A Counterintuitive Probability Game
This is a super facinating observation. At first it does seem very counterintuitive but after a little examination, the resulting probabilities make a lot of sense.<p>For sake of easy math, pick C, your threshold, to be fixed at 0, which is halfway between -inf and inf.<p>Now there are 4 scenarios. The numbers player 1 chooses, A and B, can be both above or both below C with probability 1&#x2F;4 each. In both those cases, you have a 50&#x2F;50 chance of being correct on which number is larger depending on which number you chose to see.<p>Then, with the other 1&#x2F;2 of the times, one number will be above and the other will always be below C. In those cases, the selected strategy will have you always choose the correct highest number.<p>All together you are right 1&#x2F;2 + 2 * 1&#x2F;4 * 1&#x2F;2 = 3&#x2F;4 = 75% of the time. Simple probability but totally counterintuitive from the onset.<p>Now what if C!=0 or numbers are not being selected uniformly by your opponent? You can replace the 1&#x2F;2&#x27;s with (p) and (1-p) in the right spots without making any distributional assumptions and it seems (without doing the math out fully) that things cancel nicely and show that you always have a strictly greater than 1&#x2F;2 chance of guessing right no matter what. Exercise left for the reader :P