Question 509903
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The probability of drawing a red is *[tex \Large \frac{9}{17}], hence the payout for drawing a red is *[tex \Large \frac{9}{17}\ \times\ -1.00\ =\ -\frac{9}{17}]


The probability of drawing a white is *[tex \Large \frac{5}{17}], hence the payout for drawing a white is *[tex \Large \frac{5}{17}\ \times\ 2.00\ =\ \frac{10}{17}]


The probability of drawing a blue is *[tex \Large \frac{3}{17}], hence the payout for drawing a blue is *[tex \Large \frac{3}{17}\ \times\ 3.00\ =\ \frac{9}{17}]


The overall expected payout is the sum of the three expected payouts caluclated above:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -\frac{9}{17}\ +\ \frac{10}{17}\ +\ \frac{9}{17}\ =\ \frac{10}{17}\ \approx\ $0.588].


So what does that mean?  If the owner of the game charges you $0.58 or less each time you play, then go for it (as long as the game owner can afford to run his loser, that is). On the other hand, if it costs $0.59 or more to play, walk away -- it is a loser for the player.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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