13 replies to this topic

[bonanova]

Suppose n tickets numbered 1-n are in a box, and you draw one of them

You don't know how many tickets are in the box, but you are asked to estimate how many there are.

Your ticket has the number p on it.

 

What estimate of n has the highest likelihood of being correct?

[Krishna Kutty]

Spoiler for

There are p+1 tickets for sure.

[bonanova]

Spoiler for

There are p+1 tickets for sure.

 

Since you don't know the value of n, but are asked to estimate it, you don't know that p and n are different.

[nakulendu]

Spoiler for try my luck

I picked p in my draw. Lets estimate n = p. What would have been the likelihood that I got the max number ? : 1/p
If say n = p+1. What would have been the likelihood I picked p in my draw ? : 1/(p+1)
1/p > 1/(p+1).
So given that I picked p , the estimate of n that would give the best probability of me picking p is p itself.
So best estimate of n is p

[soop]

Spoiler for rrgghh, trying to get my head around this.

Well, we know there are at least p in n, but given the possibility of us picking p is at least 1 in p, the fact that we know there are at least p-1 other numbers we didn't pick.

 

I really can't picture the math, but I think the probability that we picked the highest number straight away is lower... I'm picturing it like a bell curve, which I know it can't be.  But I'm guessing the answer is somewhere above p, if not double it.

[dark_magician_92]

Suppose n tickets numbered 1-n are in a box, and you draw one of them

You don't know how many tickets are in the box, but you are asked to estimate how many there are.

Your ticket has the number p on it.

 

What estimate of n has the highest likelihood of being correct?

Spoiler for i think

It must be p only because then the Probabability of picking out p = 1/p, for n>p, the probability decreases.

[bushindo]

Suppose n tickets numbered 1-n are in a box, and you draw one of them

You don't know how many tickets are in the box, but you are asked to estimate how many there are.

Your ticket has the number p on it.

 

What estimate of n has the highest likelihood of being correct?

 

It seems like this problem is dependent upon what is a reasonable a priori distribution for N.

Spoiler for

Let P(N) be the prior distribution for the number N. Let P(p|N) be the conditional probability for p given N; we know that P(p|N) = 1/N. We wish to compute the conditional probability P(N|p), which can be expanded using Bayes theorem

 

So, if we are given a well-defined probability function P(N), we can easily compute the number N that gives that maximum probability P(N|p).

 

The problem is that we are not given what P(N) is. One reasonable choice would be to assume that P(N) is the improper uniform distribution from p to infinity (every number has an equal chance of occuring). However, if we plug that improper distribution into the equation above, we would be able to cancel out P(N), but we'd end up with a harmonic series in the denominator, which does not converge.

 

[bushindo]

Suppose n tickets numbered 1-n are in a box, and you draw one of them
You don't know how many tickets are in the box, but you are asked to estimate how many there are.
Your ticket has the number p on it.
 
What estimate of n has the highest likelihood of being correct?

 
It seems like this problem is dependent upon what is a reasonable a priori distribution for N.

Spoiler for

Let P(N) be the prior distribution for the number N. Let P(p|N) be the conditional probability for p given N; we know that P(p|N) = 1/N. We wish to compute the conditional probability P(N|p), which can be expanded using Bayes theorem
bayesian.png
 
So, if we are given a well-defined probability function P(N), we can easily compute the number N that gives that maximum probability P(N|p).
 
The problem is that we are not given what P(N) is. One reasonable choice would be to assume that P(N) is the improper uniform distribution from p to infinity (every number has an equal chance of occuring). However, if we plug that improper distribution into the equation above, we would be able to cancel out P(N), but we'd end up with a harmonic series in the denominator, which does not converge.
 

 
 
Oh wait, the answer is much simpler than that. Turns out this is the easiest bonanova puzzle I have ever seen =)

Spoiler for

Since "n tickets numbered 1-n are in a box", if we get number p = (1-n), then it is easy to see that n = 1 - p.

[bonanova]

 

Suppose n tickets numbered 1-n are in a box, and you draw one of them
You don't know how many tickets are in the box, but you are asked to estimate how many there are.
Your ticket has the number p on it.
 
What estimate of n has the highest likelihood of being correct?

 
It seems like this problem is dependent upon what is a reasonable a priori distribution for N.

Spoiler for

Let P(N) be the prior distribution for the number N. Let P(p|N) be the conditional probability for p given N; we know that P(p|N) = 1/N. We wish to compute the conditional probability P(N|p), which can be expanded using Bayes theorem
bayesian.png
 
So, if we are given a well-defined probability function P(N), we can easily compute the number N that gives that maximum probability P(N|p).
 
The problem is that we are not given what P(N) is. One reasonable choice would be to assume that P(N) is the improper uniform distribution from p to infinity (every number has an equal chance of occuring). However, if we plug that improper distribution into the equation above, we would be able to cancel out P(N), but we'd end up with a harmonic series in the denominator, which does not converge.
 

 
 
Oh wait, the answer is much simpler than that. Turns out this is the easiest bonanova puzzle I have ever seen =)

Spoiler for

Since "n tickets numbered 1-n are in a box", if we get number p = (1-n), then it is easy to see that n = 1 - p.

LOL

That may be the first and only time I use LOL in this forum.

 

I'm not that clever. Really.

But I love it. It's a better puzzle than the one I intended.

Honorable mention.

.

[bonanova]

Spoiler for try my luck

I picked p in my draw. Lets estimate n = p. What would have been the likelihood that I got the max number ? : 1/p
If say n = p+1. What would have been the likelihood I picked p in my draw ? : 1/(p+1)
1/p > 1/(p+1).
So given that I picked p , the estimate of n that would give the best probability of me picking p is p itself.
So best estimate of n is p

 

 

 

Suppose n tickets numbered 1-n are in a box, and you draw one of them

You don't know how many tickets are in the box, but you are asked to estimate how many there are.

Your ticket has the number p on it.

 

What estimate of n has the highest likelihood of being correct?

Spoiler for i think

It must be p only because then the Probabability of picking out p = 1/p, for n>p, the probability decreases.

 

You are both correct.

 

nakulendu posted first, so I marked his answer as the solution.