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A roulette wheel has 18 red wedges, 18 black wedges and 2 green wedges. Wedges are equally likely to come up in one spin of the wheel. A gambler bets one dollar on red coming up in independent spins of the wheel. Briefly explain why the gambler shouldn’t play for too long.

I've tried it for a few times, but my answer doesn't seem to be correct. Can someone explain how it works to me? Thanks a lot!

(homework) (convergence in distribution)

Smith
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    There are many types of explanation, such as the mean loss when one plays $n$ times, or the probability one will be more than such and such in the red (pun intended) if one plays $1000$ times. What have you tried? – André Nicolas Aug 04 '13 at 05:47
  • Briefly: because the gambler will go bankrupt, almost surely (and after a time depending linearly on their initial fortune). – Did Aug 04 '13 at 09:58

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Since all wedges are equally likely, the expected value is

$$\frac{(\textrm{# of wedges where you win}) \cdot \$1 +(\textrm{# of wedges where you lose}) \cdot -\$1}{(\textrm{total # of wedges})}.$$

Can you calculate the expected value? What can you conclude?

user7530
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