Probabilit Question Question:
John and Peter play a game with a coin such that P(head) = p. The game consists of tossing a coin twice. John wins if the same result is obtained in the two tosses, and Peter wins if the two results are different. 
• (a) At what value of p is neither of them favored by the game? 
• (b) If p is different from your answer in (a), who is favored?
My Answer:
(a) p = 0.5 makes sure that John and Peter has the same chance to win or lose in the game, so nobody is favored by the game.
(b) If p > 0.5, then John has a higher overall probability to win the game (John is favored by the game. In averse situation, where p<0.5, Peter is favored by the game.
Example: p = 0.51 and pc = 0.49
P(John) = 0.51*0.51+0.49*0.49 = 0.5002
P(Peter) = 0.51*0.49+0.49*0.51 = 0. 4998

The example shows that in the case of p>0.5, John has a higher probability to win. In an averse case, Peter would be favored by the game.
Is my answer correct?
Thanks for comments.
 A: Yes your answer is correct. There are four possible outcomes:


*

*Head head

*Head tail

*Tail head

*Tail tail


When the probability of head (and tail) is $0.5$, each of these outcomes is equally likely ($0.25$ probability, which is $0.5\cdot 0.5$). If you have the probability of a head (or tail) to be greater than $0.5$, it makes the likelihood of a head tail or tail head toss less likely. 
Let's try to prove that if the game is fair then the probability of rolling head or tail is equally likely. We are trying to  the probabilities of getting either cases $1$ or $4$. That is to say,  where the probability is equal to $0.5$
$$x^2 + (1-x)^2=0.5$$
That is when the following is true
$$2x^2 - 2x + 0.5 = 0$$
$$(x-0.5)(2x - 1) = 0$$
Both solutions to this quadratic require $x = 0.5$
Now,  let's us investigate further. 
If I let $f(x)  = x^2 + (1-x)^2$ then :
$$f'(x) = 2x +2x-2$$
So there is a minimum or maximum at $0.5$. Further differentiation tells us this is a minimum,  so John has at least a probability of $0.5$ of winning,  which we have already proved is when $P\text{(Head)} = P\text{(Tails)} = $0.5$,  but this probability increases where we have an uneven coin. 
