# Probability Question for 2-Sided Coin (Verification)

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 =$$\frac{2}{3}$$;

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.

• You are correct. – Parcly Taxel Oct 29 '18 at 0:54

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 $$5$$.