Question

An individual uses the following gambling system at Las Vegas. He bets \(\$ 1\) that the roulette wheel will come up red. If he wins, he quits. If he loses then he makes the same bet a second time only this time he bets \(\$ 2\); and then regardless of the outcome, quits. Assuming that he has a probability of \(\frac{1}{2}\) of winning each bet, what is the probability that he goes home a winner? Why is this system not used by everyone?

Short answer

The individual using this gambling system at Las Vegas has a probability of \(\frac{3}{4}\) of going home a winner. However, this system is not used by everyone due to practical constraints, such as betting limits, limited personal bankrolls, and the long-term casino edge. Additionally, the assumption of a \(\frac{1}{2}\) probability of winning per bet might not be true for all gambling games.

Step-by-step solution

Understand the betting system

The individual follows a specific gambling system, which includes two rounds of betting: 1. First bet: He bets $1 on red. If he wins, he quits as a winner. 2. Second bet: If he loses the first bet, he bets $2 on the same color again, hoping for a win. Regardless of the outcome, he quits after this second bet. Each bet has a probability of \(\frac{1}{2}\) of being a winner.

Calculate the probability of winning in each round

Since the probability of winning each bet is \(\frac{1}{2}\), we can calculate the probability of winning in each of the two rounds: 1. Probability of winning in the first round (winning the first bet): \(P(Round 1) = \frac{1}{2}\) 2. Probability of winning in the second round (losing the first bet and then winning the second bet): \(P(Round 2) = P(Lose, Win) = P(Lose) \times P(Win|Lose) = (\frac{1}{2}) \times (\frac{1}{2}) = \frac{1}{4}\)

Calculate the total probability of going home a winner

Now we must find the total probability of going home a winner. This includes the probability of winning either in the first round or in the second round, so we can add the probabilities from the previous step: \(P(Winner) = P(Round 1) + P(Round 2) = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}\) The individual has a probability of \(\frac{3}{4}\) of going home a winner.

Discuss why this system is not used by everyone

Although the probability of going home a winner might seem high with this system, there are a few reasons why it's not used by everyone: 1. Casinos often have betting limits, which means that after a certain number of losses, an individual would not be able to double their bet. 2. The system assumes that the individual has an unlimited bankroll and can keep doubling their bet, which is not the case for most people. 3. The system is based on the assumption that each bet has a \(\frac{1}{2}\) probability of winning, which is not true for all gambling games. In the case of roulette, for example, the actual probability of winning a red or black bet is less than \(\frac{1}{2}\) since there are green spaces on the wheel. 4. While this system might work in the short term, it doesn't eliminate the long-term house edge present in casino games, so it's unlikely to make an individual consistently profitable. In conclusion, the individual using this gambling system at Las Vegas has a probability of \(\frac{3}{4}\) of going home a winner. However, this system is not used by everyone due to practical constraints and the long-term casino edge.