Problem 27 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

In American roulette, as described in Example 6, a player may bet on a split (two adjacent numbers). In this case, if the player bets $$\$ 1$$ and either number comes up, the player wins $$\$ 17$$ and gets his $$\$ 1$$ back. If neither comes up, he loses his $$\$ 1$$ bet. Find the expected value of the winnings on a $$\$ 1$$ bet placed on a split.

Step-by-Step Solution

Answer

The expected value of winnings for a $1 bet placed on a split in American roulette is \$0. This means that, in the long run, the player should neither expect to win nor lose. However, this is a theoretical value, and actual outcomes might vary due to randomness and luck.

Step 1: Determine the probability of winning and losing the bet

American roulette has 38 numbers(1 to 36, 0 and 00) of which we bet on 2 adjacent numbers. So, we have the following probabilities for winning and losing in a split bet: Probability of winning= \(\frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}}\) _ Probability of winning = \(\frac{2}{38}\), since we bet on two adjacent numbers (out of a total of 38). Probability of losing= \(\frac{\text{Number of unfavorable outcomes}}{\text{Total possible outcomes}}\) _ Probability of losing = \(\frac{36}{38}\), as there are 36 remaining numbers that will result in losing the bet.

Step 2: Calculate the expected value of the winnings

Expected value (EV) can be found using the formula: Expected Value (EV) = \(\text{P(winning) × Winnings} - \text{P(losing) × Loss}\) Here, winning \(17 and getting the \)1 bet back, makes the winnings $18. - P(winning) = \(\frac{2}{38}\) - Winnings = $18 - P(losing) = \(\frac{36}{38}\) - Loss = $1 (bet amount) Let's plug these values into the formula: EV = \[\left(\frac{2}{38}\right) \times 18 - \left(\frac{36}{38}\right) \times 1\]

Step 3: Solve for EV and interpret the result

Now, we will solve the equation for Expected Value (EV): EV = \[\left(\frac{2}{38}\right) \times 18 - \left(\frac{36}{38}\right) \times 1\] = \(\frac{36 - 36}{38}\) = \(\frac{0}{38}\) = $0 The expected value of the winnings for a $1 bet placed on a split is zero. This means that in the long run, the player should neither expect to win nor lose when placing split bets in American roulette. However, it's important to note that this is a theoretical value, and the actual outcomes in gambling might vary due to randomness and luck.

Key Concepts

Expected Value

Expected value is a crucial concept in probability and gambling games like roulette. It represents the average outcome you can expect over a large number of repetitions of a particular game or bet.
In mathematical terms, it is calculated as the sum of all possible outcomes, each multiplied by their respective probabilities.

For this specific exercise, the formula used was:

• Expected Value (EV) = P(winning) × Winnings - P(losing) × Loss

Here's what these terms mean:

• P(winning): the probability of winning the bet.
• Winnings: the amount you win if the bet is successful.
• P(losing): the probability of losing the bet.
• Loss: the amount you lose if the bet is unsuccessful.

In this example, with the calculation \[ EV = \left(\frac{2}{38}\right) \times 18 - \left(\frac{36}{38}\right) \times 1 \]the expected value is zero, meaning that over the long run, you neither gain nor lose money.

American Roulette

American roulette is a popular gambling game found in casinos, featuring a spinning wheel with numbered slots. The American version of the game consists of 38 slots, numbered 1 to 36, plus a single 0 and a double 00 slot.
This wheel layout slightly differs from the European roulette, which has only one 0 slot, resulting in 37 numbers overall.

When you place a bet in American roulette, you can bet on individual numbers, split bets (bet on two adjacent numbers), or even place bets on larger groups of numbers.
The inclusion of the 0 and 00 slots in American roulette increases the house edge, slightly decreasing your odds of winning compared to other versions.

• Split bet: betting on two adjacent numbers, giving a higher payout but lower probability of winning.
• House advantage: the additional 00 slot increases the casino's advantage over the player.

Probability of Winning

The probability of winning in American roulette varies depending on the type of bet you place. For a split bet, as discussed in the exercise, you are essentially betting on two adjacent numbers.
There are 38 possible outcomes on the roulette wheel, but only 2 possible favorable outcomes, those of the numbers you've selected.
This results in the probability of winning being:

• Probability of winning a split bet = \( \frac{2}{38} \)

This is a direct calculation of favorable outcomes (the 2 numbers you selected) divided by the total possible outcomes (all 38 slots on the wheel). The odds of winning are relatively low, reflecting the typical risk associated with gambling games.

Probability of Losing

Understanding the probability of losing is equally important as understanding that of winning, in games like roulette.
Since there are only 2 numbers on the wheel that can win in a split bet, this leaves 36 numbers that do not result in a win.
You'll compute the losing probability as the difference between 1, the sum of all probabilities, and the probability of winning:

• Probability of losing a split bet = \( \frac{36}{38} \)

The 36 unfavorable outcomes represent the slots that do not match your chosen numbers.
This high probability of losing emphasizes the inherent risk in gambling and the reason why a house edge typically favors the casino.