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{\bf Question}

At Las Vegas, a man with \$20 needs \$40, and hopes to raise the
money by playing roulette.  He is considering two strategies: bet
\$20 on \lq evens' all at once and stop if he wins or loses, or
bet on \lq evens' one dollar at a time until he has won or lost
\$20. Which would you advise?

(Assume the roulette wheel has numbers $0, 0, 1, 2,\ldots , 36$
all equally likely to occur).



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{\bf Answer}

If he bets \$20 all in one go the probability of winning is $\ds
\frac{18}{38} = 0.47\ldots$

If he bets \$1 at a time until either he loses or wins \$40, this
can be regarded as a gambler's ruin problem, with $$z = 20
\hspace{.2in} a = 40 \hspace{.2in} p = \frac{18}{38} \hspace{.2in}
q = \frac{20}{38}$$ So the probability that he wins is the same as
that of his adversary being ruined i.e.  $$p_z = \frac{1 - \left(
\frac{q}{p} \right)^z }{1 - \left( \frac{q}{p} \right)^a} =
\frac{1 - \left( \frac{20}{18} \right)^{20} }{1 - \left(
\frac{20}{18} \right)^{40}} = \frac{1}{1 + \left(
\frac{20}{18}\right)^{20}} = 0.108\ldots$$ So he is much less
likely to win \$40 by betting \$1 at a time than by betting all
\$20 in one go.



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