Announcement

**Dauphin** · November 26, 2002, 10:41

Originally posted by The Vagabond

It seems to me the following configuration works out nicely:

A has 2 red stamps; B -- 1 red and 1 green stamps; C -- 2 green stamps. (Or else, A has 2 green stamps while C has 2 red stamps.)

That can't be right...

When A says "no", that reveals to B that he is not GG, and therefore must have at least one R.

B can only see two Rs, therefore says "no". This reveals to C that he can not have a red. Else B would have seen 3Rs and known that he was GR and would have said "yes".

Knowing that he can't be RR or RG, C says "Yes" - he is GG

Still working on the answer...

**One_Brow** · November 26, 2002, 10:59

Originally posted by The Vagabond

We have A:Sagace, B:Vaga, and C:ZTau 3 logicians and friends.

Ming The Moderator takes a set of 8 stamps, 4 red and 4 green, known to the logicians, and loosely affixes two to the forehead of each logician so that each logician can see all the other stamps except those 2 in the moderator's pocket and the two on her own head. He asks them in turn if they know the colors of their own stamps: A: "No" B: "No" C: "No" A: "No B: "Yes"
What are the colors of her stamps, and what is the situation?

It seems to me the following configuration works out nicely:

A has 2 red stamps; B -- 1 red and 1 green stamps; C -- 2 green stamps. (Or else, A has 2 green stamps while C has 2 red stamps.)

Actually, it cold be any configuration where B has one stamp of each color.

Notation: (RR) RG RG GG refers to two reds in the pocket, one each on A, one each on B, two greens on C.

There are 21 total possible combinations.

A no: Eliminate (RR) RR GG GG, (GG) GG RR RR
B no: Eliminate (RR) GG RR GG, (GG) RR GG RR
C no: Eliminate (RR) GG GG RR, (GG) RR RR GG,
(RG) RR GG RG, (RG) GG RR RG (since if he had two of one color, A or B would have said "yes")
Note we have elimiated all cases where both A and B have two stamps of the same color.
A no: Eliminate all cases where B has two stamps of same color, since that would mean A has RG.

All of the following combinations are still possible with the given information after A's second no:
(RR) RG RG GG
(RR) GG RG RG
(RG) RR RG GG
(RG) RG RG RG
(RG) GG RG RR
(GG) RG RG RR
(GG) RR RG RG

**One_Brow** · November 26, 2002, 11:10

Originally posted by The Vagabond
A car smashes into the tree. The driver gets out, looks around, and says: "How great it's halved! Otherwise I'd be dead now".

Question: What did he mean?

Disclaimer: Don't take it too seriously. It's half joke.

Two idea, but neither seems to be a half-joke.

1) The tree was sliced vertically and half ofit removed. The driver's section struch where half the tree should have been.

2) It was a very narrow, reinforced tree. Instead of stopping the car, it split the car in two. Since the driver wasn't wearing his seat belt, he would have been killed if the cr had been stopped completely.

**Dauphin** · November 26, 2002, 11:24

On the joke angle, I figured it had something to do with his car insurance and it "killing him" to pay for repairs if he didn't have the insurance.

**Zero-Tau** · November 26, 2002, 11:43

Originally posted by Sagacious Dolphin

That can't be right...

When A says "no", that reveals to B that he is not GG, and therefore must have at least one R.

B can only see two Rs, therefore says "no". This reveals to C that he can not have a red. Else B would have seen 3Rs and known that he was GR and would have said "yes".

Knowing that he can't be RR or RG, C says "Yes" - he is GG

Bad logic here. When A says no, B learns that he is not GG. However, because C does not know what he himself has, he cannot know that B is aware that he isn't GG. So C cannot exclude GR as a possibility for himself.

One_Brow: Yep, all the combinations you listed are possible after A's second "no", but not in all of them is it possible for B to figure out his own stamp immediately afterwards. That would be a further reduction of the number of combinations.

**One_Brow** · November 26, 2002, 12:31

Zero-Tau,

The key is that when C says "no", both A and B know that at least one of them, perhpas both, is RG. On any combination where both A and B are uni-color, either A or B will know the answer or C will determine that he is RG. Since A does not know for sure that A is RG, B can not be uni-color, and so knows that he is RG.

Originally posted by Sagacious Dolphin

That can't be right...

When A says "no", that reveals to B that he is not GG, and therefore must have at least one R.

B can only see two Rs, therefore says "no". This reveals to C that he can not have a red. Else B would have seen 3Rs and known that he was GR and would have said "yes".

Knowing that he can't be RR or RG, C says "Yes" - he is GG

I assume you are referring to the case
(RG) RR RG GG.

A says no: B knows he has a red. C does not know that B knows he has a red. C knows he could be RG or GG. C does not know the difference between:

(GG) RR RG RG
(RG) RR RG GG

If the first case were true, B would not know that B had a red, for B would still see (RG) RR GG RG as a possibiltiy, and could not choose between that and the actual distribution.

**Guest** · November 26, 2002, 14:21

One Brow is right...
All the configuration with B having RG and only those are right.
Whats interesting, is that if I told you in the problem that the solution for B exists and is unique, you could have guessed immediatly that it is RG, because by symmetry, if B has RR, then if we switch all the red and greens, it could guess GG.
Now Ill leave you a few hour to make a new one...
I dont mind posting more, though then I cant answer them and in a way I feel this has become Lul's Brain Teaser Thread

(Of course I dont mind going on, if you guys like those, just tell me if I should raise the level of diff or lower, or ask more a certain kind of question, I like to ask questions I believe anyone can answer, not just those with a math or other training.)

**Dauphin** · November 26, 2002, 14:55

I've got one, that requires a bit of British Rail knowledge...

Which British train station is furthest from the area that it is named after?

**Paul Hanson** · November 26, 2002, 15:35

Waterloo?

**Dauphin** · November 26, 2002, 15:43

**Zero-Tau** · November 26, 2002, 17:19

Mmm.. time for a new one.

Take a look at the chess position below. It is white's turn. What were the last 5 moves?

Attached Files

**Dauphin** · November 26, 2002, 17:50

If black moved last, all I can say is that he didn't move his king, and unless the board is being deceptive he didn't move either of those pawns.

I guess the board is being deceptive by making you think that black started at the top as we see it.

**Zero-Tau** · November 26, 2002, 17:58

Black did start at the top, and he did move his king last. Yes, it is possible (albeit tricky).

**Dauphin** · November 26, 2002, 18:01

En Passant?

**Zero-Tau** · November 26, 2002, 18:03

Yep. Now you should be able to figure out the rest.

Announcement

Brain teaser thread

Comment

Comment

Comment

Comment

Comment

Comment

Comment

Comment

Comment

Comment

Comment

Comment

Comment

Comment

Comment