Uneven Heads against Tails

You and your friend have an equal probality of getting heads or tails on all of the first 2014 tosses. Hence in a perfect probability world you and your friend will get the same amount of tails for the first 2014 tosses. Therefore all that matters is the last toss, making the answer $\boxed {\frac {1} {2}}$ , the probability of getting tails on the last coin flip and as a result having more tails than your friend.

Let A throw $n+1$ coins, and B $n$ coins. Suppose that player A throws $a$ tails, while player B throws $b$ heads. Then $a > b \quad \Leftrightarrow \quad n+1-a < n+1-b \quad \Leftrightarrow \quad n+1-a \le n - b$ since $a$ and $b$ are integers, and so $P[a > b] \; = \; P[n+1-a \le n-b] \; = \; 1 - P[n+1-a > n-b] \;,$ so that the probability that A throws more tails than B is $1$ minus the probability that A throws more heads than B. By symmetry, the probabilities that A throw more heads than B and that A throws more tails than B must be equal, and hence both are equal to $\boxed{\tfrac12}$ .

In general, suppose you and your friend first flip $n$ (fair) coins and keep a tally of the number of tails thrown. Let $p_{a}$ be the probability that you end up with more tails, $p_{b}$ be the probability that your friend ends up with more tails and $p_{t}$ be the probability that you both end up throwing the same number of tails. As these three events are mutually exclusive we have that $p_{a} + p_{b} + p_{t} = 1$ , and be symmetry $p_{a} = p_{b}$ . This gives us that $p_{t} = 1 - 2p_{a}$ .

Now you toss your extra coin. If after each tossing $n$ coins you and your friend are tied in the number of tails thrown, then there is a probability of $\frac{1}{2}$ that after tossing your extra coin you will end up having thrown more tails. You're guaranteed of having more tails after tossing $n + 1$ coins if you are already in the lead after tossing $n$ coins, so the probability that you will end up with more tails after your extra flip is

$p_{a} + \dfrac{1}{2}p_{t} = p_{a} + \dfrac{1}{2}(1 - 2p_{a}) = \dfrac{1}{2} = \boxed{0.5}$ .