# Thread: Probability: Flipping A Coin...

1. ## Probability: Flipping A Coin...

Each day Alice and Bob flip a coin to see who buys coffee ($1.20 a cup). Bob flips and Alice calls the outcome. If the person who calls the outcome is correct, the other buys the coffee;otherwise the caller pays. (a) Determine Alice’s expected winnings if a fair coin is used and she calls the outcomes at random. Does Bob have an advantage if he decides to use a biased coin? Motivate your answer. (b) Ignoring your advice, Bob decides to cheat with a biased coin that results in heads 2/3 of the time. Suspecting this, Alice decides to change her strategy as well to one where she guesses heads 75% of the time. Over the course of 100 days, how many times do you expect Alice to win a cup of coffee? I'm not sure what direction to go in with this one! Please help!!! 2. Originally Posted by tuheetuhee Each day Alice and Bob flip a coin to see who buys coffee ($1.20 a cup). Bob flips and Alice calls the outcome. If the person who calls the outcome is correct, the other buys the coffee;otherwise the caller pays.

(a) Determine Alice’s expected winnings if a fair coin is used and she calls the outcomes at random. Does Bob have an advantage if he decides to use a biased coin? Motivate your answer.

(b) Ignoring your advice, Bob decides to cheat with a biased coin that results in heads 2/3 of the time. Suspecting this, Alice decides to change her strategy as well to one where she guesses heads 75% of the time. Over the course of 100 days, how many times do you expect Alice to win a cup of coffee?

I'm not sure what direction to go in with this one! Please help!!!
(b) I'd suggest drawing a tree diagram. The first two branches are the choices of Alice calling heads (3/4) or tails (1/4). From each branch you want two branches that give the result of the coin toss: Heads (2/3) or Tails (1/3).

Alice wins if:

1. She calls tails and tails come up: $\left( \frac{1}{4} \right) \, \left( \frac{1}{3} \right) = \left( \frac{1}{12} \right)$.

2. She calls heads and heads come up: $\left( \frac{3}{4} \right) \, \left( \frac{2}{3} \right) = \left( \frac{6}{12} \right)$.

So the probability of Alice winning on a single day is $\frac{7}{12}$ (so clearly Bob doesn't have an advantage with this biassed coin) when Alice suspects).

Now let X be the random variable number of days that Alice wins.

Then X~Binomial(p = 7/12, n = 100).

E(X) = .......

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(a) It should be clear now how to prove that:

1. The probability of Alice winning on a single day is 1/2 and therefore her expected winnings is zero (by symmetry this is an obvious result of course).

2. Bob does not necessarily have an advantage when using a biassed coin. Some obvious questions (left for you to consider) are:

1. What happens if Alice does NOT know that the coin is biassed?

2. Will a biassed coin ever give Bob an advantage? Under what circumstances?