What Is Your Expected Payoff When Buying Insider Information on a Coin Flip Bet?

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SUMMARY

The discussion focuses on calculating the expected payoff when purchasing insider information about which coin is selected in a coin flip bet scenario. The two coins have probabilities of landing heads at 0.6 and 0.3, respectively. The expected value of the payoff after buying the information is determined to be 1 - c, where c is the cost of the information. It is concluded that if c is less than 1, it is beneficial to purchase the information, as the expected payoff without the information is 0.

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Homework Statement


There are 2 coins in a bin. When one of them is flipped it lands on heads with probability 0.6 and when the other is flipped it lands on heads with probability 0.3. One of these coins is to be randomly chosen and then flipped. Without knowing which coin is chosen, you can bet any amount up to 10 dollars and you then either win that amount if the coin comes up heads or lose if it comes up tails. Suppose, however, that an insider is willing to sell you, for an amount c, the information as to which coin was selected. What is your expected payoff if you buy this information? Note that if you buy it and then bet x, then you will end up either winning x - c or -x - c (that is, losing x + c in the latter case). Also, for what values of c does it pay to purchase the information?

2. The attempt at a solution
I'm asked to find E[Y] where Y is a discrete random variable representing the payoff given that I've paid c to buy information. What are the possible values Y can take? x can be at most 10 - c so -(10 - c) - c = -10 <= Y <= (10 - c) - c = 10 - 2c.

If Y >= -c, then the coin must have landed heads. Otherwise, it must have landed tails. Thus, P{Y >= -c} equals the probability of the coin landing heads, P(H), and P{Y < -c} equals the probability of the coin landing tails, P(T).

Let p(y) = P{Y = y}. There are 11 - c values of y for which y >= -c and 10 - c values of y for which y < c. Thus p(y) = P(H) / (11 - c) for y >= -c and p(y) = P(T) / (10 - c) otherwise.

E[Y] = P(T) / (10 - c) * (-10 + ... + -c - 1) + P(H) / (11 - c) * (-c + ... + 10 - 2c)

Is this correct?
 
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Let the two coins be "heavy tail" (ht) and "light tail" (lt).

From Bayes' theorem, P(H) = P(H|ht)P(ht) + P(H|lt)P(lt) = 0.6 0.5 + 0.3 0.5 = 0.45. Your expected value if you don't buy the information is then 45 cents on the dollar.
 
I'm interested in the expected value if I buy information. I don't understand how knowing the expected value if I don't buy information helps.
 
Isn't your decision whether to buy the info a function of the expected value without it?

Maybe you can see a way to solve this not knowing the expected value of the uninformed bet. Still, it is a conditional probability question -- as per my previous post.
 
EnumaElish said:
Isn't your decision whether to buy the info a function of the expected value without it?
I don't think so. I'm asked for the expected value of the payoff given that I have the info. (which means I bought it). There is no decision to be made. The last question in the problem statement does suggest that my decision to buy the info. should be a function of c, the cost of the info.

Maybe you can see a way to solve this not knowing the expected value of the uninformed bet. Still, it is a conditional probability question -- as per my previous post.

In your previous post, you calculated the probability of heads and that I win the bet, without the info. I don't know exactly where I would apply this information.

Perhaps you can enlighten me by telling me if p(y) is correct in my first post.
 
I decided to have a look at the solution to the problem. The solution is quoted below with my comments and questions in red.

If you wager x on a bet that wins the amount wagered with probability p and loses that amount with probability 1 - p, then your expected winnings are xp - x(1 - p) = (2p - 1)x, which is positive (and increasing in x) if and only if p > 1/2. Thus, if p <= 1/2 one maximizes one's expected return by wagering 0, and if p > 1/2 one maximizes one's expected return by wagering the maximal possible bet.

Thus, if the information is that the .6 coin was chosen then you should bet 10, and if the information is that the .3 coin was chosen, then you should bet 0. Hence, your expected payoff is 1/2(2 * .6 - 1)10 + 1/2(0) - c = 1 - c.

I would actually call this the expected value of your expected winnings. But why is the c subtracted here? Shouldn't it be 1/2(2 * .6 - 1)(10 - c) + 1/2(0 - c) = 1 - 3/5c?

Since your expected payoff is 0 without the information (because in this case, the probability of winning is 0.45 < 1/2) it follows that if the information costs less than 1 then it pays to purchase it.
 

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