Math Help - Expected value ?

1. Expected value ?

Today i encountered a mathematics question...

i don't have the exact writing of the question on hand, but i try to state the question as clearly as possible.

That is, in a gambling game, which is to throw a fair dice, so the probability of getting an odd number is equal to that of even number. One can bet on either odd number or even number, the stake is $1 , if he wins, then he will win$1, while getting back $2 if stake is also counted. If he loses, then he will lose the stake so what is expected value of game? the answer was something$1 (it should be correct)

but I don't quite understand... my argument is ....

while if I lose, then I will lose the stake, then it will be -$1 if I win, then i will get back$2 in total, so it should be $2 1/2 ($2) + 1/2 (-$1) =$1/2 = $0.5 I really don't quite understand how$1 comes even though my teacher had explained to me, she was very unclear

2. Originally Posted by kenny1999
Today i encountered a mathematics question...

i don't have the exact writing of the question on hand, but i try to state the question as clearly as possible.

That is, in a gambling game, which is to throw a fair dice, so the probability of getting an odd number is equal to that of even number. One can bet on either odd number or even number, the stake is $1 , if he wins, then he will win$1, while getting back $2 if stake is also counted. If he loses, then he will lose the stake. "Getting back the stake" is a given when you win and isn't counted toward winnings. If you win you gain$1. If you lose, you lose $1. Imagine doing this 1000 times. Because "the probability of getting an odd number is equal to that of even number" you would expect to get around 500 odd numbers and 500 even numbers. That is, you would expect to win$500 and lose $500, netting$0. This is a "fair game".

so what is expected value of game?

the answer was something $1 (it should be correct) but I don't quite understand... my argument is .... while if I lose, then I will lose the stake, then it will be -$1

if I win, then i will get back $2 in total, so it should be$2

1/2 ($2) + 1/2 (-$1) = $1/2 =$0.5

I really don't quite understand how $1 comes even though my teacher had explained to me, she was very unclear If you have given this correctly, the expected value is neither$1.00 nor $0.50, it is 0. 3. Hello, Kenny! Your description is not clear . . . the "one" and "he" are confusing. Here is my interpretation of the game. That is, in a gambling game, which is to throw a fair dice, so the probability of getting an odd number is equal to that of even number. You can bet on either odd number or even number. If you win, then you win$1.
If you lose, then you lose $1. What is expected value of game? Ignore the reference to the "stake"; you do NOT get$2 for a win.

. . If you win, he hands you a dollar.
. . If you lose, you give him your dollar.
And that is how the money changes hands.

We have: . $\begin{array}{ccc}P(\text{win }\1) &=& \frac{1}{2} \\ \\[-4mm] P(\text{lose }\1) &=& \frac{1}{2} \end{array}$

Therefore: . $EV \;=\;(+1)\left(\tfrac{1}{2}\right) + (-1)\left(\tfrac{1}{2}\right) \;=\;\tfrac{1}{2} - \tfrac{1}{2} \;=\;0$

The game is "fair".
Playing repeatedly, you can expect to break even.

4. thanks all for the answers.

but the answer of the expected value was certainly $1 was it required to find the expected value of "what is finally getting back" then the result should be the sum of the stake ($1) and the expected value of winning), right?