Probability Question for 2-Sided Coin (Verification)

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Suppose you and your friends have a two-sided coin each. Your coin lands Heads with probability $\frac{1}{6}$, while your friend's coin lands Heads with probability $\frac{3}{4}$. The two coins are independent of one another. Suppose you play a game where you both flip your coins once, and if they both land on the same side (i.e., both Heads or both Tails) you get $x from your friend, but if they land on different sides, then your friend gets 2 dollars from you.

What is the minimum integer value of $x$ for which your expected total winnings after 3 rounds of this game are positive (i.e., you are expected to make money rather than lose some)?

Hint: Let $W$ denote your total winnings after 3 rounds of this game. What values of $W$ are possible?

This is what I have so far:

Your Coin: Probability of Heads P(H1) = $\frac{1}{6}$ ; P(T1) = $\frac{5}{6}$

Your friends coin: Probability of Heads P(H2) = $\frac{3}{4}$ ; P(T2) = $\frac{1}{4}$

For a given round:

$W$= you win : Both heads or both sides

$L$= you lose : landing on different side

P(W) = Probabilty of Both heads or Both tails = P(H1H2) + P(T1T2) = P(H1)P(H2) + P((T1)P(T2) = $\frac{1}{6}$$\frac{3}{4}$$\frac{5}{6}$$\frac{1}{4}$= $\frac{1}{3}$

P($L$) = Probability of landing different sides = 1- Probabilty of Both heads or Both tails = 1-$\frac{1}{3}$ = $\frac{2}{3}$

P($W$) = $\frac{1}{3}$ P($L$) = $\frac{2}{3}$

The game is played for three rounds : Let $Y$ be the number of rounds you win

Then possible values of $Y$= 0,1,2,3;

$W$ : Total winning after three rounds

when $Y$ = 0 ; You lose 3 rounds and lose 3*2 = 6$ ; W = 0x – 6 = -6

Y=1 ; You win one round and lose 2 round ; W = 1x – 4 = x – 4

Y=2 ; You win 2 rounds and lose one round : W= 2x -2

Y=3 ; You win all three rounds W = 3x – 0 = 3x

If Y is random variable representing number of wins in 3 rounds with probability of winning a round: p=$\frac{1}{3}$ and q =1 -p =$\frac{2}{3}$;

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If expected value of winning after 3 rounds to be positive [ex) x – 4 > 0 = x > 4]

Then, when x > 4 then expected value of winning after 3 rounds to be positive

when value of x = 5 ; then expected value of winning is x-4 ex) 5-4 = 1

Minimum integer value of x > 4 is 5

Minimum integer value of x = 5 for which you expected total winnings after 3 rounds of this game are positive.

Best Answer

Alternatively, let $W_i$ be the expected winning on day $i$.

$$E[W_i] = \frac13 x - \frac23\cdot 2$$

By linearity of expectation, expected winning over $3$ days would be $E[W_1+W_2+W_3]=3E[W_1]=x-4$

Hence the answer for the smallest integer satisfying $x>4$ would be $5$.

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