2
$\begingroup$

Say you pay 10 dollars to play a game, and if you win, you get a certain sum back. If you lose, nothing is returned.

For example, you roll two dice, getting numbers $2–12$. Only upon obtaining odd numbers do you get $x$ dollars back, otherwise, ten dollars lost (cost of playing the game).

As the generous owner of such a game, I strive to make the expected value, the dollars overall returned, $9$.

So, my goal is to find the value of $x$.

What I currently have is:

$$9= (x-10)(\frac{18}{36}) + (-10)(\frac{18}{36})$$

$$9= (x-10)(\frac{1}{2}) + (-10)(\frac{1}{2})$$

$9$ is the expected value, $(x-10)$ is the return minus the cost, and the $(-10)(\frac{1}{2})$ represents the scenario where the player loses.

$$9\times2=x-10-10$$

$$18+20=x$$

$$x=38$$

Does this mean that if I want to have an overall return of $9, the reward for getting an odd number needs to be thirty-eight dollars? Looking at it from a logical standpoint, that number seems a bit high (I feel like it should be in the twenties). Did I do something wrong somewhere? Or is it all correct and I'm just mistaken?

$\endgroup$
3
  • $\begingroup$ You might found this interesting en.wikipedia.org/wiki/St._Petersburg_paradox. It's a game that seems like its payoff isn't worth much but the expectation of the game is infinity! $\endgroup$
    – BigbearZzz
    Commented Jan 14, 2016 at 2:32
  • 1
    $\begingroup$ If your goal is an expected profit of \$9 for the player each game, \$38 makes sense to me. If you play twice and win once and lose once, that's an average return of \$19 per game. Minus the \$10 you paid to play equals a profit of \$9 per game. $\endgroup$ Commented Jan 14, 2016 at 2:56
  • 1
    $\begingroup$ Your analysis is correct provided that when you roll an odd number you earn $\$38$ less the $\$10$ paid to play the game in the first place. Then your expected net earnings is $\frac{38}{2}-10 = 9$ dollars. $\endgroup$
    – Brian Tung
    Commented Jan 14, 2016 at 2:58

1 Answer 1

3
$\begingroup$

It's all correct and you're mistaken.

Suppose the reward is 10 dollars: after $n$ games you will pay $10n$ dollars but you'll only win about half the time, winning about $5n$ total. And who would really play this game? This game is "Flip a coin, if it's heads then you give me $10 and if it's tails you get to keep your money." Sounds like extortion.

Suppose the reward were 20 dollars: now the game is symmetric: you pay $10n$ dollars and win about $10n$ back, so on average you break even.

Suppose the reward were 40 dollars: now you pay $10n$ but win about $20n$ back, so asymptotically your winnings per game total $10/game.

The \$9/game sum is a little under the \$10/game one. Possibly you are confused because the 10 dollars is being deducted for every game, not just for the ones where you lose?

$\endgroup$
2
  • $\begingroup$ Ah yes, a bit of a foolish misinterpretation on my part. Thanks to you and the comments! $\endgroup$
    – john2546
    Commented Jan 14, 2016 at 3:08
  • $\begingroup$ For the 40 dollar case, I think it should read, "... but win about $30n$ back", not $20n$. $\endgroup$
    – Théophile
    Commented Jan 14, 2016 at 3:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .