A Gambling Game

You're at your favorite casino when you spot a new machine with a game that you've never played before. The machine works like this: There is a container containing some unknown number of balls. You cannot see the container (the machine is mounted on a wall with the container on the other side) and so you cannot judge its size. The balls inside the container are labelled 1 through n (where n is the number of balls in the container). When you press a button on the machine, a randomly selected ball pops out, so that you can see the number on that ball. At this point, you bet on how many balls are in the container.

Q1: Suppose you press the button for the first time and the ball labelled '12' pops out. How many balls should you bet are in the container?

Q2: There is another machine at a competing casino which is exactly the same except that the container with the balls is visible, so you can estimate the maximum number of balls that can be contained within it. Can you take advantage of this fact to win more money?

I'll post answers in a day or two.

A1: 23.

A2: Yes.

incorrect

You said the container is visible is part 2...
Are the balls visible or semi-visible, or is the container full?

I guess even if not, you could at least get an upper bound of what a reasonable estimate would be.

I don't see anything wrong with Thomaster's answer

*edit* Actually, I can see why guessing 2x-1 would yield poor results if there are an even number of balls in the container.

A guess of x would actually be more effective if your answer needs to be exact.

Q1: Gut says '12' (it's the n for which a bet on ball '12' would have the highest probability of winning), but I think you need to specify a prior of some sort on the distribution of n to have an answer.

Q2: Do you mean that we should specify a prior for n that is uniform on 12 to nmax? In that case, I'd say 12 again.

This '12' answer could be completely wrong, though. I have to think it through.

Originally posted by forkedknight
You said the container is visible is part 2...
Are the balls visible or semi-visible, or is the container full?

I guess even if not, you could at least get an upper bound of what a reasonable estimate would be.

I don't see anything wrong with Thomaster's answer

*edit* Actually, I can see why guessing 2x-1 would yield poor results if there are ...[text shortened]... container.

A guess of x would actually be more effective if your answer needs to be exact.

Edited out due to stupidity.

forkedknight,

in Q2 the balls are not visible and the container is not necessarily full (though of course, it could be full by chance)

Originally posted by Palynka
Q1: Gut says '12' (it's the n for which a bet on ball '12' would have the highest probability of winning), but I think you need to specify a prior of some sort on the distribution of n to have an answer.

Q2: Do you mean that we should specify a prior for n that is uniform on 12 to nmax? In that case, I'd say 12 again.

This '12' answer could be completely wrong, though. I have to think it through.

I thought 12, but the question is "How many balls should you bet are in the container?" and you bet after the 12 ball comes out. So it would be 11. But it's probably just an imprecise question, not a trick question.

I think the second question needs clarification on whether "you can estimate the maximum number of balls that CAN BE contained within it" means you can estimate the number of balls that ARE contained within it. That is, can you see how full it is.

Originally posted by JS357
I thought 12, but the question is "How many balls should you bet are in the container?" and you bet after the 12 ball comes out. So it would be 11. But it's probably just an imprecise question, not a trick question.

I think the second question needs clarification on whether "you can estimate the maximum number of balls that CAN BE contained within it" means ...[text shortened]... mate the number of balls that ARE contained within it. That is, can you see how full it is.

Ah, yes, 11 of course. I thought how many balls were in the container.

The second just means how many can be fitted physically in the container (nmax, if you will). Like a 25*25cm opaque box that can fit 10k balls if full, for example.

If n=12, there's a chance of 1/12 of number 12 popping out, if n=23, there's a chance of 1/23 of number 12 popping out.

That would also imply an upper bound is irrelevant.

I'm sure that this is going to rely on "equal probability" for every n from 1 to infinity and I'm going to be upset by people dividing by zero. 😠

Originally posted by Thomaster
That would also imply an upper bound is irrelevant.

No! It solves the problem I mention above, although nobody will care as usual.

Originally posted by Palynka
No! It solves the problem I mention above, although nobody will care as usual.

If I'm guessing the number that pops out, I don't need to know the upper bound. I would not influence my guess.

Originally posted by Thomaster
If I'm guessing the number that pops out, I don't need to know the upper bound. I would not influence my guess.

You guess the number that pops out because there is an implicit assumption about a uniform distribution for n.

Palynka,
Yes, I'm assuming uniform distribution.

JS357,
As to your first point, you are right, it wasn't a trick, just imprecision.
For the second, yes, I meant "can be" as in the maximum capacity of the container.