Matti

]]>oh, you theorists… We need to specify even the goriest detail lest you’ll argue the solution is undetermined. Of course the kids will try to balance the overall weight the best they can!

If I have time today I will try to solve the case of a flat prior weight distribution.

Cheers,

T.

By intuition (the only thing still working at this time in my brain) the most profitable bias in the seesaw balance is the average difference between two children’s weights. Am I right ?

Cheers,

T.

You made me remember a thought I had a week ago, while looking at my two kids playing on a see-saw.

Imagine the see-saw is perfectly balanced: two kids of uneven weight will unbalance it. Now, given a gaussian distribution of

children weights, of mean M and width W, what is the most profitable unbalance the see-saw

should have when unloaded, in order to minimize on average the

unbalance once two random kids play on it ?

The problem is easy to solve, but even easier it is to know that there is indeed a unique solution, for a given single-modal distribution of weights. Or at least, that is what my intuition tells me.

Cheers,

T.