- #1

twoflower

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Hi all!

I have a problem solving this exercise:

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.

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).

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:

[tex]

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

[/tex]

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)

[tex]

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}

[/tex]

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

[tex]

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)

[/tex]

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

Thank you for any response.

Standa

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:

[tex]

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

[/tex]

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)

[tex]

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}

[/tex]

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

[tex]

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)

[/tex]

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

Thank you for any response.

Standa

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