Situations with doing a same event n times

[10may]

While calculating the expectation value of problems in which an event is done n times, like rolling a dice n times, playing roulette n times, etc., we simply multiply expectation of event being done once with n.
But shouldn't we calculate the Expectation by proper method? The answer is both of those scenario isn't matching for me.

Let's take the specific example of rolling a dice 2 times. We need to calculate the expected value of the sum
What I've seen people doing is - E(sum of 2 dice) = 2*E(sum of one dice)

But if I calculate E properly :-
E(sum of 2 dice) = sum(xf(x)) = sum(P(sum is x)*(value of x))

This does not match with the answer above.
Please see the attached image.

Why is this so?

 

[BigBeachBanana]

While calculating the expectation value of problems in which an event is done n times, like rolling a dice n times, playing roulette n times, etc., we simply multiply expectation of event being done once with n.
But shouldn't we calculate the Expectation by proper method? The answer is both of those scenario isn't matching for me.

Let's take the specific example of rolling a dice 2 times. We need to calculate the expected value of the sum
What I've seen people doing is - E(sum of 2 dice) = 2*E(sum of one dice)

But if I calculate E properly :-
E(sum of 2 dice) = sum(xf(x)) = sum(P(sum is x)*(value of x))

This does not match with the answer above.
Please see the attached image.

Why is this so?

You haven't considered all of the possibilities an outcome could happen. For example, to get a sum of 3, you can get either (1 and 2) or (2 and 1).
You've only considered the first scenario but not the second.

If you know a bit about combinatorics, the binomial coefficient will help you count, otherwise, creating a matrix can help you see all of the possible outcomes.

 

[blamocur]

What I've seen people doing is - E(sum of 2 dice) = 2*E(sum of one dice)

For any two random variables [imath]X,Y[/imath] we always have [imath]E(X+Y) = E(X) + E(Y)[/imath]

 

[10may]

You haven't considered all of the possibilities an outcome could happen. For example, to get a sum of 3, you can get either (1 and 2) or (2 and 1).
You've only considered the first scenario but not the second.

If you know a bit about combinatorics, the binomial coefficient will help you count, otherwise, creating a matrix can help you see all of the possible outcomes.

View attachment 35139

Hey @BigBeachBanana I did considered all the possibilites. If you see my uploaded image, the 1/18 factor is there due to this only (2*1/6*1/6)

 

[BigBeachBanana]

Hey @BigBeachBanana I did considered all the possibilites. If you see my uploaded image, the 1/18 factor is there due to this only (2*1/6*1/6)

Check 7. You got 2+5 and 3+4, but where's 1+6?
Also, wrote 1+4 = 4 which isn't true.
My point is you did consider it to some extent but not all. Listing them using the matrix above keeps things straight.