Probability with coins I'm self learning and I stumbled upon the following task, but I struggle to find the solution:
Two players flip coins. The first player flips 3 coins, the second player flips 2 coins. The player that gets most tales wins 5 coins. If both players get the same amount of tales, the game starts over.


*

*What is the probability of the first player to win on the first attempt?

*What is the probability of the first player to win the game?

*How is the prize distributed?


My solution:
if H=heads, T=tails then on the first attempt the following outcomes are possible:
{(HHH, HH), (HHH, HT), (HHH, TH), (HHH, TT),
(HHT, HH), (HHT, HT), (HHT, TH), (HHT, TT),
(HTH, HH), (HTH, HT), (HTH, TH), (HTH, TT),
(THH, HH), (THH, HT), (THH, TH), (THH, TT),
(HTT, HH), (HTT, HT), (HTT, TH), (HTT, TT),
(THT, HH), (THT, HT), (THT, TH), (THT, TT),
(TTH, HH), (TTH, HT), (TTH, TH), (TTH, TT),
(TTT, HH), (TTT, HT), (TTT, TH), (TTT, TT)}
Total cases: 32; First player wins in 16; Second player in 6; Game is repeated in 10.


*

*The probability of the first player to win the game on the first attempt is $\frac {16} {32} = \frac 12 $.

*The probability of the first player to win the game is $\frac {16}{32}\frac {10}{32} = \frac {5}{32}$ ?? 

*I have no idea about the third question. Where can I read more about this?
I'm not very sure if the second is correct. Is it right to conclude that if the game is repeated $n$ times the chance of the first player to win is the same as if the game is repeated 1 time?
 A: About the second part, you can think this way:
Firstly, in each trial the probability that the first player wins is $\frac{1}{2}$, as you have calculated. The probability of the second person to win a trial is $\frac{3}{16}$. The probability of a draw is $1-\frac{1}{2}-\frac{3}{16}=\frac{5}{16}$.
Having the probabilities for a single trial, the probability that the first person wins, in total, is calculated considering the probabilities of the following scenarios:
1- the first person wins in the first trial ($\frac{1}{2}$)
2- the first trial ends in a draw and in the second trial, the first person wins ($(\frac{5}{16})(\frac{1}{2})$)
3- in general, we need to have $n$ draws and one win (for the first person) at the end, which happens with the probability $(\frac{5}{16})^n(\frac{1}{2})$
Since the mentioned scenarios are disjoint, they can be added up to give the final answer
$\frac{1}{2}\sum_{i=0}(\frac{5}{16})^i=\frac{1}{2}\frac{1}{1-\frac{5}{16}}=\frac{8}{11}$
For the third part, I think it should be noted what prize distribution is.
