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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.

  1. What is the probability of the first player to win on the first attempt?
  2. What is the probability of the first player to win the game?
  3. 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.

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

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

  3. 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?

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    $\begingroup$ You've got to be careful. For the first player, $HHT$ is a different throw from $HTH$, and they need to be separate entries. If they aren't treated as separate, then the probabilities aren't uniform, they go like this: $P(3H) = 1/8, P(2H) = 3/8, P(1H) = 3/8, P(0H) = 1/8$. $\endgroup$ – Arthur Jun 15 '16 at 12:27
  • $\begingroup$ Thanks! Modified my question. The cases should be correct now, but is it so with my answer? $\endgroup$ – Ivan Prodanov Jun 15 '16 at 12:39
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    $\begingroup$ You do not need to list all solutions. Player 1 and player 2 are independently distributed, meaning, the outcome of player 1 does not affect the probability of the outcomes of player 2, and conversely. Player 1 plays a Binomial distribution with $n=3$ attempts and probability of success $p=\frac{1}{2}$. Player 2 plays also a binomial distribution with $p = \frac{1}{2}$, but with $n=2$ attempts. $\endgroup$ – Lærne Jun 15 '16 at 12:46
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    $\begingroup$ Using binomial distribution for the first answer looks interesting. So in order for the first player to win on the first trial it would be $\binom 3 3p^3(1-p)^0 + \binom 3 2p^2(1-p)^1(1-\binom 2 2p^2(1-p)^0) + \binom 3 1p(1-p)^2\binom 2 0p^0(1-p)^2$ $\endgroup$ – Ivan Prodanov Jun 15 '16 at 13:13
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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.

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  • $\begingroup$ Thanks! The third part I believe is related to probability distribution. Any clue on this one? $\endgroup$ – Ivan Prodanov Jun 15 '16 at 13:26
  • $\begingroup$ We need to have a random variable defined first, so we can calculate the probability distribution. So, the distribution of prize is not well defined. $\endgroup$ – Med Jun 15 '16 at 13:36

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