1. Conduct the following experiment 10 times: first, toss a coin 10 times in a row and record the number of heads, then toss the coin 9 times in a row and also record the number of heads. We will call the experiment successful if the number of heads in the first case is greater than in the second. After conducting a series of 10 such experiments, record the number of successful and unsuccessful experiments. Present the collected statistics in the form of a table.

a) Vanya tosses a coin 3 times, and Tanya - two. What is the probability that Vanya has more heads than Tanya?

b) Vanya tosses a coin \( n+1 \) times, and Tanya - \( n \) times. What is the probability that Vanya has more heads than Tanya?
# Solution.

a) The problem is a particular case of the more general problem from part b).

Answer: $\frac{1}{2}$.

b) Let Vanya first toss the coin $n$ times, just like Tanya. After this, the following events arise:

$$
\begin{aligned}
A= & \{y \text { Vanya has more heads }\} ; B=\{y \text { Tanya has more heads }\} ; \\
& C=\{y \text { Vanya and Tanya have the same number of heads }\} .
\end{aligned}
$$

It is clear that $\mathrm{P}(A)+\mathrm{P}(B)+\mathrm{P}(C)=1$ and that due to symmetry (the players can be swapped without changing the situation) $\mathrm{P}(A)=\mathrm{P}(B)$, where $\mathrm{P}(A), \mathrm{P}(B)$, $\mathrm{P}(C)$ are the probabilities of the respective events.

Now let Vanya toss the coin one more time, the $(n+1)$-th time. If events $A$ or $B$ have already occurred, the outcome of the last toss will not affect the final result: if $A$, Vanya will win, if $B$ - he will not win. Vanya will have more heads in the case of $C$ if a head appears in the $(n+1)$-th toss (the probability of this event is $\frac{1}{2}$). Therefore, the probability of Vanya winning is:

$$
\mathrm{P}(A)+\frac{1}{2} \mathrm{P}(C)=\mathrm{P}(B)+\frac{1}{2} \mathrm{P}(C)=\frac{1}{2}
$$

Answer: $\frac{1}{2}$.