How to guess the bigger number in two hands?

Today morning, I received a question from one of my old friends

Alice writes respectively one number in two hands, you can see one of them, then is there a strategy to guess in which hand is placed the bigger number with a probability bigger than 50%?

His intuition tells my friend that nothing will change even though we know one number because the other has always possibility to change. But the maths tell me that it is not the fact. (Thanks to Polytechnique, I remember that I have done this question in PC but not in the form to try to get the design)

First, we recall that for a fair game, we should have the possibility to see left or right hand. If not, the "cheat strategy" is that Alice shows always the left hand with a smaller number but hides the bigger in the right, then we have no chance to win.

The strategy is simple: we require left hand or right hand randomly with possibility of 0.5, then if it is bigger than 10, we guess this number is bigger. Else we guess that it is small. We neglect the situation that the two number are equal. Moreover, 10 is not essential, we can use any number as a criteria. 

What happens? We note the two number as random variable X, Y. As we require the two number randomly, we can suppose that they have a symmetry distribution. i.e p(x,y) = p(y,x). The domain that our strategy does not work is {Y>X>10} and {10>X>Y}. But our strategy works is the area {X>Y>10},{10>Y>X},{X>10>Y},{Y>10>X}. The first two compensate the negative situation. The last two situation we win.

If you do not believe in it, a simple MATLAB simulation will show our proof is correct.

We can also prove by contradiction. If as what we suppose, nothing has changed. Then, the number showed should be the mean number of the distribution. Then, any number is the mean? This is obviously wrong.

In conclusion, to figure out this question, we should has a basic knowledge about the base of probability. That is, what is the experimental space and what is measurable, what is probability. This also underlies the significance of the information and conditional probability.

PS: If I play this game with my naught cousin and I hope to win as many as possible, what should I do? Firstly, I will use the statistic method to approximate the mean of the distribution, given that we believe the number in two hands are independent. Secondly, I will make the choice by two hands as randomly as possible, at least, he should not know how I guess, or I will lose the game.