Jump to content
BrainDen.com - Brain Teasers
  • 0
Sign in to follow this  
Kradec na kysmet

The Pebbles

Question

We are in a TV game. We have 8 same looking boxes, each of which has 2 pebbles. Each pebble could be either precious or not precious. We choose a box and the host, without looking inside the box, pulls out a pebble. It comes out to be precious. The TV host then declares that we have exactly 50% chance that the 2nd pebble inside the box is precious, too.
If we know that, in the beginning of the game, the count of precious pebbles is no less than the count of not precious pebbles, which of the statements MUST be true?
We assume that the TV host knows the pebbles distribution in the boxes - in what amount of boxes the precious pebbles are 0,1 or 2. The TV host, however, doesn't know the kind of pebbles in each specific box, as he himself doesn't differ the boxes from one another.

1. If we swapped the boxes, we would have higher chance of picking up precious pebble next.

2. If we swapped the boxes, we would have same chance of picking up precious pebble next.

3. If we swapped the boxes, we would have lower chance of picking up precious pebble next.

4. In the beggining of the game, there've been exactly 2 boxes with 2 precious pebbles each.

5. In the beginning of the game, half the boxes have had 1 precious and 1 not precious pebble.

6. In the beginning of the game, the precious and not precious pebbles have been equal in count.

7. In the beginning of the game, the amount of boxes with 2 precious pebbles is equal to the amount of boxes with 2 not precious pebbles.

8. The TV host has gone wrong. There's no way that there's 50% chance for 2nd precious pebble in our box, as there is unequal amount of pebbles left in the game.

9. None of the above must be true.

  • Like 1

Share this post


Link to post
Share on other sites

4 answers to this question

Recommended Posts

  • 0

Kradec, here’s my argument:


The host says we have exactly 50% chance of getting P after getting P.

How many ways did we get P?
If we were in PN-a, there was one possibility: he pulled the P.
Ditto for the PN-b.
If we were in PP, there were two cases: (1) he picked P1 and (2) he picked P2
In these four cases:
PN-a: the chance of getting another P is 0
PN-b: the chance of getting another P is 0.
PP(1): the chance of getting another P is 1.
PP(2): the chance of getting another P is 1.

the expected value is 2 out of 4 = 1/2

Share this post


Link to post
Share on other sites
  • 0


my answer:

 

3567 are True.

Argument:

 

The only distribution that has at least 4 Precious, and yields exactly 50% probability of finding a second P is (PP,PN,PN,NN). 

There are 4 possible situations:
We picked P1 from PP
We picked P2 from PP
We picked P from PN
We picked P from other PN

If we swap boxes, then
* if in PP (half the cases), our chance of picking P is 1/2 * 2/3 = 1/3, as we might swap to NP, NP, or NN
* if in PN (half the cases), we will get PP in 1/3 cases, PN in 1/3 cases, and NN IN 1/3 cases, so 1/3 * (1 + .5 + 0)= 1/2

So probability of picking P after swapping is (1/3 + 1/2)/2 = 5/12

Therefore 1 and 2 are false, 3 is true.

The rest of the statements are self-evident given this distribution.

 

 

 

 

Edited by CaptainEd
Correct brsces

Share this post


Link to post
Share on other sites
  • 0
6 hours ago, CaptainEd said:


my answer:

  Hide contents

3567 are True.

Argument:

  Hide contents

The only distribution that has at least 4 Precious, and yields exactly 50% probability of finding a second P is (PP,PN,PN,NN). 

There are 4 possible situations:
We picked P1 from PP
We picked P2 from PP
We picked P from PN
We picked P from other PN

If we swap boxes, then
* if in PP (half the cases), our chance of picking P is 1/2 * 2/3 = 1/3, as we might swap to NP, NP, or NN
* if in PN (half the cases), we will get PP in 1/3 cases, PN in 1/3 cases, and NN IN 1/3 cases, so 1/3 * (1 + .5 + 0)= 1/2

So probability of picking P after swapping is (1/3 + 1/2)/2 = 5/12

Therefore 1 and 2 are false, 3 is true.

The rest of the statements are self-evident given this distribution.

 

 

 

 

How is the chance 50% for 2nd precious if precious is picked in this case.....

Share this post


Link to post
Share on other sites
  • 0
6 hours ago, CaptainEd said:

Kradec, here’s my argument:

  Reveal hidden contents


The host says we have exactly 50% chance of getting P after getting P.

How many ways did we get P?
If we were in PN-a, there was one possibility: he pulled the P.
Ditto for the PN-b.
If we were in PP, there were two cases: (1) he picked P1 and (2) he picked P2
In these four cases:
PN-a: the chance of getting another P is 0
PN-b: the chance of getting another P is 0.
PP(1): the chance of getting another P is 1.
PP(2): the chance of getting another P is 1.

the expected value is 2 out of 4 = 1/2

 

 

My bad, but you missed 4 ( just to add )

Share this post


Link to post
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Answer this question...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...
Sign in to follow this  

  • Recently Browsing   0 members

    No registered users viewing this page.

×
×
  • Create New...