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If there are 9 white balls and one black ball in a bag. The white balls are valued at 10 dollars and black ball at 100 dollars. How much are you willing to pay for each pull from the bag (only one ball per pull) but you can pull as many times as you wanted.

Before you start pulling balls you must choose how much you are willing to pay for each pull. After each pull you keep the ball (it doesn't go back in the bag).

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    $\begingroup$ Do we know that there are no poisonous snakes in the bag? $\endgroup$ – Will Jagy Jul 5 '15 at 17:13
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    $\begingroup$ Could you clarify: is the ball that you pull put back in the bag before you pull again? $\endgroup$ – Mike Pierce Jul 5 '15 at 17:28
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So you have a $9/10$ chance of winning $10$ and a $1/10$ chance of winning $100$. Thus your expected outcome is

$$\frac{9}{10} \cdot 10 + \frac{1}{10}\cdot 100 = 19$$

If this is how much you expect to win per pull, you can answer how much you would be willing to pay per pull. After the first pull, you either have 8 white balls and 1 black, or 9 white and 0 black, and need to reevaluate.

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Since the bag contains 190 dollars, I'll eagerly pay $19 per pull. At the worst, I'll break even by continuing till the end !

Since the $100 bill can come equi-probably on any draw, if the operator wants to break even,

I will get $\frac{1}{10}\cdot [100 +110 +120 + 130 + ...+190] = 145$ dollars

against $\frac{1}{10}\cdot[x+2x+3x+4x+5x+6x+7x+8x+9x+10x] = 5.5x$ dollar charge

So the fair amount (which I could pay) = 145/5.5 = $\approx$ 26.36 dollars

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Assume I have to pay $r$ dollars per pull. I shall pull until I obtain the black ball. The expected number of pulls for that is $5.5$. Therefore the expected revenue per game will be $5.5(10-r)+90$ dollars. If this is positive the game is in my favor. The break-even point is $r_0={145\over5.5}\doteq 26.35$ dollars.

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