I figure that people usually offer bets to make money, so if I assume that you are a selfish actor, your coins will come up tails practically all the time—you have no incentive that I can think of to ever make them come out heads. Expected payout is $10 each time, so I will pay a maximum of $10 to play the game each time.
Or is your intent for this to be an “Omega tells you...” thought-experiment?
I have clarified my post to specify that for each flip, I pick a coin from this infinite pool at random. Suppose you also magically know with absolute certainty that these givens are true. Still $10?
The properties of the pool are unknown to you, so you have to take into account the possibility that I’ve tuned them somehow. But you do know that the 100 coins I drew from that pool were drawn fairly and randomly.
If I were trying to make a profit then I’d need to know how much to charge for entry. If I could calculate that then yes, I’d offer the bet regardless of how many heads came out of 100 trials.
But this is entirely beside the point; the purpose of this thought experiment is for me to show which parts of bayesianism I don’t understand and solicit some feedback on those parts.
In particular, a procedure that I could use to actually pick a break-even price of entry would be very helpful.
This is actually essential to the problem. If you would only bet me if the coins came up heads, you could make all the coins heavily biased toward tales. In the rare scenario that 100 coins happened to come up anyways, you could show this to me to try to trick me into accepting, when you know that the next coin, like all the others, is guaranteed to be biased toward tales.
For actually solving the problem, I am no expert, but I think Laplace’s law of succession applies here. Laplace’s law states that when the only information that you have is the fact that there are only two possible results to a process (such as heads and tales) and the results to a number of trials that have already been done, the probability that the next result turns out a specific way is (s+1)/(n+2), where s is the number of times that is happened that way in the past and n is the total number of trials so far. I am not sure if that applies here because we may be working with a bit more information than this, but it might be correct.
This is actually essential to the problem. If you would only bet me if the coins came up heads, you could make all the coins heavily biased toward tales. In the rare scenario that 100 coins happened to come up anyways, you could show this to me to try to trick me into accepting, when you know that the next coin, like all the others, is guaranteed to be biased toward tales.
I figure that people usually offer bets to make money, so if I assume that you are a selfish actor, your coins will come up tails practically all the time—you have no incentive that I can think of to ever make them come out heads. Expected payout is $10 each time, so I will pay a maximum of $10 to play the game each time.
Or is your intent for this to be an “Omega tells you...” thought-experiment?
I have clarified my post to specify that for each flip, I pick a coin from this infinite pool at random. Suppose you also magically know with absolute certainty that these givens are true. Still $10?
You still decide what coins are in the infinite pool, right?
The properties of the pool are unknown to you, so you have to take into account the possibility that I’ve tuned them somehow. But you do know that the 100 coins I drew from that pool were drawn fairly and randomly.
Okay. I retract my permanent answer of $10.
Would you have tried to bet us anyways if you had not landed heads 100 times?
If I were trying to make a profit then I’d need to know how much to charge for entry. If I could calculate that then yes, I’d offer the bet regardless of how many heads came out of 100 trials.
But this is entirely beside the point; the purpose of this thought experiment is for me to show which parts of bayesianism I don’t understand and solicit some feedback on those parts.
In particular, a procedure that I could use to actually pick a break-even price of entry would be very helpful.
This is actually essential to the problem. If you would only bet me if the coins came up heads, you could make all the coins heavily biased toward tales. In the rare scenario that 100 coins happened to come up anyways, you could show this to me to try to trick me into accepting, when you know that the next coin, like all the others, is guaranteed to be biased toward tales.
For actually solving the problem, I am no expert, but I think Laplace’s law of succession applies here. Laplace’s law states that when the only information that you have is the fact that there are only two possible results to a process (such as heads and tales) and the results to a number of trials that have already been done, the probability that the next result turns out a specific way is (s+1)/(n+2), where s is the number of times that is happened that way in the past and n is the total number of trials so far. I am not sure if that applies here because we may be working with a bit more information than this, but it might be correct.
In this case:
P(heads) = 101⁄102
EV(heads) = 101/102*V(heads) + 1/102*V(tails) = (101*$1000+1*$10)/102 = $101010/102 = $990 5⁄17
You can read more about Laplace’s law, including a derivation, in chapter 6 of Probability Theory: the Logic of Science by Edwin T. Janes.
This is actually essential to the problem. If you would only bet me if the coins came up heads, you could make all the coins heavily biased toward tales. In the rare scenario that 100 coins happened to come up anyways, you could show this to me to try to trick me into accepting, when you know that the next coin, like all the others, is guaranteed to be biased toward tales.