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You only pay a fixed sum at the beginning to play. The question is, what is the upper bound on that fixed sum, and why is it so much smaller than your expected
by mgraczyk 11y ago
You only pay a fixed sum at the beginning to play. The question is, what is the upper bound on that fixed sum, and why is it so much smaller than your expected winnings?
- ajmurmann 11y agoYou pay the fixed sum before every game. If you go with Ian Hacking's suggested low price of $25, I can easily see losing several thousand dollar before you get lucky and hit one of the rare, high value events.
- danbruc 11y agoYou have to pay for every game. Half of the time you win $2 which means half of the time you lose $3 if a game costs $5.
- logicallee 11y ago(you've misunderstood the description.)
- justinsaccount 11y agoCorrect, half the time you win $2, but what happens the other half of the time? With the other half... half of the time you win at least $4. With the other half of that... half of the time you win at least $8. With the other half of that... half of the time you win at least $16. So, if you simulate the average winnings in a single round, you get data like this: [8, 2, 8, 2, 8, 32, 8, 8, 4, 4, 2, 4, 4, 16, 2, 2, 16, 4, 8, 2] Which even with losses of $3 and $1 for most games still works out to a $44 profit at $5 game. However, and this is the point of the paradox, if you run the numbers for 100 rounds, you start to see average winnings per round like: [6.48, 6.6, 15.68, 10.02, 10.26, 17.04, 5.86, 11.96, 8.92, 7.34, 6.56, 17.14, 9.92, 9.64, 11.48, 12.44, 19.64, 171.42, 12.82, 5.9] So if you had played those 20 times, at $5/game * 100 rounds you would have a $27,000 profit on $10000.