# Expected number of 5 in n tosses of die

1. Aug 25, 2014

1. The problem statement, all variables and given/known data
Show that the expected number of 5's in n tosses of a die is n/6.

2. Relevant equations
x = random variable representing the number of 5's

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

3. 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!

2. Aug 25, 2014

### gopher_p

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?

3. Aug 25, 2014

### Ray Vickson

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.

4. Aug 25, 2014

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!

5. Aug 25, 2014

### Ray Vickson

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
$$I_i = \begin{cases} 1, \text{ if toss }i = 5\\ 0 , \text{ otherwise} \end{cases}$$
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 $EI_1 = EI_2 = \cdots = E I_n = 1/5.$

6. Aug 26, 2014

### dirk_mec1

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

7. Aug 26, 2014

### MrAnchovy

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.

8. Aug 26, 2014

### Ray Vickson

Yes, of course.

9. Aug 26, 2014

### Ray Vickson

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

10. Aug 26, 2014

### phinds

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

11. Aug 26, 2014

### MrAnchovy

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.

12. Aug 26, 2014

### Ray Vickson

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.

13. Aug 26, 2014

### MrAnchovy

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

14. Aug 26, 2014

### Ray Vickson

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