Probability - Roulette

Question

A gambling book recommends the following strategy for the game of roulette. It recommends that a gambler bets 1 euro on red. If red appears (with probability 18 38 ), then the gambler should take his or her profit and quit the game. If the gambler loses his or her bet (which has probability 20 38 of occurring), he should make additional 1 euro bets on red on each of the next two spins of the roulette wheel and then quit. Let X denote the gambler’s winnings when he quits. (a) Find P(X >0). (b) Is this a winning strategy? Explain your answer. (c) Compute E(X).

Answer 1

Scenarios: 1) he wins the first bet. Net + 1

2) he loses the first bet and wins bets 2 and 3. Net + 1

3) he loses the first bet and splits bets 2 and 3. Net -1

4) he loses all 3. Net -3

$P(1) = \frac {18}{38}\\ P(2) = \frac {20}{38}(\frac {18}{38})^2\\ P(3) = \frac {20}{38}(2\cdot \frac {18}{38}\frac {20}{38})\\ P(4) = (\frac {20}{38})^3$

If I set it up correctly we have covered all of the possibilities, and $P(1) + P(2) + P(3) + P(4) = 1$

$P(X>0) = P(1) + P(2)$

b) Roulette has a house edge on every bet placed of $\frac {2}{38}.$ This edge does not go away under any betting strategy.

c) $E(X) = P(1) + P(2) - P(3) - 3P(4)$