A variation on an old paradox
The infamous "St. Petersburg Paradox", analyzed by the Bernoulli family, is as follows:
I flip a coin. If it first comes up heads on the first toss I pay you $2, on the second I pay you $4, on the third $8, and so on. How much is a fair price for me to charge you to play?
If we calculate the mathematical expectation, your expectation is ( $ 2 ∗ 1 / 2 ) + ( $ 4 ∗ 1 / 4 ) + ( $ 8 ∗ 1 / 8 ) . . . = $ 1 + $ 1 + $ 1 . . . . out to infinity, or in other words, infinite. So no amount you can pay me is fair.
Now let's modify the situation as follows. If the first head appears on the first or second flip I pay you $2, on the third or fourth flip I pay you $4, on the 5th or 6th flip I pay you $8, and so forth. What is a fair price for me to charge you, so that the price is exactly equal to your mathematical expectation? Give answer in dollars.
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1 solution
i don't understand tell me again plse
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On the first toss, you are expected to get $ 2 with a probability of 2 1 . This makes the expected money gained to be $ 2 × 2 1 = $ 1 .
Furthermore, the probability of getting $ 2 more dollars happens when you make two heads tosses, with a probability of 2 1 × 2 1 = 4 1 . In this case, the expected amount of money gained is $ 2 × 4 1 = $ 2 1 .
Thus, the expected amount of money gained in the first two tosses is $ 1 + $ 2 1 = $ 2 3 .
You can do a similar calculation to find the expected amount of money gained for the third and fourth tosses to be $ 4 3 , then 5th and 6th tosses to be $ 8 3 , and so on.
So the total expected money gain is $ 2 3 + $ 4 3 + $ 8 3 + ⋯ = 1 − 2 1 $ 2 3 = $ 3 by using the Infinite Geometric sum formula.
Under the new conditions, your mathematical expectation is [$2 * (1/2 + 1/4)] + [$4 * (1/8 + 1/16)] + [$8 * (1/32 + 1/64)]... = 3/2 + 3/4 + 3/8 +... = 3 (1/2 + 1/4 + 1/8 +...) = 3 * 1 = 3