A gambling book recommends the following "winning strategy" for the game of roulette. It recommends that a gambler bet $1 onred. If red appears (which has probablity 18/38), then the gamblershould take her $1 profit and quit. If the gambler loses this bet (which has probablity 20/38 of occurring), she should make additional $1 bets on red on each of the next two spins of the roulette wheel and then quite. Let X denote the gambler's winnings when she quites. (a) Find P{X > 0}. (b) Are you concinved that the strategy is indeed a "winning" strategy? Explain your answer. (c) Find E[X].

A gambling book recommends the following "winning strategy" for the game of roulette. It recommends that a gambler bet $1 onred. If red appears (which has probablity 18/38), then the gamblershould take her $1 profit and quit. If the gambler loses this bet (which has probablity 20/38 of occurring), she should make additional $1 bets on red on each of the next two spins of the roulette wheel and then quite. Let X denote the gambler's winnings when she quites. (a) Find P{X > 0}. (b) Are you concinved that the strategy is indeed a "winning" strategy? Explain your answer. (c) Find E[X].

Question
A gambling book recommends the following "winning strategy" for the game of roulette. It recommends that a gambler bet $1 onred. If red appears (which has probablity 18/38), then the gamblershould take her $1 profit and quit. If the gambler loses this bet (which has probablity 20/38 of occurring), she should make additional $1 bets on red on each of the next two spins of the roulette wheel and then quite. Let X denote the gambler's winnings when she quites.
(a) Find P{X > 0}.
(b) Are you concinved that the strategy is indeed a "winning" strategy? Explain your answer.
(c) Find E[X].

Answers (1)

2020-12-07
a) Remember that when you bet $1 and lose the bet,you lose the $1. Therefore, you have to earn $2 just to walk out with a net gain of $1. So the answer is:
\(\displaystyle{P}{\left\lbrace{X}{>}{0}\right\rbrace}={\left({\frac{{{18}}}{{{38}}}}\right)}+{\left({\left({\frac{{{20}}}{{{38}}}}\right)}\cdot{\left({\frac{{{18}}}{{{38}}}}\right)}\cdot{\left({\frac{{{18}}}{{{38}}}}\right)}\right)}={.5918}\)
b) To get this answer, you really need to do part first. I'll come back to this after that.
c) Remember that there is a chance that you can lose some dollars when playing. So the answer is as follows:
\(\displaystyle{E}{\left[{X}\right]}=-{3}{\left[{P}{\left(-{3}\right)}\right]}+-{2}{\left[{P}{\left(-{2}\right)}\right]}+-{1}{\left[{P}{\left(-{1}\right)}\right]}+{0}{\left[{P}{\left({0}\right)}\right]}+{1}{\left[{P}{\left({1}\right)}\right]}\)
\(\displaystyle=-{3}{\left({\left({\frac{{{20}}}{{{38}}}}\right)}\cdot{\left({\frac{{{20}}}{{{38}}}}\right)}\cdot{\left({\frac{{{20}}}{{{38}}}}\right)}\right)}+{\left(-{2}\right)}{\left({0}\right)}+{\left(-{1}\right)}{\left[{\left({\left({\frac{{{20}}}{{{38}}}}\right)}\cdot{\left({\frac{{{18}}}{{{38}}}}\right)}\cdot{\left({\frac{{{20}}}{{{38}}}}\right)}\right)}+{\left({\left({\frac{{{20}}}{{{38}}}}\right)}\cdot{\left({\frac{{{20}}}{{{38}}}}\right)}\cdot{\left({\frac{{{18}}}{{{38}}}}\right)}\right)}\right]}+{0}{\left({0}\right)}+{1}{\left[{\left({\left({\frac{{{20}}}{{{38}}}}\right)}\cdot{\left({\frac{{{18}}}{{{38}}}}\right)}\cdot{\left({\frac{{{18}}}{{{38}}}}\right)}\right)}+{\left({\frac{{{18}}}{{{38}}}}\right)}\right]}\)
\(\displaystyle=-{3}{\left({.1458}\right)}+{0}+{\left(-{1}\right)}{\left({.2624}\right)}+{0}+{1}{\left({.5918}\right)}\)
\(\displaystyle=-{.108}\)
Now, going back to part b. This strategy is NOT a"winning strategy" because the expected winnings is a negative value, hence you really have an expected loss.
0

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