Solving a Probability Exercise with Peter and Paul

[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:

<br /> EZ = \sum z_{i} p_{i} = 1*0.5 + (-1)*0.5 = 0<br />

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)

<br /> 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}<br />

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

<br /> P(W &gt;= 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)<br />

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

Thank you for any response.


Standa

 

[AlephZero]

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

 

[twoflower]

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

<br /> \lim_{n \rightarrow \infty} n\ p_{n} = \lambda \in (0,\infty)<br />

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?

 

[AlephZero]

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.

 

[twoflower]

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)

<br /> P (W &gt;= 5) = P (X &gt;= 15) = 1 - P(X &lt;= 15) = 1 - P\left(\frac{X - EX}{\sqrt{var X}} &lt;= \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}<br />

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

 

[twoflower]

Of course I had it wrong :)

P (W &gt;= 5) = P (X &gt;= 15) = 1 - P(X &lt;= 15) = 1 - P\left(\frac{X - EX}{\sqrt{var X}} &lt;= \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})}} &lt;= \frac{15 - 25.\frac{1}{2}}{\sqrt{25.\frac{1}{2}(1 - \frac{1}{2})}}\right)<br />

<br /> = 1 - P\left(\frac{X - 17.5}{\frac{5}{2}} &lt;= \frac{-\frac{5}{2}}{\frac{5}{2}}\right) \approx 1 - \Phi(-1)<br />