Question # A gambling book recommends the following “winning strategy” for the game of roulette: Bet $1 on red. Probability ANSWERED asked 2021-08-23 A gambling book recommends the following “winning strategy” for the game of roulette: Bet$1 on red. If red appears (which has probability $$\displaystyle\frac{{18}}{{38}}$$),then take the $1 profit and quit. If red does not appear and you lose this bet ( which has probability $$\displaystyle\frac{{20}}{{38}}$$ of occurring), make additional$1 bets on red on each of the next two spins of the roulette wheel and then quit.
Let X denote your winnings when you quit.
(a) Find P{X>0}.
(b) Are you convinced that the strategy is indeed a “winning” strategy?
(c) Find E[X]. 2021-08-24
(a)
Observe that possible outcomes of X are 1,-1,-3.
The gambler will win 1$if he wins in the first bet or he loses the first bet, but he have success in the second and the third. He will win -1$ if he loses the first and the second bet, but wins the third or loses the first, but wins the second.
Finally, he will win -3\$ if he loses all the bets. Hence $$\displaystyle{P}{\left({X}{>}{0}\right)}={P}{\left({X}={1}\right)}=\frac{{18}}{{38}}+\frac{{20}}{{38}}\cdot{\left(\frac{{18}}{{38}}\right)}^{{2}}=\frac{{4059}}{{6859}}\approx{0.5918}$$
(b)
No, the strategy is not the winning one.
Even though that the probability that the gambler wins in this system is greater that 1/2, we will show in (c) that the expected winnings is less than zero since there is a big risk of losing all the bucks.
(c)
See that
$$\displaystyle{P}{\left({X}=-{1}\right)}={2}{\left(\frac{{20}}{{38}}\right)}^{{2}}\cdot\frac{{18}}{{38}}=\frac{{1800}}{{6859}}$$
$$\displaystyle{P}{\left({X}=-{3}\right)}={\left(\frac{{20}}{{38}}\right)}^{{3}}=\frac{{1000}}{{6859}}$$
so, using definition of expectation, we nave that
$$\displaystyle{E}{\left({X}\right)}={1}\cdot\frac{{4059}}{{6859}}-{1}\cdot\frac{{1800}}{{6859}}-{3}\cdot\frac{{1000}}{{6859}}\approx-{0.11}$$
which says that the gambler is expected to lose 11 cents.