bonanova

A sealed, opaque and weightless box contains marbles of uniform weight and no more than 61 in number. I give you a precision two-pan balance and offer you $50 to tell me how many marbles the box contains. You may purchase any number of calibrated weights at $10 each and in units of your choice to assist you. Postage and handling for the calibrated weights is a flat $2 charge. Will you accept the offer, and what is your reasoning?

Agree. Better simply to say simply that in any large ensemble of N outcomes, where 60% will be H, the outcomes can be partitioned into groups that are totally fair (H=T) and totally biased (all H.). The fair partition will be N x 2 p(T) in size.

T T T T H H H H - H H is a result that is representative of p(H) = 0.6. That does not mean that in every ten flips it will happen. There can be (arbitrarily long) strings of T, so long as p(T) is not zero. So T T T T T T T T T T can happen. But p(H) = 0.6 says that, on average, whenever there are 10-T strings there will also be 10-H strings, and they will occur in the following proportions: T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H ------------------- H H H H H H H H H H H H H H H H H H H H and the hyphens indicate how to partition them.

Agree with SP. This is a classic.

It may not be clear whether mean (median) applies to the speeds of the lanes or the ensemble of cars. My assumption was the former; as I think of it now, the latter makes a better puzzle. Edit: and it is what the OP says.

Nice segue to a cute poem from the late Martin Gardner: Pi goes on and on and on ... And e is just as cursed. I wonder: Which is larger When their digits are reversed?

Once bias exists, infinity won't guarantee a return to equality. The probability of equality decreases with bias, but it's never zero unless one of the outcomes never occurs at all. Note that saying p(H=T) < 1 doesn't prohibit H=T, it only says H=T is not a certain eventual outcome.

On further thought, I think my answer gives only a lower bound.

Point is you can change cars passed or passing by adjusting their density independent of Alice's speed. She just can't be the slowest or fastest car. At any speed other than slowest or fastest the passed and passing cars can be equal in number.

The guy with the brunette would immediately say brunette if you had the other blonde.

B knows A's 3-bit sequence was sent. Does B also know what it is?

I think harey has it.

By "not on top of" or below, you mean "not immediately on top of" or below; that is, "not touching". That is, the spade 10 cannot touch the heart 10 (or any other 10), and also the spade 10 cannot touch the spade King (or any other spade.) Is that the sense of the question?

So I would calculate L and subtract that from my secret number. Cool. Thx.