It’s your move

I’m thinking of two numbers between 0 and 1. Your goal is to guess a number which falls in between my two numbers. Each guess costs you $1, and if you guess correctly you win the reciprocal of the length of my range (ie if I am thinking of 0.2 and 0.4, a correct guess wins you $5). At any time you may request that I choose of a new pair of numbers, and of course I will pick a new pair of numbers whenever you win.

What’s your strategy? Under certain conditions, this game is fair. How might you be able to have a positive expectation?

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This entry was posted on Wednesday, June 2nd, 2010 at 12:36 pm and is filed under feature, games, probability, stats. You can follow any comments to this entry through the RSS 2.0 feed. Both comments and pings are currently closed.

2 comments

  1. Martin Gillette
    June 3, 2010 at 7:22 pm

    The best strategy would be to pick the midpoint, 0.5, every time. So long as the numbers are chosen randomly each time that’s your best bet.

  2. David
    July 16, 2010 at 4:26 pm

    I can only think of directions in regards to the “Fairness” part of the challenge.
    Well, if , c, the mean of your two numbers [a, b], has a small variance from the centre, than I’ll go with the previous comment – and always pick 0.5.
    If such a variance is imposed (or viewed), then you actually need to always go with [0, 1]. Giving me a single $ each time – and costing me the same. Hence, E=0.
    So imposing small Var on your mean value, is a recipe for a “Fair” game.
    For that matter, if YOU have to meet such a constraint, then it hardly matters around which point are your two #s revolving.
    If all I know, is the above constraint, I am taking a binary-search to find your mean using say a kind of Win-Stay-Lose-Shift on the values I’m raffling.
    For large enough N (considering the constraint is still on, and has remained the same throughout) I will eventually find your mean, and will always gamble on it.
    This is only fair in the large N sense. Asymptotically fair. I still lose money during “excavation”.
    Again, since for large enough N I’m statistically winning – your best bet is to lower my income from every single hit – again choosing 0,1 as your two #s.
    So having a small variance from a (secretly) pre-chosen, fixed point, is a recipe for us both playing nice – Keeping our Dollars (or Shekels) in our pockets. Besides, we’ve got 8-legged Paul for such coin-flips 🙂