# Solving a Probability Exercise with Peter and Paul

• twoflower
In summary, the conversation discusses a card game where Peter and Paul play and the winner gains one dollar while the loser pays one dollar. The conversation then delves into finding the distribution and expected value of a random variable Z representing Peter's gain from a single round, as well as the probability of Peter gaining at least five dollars within 25 rounds. The conversation also mentions using the Central Limit Theorem and Normal approximation to calculate the probability, with a final result of 0.8413.
twoflower
Hi all!

I have a problem solving this exercise:

## Homework Statement

Peter and Paul play a card game and after each round the winner gets (and the defeated pays) one dollar. Both of them have equal chance to win in the round, the round can't end in a tie.

## Homework Equations

a) Find out the distribution and expected value of a random variable Z which represents Peter's gain from one single round. (Losing means negative gain).

b) Compute at least approximate probability that Peter will gain at least 5 dollars within 25 rounds (the rounds are independent).

## The Attempt at a Solution

a) I think it's something like an alternative distribution Alt(p), where p = 0.5 Because the random variable Z is discrete, its expected value can be computed this way:

$$EZ = \sum z_{i} p_{i} = 1*0.5 + (-1)*0.5 = 0$$

The distribution is:

Z is 1 with probability 0.5
Z is -1 with probability 0.5

b) To win at least five dollars in 25 rounds, Peter has to win at least 15 rounds. He can also win 16, 17, ..., up to 25 rounds. If the random variable W denotes count of wins during 25 rounds, I think it's true that (from properties of binomial distribution)

$$P(W = k) = \left( \begin{array}{c} 25 \\ k \end{array} \right) \left(\frac{1}{2}\right)^{k} \left(\frac{1}{2}\right)^{25-k} = \left( \begin{array}{c} 25 \\ k \end{array} \right) \left(\frac{1}{2}\right)^{25}$$

So the probability that Peter will gain at least five dollars during 25 rounds will be

$$P(W >= 5) = \sum_{k = 15}^{25} \left( \begin{array}{c} 25 \\ k \end{array} \right) \left(\frac{1}{2}\right)^{25} = \frac{1}{2^{25}} \sum_{k = 15}^{25} \left( \begin{array}{c} 25 \\ k \end{array} \right)$$

Is it correct? And if it is, how to get at least the approximate result?

Thank you for any response.

Standa

Last edited:
The number of rounds Peter wins is a Binomial distribution. What other distribution can you use to approximate the Binomial?

Thank you for hint. As far as I can remember, I can use Poisson distribution to approximate the binomial distribution. Anyway, to be allowed to do that, it would have to be true that

$$\lim_{n \rightarrow \infty} n\ p_{n} = \lambda \in (0,\infty)$$

But in my case $p_{i}$ is constant for every $i$ ($p_{i} = 0.5$) and I'm afraid I can't use the approximation with Poisson distribution in this case...

Am I right?

Right, the Poisson approximation isn't a good idea here, but there is also the Normal approximation (a special case of the Central Limit Theorem). The usual rule of thumb is that the Normal approx. is OK when np >= 10, so that applies here.

Last edited:
AlephZero said:
Right, the Poisson approximation isn't a good idea here, but there is also the Normal approximation (a special case of the Central Limit Theorem). The usual rule of thumb is that the Normal approx. is OK when np >= 10, so that applies here.

Thank you, but I have trouble applicating the Central Limit Theorem in this case. I have ($X$ is random variable describing number of Peter's wins in 25 rounds)

$$P (W >= 5) = P (X >= 15) = 1 - P(X <= 15) = 1 - P\left(\frac{X - EX}{\sqrt{var X}} <= \frac{15 - EX}{\sqrt{var X}}\right) \approx \int_{-\infty}^{\frac{15 - EX}{\sqrt{var X}}} \frac{1}{\sqrt{2\pi}}\exp^{-\frac{t^2}{2}}\mbox{dt}$$

But I think $\mbox{var} X = 25.\mbox{var} X_{i} = 0$ and thus I got zero in the denominator

Last edited:
Of course I had it wrong :)

$$P (W >= 5) = P (X >= 15) = 1 - P(X <= 15) = 1 - P\left(\frac{X - EX}{\sqrt{var X}} <= \frac{15 - EX}{\sqrt{var X}}\right) = 1 - P\left(\frac{X - 25.\frac{1}{2}}{\sqrt{25.\frac{1}{2}(1 - \frac{1}{2})}} <= \frac{15 - 25.\frac{1}{2}}{\sqrt{25.\frac{1}{2}(1 - \frac{1}{2})}}\right)$$

$$= 1 - P\left(\frac{X - 17.5}{\frac{5}{2}} <= \frac{-\frac{5}{2}}{\frac{5}{2}}\right) \approx 1 - \Phi(-1)$$

Last edited:

## 1. How do Peter and Paul approach solving probability exercises?

Peter and Paul typically use a combination of theoretical and computational methods to solve probability exercises. They start by understanding the problem and identifying any known information, then use mathematical formulas and equations to calculate the probability of an event occurring. They may also use simulations or experiments to test their results.

## 2. What are some common mistakes people make when solving probability exercises?

Some common mistakes include misinterpreting the question, using incorrect formulas or equations, and not considering all possible outcomes. It's also important to be careful with decimal and percentage conversions, and to remember to account for any given information or conditions in the problem.

## 3. Can you provide an example of a probability exercise with Peter and Paul's solution?

Sure, here's an example: "If a fair six-sided die is rolled twice, what is the probability of rolling two even numbers?" Peter and Paul would start by identifying the total number of possible outcomes (36), then determining the number of favorable outcomes (18). They would use the formula P(A) = number of favorable outcomes / total number of outcomes to calculate the probability of rolling two even numbers as 18/36, or 1/2.

## 4. How can I improve my skills in solving probability exercises?

One way to improve is to practice regularly with a variety of problems. You can also study and learn different formulas and techniques for solving probability exercises, and seek guidance from experienced mathematicians or resources such as textbooks or online tutorials.

## 5. Are there any real-world applications for solving probability exercises?

Absolutely! Probability is used in many fields such as finance, engineering, and statistics to make predictions and informed decisions. For example, insurance companies use probability to determine rates and assess risk, while manufacturers use it to predict the likelihood of defects in their products. Understanding and being able to solve probability exercises is an important skill for many careers.

Replies
5
Views
763
Replies
6
Views
824
Replies
3
Views
418
Replies
11
Views
1K
Replies
6
Views
362
Replies
6
Views
1K
Replies
3
Views
788
Replies
6
Views
1K
Replies
14
Views
2K
Replies
0
Views
489