**Edited by cheyenne 09, 26 October 2006 - 10:03 AM.**

# Riddle Me This

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#211
Posted 26 October 2006 - 09:57 AM

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#212
Posted 26 October 2006 - 10:29 AM

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#213
Posted 26 October 2006 - 10:46 AM

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#214
Posted 26 October 2006 - 01:41 PM

the changing of pitch will make everyone know, if they are able to discuss for instance yell the color of your hat if the persons hat in front of you is white, vice versa for black, and the person in teh back will just tell the person in front of them what color hat they are wearin as a sacrifice

isnt that almost exactly what i said? :-s

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#215
Posted 26 October 2006 - 01:47 PM

***dsenette**suggests a "no more what color hat am i wearing riddle" rule...that's like the 4th one in this thread alone...the hat riddles always get SOOO frustrating and are often either misposted...poorly worded...or otherwise almost impossible to accurately solve...after this one...please...no mor hat riddles

**Edited by dsenette, 26 October 2006 - 01:48 PM.**

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#216
Posted 26 October 2006 - 03:29 PM

so we can get back to posting some good Riddles and some Funny ones and maybe some Different ones as well Thank you to everyone for there patients.

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#217
Posted 26 October 2006 - 03:54 PM

Simple, just have each person, starting from the back, recite the color of the guy in front of them. I.E. back guy tells second from back he has a white hat, etc. The only one unable to inform anyone of a color is the man in front, but he receives his color from the man behind him.

Seems pretty straightforward to me. The only one left guessing is the first man and good luck to him.

(If I got this right, I pass my turn to do a riddle to Cheyenne, since I don't check here as often as I should)

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#218
Posted 27 October 2006 - 12:54 AM

**Edited by c7ng23, 27 October 2006 - 01:12 AM.**

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#219
Posted 27 October 2006 - 07:39 AM

they can only say one word...so if they call out the color of the hat in front of them....they just killed themselves because they didn't say the color of THEIR hat...they just said the color of someone elses hat (i.e. guy in the back sees a black hat so he says black...he's dead if his hat isn't black too....2nd guy now knows his hat is black...and let's say..the guy in front of him has a white hat...well...if he says white...then h e dies because he's got a black hat on...nto white hat...etc..)Simple, just have each person, starting from the back, recite the color of the guy in front of them. I.E. back guy tells second from back he has a white hat, etc. The only one unable to inform anyone of a color is the man in front, but he receives his color from the man behind him.

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#220
Posted 27 October 2006 - 07:45 AM

Suppose 100 prisoners are in a line. Each prisoner is given a hat, but are not allowed any form of communication after they are given the hats (no blinking morse code or anything like that). They are standing in a line facing front and can see all the hats of the others in front of them, but not their own or anyone behind them. They are told that the probability of the hat they are wearing being black is 2/3, and that the remaining hats are white and that each hat was independently randomly chosen with these odds. Now each prisoner will be called up in the order of the line starting with the one in the back (who can see 99 other hats in front of him/her), one by one is forced to guess which color the hat on their own head is. If they guess correctly they live, if they guess incorrectly they are executed on the spot right then and there. As the procedure progresses the others can hear the guesses of those behind him/her and whether or not they guessed correctly.

Ok, assuming that the prisoners are allowed to collude and strategize before they are given their hats (they know of the pending situation before it takes place, but cannot communicate at all once the hats are given out), what should they do to maximize their probability of their own survival? What should they do to maximize the number of surviving prisoners (is this the same question)? And assuming there is a difference in the two strategies, what if some of the prisoners (with probability p) decides to be selfish and go with a selfish strategy instead of strategy which maximizes the number of survivors?

- In the selfish situation, obviously, they should just pick black. They have a better chance of being correct. So the real question is how to the try to increase the odds of as many following them as possible to survive.

- If a prisoner were being selfless for the benefit of others, he/she could use their guess as a binary communications mechanism instead of a genuine guessing mechanism.

- If each guess only told exactly enough information for the next person to know their hat color, then this strategy could only be used once (one person projects the information forward, and one person receives the information).

- Some set of prisoners must give information about all the prisoners.

- What is exactly the difference in knowledge about all the prisoners between the first and second prisoner to have to guess? What is the most minimum way of encoding this difference in knowledge, in a way that the *difference* between these two sets of knowledge is known to all the prisoners who hear guesses?

- Suppose the first prisoner asked simply gives information whether or not a count of the number of white hats he/she sees is even or odd? Note that this strategy is independent of the actual number of prisoners.

- This second strategy will save everyone deterministically except possibly the first person, who basically has a 50% chance of guessing wrong and dying; so a 99.5% success rate. The greedy strategy basically has a 66.6% success rate. The optimal strategy (from either the selfless or selfish point of view) is to be part of the second strategy, if possible, though it might be out of your control. So lets consider the third case where some prisoners might be selfless, and some selfish.

- As soon as a prisoner can established that some deterministic previous prisoner has given the parity of the number of white hats he/she sees ahead, they can assume they the other prisoners before him/her knows at least as much, and have switched to the selfless strategy thus guaranteeing their own survival.

- How would any given prisoner interpret the possibility that a previous prisoner guessed their color as white and survived? Does a sequence of black guessers and surivivors tell them anything? What if someone prior to them has guessed white but guessed wrong?

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#221
Posted 27 October 2006 - 08:01 AM

there's a bit of a combination of all of the previouisly suggested techniques

As the first prisioner will not be given any indication the color of his hat, whatever color he says will only give him a 50% chance of avoiding the alligators. He can, however, offer a 100% chance of survival to the two prisoners in front of him. If the two prisoners have like colored hats, he can say "Red hat", and if they have different colored hats he can say "Blue hat". The next prisoner can then know the color of his own hat by knowing the color of the prisoner in front of him; and, when he calls out the color of the hat he has deduced his own to be, the prisoner in front of him will be able to deduce his own hat color. This pattern can work for every set of 3 prisoners. If the last set of prisoners is two, then the backmost prisoner of that set can simply identify the color of the hat of the prisoner in front of him. This will give N prisoners, where N is greater than 1, an average survival rate of at least 75% (around 83%, if N/3 is modulo zero).

If the prisoners were able to alter their tones, like having ability to say RED/BLUE and HAT in either high pitch or low pitch. The information the first person can provide to the prisoners in front of him can give them a 100% survival rate. Only the first person risks a 50% chance of being an alligator appetizer.

Depending upon the color of the prisoner's own hat, he can alter the tones to inform the (minimum) two prisoners ahead of him the color of their hat, and do this without compromising his own ability to correctly call out the color of his own hat. An example of this would be, if he knew his own hat were red, he could just call out his hat color. An alternating high-low pitch, "redHAT", could be used to tell the individuals in front of him that their hats are the same color; and a low-high pitch, "REDhat", could provide those two ahead that their hats are different in color. Each prisoner can use this technique to inform the two prisoners ahead.

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#222
Posted 27 October 2006 - 02:23 PM

and....Please Dont Post Riddles that you dont have the Answer to.....as the Point of most Riddles is to solve them or at least Try to solve them and there are Kids using this as well so Please try and post some easy ones from Time to Time. Thank you Everyone

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#223
Posted 28 October 2006 - 01:27 PM

Three men rent a hotel room. The manager tells the men that the total cost is $30. The men each pay $10. They go into their room.

The manager remembered that there was a special that week, and that it was only supposed to be $25. He hands the bell-boy $5 and tells him to deliver the men their change.

The bell-boy could not figure out how to split $5 three ways., so he decides to keep $2 and give $1 back to each of them.

after receiving their change, each of the men have now paid $9 total.....

9x3=27 The bellboy only kept two dollars, so that brings it to a grand total of $29. Where is the extra dollar?

**note, yes, you do have to take into consideration that the men have paid $9 each. **

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#224
Posted 28 October 2006 - 02:02 PM

divide $25.00 by 3

the answer is not 9

we need to find a new source of riddles

<<<EDIT>>> not a slur against your riddle silenced, just stating we need new ideas

**Edited by harrythook, 28 October 2006 - 02:05 PM.**

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#225
Posted 28 October 2006 - 02:04 PM

so out of the 30 dollars 3 went to the men=27

so the 3 dollars isn't missing it is in the 3 mens hands

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