is the EV for this game really infinity?

[cabiness42]

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Well, if you read my post, I didn't just pull 2 out of the air. I calculated an expected number of flips the game would last based on the probabilities of each individual number of flips. The result was a series that converged to 2.

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I'm sorry, I didn't see your first post there. My mistake.

Anyway, as others have pointed out already, that method does not produce accurate results to answer the question that is being asked. You have to consider the value placed on each result.

For example, suppose I offer you the same game... except if you manage to flip 4 tails in a row, you win a million dollars.

The game will still last an average of 2 flips, the payout for which is $4. But I assume you would be willing to pay more than $4 for a 1/16 chance at a million dollars.

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Sure, but now you've changed the game. In the original game, the (probablilty*payout) was the same for each number of tosses. Now that's different so it doesn't work.

Yes, mathematically the EV is infinite, but that also assumes that the game can go on infinitely, when in reality it can't.

What I am saying is that if you play the game 1 time, $4 is a fair amount to pay.

[pococurante]

Whether I change the game or not, the point is still the same. You can't get the EV by calculating the average # of flips.

Let's play the game 1024 times. Results:

512 times the game will last 1 flip = 512x$2 = $1024
256 times the game will last 2 flips = 256x$4 = $1024
128 times the game will last 3 flips = 128x$8 = $1024
64 times the game will last 4 flips = $1024
32 times the game will last 5 flips = $1024
16 times the game will last 6 flips = $1024
8 times the game will last 7 flips = $1024
4 times the game will last 8 flips = $1024
2 times the game will last 9 flips = $1024
1 time the game will last 10 flips = $1024
(final game with infinite value is being ignored here)

Results: In 1,024 games, a total of $10,240 was paid out.

Average number of flips per game: 2
Average payout per game: $10