Balain

2013-07-25, 09:38 PM

I am taking an intro course to stats and been studying for exam. One question I am having problems with.

The previous question we are told we are playing a game with 9/26 chance of winning, 17/26 chance of losing. If we lose we pay $20. The expected amount to win if one game is played is $5. We have to figure how much money we earn if we win.

So we end up with:

X = The amount of money we win or lose.

X 52.22 -20

P(x) 9/26 17/26

E(x) = 5

All of that is fine for me.

Then we are asked. If we play the game 37 times what is the chance we lose money. (here is his solution which I Understand where the numbers come from but the logic is eluding me)

let x = number of wins, y = number of loses, so x+y=37

We want an expectation of $0, that is E(x) = 0

E(x) = sum(x*p(x))

E(x) = 52.22*x + (-20)*y = 0

0 = 52.22*x + (-20)*(37-x)

0 = 52.22*x - 740 +20*x

x = 10.22

Therefore to break even we need to play 10.22 games. So playing 10 games we should lose money and playing 11 or more games we should earn money.

therefore P(losing money) = p(x=0)+p(x=1)+p(x=2)+...+p(x=10)

or 1-P(earning money) = 1 - p(x=11)+...+p(x=37)

both way to long so use normal approximation.....etc etc etc

My problem is the whole E(x)=52.22*x + (-20)*y = 0, 52.22*x + (-20)(37-x) = 0

x,y are suppose to be the probabilities of winning and losing. Is formula is just swapping probabilities with number of games played. which is just changing the game or the essence of the game. So then I thought well change it to 52.22*x =(-20)*(1-x) = 0. then take x * 37 = 10.3 or something.... Still getting 10 games to lose money. I still have a problem with this cause it is still changing the probabilities and thus the game.

Can anyone help me clear up my thinking?

The previous question we are told we are playing a game with 9/26 chance of winning, 17/26 chance of losing. If we lose we pay $20. The expected amount to win if one game is played is $5. We have to figure how much money we earn if we win.

So we end up with:

X = The amount of money we win or lose.

X 52.22 -20

P(x) 9/26 17/26

E(x) = 5

All of that is fine for me.

Then we are asked. If we play the game 37 times what is the chance we lose money. (here is his solution which I Understand where the numbers come from but the logic is eluding me)

let x = number of wins, y = number of loses, so x+y=37

We want an expectation of $0, that is E(x) = 0

E(x) = sum(x*p(x))

E(x) = 52.22*x + (-20)*y = 0

0 = 52.22*x + (-20)*(37-x)

0 = 52.22*x - 740 +20*x

x = 10.22

Therefore to break even we need to play 10.22 games. So playing 10 games we should lose money and playing 11 or more games we should earn money.

therefore P(losing money) = p(x=0)+p(x=1)+p(x=2)+...+p(x=10)

or 1-P(earning money) = 1 - p(x=11)+...+p(x=37)

both way to long so use normal approximation.....etc etc etc

My problem is the whole E(x)=52.22*x + (-20)*y = 0, 52.22*x + (-20)(37-x) = 0

x,y are suppose to be the probabilities of winning and losing. Is formula is just swapping probabilities with number of games played. which is just changing the game or the essence of the game. So then I thought well change it to 52.22*x =(-20)*(1-x) = 0. then take x * 37 = 10.3 or something.... Still getting 10 games to lose money. I still have a problem with this cause it is still changing the probabilities and thus the game.

Can anyone help me clear up my thinking?