Probability and Stats (loaded coin)
Smith is offered the following gamble: he is to choose a coin at random from a large collection of coins and toss it randomly.The proportion of the coins in the collection that are loaded towards a head is $p$. If a coin is loaded towards a head, then when the coin is tossed randomly, there is $\frac{3}{4}$ probability that a head will turn up and a $\frac{1}{4}$ probability that a tail will turn up. Similarly, if the coin is loaded towards tails, then there is a $\frac{3}{4}$ probability that a tail will turn up and a $\frac{1}{4}$ probability that a head will turn up. If Smith tosses a head, he loses \$$100$, if he tosses a tail, he wins \$$200$. Find the probability value $p$ for which Smith's gain is $0$ when taking the gamble.
$\endgroup$ 21 Answer
$\begingroup$I'm assuming the coins that are not loaded towards heads are loaded towards tails. Write out the expected gain in terms of p, set it to 0, and solve.
Compute the expected gain in terms of P(heads) and P(tails), the overall probability of flipping heads, or tails, respectively.
Expected gain = P(heads)(-100) + P(tails)(200).
We can then compute P(heads) and P(tails) in terms of p, the percentage of coins biased towards heads.
P(heads) = (3/4)p + (1/4)(1-p)
P(tails):
P(tails) = (1/4)p + (3/4)(1-p)
Now just solve for p:
$\endgroup$0 = (-100)[(3/4)p + (1/4)(1-p)] + 200[(1/4)p + (3/4)(1-p)]
