# Not so obvious probability puzzle.

You are given two boxes with a number inside each box. The two numbers are different but you have no idea what they are. You pick one box to open; read the number inside; and then guess if the number in the other box is larger or smaller. You win if you guess correctly, and lose otherwise.

Is there anyway that you can win the game with more than $50$ % chances no matter what the two numbers are?

I cannot think of a way. Suppose you get $25$ from first box, now probability of you winning is undoubtedly $1/2$. Either the number in box $2$ is greater than 25 or less than.

• So the numbers inside are chosen from 1-50? – Jason Chen May 14 '15 at 5:10
• I guess numbers have no bound on them,except assuming positive integers. Nothing was mentioned in the question atleast – Bhaskar Vashishth May 14 '15 at 5:11
• If there is no bound, always guess that the second number is greater. There are infinitely more positive integers greater than the first number than there are smaller numbers. – Jason Chen May 14 '15 at 5:12
• Choose a random number $X$ before opening the box, from any continuous distribution over the positive reals that you like. If the revealed number is greater than $X$, guess that the other number is smaller, and vice versa. Remarkably, this gives you a greater than 50% chance of winning. – mjqxxxx May 14 '15 at 5:14
• To answer the question we need to know the probability distribution. And there is no probability distribution on the natural numbers that assigns equal probability to all $n$, – André Nicolas May 14 '15 at 5:14

To elaborate on mjqxxxx's answer: let's say the two numbers are $a,b$ with $a<b$.
Strategy: pick the box randomly (so that the number you see - denoted by $Y$ - is equally likely to be $a$ or $b$). Also, pick a random threshold $X$ - let's say from the Gaussian distribution on the reals.
If $X>Y$, then guess that the other number is larger. If $X<Y$, then guess that the other number is smaller.
The probability of winning is \begin{align} &\Pr(Win|X\in[a,b])\cdot \Pr(X\in[a,b])+\Pr(Win|X\notin[a,b])\cdot \Pr(X\notin[a,b])\\ =& 1\cdot \Pr(X\in[a,b]) + (1/2)\cdot \Pr(X\notin[a,b])\\ =&\frac{1}{2}+\frac{\Pr(X\in[a,b])}{2}\\ >& \frac{1}{2} \end{align}, since the Gaussian distribution assigns positive probability to every interval $[a,b]$. Note that we cannot provide a lower bound on the gap from 1/2.