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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?

Razanir
2013-07-25, 11:24 PM
Okay. Let W = Money won for winning and L = Money lost for losing.

E(Profit) = P(win)*W + P(lose)*L
\$5 = 9/26 * W - 17/26 * 20

Solve for W = \$102.81

---

Now for 37 games, the most money we can lose is \$740 (\$20 * 37 games). Each win will decrease our losses by \$82.81 (\$102.81 - \$20). 720/82.81 is just under 9 (8.93), so we'll declare that you need 9 wins in order to not lose money. Therefore, we just need to find P(b(37,9/26)<9), where b(n,p) is a binomial distribution with n samples and probability p. Plugging this into a TI-84, I get the answer to be .06460.

Jay R
2013-07-26, 10:32 AM
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.

Here's the mistake. To break even we must win 10.22 games of the 37 we play. If we win ten games and lose 27, the payoff is 10 * 52.22 - 27 * 20 = -\$17.78. That's a net loss.

But if we win 11 games and lose 26, the payoff is 11 * 52.22 - 26 * 20 = \$54.44. That's a net gain.

Therefore you lose if you win ten games or fewer.

The probability of losing all 37 games is (17/26)^37.

The probability of winning exactly one game is 10 * (17/26)^36 * 9/26.

...

The probability of winning exactly ten games is (37! / (10! * 27!)) * (17/26)^27 * (9/26)^10.

Those are all the ways to lose money, so the sum of all of those probabilities is the probability of losing money

Razanir
2013-07-26, 02:43 PM
Yeah... I actually thought to check my work now that it's late afternoon. The \$52.22 number is right, not sure how I got twice that for a win. (I blame it on the fact I was attempting to put it into one line in Wolfram Alpha)

P(b(37,9/26)<11) = .2147

Balain
2013-07-27, 01:30 AM
Thanks Guys