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There are $N$ dollars in a box. You have a probability $P$ (fixed) to win everything in the box and $1-P$ to pay a fee of $200$ dollars ($100$ go in the box, the other half is lost). You can try it as much as you want.

I've calculated mathematical expectation $E$ if one tries one time, two times or three times and solved the equation saying that $E\gt0$ to know when it is worth it to steal the box (depending on $N$).

I'd like to solve the problem for $n$ times trying to steal the box (stopping on a first success).

Here are the first iterations :

  • Trying one time : $E = P\times N-200\times (1-P)$ (it is worth to steal the box when $E\gt0 \Leftrightarrow N>128$)
  • Trying two times (stopping if succeed on first attempt) : $E = P\times N-P\times(1-P)\times(N-100)-400\times(1-P)^2$ (it is $N-100$ and not $N-200$ because 100 dollars go in the box if you fail the first time). It is worth it if $E\gt0 \Leftrightarrow N>99$.
  • Third iteration : Becomes really complicated, I've solved it using Soulver (numerical method)

Is it possible to generalize the problem to $n$ tries ?

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1  
Regardless to the mathematical answer, for $n>4$ it is worth to steal the box just for the psychological satisfaction of finally having the prize! :-) –  Asaf Karagila Feb 23 '12 at 12:27
2  
$E=(\sum_{k=0}^{n-1} P(1-P)^k(N-100k))-200n(1-P)^n$ –  Angela Richardson Feb 23 '12 at 13:02
    
The formula works, thank you ! –  Skydreamer Feb 23 '12 at 13:28

1 Answer 1

up vote 3 down vote accepted

Angela Richardson has given you a formula in her comment, but it is easier to find the overall optimal strategy.

  1. If it is worth playing a given finite number of turns then it is worth playing a potentially unlimited number of turns until you win, as there will be more money in the box later.
  2. If you will play a potentially unlimited number of turns until you win then you can ignore the 100 dollars that stays in the box, as you will get it back.
  3. So you need to compare the $N$ dollars you will gain against the 100 dollars you may lose several times which does not stay in the box.
  4. The expected number of times you lose before winning is $\frac{1-P}{P}$ [this is a negative binomial distribution] so you expect to lose $100\left(\frac{1-P}{P}\right)$ dollars before winning $N$ dollars.
  5. Your expected gain if you play until you win is $N-100\left(\frac{1-P}{P}\right)$ dollars.
  6. So if you have a linear utility of money and no attitude to risk you should play until you win if $N \gt 100\left(\frac{1-P}{P}\right)$ or equivalently if $P \gt \frac{100}{100+N}$, not play at all if $N$ or $P$ are smaller than these bounds, and either play until you win or do not play at all if there is equality.
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+1, nice comprehensive answer! –  joriki Feb 23 '12 at 21:01
    
Nice approach to the problem ! Thank you ! –  Skydreamer Feb 23 '12 at 21:45

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