Problem 3 - A First Course in Probability

Question

Gambles are independent, and each one results in the player being equally likely to win or lose 1 unit. Let \(W\) denote the net winnings of a gambler whose strategy is to stop gambling immediately after his first win. Find (a) \(P\\{W>0\\}\) (b) \(P\\{W<0\\}\) (c) \(E[W]\)

Step-by-Step Solution

Answer

(a) \(P\\{W>0\\} = \dfrac{1}{2}\) (b) \(P\\{W<0\\} = \dfrac{1}{4}\) (c) \(E[W] = -\dfrac{7}{2}\)

Step 1: Understanding the question

We have a gambler who will stop gambling immediately after his first win. Now, we need to find the following probabilities and the expected value: (a) \(P\\{W>0\\}\) which represents the probability of having net winnings higher than zero, (b) \(P\\{W<0\\}\) is the probability of having net winnings below zero, (c) \(E[W]\) is the expected value of net winnings.

Step 2: Probability of having net winnings higher than zero (Finding \(P\\{W>0\\}\))

To have net winnings higher than zero, the gambler must win the first turn. Since the gambler is equally likely to win or lose, the probability of winning is 1/2. Therefore, \(P\\{W>0\\} = \dfrac{1}{2}\).

Step 3: Probability of having net winnings below zero (Finding \(P\\{W<0\\}\))

To have net winnings below zero, the gambler must lose the first game and win the second. After losing the first game, with the same reason as the previous step, the probability of winning the next game is also 1/2. Hence, \(P\\{W<0\\} = P\\{Lose_1\} * P\\{Win_2 | Lose_1\\} = \dfrac{1}{2} * \dfrac{1}{2} = \dfrac{1}{4}\).

Step 4: Using Geometric Distribution to find the net winnings probability

Let \(X\) be a random variable representing the number of trials required for the first win. Since the gambler will stop after the first win, we have a scenario for the Geometric Distribution. Therefore, for \(n\) trials, the probability function is given by: \(P(X = n) = p(1-p)^{n-1}\), where \(p = \dfrac{1}{2}\) is the probability of winning

Step 5: Finding Expected Value \(E[W]\)

Now, since the gambler loses one unit when he loses, and gains one unit when he wins, we can wright expected value formula as: \(E[W] = \sum ^{\infty}_{n=1} (p(1-p)^{n-1})[2n-3]\) where, for each of the first \(n\) turns, we add the probability of that event happening multiplied by the net gain (or loss) in that event. Substitute the value of \(p\) and solve it: \(E[W] = \sum ^{\infty}_{n=1} (\dfrac{1}{2}(\dfrac{1}{2})^{n-1})[2n-3]\) now, let's calculate E[W] \(E[W] = \dfrac{1}{2} \sum ^{\infty}_{n=1} (2n-3)(\dfrac{1}{2})^{n-1}\) \(E[W] = \dfrac{1}{2} (1-2\sum ^{\infty}_{n=1}n(\dfrac{1}{2})^n)\) We know that the sum of the geometric series is given by: \(\dfrac{a}{1-r}\), where \(a\) is the first term and \(r\) is the ratio. Now this series(sum) is differentiated w.r.t r, then we get the sum of n terms of geometric series: \(\sum n ar^{n-1} = \dfrac{a}{(1-r)^2}\) Using this formula: \(\sum ^{\infty}_{n=1} n(\dfrac{1}{2})^n = a(\dfrac{1}{(1-r)^2}\) with \(a = 1\) and \(r = \dfrac{1}{2}\), \(\sum ^{\infty}_{n=1} n(\dfrac{1}{2})^n = \dfrac{1}{(\dfrac{1}{2})^2} = 4\) Then the expected value is given by: \(E[W] = \dfrac{1}{2}(1-2(4)) = -\dfrac{7}{2}\) #Results# Thus, we found: (a) \(P\\{W>0\\} = \dfrac{1}{2}\) (b) \(P\\{W<0\\} = \dfrac{1}{4}\) (c) \(E[W] = -\dfrac{7}{2}\)

Key Concepts

Probability of Winning

In probability theory, understanding the probability of a specific outcome is foundational. In the context of this gamble, we look at the probability of the gambler ending up with net winnings greater than zero. The gambler stops playing after their first win, making it crucial to understand the probability of winning in a single round.

We know each gamble is independent, and each outcome (win or loss) holds the same likelihood. Therefore, the probability of winning in a single round is:

• Win: 1/2 (probability of getting the desired outcome).
• Stop: Occurs right after a win, as per strategy.

When focusing solely on the probability of getting at least a net zero outcome, we calculate the chance of winning on the first try because that's the only scenario which results in a positive net:

• \[ P\{W > 0\} = \frac{1}{2} \]

This probability reflects the chance of the first gambling instance succeeding, leading to a positive winning immediately.

Expected Value

The concept of expected value in probability provides insight into what outcome to expect on average when an experiment is repeated many times. For this problem, we're curious about the expected net winnings of the gambler.

When the strategy is to stop at the first win, expected value calculations factor in the probability of sequences of losses and eventual winnings, modeled by the geometric distribution:

• Lose repeatedly before the first win
• Win the first time afterwards (stop strategy)

The calculation of expected value involves a more complex expression than simply adding probabilities:

• If \( X \) is the number of trials until the first success:\[ E[W] = \sum ^{\infty}_{n=1} (p(1-p)^{n-1})[2n-3] \]

Here, the term \((2n-3)\) reflects the gain or loss per trial after losing several rounds before a win. Simplifying, using geometric series properties, we find:

• \[ E[W] = -\frac{7}{2} \] for this specific scenario.

A negative expected value implies a potential loss over time with repeated instances of the game, highlighting the downside risk inherent in gambling.

Independent Events

Understanding independent events is central in probability theory, especially in gambling scenarios. Two events are considered independent when the outcome of one does not affect the outcome of another.

In this gamble:

• Each turn (each game round) is independent.
• Whether the gambler wins or loses doesn’t alter the 50% chance of winning or losing on the next round.

This assumption allows us to calculate probabilities and expected values straightforwardly. For instance:

• The probability of successive outcomes such as losing once then winning (a combination of separate events) can be expressed as the product of their separate probabilities.

If a loss followed by a win is the outcome we're analyzing:

• \[ P\{Lose_1\} \cdot P\{Win_2 | Lose_1\} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \]

Thus, using the concept of independent events, the calculations provide an ability to assess and strategize gameplay under probabilistic conditions accurately.