daily puzzle: the shady minter

The chance of getting away with each box is independent of the other boxes and is 99/100. (99/100)^n gives you the probability of "getting away with it" for n boxes, so 0.3660323412732 (approximately) for 100 boxes.

The probability tends to 0 as the number of boxes tends to infinity unless he cuts the treasurer in on the deal.

What does "plugs" mean in this context?

Fake coins.

The King's treasurer is unfortunately far too dim to solve the problem efficiently. All you need is a balance and a bit of patience and a binary chop procedure to isolate the one coin with the wrong mass.

It is highly unlikely that the plug is the same mass as a gold coin of those dimensions although one made of tungsten might pass the density test it wouldn't pass the bite test...

Too boring to be worth bothering to answer.

so 0.366032 (approximately) for 100 boxes.

That's easy.

Massive fail. There are n coins per box, and n boxes get are sampled.

Yeah, I didn't notice that n was also increasing the size of the box. So it would be ((n-1) / n)^n or

(n-1)^n / n^n

Yeah, not sure. The ratio itself tends to one as n increases, but taking the power of n makes it smaller as n increases. So not sure if it goes to 1 or 0. I'm wondering if (n-1)^n forms a well known power series. Not in a math mood at the moment.

You chose a very unfortunate way to write it down. You should recognise this classic standard form.

lim n-> inf (1-1/n)^n

(although it is usually written with the opposite sign)

It is a variant of expanding (1+x)^n with x = -1/n in this case

the probability of winning ( not being dioscovered) is (1-(1/n))^n

or / n-1 \ n | --- | \ n / or (n-1)^n ------- n^n

That tends asymptotically to 1/e