2
$\begingroup$

Say I have a bag of 100 balls, 50 red and 50 blue. 100 people pick one ball each including me. I want to pick a red ball. I get to choose when I pick.

Clearly there is a high probability that at some point in the process there will be more reds than blue. My question is, at the very start is there a higher probability of me winning by waiting on the chance of the odds turning in my favour, even if it’s only slightly? As in does the probability of that happening combined with the probability of then picking red outweigh the 50% chance you have at picking red at the very start?

$\endgroup$
  • 1
    $\begingroup$ Would you be able to see what the others have chosen? $\endgroup$ – Rhys Hughes May 17 '18 at 19:00
  • $\begingroup$ Yes you would be able to see what others have chosen $\endgroup$ – Christian T May 17 '18 at 21:58
3
$\begingroup$

No, it makes no difference.

Number the balls according to when they will be drawn out of the bag, so that whoever chooses $4$th gets ball number $4$, etc. Now suppose you have a strategy for deciding when to pick that gets red more often than not. Suppose for a particular ordering you decide to pick at time $r$, so you get the $r$th ball. You must do exactly the same for the ordering which is the same for the first $r-1$ balls, but then is reversed after that (because you have exactly the same information at the point you decide to act). Thus, if you follow the strategy your chance of picking a red ball doesn't change if, instead of getting the next ball when you pick, you always get the $100$th ball when you pick. But in that case it makes no difference what you do (since you always get the same ball), and the chance of the $100$th ball being red is $50\%$.

This is counterintuitive because it seems like you should be able to wait until there is one more red ball than blue left, for more than $50\%$. But the point is that this isn't guaranteed to happen. When it does, you gain a small extra chance of getting a red ball, but on the rare times it doesn't, you lose much more.

[For those who are familiar with the concept, what's really going on is that the probability of the next ball being red is a martingale, and the optional stopping theorem applies.]

$\endgroup$
  • $\begingroup$ So does it work out that the probability of winning is always exactly 50%, no matter what your condition for when you choose is? $\endgroup$ – Christian T May 20 '18 at 7:36
  • $\begingroup$ @ChristianT Yes, for any (possibly randomised) strategy that only depends on what you've seen so far. $\endgroup$ – Especially Lime May 20 '18 at 10:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.