Ana and Bastian

Since Bastian only flips one more coin than Ana, he cannot have both more heads and more tails, but must have more of one. So $P(\text{Bastian flips more heads}) + P(\text{Bastian flips more tails}) = 1$

But heads and tails are equally likely results for both players, the probability of Bastian getting more heads than Ana must be the same as the probability of Bastian getting more tails than Ana. Thus, the probability of either is $\boxed{\dfrac{1}{2}}$ .

Let $p$ be the probability that after both Ana and Bastian have tossed their coins 50 times, they have the same number of heads. Then the probability that Bastian already has more heads is $\dfrac{1}{2}(1 - p)$ , (the same probability that Ana already has more heads), in which case he will have more heads than Ana regardless of the result of his 51st toss. If they are tied after 50 tosses, however, then he can still end up with more heads if his 51st toss is a head, which would occur with probability $1/2$ . So the total probability that Bastian ends up with more heads after his 51st toss is

$\dfrac{1}{2}(1 - p) \times 1 + p \times \dfrac{1}{2} = \dfrac{1}{2} = \boxed{0.5}$

Consider these two eventualities:

$E_1$ : B throws more heads than A

$E_2$ : B throws more tails than A

E1 and E2 are mutually exclusive. Suppose Ana had H heads and T tails, where H+T=50. B would need or more 52 tosses to throw H+1 or more heads and T+1 or more tails. He only has  51 tosses.

Either E1 or E2 has to be true. If both were false, the B must throw no more than H heads and no more than T tails. That is only possible with 50 or less tosses.

Combined, $E_1$ and $E_2$ must be complementary, meaning that $P(E_1)+P(E_2)=1$ .

By symmetry, for fair coins $P(E_1)=P(E_2)$ . From which the answer $P(E_1)=\frac{1}{2}$ follows directly.

The situation needn't be more complex. If Bastian and Ana has 50 tosses each, the probability of Bastian and Ana having equal chances of heads and tails is same. If then Bastian is given an extra toss, this toss has no memory of the previous 50 tosses and since the previous 50tosses cancels outs Ana's then there is exactly 1/2 chance of Bastian having a head, as there should be if that was the first toss.

$\frac{\sum _{k=0}^{51} \binom{51}{k} \sum _{j=0}^{k-1} \binom{50}{j}}{2^{101}} = 1/2$

Although the problem looks hard, because 50 and 51 coin tosses are a large amount, let's reduce it to 0 and 1 coin tosses.

Here, the chances of Ana getting 0 heads is 100%, because she have no coin tosses. The chances of Bastian getting more than 0 heads is obviously 50%, as he only have 1 coin toss.

So the answer must be 50%, or 0.5.

The expected value for the number of heads for Bastian is 25.5. The probability of having more heads than the expected value is 0.5. That means that the probability of throwing 26 or more heads is 0.5.

$1 = Prob(H_b > H_a) + Prob(H_b = H_a) + Prob(H_b < H_a)$ , where $H_b =$ Brian's number of heads and $H_a =$ Ana's number of heads.

Note that $Prob(T_b > T_a) = Prob(H_b = H_a) + Prob(H_b < H_a)$ . That is, for Brian to have more tails than Ana does, he must have fewer or the same number of heads, and vice versa.

Therefore $1 = Prob(H_b > H_a) + Prob(T_b > T_a)$ .

Due to the symmetry between head and tail, $Prob(H_b > H_a) = Prob(T_b > T_a)$ . That is, the likelihood that Brian has more heads than Ana is equal to the likelihood that he has more tails than she does.

Therefore $1 = 2 \cdot (Prob(H_b > H_a)$ .

(Updated 16/12/2018). After 50 tosses there are 3 possible outcomes: A has more than 50 heads, B has more than 50 heads, or they have 25 each. The probabilities are $p_1$ , $p_2$ and $p_3$ ; $p_1+p_2+p_3=1$ . Due to symmetry $p_1=p_3$ and $2p_1+p_2=1$ . At this point B is already winning with probability $p_1$ and A is loosing with the same probability. When B tosses the coin again, his chance of winning increases by $p_2*(1/2)$ to a total of $p=p_1+p_2/2=1/2$ .

Very simple: Both toss their coin 50 times. They both expect to get the same number of heads. Now, Bastian can toss his coin once again. His chances to get another head are 50% (1/2). This is also the probability of getting more heads than Ana.

An unbiased coin has been tossed 101 times. It is not possible for the number of heads and tails to be equal.Thus the only two possible outcomes are the number of heads is greater than the number of tails or the number of tails is greater than the number of heads. The symmetry of the outcomes provides that each of these two outcomes are equally likely..so the probability that either tosses more heads than the other is 1/2.

They all flip 50 coins. They have an equal chance of getting heads or tails. Now the other person flips one more coin. There you go. That’s the deciding factor. 50% chance.

Their probabilities of getting heads or tails are the same for their first 50 flips.. until Bastian flips another coin which has a 50/50 chance of getting a head or tail.

Imagine that they'll only do 50 flips each, they get the same probabilities right? Until Bastian gets another shot of doing a flip

I don't know if I'm oversimplifying it but this is how I look at that question.

Since they both flipped the coin 50 times, they both have the same probabilities profile (to get any number or head or tails).

So what happens next has nothing to do with what happened previously.

On the 51st coin toss, it's as if Ana had never tossed a coin and Bastian is flipping it for the first time.

He has a 1/2 chance to get a head.

So he has 1/2 chance to get more heads than Ana.