View Full Version : An interesting probability problem

06-09-2003, 01:11 AM
You are one of three prisoners. You cannot communicate with your compatriots in any way, but may assume that the other two are brilliant, i.e. they will do the right thing. You are each outfitted with a hat - which you cannot see. The warders have an infinite supply of red and black hats, so you know your hat must be one of those two colors. The three of you are led to a courtyard, to view each other's hats (sort of like that poker game where everyone holds a card to their own forehead). You are led back to your cells, and asked to state what color hat you are wearing. You may choose red, black, or "abstain." Warning, if all three abstain you will all be killed. Also, if any one of you guesses your color incorrectly, you will all be killed. You do not know beforehand the purpose of going to the courtyard, and you are all questioned simultaneously while separated from one another. What is the best strategy?


06-09-2003, 02:38 PM
I'll take a stab at this. If you see your fellow prisoners have different color hats you abstain. If you see 2 of them with the same color hat (such as black) you guess the other color (red).

06-09-2003, 03:07 PM
I must be missing something. If you are not given the game until after you return from your walk, the info is basically useless. The fact that there is an infinite supply of each color means the probability does not change based on the color hats they are wearing.

Now I do not think you can do better then have me guess (always black, always red or flip a coin) and have the other two always abstain. That will give us a 50% chance of living. However, this is not really plausible. How genius are the other two. Do they know they should abstain so that I can make the guess? If they do not you may all want to guess which means that you live only (1/8) of the time.

I am sure that this makes no sense but what can I say.

06-09-2003, 03:37 PM
Congrats, Jimbo! If I ever get thrown in the clink I want you there with me, we might get out alive. I will leave it open as to why all the prisoners should logically reach the same conclusion.

06-09-2003, 03:43 PM
gilly, sorry but they cannot confer and decide to make you the guesser. They can however come up with a better strategy through logic. See Jimbo's reply and you may figure out why.

06-09-2003, 05:37 PM
I liked that one. Got any more like it?

Oh, and by following Jimbo's solution, the prisoners have a 75% chance of living, which is the best they can get. They're screwed if they all have the same color though.

06-09-2003, 07:09 PM
Hello. Your puzzle is very interesting and I admit I cannot solve it. However, I know one that is very similar.

A king has three genius advisors and wants to marry one of them off to his daughter, who is ridiculously hot and tolerant of boyfriends spending large amounts of time playing poker with their buddies. The three advisors have their eyes blindfolded and hats put on their heads. They are told that their hats were chosen from a group of two red and three black hats, and then their blindfolds are removed. The first one to correctly guess the color of his hat gets the girl, but an incorrect guess means death. After a while, the first advisor says he passes, followed closely by the second one. Immediately the third advisor says "I'm wearing a black hat". He was correct and was able to explain why he was certain and a date was set. How did he know?

06-09-2003, 07:30 PM
That is a good puzzle but I have heard it before, so in the spirit of fun I will leave it for someone else to solve. I do have another puzzle that is somewhat difficult and that took a bit of time for me to solve that I will put up soon.


06-09-2003, 07:33 PM
LB, right you are. I do have another one that drove me crazy for a couple of hours - I will post it soon.


rusty JEDI
06-11-2003, 02:51 AM
I will leave it open as to why all the prisoners should logically reach the same conclusion.

Who do you think you are? Sklansky?


06-11-2003, 01:46 PM
Of course the other 2 were wearing red hats.

06-11-2003, 03:16 PM
Dragon and a knight get bored of each other and decide for entertainment reasons they'll try to kill each other, hence kill their boredom or boredom kill them.

There are 7 wells, all numbered, which all contain magic water. If you drink from a well then drink from a well with a lower number you die. If you drink from a well then drink from one with a higher number you live.


Drink from 1, drink from 2 and you live.
Drink from 7 then 6 and you die.

They agree to use this system as their fight, they both bring water and then they go away to drink again.

Because the dragon can spit fire and the knight can't argue with that only the dragon can use well 7.

How can the knight live and kill the dragon?

06-11-2003, 05:22 PM
No, that's not it. If that were the case, he wouldn't have had to wait for the other two to pass. I'm sorry, but I should have made that clear from the get go. In fact, to make that point clear, the puzzle is sometimes told with the third advisor being blind and thus unable to see the hats at all. But he still guesses correctly, and by using the same logic.

06-12-2003, 06:51 PM
Here's what I have:

1) If two read hats are out, then one of the first two guessers will most definitely say black for obvious reasons. Therefore, 1&3 or 2&3 are not wearing red hats together. BUT, this doesn't preclude 3 from wearing a red hat while the other two wear black (see the following points as a rebuttal).

2) If 3 wears the red hat, then 1 would have to pass because he sees one red, one black. BUT, the second guesser has to assume he wears a black hat because the first guesser passed (see point #1). Moreover, by passing, he reveals that the first guesser has a black hat assuming 3 wears the red hat. Therefore, it is likely that player three can infer that he is wearing a red hat. But he would never had a chance to guess because player two would have guessed black.

3) If player 2 wears a red hat, then the situation is questionable. Here, player one passes because he sees one red one black. Player 2 can guess red, but he could easily guess black; no information was revealed by player 1. But by passing he reveals considerable amounts of information to player 3 (see point #2).

4) Similar thinking applies when no red hats are worn. Player one passes because there is still one black that could be worn as opposed to two red hats. Player two also passes using the same logic. But as before, player 2 revealed significant amounts of information to player 3.

The information revealed by player two (pts #3 & #4) "tells" player 3 that he wears a black hat because he actually has a chance to guess. Therefore, player 3 guesses correctly, gets the girl, and has a wild night...

06-12-2003, 07:01 PM
If you drink from a well then drink from a well with a lower number you die. If you drink from a well then drink from one with a higher number you live.

Must you drink from two different wells? What happens if you drink from the same well twice?

They agree to use this system as their fight, they both bring water and then they go away to drink again.

Can you clarify? I read this as bring one "cup", sit and drink, get another "cup", sit and drink, etc...

OR, is it bring a "cup" for your opponent, they drink, then you get another cup for your opponent...

06-15-2003, 08:15 PM
Drink the same well you die.

You bring water for the other person then they drink from whichever they wish.

06-15-2003, 10:30 PM
I just don't see an answer for this one. From how I understand the game, the knight and dragon bring water for the other. After drinking that water, each goes back to the wells for a second drink. If this was true, then the dragon would bring water from well 7 for the knight and always drink from well 7 as a second drink. Therefore, the knight will always die and the dragon will always live.

If my understanding of the game is incorrect, I'd like to know. This is an intriguing problem...

06-16-2003, 10:40 AM
Your understand is prefectly right. This problem is better spoken because when i was typing i had to be pretty sure of what i was typing so people can't go back and quote and say how badly worded i did it.

So there is your clue.

Code Red
06-17-2003, 12:32 AM
By looking into the reflection in the eyes of his opponents he was able to determine the color of his hat. Or both of his opponents hats are red, making his black automatically.

06-17-2003, 07:39 AM
After several moments of thought, my guess is going to be this is less of a probability riddle.

Now this is only a guess and I could be wrong. You said they "go away". While they are "away" the knight brings a cup of "unmagical" water for the dragon to drink, and the dragon obviously gives the Knight the magic water #7, considering theres no higher well, and expecting the knight to die because he cannot be cured.

the dragon proceeds to drink the unmagical water, then drink from well #7 to "cure" himself.

if the knight drinks at all (?) he must first drink from a lower #'d well, and then he can cure himself with #7 but given the dragon will soon be dead, i dont think the knight has to drink anything?

06-17-2003, 12:32 PM
Yep. That's right. Knight drinks from any 1 - 6 well before going, and gives the dragon normal water.