chinese emperor

[WorkHorse]

chinese emperor aha:!


A chinese emperor had to choose a new adviser amongst 3 sages, all of them equally wise. He placed a problem to them: "To choose one of you, you'll play a simple and fair game: In this sack there are 3 white balls and 2 black balls. Each of you will be blindfolded and will pick one ball and place it on your head. After that, the blindfolds will be removed and each one in turn will try to guess the colour of the ball upon his head, by observation of the other picked balls. However, beware. You may pass your turn in guessing, but if you state a colour and fail, you're disqualified. This way I'll learn which one is the most intelligent amongst you" The sages talked briefly to each other and promptly refused: "Dear lord, it's of no use, since the game is not fair. The last one of us to guess in the first round will know the answer." and the sages promptly demonstrated this to the emperor, who was so amazed by their wits that he appointed all 3 has his advisers. Could you demonstrated it ? NOTE: If the emperor had any wits at all he would have named them all advisers in the first place... maybe spending reduction ?

[jemidiah]

A picks white
B picks white
C picks black

A guesses first, tries black, is disqualified
B guesses next, tries black, is disqualified
C guesses next, tries white, is disqualified

They all lost. Did I miss something?

[WorkHorse]

Originally posted by jemidiah
Did I miss something?

Yes. I think we need to make the assumptions that 1) All the sages want to be advisors. 2) They will answer only if they are certain that they are correct--otherwise they pass. I'm not sure if that is how the question really works, but I think that it can be worked and it makes it more interesting. I'll let people play with it a bit before I give the solution I came up with.

[jemidiah]

Hmm. Can they pass to the point of starting the game over?

[WorkHorse]

Originally posted by jemidiah
Hmm. Can they pass to the point of starting over?

Nope. The guesses go to Sage 1, then Sage 2, then Sage 3. When Sage 3 has answered (if one of the Sages has not already won), one sage will have answered correctly with 100% confidence.

[jemidiah]

The only thing I've thought of is that two of them must pick black so that the third will know his must be white, though I'll noodle it for a while...

[WorkHorse]

A Sage will pass if he is not 100% confident that he is correct. No guessing involved.

[Something Else]

If the 2 black are picked, then the 1 with white on his head will immediately know he's got white on his head.

Hmm, thats about as far as I got right now.

hmm:

The last one of us to guess in the first round will know the answer

all the balls are picked, the blindfolds are removed, and contestent #1 comes up to the podium...
Regis, I'm not really sure, but since #2 has black and #3 has black, then the odds are I've got white on my head???



Hmmm, I think that disqualifies him, even if he got it right?

How about:
The Case:
Contestants #2 and #3 both have white on their heads.
Who cares about what Contestant #1 has on his head at this time.

all the balls are picked, the blindfolds are removed, and contestent #1 comes up to the podium...
Regis, I'm not really sure, but thinking to himself, since #2 has white and #3 has white, the odds are I've got Black on my head, so I'll say: is there a black ball on my head?

Now, at this time, #2 and #3, knowing #1 would guess the ball that is most likely, knows that they don't both have black balls on their heads. They must have some arrangement of 1 black and 1 white, or both white. If they have 1 black and 1 white, then that would leave 1 black and 2 white. If they were contestant #1, they would know it would be more likeley that #1 had a white on his head, if #2 and #3 had a white and a black, but he guessed black instead, which means they both have white on their heads, which would have left for num1 2 black and 1 white, which does make it more likeley that theirs a black on his head.

So, if Num1 is wrong, then the first one chosen to answer between #2 and #3 will answer correctly.

Hmm, so:
Case #1: Contestant #1 Sees 2 Black Balls, Guesses White, Wins.
Case #2: Contestent #1 sees 2 white balls, Guesses Black, either
wins, or loses, and Contestent #2 wins. {See Above}

Case #3: Contestent #1 sees 1 white and 1 black. That leaves 2 white and 1 black. Guesses white. Either wins or loses. If lost, #2 knows what #3 and #1 have on their head, and knows the guess so far was white. Ok, so at this time, #2 is up, #1 has black on his head, #2 sees what #3 has on his head. If #3 Has Black on his head, #2 Knows he's got white on his head. If #3 Has white on his head, then that leaves 2 white, 1 black. So the odds are #2 has white on his head. However, that would mean #1 should have guesses... wait a sec, I've lost it. is it 3 white and 2 black, or is it 3 black and 2 white? Darn this edit mode!

[jemidiah]

Nice reasoning, though it was said that none would guess. Hmm, what could it be???

[NotLKH]

Originally posted by jemidiah
Nice reasoning, though it was said that none would guess. Hmm, what could it be???

The last one of us to guess in the first round...

Are you sure?
Doesn't this sound like guessing was allowed?

[WorkHorse]

Sorry. I thought about it some more and it the Sages can guess. In fact, being Sages they WOULD guess because there is no penalty involved with being wrong and they know that by the end of the round a Sage will have been able to answer correctly. So if a Sage doesn't know for sure, he might as well take a guess, but unless he gets lucky and gets it right (which would end the game) it doesn't matter whether he guesses wrong or passes.

I think we need to assume though that the Sages are smart enough to always guess as best they can. And no fair with an aha!!!!! answer like "there are only two black balls so one must be white, so very Sage answers white and therefore one must be right." As I said, I think we must assume they want the job and are smart.

I'll look at Lou's answer to see if it is the same as mine in different words.

[WorkHorse]

I think I like it better with my previous requirement "sage only answers if he knows for sure". It's still possible with that limitation. But, Lou's idea is neat too. I'm pretty sure that if we use the "sage always guesses based on probability" (which makes sense considering they are sages) that by the end of the second ANSWER one sage would have guessed correctly. Sage three doesn't have a chance. Them sages may not have been such smarty pants after all.

[WorkHorse]

Nope. Lou, your solution doesn't work. Here's the flaw:

Originally posted by Something Else
Regis, I'm not really sure, but thinking to himself, since #2 has white and #3 has white, the odds are I've got Black on my head, so I'll say: is there a black ball on my head?

This is one of those things where it SEEMS more likely to be black, but it isn't. I think it is like the "gold behind one door, pick a door, a door without the gold is opened, now guess the correct door" thing where it SEEMS like you end up with 50/50 odds, but you don't. If Sage 1 does not see two black balls, that option is eliminated from from the probability of remaining balls, which makes it equally likely that the ball on his head is either black or white.

If Sage 1 does not see two black balls (if he did, he would know for sure that his is white) then he can only make a 50/50 guess--which means that subsequent Sages can't base their guess on his because it was only a blind guess.

And it does work with a "stupid Sage", provided that the sage knows enough to always answer correctly when he knows for sure what is on his head (in other words, a Sage can't "bluff out" another sage by giving an intentionally "against the odds" answer when he doesn't know for sure). It doesn't matter what answer another sage gave. If a Sage does not "know for sure", he's just making a blind guess with even odds.

[Something Else]

However, its not wether the first, or even the second, get it right. its that, given that two have guessed before the third, that, being the third, and given the observations that he has, in combination with the guesses based on probability that the previous two have given, that there is no way an educated sage will even be guessing, but will give THE correct answer, being in position #3.

In Other words, more than enough info has been given, even if the first two were wrong in their "guesses", that there is no way that the third sage could be wrong.

I'll see about my guessing technique in the morning, se if I get it straightened out before I go off to work.

-Lou

[Something Else]

Originally posted by WorkHorse

meOriginally posted by Something Else
Regis, I'm not really sure, but thinking to himself, since #2 has white and #3 has white, the odds are I've got Black on my head, so I'll say: is there a black ball on my head?

Nope. Lou, your solution doesn't work. Here's the flaw:



This is one of those things where it SEEMS more likely to be black, but it isn't.

Hmm. Ok. Lets see. 3 white, two black. we see 2 white have been drawn. that leaves 1 white and 2 black. Hmm. The odds are, given 2 remaining black and 1 white, 2/3rds chance that I have a black.

I think 2/3rds black is more likely than 1/3 white. How is this wrong?

[WorkHorse]

Originally posted by Something Else
I think 2/3rds black is more likely than 1/3 white. How is this wrong?

I think the error there is that your probability includes the possibilty of 2 black on the others heads, which would have already been excluded. But I think I was in error saying that that results in equal probability. It is 3/5 more likely to be white. (9/15 compared to 10/15 (2/3) because the double black option is not excluded only from the possible combinations, but from the total number of possible combinations). So I don't think you get 2/3, but you do get a more probable option. So I'm going back to your solution, which I think means that by Sage 2 the correct anwer willbe given (the non-smarty-pants option). Although, it is far more complex and less elegant than really needs to be given to the original problem (but perhaps more brilliant).

[kedaman]

Possible ordered outcomes:
WWWBB
WWBWB
WWBBW
WBWWB
WBWBW
WBBWW
BWWWB
BWWBW
BWBWW
BBWWW
Lets take the three first and give then to Sage 1, Sage 2 resp Sage 3.

3/10 Sage 1 sees WW chooses B
..2/3 Success
..1/3 Die -> Sage 2 chooses W Success (Knowns sage 1 Saw 2 White)
6/10 Sage 1 sees WB chooses W
..2/3 Success
..1/3 Die Sage 2 chooses opposite colour to Sage 3 Success
1/10 Sage 1 sees BB chooses W Success
Sage 1 skips in case WW and WB:
WWWBB (outcomes ordered in Sage 2 POV)
WBWWB
WBWBW

WWBWB
WWBBW
BWWWB
BWWBW
BBWWW

BWBWW
..1/3 Sage 2 sees WW Chooses B
....2/3 Success
....1/3 Die -> Sage 3 chooses W Success
..5/9 Sage 2 sees WB Chooses W
....4/5 Success
....1/5 Die -> Sage 3 Chooses W Success
..1/9 Sage 2 sees BB Chooses W Success
..Sage 2 skips seeing WW -> Sage 3 chooses W Success
..Sage 2 skips seeing WW and WB:
WWWBB (outcomes ordered in Sage 3 POV)
WWBWB
WWBBW

BWWWB
BWWBW
WBWWB
WBWBW
BBWWW
....3/8 Sage 3 sees WW chooses B
......2/3 Success
......1/3 Die -> Sage 1 chooses W Success
....5/8 Sage 3 sees WB or BB chooses W Success
....Sage 3 skips seeing WW -> Sage 1 chooses W Success

Lets backtrack to see how smart it is for Sage 2 to skip WW and WB:
Sage 2 has no chance of winning by skipping, thus no skipping WW and WB
Lets backtrack to see how smart it is for Sage 2 to skip WW:
Sage 3 wins, stupid, no skipping.
Sage 2 will thus not skip:
..1/9 + 1/3*2/3 + 5/9*4/5 =7/9 Sage 2 Success
..2/9 Sage 3 Success
The Game cannot go further from Sage 3.
QED.

[kedaman]

btw, Sage1 will not skip either for the same reason, and thus the game will end with Sage1 either knowing one of ten times, guessing correct nine of fifteen, Sage 2 winning nine of thirty times. Very unfair. Sage 3 will know this of course, but oh well..

[fundu]

The method is very simple if we assume that no one will guess. That is they will pass if they cannot make out by the information they already have by then. Thus even if the 1st two cannot make out the 3rd will do certainly.

I am running short of time so will write the method in detail within next 6 hours

[wossname]

I think we are missing the fundamental question...

Who was the poor white guy who had his 3 balls removed?

I dare say the black dude wasn't too chuffed either.

[fundu]

Lets suppose that no one will guess and all the three sages are very intelligent. Now lets start our approach to the problem.

1st Sage's Turn:
If the other two sages have black balls then as the total number of black balls is two, sage1 can certainly tell it white & game over.
If not, what are the possibilities, either both have white balls or one has white and one has black ball. I hope you all will agree to that. In these two cases the 1st sage will have 2 pass.

2nd Sage's Turn:
As the 1st sage passed and all these sages know that everyone is intelligent and loyal, he can certainly conclude the result in the last para above that is either WW or WB combination.
Now suppose he sees that the 3rd sage has a black ball, then with the info that 1st sage passed and so oth of them have atleast one white ball he will conclude that he has a white ball. Note in this case even there is no need for him to look the color of the ball of the 1st sage. I agree if it is also black then no need of this but even if it is white, he can conclude the same result.
Thus the problem comes only if the 3rd sage has a white ball. In this case he can have either black or white. Even looking the colour of the 1st sage's ball won't help him. Thus he will have to pass.

3rd Sage's Turn:
He will get a turn only if the 1st two will pass. Now as he is also intelligent as others, he will conclude that the 1st sage passed mean either WW or WH combination. The 2nd passed means he doesn't have a black, that means he has white ball. So he will say that he has a white ball & will win.

I think this is pretty OK & explanatory solution.


The problem was as simple as cutting an apple with a knife.

[WorkHorse]

Yeah fundu! That was the same answer I originally came up with.

There are seven possibilities:


1234567
wwwwbbb
wwbbwwb
wbwbwbw

Because there are only two black balls, if a Sage sees two black balls on the others heads, his must be white. If a Sage sees one black ball when another sage has passed, the Sage can deduce that his ball is not black, otherwise the previous sage would have seen two black balls and known his was white.

Sage 1:

If both others are black, then Sage 1 is white. Possibility 4 is eliminated.

Sage 2:

If both others are black, then Sage 2 is white. Possibility 6 is eliminated.
If Sage 3 is black, Sage 2 can’t be black or Sage 1 would have guessed white. Therefore, Sage 2 is white. Possibility 2 is eliminated.

Sage 3:

If both others are black, then Sage 3 is white. Possibility 7 is eliminated.
If Sage 1 is black, then Sage 2 would have guessed white if Sage 3 is black. Therefore, Sage 3 is white. Possibility 5 is eliminated.
If Sage 2 is black, then Sage 1 would have guessed white if Sage 3 is black. Therefore, Sage 3 is white. Possibility 3 is eliminated.
The only possibility left is 1, in which case all are white. Therefore Sage 3 is white. Possibility 1 is eliminated.

Any Sage that wins must answer white. A Sage can never answer black because there are more at least as many Sages as there are white balls. So seeing a white ball on the next sage is never helpful. Therefore, Sage 3 can always answer white--because only if his were black could the other sages have known for sure the correct answer.

[WorkHorse]

chinese emporer2

I put this problem in a new thread http://www.vbforums.com/showthread.p...hreadid=275136

[kedaman]

that fundu and WorkHorse is shortsighted, if sage1 knows all outcomes of intelligent choice by the others, why would he skip in the first place? I think Sage3 will never get a chance because Sage2 will end up knowing based on Sage1's guess.

[alkatran]

Ok, we start with sage 1. He can see 1 of 3 different things on the other 2 heads.

bw, ww, or bb
If he sees BB he knows he has white, anything else he passes.

If we reach sage 2, he KNOWS that either he or sage 3 has a white ball. If he sees a black ball on sage 3's head...

But if he sees a white ball on sage 3's head he's not sure what is on his head, because it could have been WW that sage 1 saw, or BW.

So if sage 2 sees white he passes, and sage 3 KNOWS he has white.



I think that covers all the possibilities.

[fundu]

I completely agree with alkatran.

Thats what I said but was trying 2 explain more as some people really need lots of them.

I can't understand why others think that it will finish before the 3rd sage gets turn

[Nucleus]

I think ked is onto something there.

You can't assume than a sage will skip if he doesn't know the answer. So then picking the best sage would be the one who weighs up the probabilities in his circumstances and makes the best choice, not necessarily the one who actually wins the game.

A better way to work this out would be to ask each sage, what they would do under all the possible senarios.

[WorkHorse]

Hi Nucleus. Haven't seen you around for a while.

I was just posting some puzzles that I didn't think were very good. IMHO, the question should have had a condition where the sage is fed to the lions if guessing correctly. Perhaps the key is that the sages had to explain this to a king. If you have ever worked for a "non-technical" boss, I think you would agree that the solution that requires skipping but still results in an answer by the third guess is the "easier" solution.

Actually, they could have just explained that the contest is unfair because the first sage has a 50% chance just by guessing, which means that Sage 1 has a great advantage. In fact, if they all bluffed, Sage 1 could always just answer the opposite of his first guess. So it has to be answered by guess 4, in which case Sage 1 always wins. Unless they are all completely daft, in which case it comes down to just random chance. Might as well draw straws.

Furthermore, they could have explained that the test of any sage would be based on the "guessing sage" making the assumtion that the prior sage had made the best possible guess. That is, that the prior sage is as smart as the "guessing sage". Which means it only works if all the sages are equally smart. Therefore, th challenge begs the question of who is the smartest sage. Any selection is really only based on sage-guessing-position and luck of the draw. The question cannot determine the intellect of the sage. You supid king.