Consider two players P1 and P2

Question

Consider two players P1 and P2:

• P1 has one fair coin. 
• P2 has two coins. One of them is fair, whereas the other one is 2-headed (Her Majesty is on both sides of this coin).

The two players P1 and P2 play a game in which they alternate making turns: P1 starts, after which it is P2's turn, after which it is P1's turn, after which it is P2's turn, etc.

• When it is P1's turn, she flips her coin once.

• When it is P2's turn, he does the following:

  • P2 chooses one of his two coins uniformly at random. Then he flips the chosen coin once.
  • If the first flip did not results in heads, then P2 repeats this process one more time: P2 again chooses one of his two coins uniformly at random and flips the chosen coin once

The player who flips heads first is the winner of the game.

• Determine the probability that P2 wins this game, assuming that all random choices and coin flips made are mutually independent. Justify your answer.

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Just need a little help figuring out the answer.

I get that when it is P2's turn he has two chances of getting heads, and if he gets the 2-headed coin he wins and if he gets the fair coin he has a 50-50 chance of getting heads.

But what I can't figure out is how to actually calculate his probability of winning.

Answer 1

I'm not really sure but it may be something like this.

Player2 has probability $\frac{3}{4}$ of getting heads in a single coint toss. Let $E_{1}$ be the event that player 2 wins in the first turn, so $E_{1} = \frac{1}{2} \frac{3}{4} + \frac{1}{2} \frac{1}{4} \frac{3}{4}$ (this is because there are two ways for player 2 to win in the first round: when he gets heads at his first coin toss and when he gets heads at his second coin toss).

Following the same reasoning we have that if $E_n$ is the event where player 2 wins after n repetitions of the experiment then $E_n = (\frac{1}{2})^{n} (\frac{1}{4})^{2n-2} (\frac{3}{4}) + (\frac{1}{2})^{n} (\frac{1}{4})^{2n-1} (\frac{3}{4})$

Then if $E$ is the event where player 2 wins, we have $E = \sum\limits_{i=1}^{\infty} E_{i}$ which is the sum of the right sides of the $E_{i}$'s plus the left sides, that is $E = \sum\limits_{i=1}^{\infty} (\frac{1}{2})^{i} (\frac{1}{4})^{2i-2} (\frac{3}{4})$ $+\sum\limits_{i=1}^{\infty}(\frac{1}{2})^{i} (\frac{1}{4})^{2i-1} (\frac{3}{4})$

So $\sum\limits_{i=1}^{\infty} (\frac{1}{2})^{i} (\frac{1}{4})^{2i-2} (\frac{3}{4}) = \frac{1}{2} \frac{3}{4}\sum\limits_{i=1}^{\infty} (\frac{1}{2})^{i-1} (\frac{1}{4})^{2(i-1)} = \frac{3}{8} \sum\limits_{n=0}^{\infty} (\frac{1}{32})^{n}$ and using the geometric series this is $\frac{3}{8}\frac{32}{31} = \frac{12}{31}$

Using a similiar procedure for the other sum we get $\sum\limits_{i=1}^{\infty}(\frac{1}{2})^{i} (\frac{1}{4})^{2i-1} (\frac{3}{4}) = \frac{3}{8} \frac{1}{4}\frac{32}{31} = \frac{3}{31}$ and finally $E = \frac{12}{31} + \frac{3}{31} = \frac{15}{31}$

Hope that helps.