# trying to understand variance

Discussion in 'Probability and Statistics' started by solarchick, Apr 7, 2022.

1. ### solarchick

Joined:
Apr 7, 2022
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Hello everyone,

i have this problem in my textbook:

An example may help to make this clear. Let us modify the parameters of the die roll game we
discussed before. A die is thrown, and the player loses 1 unit on 1-5, but wins 6 units on a roll of 6.
We'll call this new game D.
The expectation of the game is:

< D > = ( 5/6 )(-1) + ( 1/6 )(6)
< D > = 1/6 units /trial

When the player wins, he gains six units. Subtracting the mean value of 1/ 6 from this outcome,
we obtain:

Vwin = (6 - 1/ 6 )^2
Vwin = ( 35/6 )^2

Likewise, when he loses he loses one unit. Subtracting the mean value of 1 6 from this, we have:
Vlose = (-1 - 1/6 )^2
Vlose = (− 7/6 )^2
The variance of the game is:
V D = p(win)(V win ) + p(lose)(V lose )
V D = ( 1/6 )( 35/6 )^2 + ( 5/6 )(- 7/6 )^2
V D ≈ 6.806 units^2 /trial^2

Suppose we toss the die 200 times. What is the chance that the player will win overall in 200
tosses? Win 40 units or more? Win 100 units or more?

My question is this: What is the red 1/6? is it the summation of all the expected value of all the dice rolls?

solarchick, Apr 7, 2022