Problem 20 - A First Course in Probability

Question

A gambling book recommends the following "winning strategy" for the game of roulette: Bet \(\$ 1\) on red. If red appears (which has probability \(\frac{18}{38}\) ), then take the \(\$ 1\) profit and quit. If red does not appear and you lose this bet (which has probability \(\frac{20}{38}\) 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? Explain your answer! (c) Find \(E[X]\)

Step-by-Step Solution

Answer

The probability of having positive winnings (X > 0) using the given roulette strategy is approximately 72.32%. However, the expected value of winnings (E[X]) is approximately -\$0.0026, which indicates that on average, you will lose about $0.0026 per game. Therefore, despite the higher probability of positive winnings, the strategy is not a "winning" strategy in the long run since the losses outweigh the potential winnings.

Step 1: Winning on first bet

Bet \(1 on red, and the probability of winning this bet is given by \[\frac{18}{38}\]. If you win in the first bet, your winning is \)\$1$, and you stop playing.

Step 2: Winning after additional bets

If you lose the first bet, you keep betting $1 on red for two more spins. You have the following winning scenarios: - Lose the first bet, win the second bet: probability: \[\frac{20}{38} × \frac{18}{38}\], winning: \(-\$1 + \$1 = \$0\) - Lose the first bet, lose the second bet, win the third bet: probability: \[\frac{20}{38} × \frac{20}{38} × \frac{18}{38}\], winning: \(-\$1 - \$1 +\$1 = -\$1\)

Step 3: Calculate P{X>0}

To find the probability of positive winnings (X > 0), we just need to account for the winning scenarios from Steps 1 and 2, which are: - Winning in the first bet - Losing the first bet, winning in the second bet So, we calculate the probability as follows: \[P(X>0)= \frac{18}{38} + \frac{20}{38} × \frac{18}{38} = \frac{18}{38} (1 + \frac{20}{38}) = \frac{18}{38} × \frac{58}{38}\] \[P(X>0) = \frac{18}{38} × \frac{29}{19} = \frac{522}{722} ≈ 0.7232\]

Step 4: Assessing the strategy

Even though the probability of having positive winnings is about 72.32%, this alone does not guarantee that the strategy is a "winning" strategy in the long run, as we need to consider the expected winnings to measure the performance of the strategy.

Step 5: Calculate E[X]

To calculate the expected value of winnings (E[X]), we multiply the winnings with their respective probabilities and sum them: \[E[X] = (\$ 1)P(X > 0) + (\$ 0)P(X = 0)+(-\$ 1)P(X < 0)\] We know that \(P(X>0) ≈ 0.7232\). In Step 2, the second scenario has a probability of winning $0, so: \[P(X=0) = \frac{20}{38} × \frac{18}{38}\] And: \[P(X<0) = 1 - P(X>0) - P(X=0)\] Now, we find E[X]: \[E[X] = (\$ 1)(0.7232) + (\$ 0)(\frac{20}{38} × \frac{18}{38}) + (-\$ 1)(1 - 0.7232 - \frac{20}{38} × \frac{18}{38})\] \[E[X] ≈ 0.7232 - (1 - 0.7232 - 0.2789) = -\$ 0.0026\] Based on the computed expected value of winnings (E[X]), the strategy is not a "winning" strategy since, on average, you will lose about $0.0026 per game. Even though you have a higher probability of winning, the losses in the losing scenarios outweigh the potential winnings.

Key Concepts

Expected Value

The concept of "Expected Value" in probability theory helps determine the average outcome of a series of random events. It is crucial in evaluating strategies, especially in gambling or games of chance like roulette. Essentially, the expected value is the long-term average result if an experiment is repeated many times.

For the roulette strategy in question, our goal is to calculate the expected value of the winnings after following a series of bets. We start by identifying the potential outcomes and their respective probabilities. Let's break it down:

• Winning on the first bet of \(1 with a probability of winning, identified as \(P(X>0)\).
• Net winnings of \)0 after losing the first bet but winning the second. This has its own combined probability.
• Finally, a net loss if all three consecutive bets are lost.

By assigning each event's value and multiplying it with the probability of that event, then summing all these products, we obtain the expected value, \(E[X]\). If the expected value is positive, you'll win money over the long term. If negative, you can expect an overall loss.

In this example, despite a high probability of positive winnings, the expected value is negative, illustrating the importance of considering both likelihood and potential loss.

Roulette Strategy

Roulette is a popular casino game, where strategizing revolves around types of bets and their odds. In the strategy presented here, the gambler is suggested to bet initially on red. The wheel contains 38 numbers — 18 red, 18 black, and 2 green (zero and double zero). Thus, the probability of landing on red is \(rac{18}{38}\).

The advised strategy involves placing a \(1 bet on red and stopping if red wins, providing immediate profit. If red doesn't win on the first attempt, you proceed with placing \)1 bets on the next two spins. This is based on the premise that catching a win after a loss will ideally break even or minimize losses.

However, analyzing the strategy reveals pitfalls:

• The strategy assumes maintaining a consistent winning probability, overlooking that each roulette spin is independent with reset probabilities.
• The potential loss from initial losses adds up, which may exceed occasional red wins.

The underlying idea is to exploit short sequences but unfortunately, this doesn't change the inherent house edge.

Probability Calculations

Probability calculations are instrumental in formulating expectations from gambling strategies. Calculating the probability of outcomes involves understanding the likelihood of winning and losing sequences.

Let's revisit how probabilities are combined here:

• Winning immediately on the first bet has a probability of \(rac{18}{38}\).
• Winning after a single loss results in \(rac{20}{38} imes rac{18}{38}\) since each spin is independent and multiplication rules for probabilities apply.

To find the probability of winning anything (\( P(X>0)\)), we add these probabilities. It's reflected in the formula we used:\(P(X>0) = \frac{18}{38} + \frac{20}{38} imes \frac{18}{38} = \frac{522}{722} \approx 0.7232.\)This indicates a 72.32% chance of positive outcomes based on the suggested bet path. However, these calculations don't account for the broader risk, as shown when further metrics like expected value reflect long-term trends. Analyzing probabilities stresses caution when adopting strategies based merely on estimated positive dumping scenarios without accounting for complete risk factors.