What is the Mean When Tossing a Dice Three Times?

[Niles]

Homework Statement

Hi all.

I thought about this earlier today. Let's say I toss a dice three times, and the outcome is 2, 4 and 5. Is the mean (2+4+5)/3 or (2+4+5)/6?

The reason why I am asking is that we can look at the mean as a weighted average of the probability of getting each outcome. So in our case, the probability for getting {1, 2, 3, 4, 5, 6} is the same, so that is why one might be inclined to say that the mean is (2+4+5)/6.

But then again, only 3 tosses has been made. So my question is:

1) Am I looking at two different things here?

2) If yes (which is probably the correct answer to #1), then is (2+4+5)/6 the expectation value (i.e. the average when tossing the dice infinitely many times), and (2+4+5)/3 the average of that sample?

Thanks in advance. I really appreciate your help.


Niles.

 

[HallsofIvy]

Homework Statement

Hi all.

I thought about this earlier today. Let's say I toss a dice three times, and the outcome is 2, 4 and 5. Is the mean (2+4+5)/3 or (2+4+5)/6?

The reason why I am asking is that we can look at the mean as a weighted average of the probability of getting each outcome. So in our case, the probability for getting {1, 2, 3, 4, 5, 6} is the same, so that is why one might be inclined to say that the mean is (2+4+5)/6.

There is no need to "weight" here since the probabilities of rolling 2, 4, 5 with a single die ("dice" is the plural) are the same. If you did that, you have to take into account the fact that those are not the only possibilities. The probability of rolling one of 2, 4, or 5, as opposed to 1, 3, or 6, is 3/6= 1/2, You would have to find the "weighted" sum (1/6)(2)+ (1/6)(4)+ (1/6)(5) divided by 1/2 (the overall probability). That gives (2+4+5)/6 times 2= (2+4+5)/3.

But then again, only 3 tosses has been made. So my question is:

1) Am I looking at two different things here?

Not really, but you forgot that 1/6+ 1/6+ 1/6 is not 1: the total probability here is 1/2 not 1.

2) If yes (which is probably the correct answer to #1), then is (2+4+5)/6 the expectation value (i.e. the average when tossing the dice infinitely many times), and (2+4+5)/3 the average of that sample?

Done correctly either way, the answer is (2+4+5)/3.

Thanks in advance. I really appreciate your help.


Niles.

 

[Niles]

Thanks for helping.