Opening boxes consecutively to find 1 prize

[JohnnyGui]

I've seen a simple guessing show but I'm a bit surprised by its probability results and don't know what I'm doing wrong.

A prize is hidden in 1 of 7 boxes and a person has to guess in which box it is.
He opens a box one after the other and I questioned myself what the probability might be for finding the prize at the $n$th box.
To find the prize at the 4th box for example, I would reason that the probability is:
$$P(n=4) = \frac{6}{7} \cdot \frac{5}{6} \cdot \frac{4}{5} \cdot \frac{1}{4} = \frac{1}{7}$$
Furthermore, the host said that on average, people would find the prize on the 3rd or 4th box.

However, I keep calculating the exact same probability of $\frac{1}{7}$ regardless of the number of boxes. This would mean that it's equally probable for the person to find the prize in the first box as in the last (7th) box, which seems weird to me.

What am I missing here? And how does the 3rd or 4th box have the highest probability on average according to the host?

 

[Dale]

What am I missing here?

Nothing. 1/7 is right

 

[JohnnyGui]

Nothing. 1/7 is right

How did the host come up with saying that on average the prize is found in the 3rd or 4th box?

 

[Dale]

How did the host come up with saying that on average the prize is found in the 3rd or 4th box?

Well, it isn’t the 3rd or 4th, it is the 4th. That is just the average of 1 through 7.

 

[JohnnyGui]

Well, it isn’t the 3rd or 4th, it is the 4th. That is just the average of 1 through 7.

But would that mean that when a lot of people do this guessing game, the majority would find the prize in the 4th box? I find that hard to understand if the fact that finding the prize at each nth box is equally probable.

 

[Dale]

But would that mean that when a lot of people do this guessing game, the majority would find the prize in the 4th box?

No. It is the mean, not the mode.

 

[JohnnyGui]

No. It is the mean, not the mode.

Thanks! This now makes sense to me.

 

[Ray Vickson]

I've seen a simple guessing show but I'm a bit surprised by its probability results and don't know what I'm doing wrong.

A prize is hidden in 1 of 7 boxes and a person has to guess in which box it is.
He opens a box one after the other and I questioned myself what the probability might be for finding the prize at the $n$th box.
To find the prize at the 4th box for example, I would reason that the probability is:
$$P(n=4) = \frac{6}{7} \cdot \frac{5}{6} \cdot \frac{4}{5} \cdot \frac{1}{4} = \frac{1}{7}$$
Furthermore, the host said that on average, people would find the prize on the 3rd or 4th box.

However, I keep calculating the exact same probability of $\frac{1}{7}$ regardless of the number of boxes. This would mean that it's equally probable for the person to find the prize in the first box as in the last (7th) box, which seems weird to me.

What am I missing here? And how does the 3rd or 4th box have the highest probability on average according to the host?

Think of the permutations of the numbers from 1 to 7, with "1" being the prize. In 1/7th of the permutations, "1" is in position 1, in 1/7th of the permutations the "1" is in position 2, etc., etc. So, for any postion, the chance of winning in that position is 1/7.

 

[JohnnyGui]

Think of the permutations of the numbers from 1 to 7, with "1" being the prize. In 1/7th of the permutations, "1" is in position 1, in 1/7th of the permutations the "1" is in position 2, etc., etc. So, for any postion, the chance of winning in that position is 1/7.

That's an intuitive way to look at it actually. Although in that way of thinking, I come across a different problem:

Say the prize is at the 4th box. For 1 permutation, taking the probability of each empty box individually into account, the chance would be:
$$\frac{1}{7} \cdot \frac{1}{6} \cdot \frac{1}{5} \cdot \frac{1}{4}$$
The first 3 probabilities represent the chance of each of the 3 empty boxes before the prize, right? Those 3 empty boxes can be opened in a different order, such that the number of possible permutations for these 3 empty boxes is $3!$. Multiplying $3!$ by the probability for one specific permutation does not give me 1/7th.

However, I can see that it does give me 1/7th if I include the 4th box with the prize in the number of possible permutations; being $4!$. But the box with the prize should stay fixed in the same place and not be "shuffled" with the 3 empty boxes. What is wrong in my way of thinking?

EDIT: Never mind, it should not be $3!$ but $6 \cdot 5 \cdot 4$ since there are 6 empty boxes available to pick from for the first 3 empty boxes.

 

[Ray Vickson]

That's an intuitive way to look at it actually. Although in that way of thinking, I come across a different problem:

Say the prize is at the 4th box. For 1 permutation, taking the probability of each empty box individually into account, the chance would be:
$$\frac{1}{7} \cdot \frac{1}{6} \cdot \frac{1}{5} \cdot \frac{1}{4}$$
The first 3 probabilities represent the chance of each of the 3 empty boxes before the prize, right? Those 3 empty boxes can be opened in a different order, such that the number of possible permutations for these 3 empty boxes is $3!$. Multiplying $3!$ by the probability for one specific permutation does not give me 1/7th.

However, I can see that it does give me 1/7th if I include the 4th box with the prize in the number of possible permutations; being $4!$. But the box with the prize should stay fixed in the same place and not be "shuffled" with the 3 empty boxes. What is wrong in my way of thinking?

I think you have it backwards and are just confusing yourself. We generate a permutation of {1,2,3,...,7} at random; then we fix that permutation and start opening boxes. It may be we are dealing with a permutation in which the prize is in box 4. Opening other boxes will not move the prize: it started inbox 4 and will just stay there. Opening other boxes will not change the contents of those boxes either: whatever was in the a box before opening any boxes will, in fact, be what we see in that box when we finally open it.

What is true that before we start opening boxes we do not know the permutation, and when we start opening boxes we learn some things about the permutation. So, when we open box 1 and find no prize in it, we now know that the prize must be in one of boxes 2--7, so now only 6! permutations remain, and the posterior probability that the prize is in box 4 is now 1/6, etc., etc. However, none of that changes the fact that before we started opening boxes there is a 1/7 chance it will eventually found to be in box 4, and opening other boxes will not move the prize. It will not change the 1/7 either, because that is a prior probability that applies before we open boxes.

 

[JohnnyGui]

I think you have it backwards and are just confusing yourself. We generate a permutation of {1,2,3,...,7} at random; then we fix that permutation and start opening boxes. It may be we are dealing with a permutation in which the prize is in box 4. Opening other boxes will not move the prize: it started inbox 4 and will just stay there. Opening other boxes will not change the contents of those boxes either: whatever was in the a box before opening any boxes will, in fact, be what we see in that box when we finally open it.

What is true that before we start opening boxes we do not know the permutation, and when we start opening boxes we learn some things about the permutation. So, when we open box 1 and find no prize in it, we now know that the prize must be in one of boxes 2--7, so now only 6! permutations remain, and the posterior probability that the prize is in box 4 is now 1/6, etc., etc. However, none of that changes the fact that before we started opening boxes there is a 1/7 chance it will eventually found to be in box 4, and opening other boxes will not move the prize. It will not change the 1/7 either, because that is a prior probability that applies before we open boxes.

I did not see it that way. What I tried to do is calculate the probability of 1 specific permutation in which the prize is at a fixed position, and then multiply that probability by the number of possible permutations for the empty boxes before the box with the prize. I edited my last post and showed that the number of permutations for the empty boxes is not $3!$ but $6 \cdot 5 \cdot 4$ if the prize is in the 4th box, which indeed gives me a probability of 1/7.

 

[WWGD]

I am a bit confused. Isn't this a geometric distribution with p=1/7?

 

[JohnnyGui]

I am a bit confused. Isn't this a geometric distribution with p=1/7?

Yes, but we were also trying to deduce it using the method of permutations.

EDIT: No, see the next post.

 

[StoneTemplePython]

I am a bit confused. Isn't this a geometric distribution with p=1/7?

Yes, but we were also trying to deduce it using the method of permutations.

No.

This problem involves sampling without replacement. A redflag is given by the fact that the expected number of cards drawn is $= 3.5 =\frac{7}{2} \neq \frac{1}{p} = 7$

 

[JohnnyGui]

No.

This problem involves sampling without replacement. A redflag is given by the fact that the expected number of cards drawn is $= 3.5 =\frac{7}{2} \neq \frac{1}{p} = 7$

Wasn't he asking about the probability itself, which is indeed 1/7? The $3.5$ appears to be the mean and has nothing to do with the probability as cleared by post #6. After that we were deriving the probability that WWGD was asking about using permutations, which indeed also shows that $p=\frac{1}{7}$.

 

[StoneTemplePython]

Wasn't he asking about the probability itself, which is indeed 1/7? The $3.5$ appears to be the mean and has nothing to do with the probability as cleared by post #6. After that we were deriving the probability that WWGD was asking about using permutations, which indeed also shows that $p=\frac{1}{7}$.

A probability parameter $p$ completely characterizes the geometric distribution. The issue is: this ain't a geometric distribution. If we were sampling with replacement it would be but we aren't. The fact that the 1st moment of this problem doesn't agree with the first moment of the geometric distribution is... a giant red flag.

 

[JohnnyGui]

A probability parameter $p$ completely characterizes the geometric distribution. The issue is: this ain't a geometric distribution. If we were sampling with replacement it would be but we aren't. The fact that the 1st moment of this problem doesn't agree with the first moment of the geometric distribution is... a giant red flag.

My bad, I misunderstood the definition of geometric distribution. I have edited my answer to direct to yours.