Expected number of 5 in n tosses of die

[eprparadox]

Homework Statement

Show that the expected number of 5's in n tosses of a die is n/6.

Homework Equations

x = random variable representing the number of 5's

E\left( x\right) =\sum _{i}p_{i}x_{i}

The Attempt at a Solution

The probability of getting i 5's in n tosses is:

\left( \dfrac {1} {6}\right) ^{i}\left( \dfrac {5} {6}\right) ^{n-i}

So the expectation value of x is:

E\left( x\right) =\sum _{i=0}^{n}i\left( \dfrac {1} {6}\right) ^{i}\cdot \left( \dfrac {5} {6}\right) ^{n-i}

I typed this into wolframalpha and I don't get n/6 as I thought I would.

I'm not sure where I went wrong. Any ideas would be greatly appreciated. Thanks!

 

[gopher_p]

Homework Statement

Show that the expected number of 5's in n tosses of a die is n/6.

Homework Equations

x = random variable representing the number of 5's

E\left( x\right) =\sum _{i}p_{i}x_{i}

The Attempt at a Solution

The probability of getting i 5's in n tosses is:

\left( \dfrac {1} {6}\right) ^{i}\left( \dfrac {5} {6}\right) ^{n-i}

So the expectation value of x is:

E\left( x\right) =\sum _{i=0}^{n}i\left( \dfrac {1} {6}\right) ^{i}\cdot \left( \dfrac {5} {6}\right) ^{n-i}

I typed this into wolframalpha and I don't get n/6 as I thought I would.

I'm not sure where I went wrong. Any ideas would be greatly appreciated. Thanks!

You're claiming that $p_i=\left( \dfrac {1} {6}\right) ^{i}\left( \dfrac {5} {6}\right) ^{n-i}$ is the probability of $i$ $5$s from $n$ tosses.

What should be true of $\sum\limits_{i=0}^np_i$? Is that, in fact, true?

 

[Ray Vickson]

Homework Statement

Show that the expected number of 5's in n tosses of a die is n/6.

Homework Equations

x = random variable representing the number of 5's

E\left( x\right) =\sum _{i}p_{i}x_{i}

The Attempt at a Solution

The probability of getting i 5's in n tosses is:

\left( \dfrac {1} {6}\right) ^{i}\left( \dfrac {5} {6}\right) ^{n-i}

So the expectation value of x is:

E\left( x\right) =\sum _{i=0}^{n}i\left( \dfrac {1} {6}\right) ^{i}\cdot \left( \dfrac {5} {6}\right) ^{n-i}

I typed this into wolframalpha and I don't get n/6 as I thought I would.

I'm not sure where I went wrong. Any ideas would be greatly appreciated. Thanks!

You have the wrong formula for the probability of i 5's in n tosses. Try writing out the "event" for small i and n, say for n = 4 and i = 1 (where the outcomes are 1 '5' and 3 'o's, o = 'other' = non-five.

 

[eprparadox]

Ah, thanks so much. The sum of the probabilities has to be 1. And what I forgot was the number of ways that each number of 5's could appear which is C(n, i).

With that addition, the expectation then does come out to n/6.

Thanks again!

 

[Ray Vickson]

Ah, thanks so much. The sum of the probabilities has to be 1. And what I forgot was the number of ways that each number of 5's could appear which is C(n, i).

With that addition, the expectation then does come out to n/6.

Thanks again!

OK, so now that you have it I am willing to show you a much more general and much easier way, using 'indicator variables". Let
I_i = \begin{cases} 1, \text{ if toss }i = 5\\<br /> 0 , \text{ otherwise}<br /> \end{cases}
The number of 5s in $n$ tosses is $N = \sum_{i=1}^n I_i$. Using linearity of expectation we have
E N = \sum_{i=1}^n E I_i = n/5,
because $EI_1 = EI_2 = \cdots = E I_n = 1/5.$

 

[dirk_mec1]

@Ray don't you mean EI_i = 1/6?

 

[pbuk]

That is a stupid question. In order to calculate the answer you have to assume that the die is fair. What is the definition of fair? That the expected value of the number of 5s (or 1s, 2s etc) in N trials is N/6.

 

[Ray Vickson]

@Ray don't you mean EI_i = 1/6?

Yes, of course.

 

[Ray Vickson]

That is a stupid question. In order to calculate the answer you have to assume that the die is fair. What is the definition of fair? That the expected value of the number of 5s (or 1s, 2s etc) in N trials is N/6.

Why do you think it a stupid question to ask for a proof of
\sum_{i=0}^n i {n \choose i} (1/6)^i (5/6)^{n-i} = n/6\:?

 

[phinds]

In order to calculate the answer you have to assume that the die is fair.

That is ALWAYS the assumption in such problems unless otherwise stated.

 

[pbuk]

Why do you think it a stupid question to ask for a proof of
\sum_{i=0}^n i {n \choose i} (1/6)^i (5/6)^{n-i} = n/6\:?

But that is not what the question asked for. It's as if the question asked "show that the distance from any point on the circumference of a circle to the centre is constant" and you proceeded to "prove" it by saying "well the equation for a circle is $\sqrt{x^2 + y^2} = r$ so...", the equation that you are using is derived from the proposition in the question and it is valid if and only if the propostion is true - you can't hoist yourself by your bootstraps and use this to prove the proposition.

That is ALWAYS the assumption in such problems unless otherwise stated.

That is exactly my point: having made that assumption there is nothing further to show.

 

[Ray Vickson]

But that is not what the question asked for. It's as if the question asked "show that the distance from any point on the circumference of a circle to the centre is constant" and you proceeded to "prove" it by saying "well the equation for a circle is $\sqrt{x^2 + y^2} = r$ so...", the equation that you are using is derived from the proposition in the question and it is valid if and only if the propostion is true - you can't hoist yourself by your bootstraps and use this to prove the proposition.


That is exactly my point: having made that assumption there is nothing further to show.

You say "But that is not what the question asked for". On the contrary, that is exactly what the question asked for. However, I do not wish to debate this issue further.

 

[pbuk]

Why do you think it a stupid question to ask for a proof of
\sum_{i=0}^n i {n \choose i} (1/6)^i (5/6)^{n-i} = n/6\:?

I think "verification" would be better than "proof", but I agree that this would not be a stupid question.

 

[Ray Vickson]

I think "verification" would be better than "proof", but I agree that this would not be a stupid question.

Right: I can live with "verification", which is more-or-less how I interpret it anyway.