toss is $0.52806$ in the case of $n = 2$ and $0.54885$ in the case of $n = 3$ . A general formula for $P(n)$ is in the works; it would probably be worthwhile to introduce the variable $k$ as well for the number of coins the professor starts with as well, (giving you $k + n$ coins). So now I'm looking for an equation for $P(k;n)$ ..... Or maybe I'll stop being so obsessive about this..... sigh.