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![[parent]](http://images.planetmath.org:8080/images/uparrow.png) Viewing Message
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| ``Re: New Year Ponder This!!!''
by rm50 on 2009-01-05 21:46:21 |
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| Here is a proposed solution, courtesy of a friend of mine who is an economist and knows about these things...
Let v(c) be the expected value of playing the (first) game given that the cost of playing is c. Assuming you play at least once, you pay c and get a number. If that number is less than v(c), you should play again (the initial cost of c is a sunk cost and is irrelevant to the decision as to whether or not to continue). The expected value of that further play is then v(c). If the number you originally get is greater than v(c), then you should stop; since the probability distribution is uniform, the expected value is (1+v(c))/2. So we get the equation
v(c) = -c + v(c)^2 + (1-v(c))(1+v(c))/2
Solving for v(c), one gets v(c) = 1 - sqrt(2c)
For the edge cases, this makes intuitive sense as an answer: if c>1/2, then you won't play at all, while if c=0, you'll play until you get 1 (which will never happen, but the limit behavior is appropriate).
For the second game, there is no difference. This is because there is a clear decision procedure for choosing whether to continue or not, and having more information cannot increase the expected value of the game based on that procedure. |
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