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Tue 7 Jun, 2011 02:23 pm
I will post a new riddle once the previous one has been answered. These riddles are tough, but once solved, will be appreciated much more.
There is line of 20 prisoners and each will be given a hat. The hat color is either red or black and the number of each color hat is random meaning there can be 20 red hats, 0 blacks, vise versa and anywhere in between (1r 19b, 2r, 18b...and so on). The last prisoner (#20) can see everyone's hat in front of him but cannot see his own. The 19th guy can see 18 hats and so on. The first guy sees no ones hat. When asked what hat color he has, the prisoner must answer in 3 seconds and can only say red or black without varying his voice, giving any sort of verbal or non verbal cue (coughing, sneezing, raising his voice etc.) If he answers incorrectly, he will be shot immediately. #20 is first to be asked, and everyone can hear what each prisoner answers. Also, the prisoners are given 1 hour beforehand to discuss a strategy. How can you guarentee the survival of 19/20 prisoners.
Assuming that they go along the line, starting from the back (you didn't specify an order so I'm assuming) the guy at the back may or may not get shot as he's guessing his own colour. However, the one in front of him is wearing a red hat so the back guy answers "red", giving the next guy his answer, and so on and so on.
@Old Goat,
Then what happens if guy #20 says red, because that's the color of #19's hat, and #19, wanting to live, say red, but #18 hat is black. #18 may say the wrong color now.
@chai2,
Bugger!
Back to the drawing board.
@Old Goat,
GOT IT!
They smuggle viagra in to the prison an hour before the test, and give one subtle prod for red and two for black.
Yes?
@Old Goat,
seems to me the person in from of that one might be risking a shot of something anyway.
in any event, it doesn't say that all the prisoners are men, does it?
yes, subtle thrusting motions.
How about if during the hour of strategy, they form 2 lines facing each other, then someone will ask "If the person facing you is wearing red hat, look down.
Then you would know the color of your hat.
@youlikeit,
Why not make it harder for the miscreants by putting a number from one to twenty on the hats and making then specify the colour of the hat and whether the sum of the numbers on the remaining hats is odd or even?
They look at their hat during their strategy hour.
@laughoutlood,
Some girlfriends and I once tried to play backgammon with dungeons and dragons dice.
Didn't work.
@youlikeit,
simple, you said 19/20 so you only need to guarentee 19, so if you start with 20 he just needs to see the hat in front of him. That would give you 19 at the end.
@youlikeit,
The strategy involves parity (even vs. odd), and everybody after #20 is guaranteed to survive. #20 has a 50/50 chance. Let's say, for instance, that #20 will say red if he sees an even number of red hats in front of him, black if he sees an odd number of red hats.
If #20 says red, #19 says black if he sees an even number of reds, and red if he sees an odd number of reds. #18 now knows the parity of red hats on prisoners 1-18 by taking into account what #20 and #19 said. It continues like this with each prisoner having to keep track of all that goes on before him.
@markr
good job, that's correct. someone told me there was a way to get 20/20 but I dont see how thats possible. anyhow, heres the next riddle.
You have 12 marbles and a scale (with 2 sides, the heavier side goes down and the side with less weight rises). 11 of 12 marbles are identical in weight, and all 12 are identical in appearance. You can use the scale 3 times (for example,weighing 6 against 6, 3 against 3, and finally 1 against 1). The one marble of different weight is EITHER heavier OR lighter than the others. Under these conditions, how can you determine which marble is of different weight, and whether it is heavier or lighter.
@youlikeit,
I solved this years ago with a decision tree (successive weighings were based on the outcome of the previous weighing), but here's a way to do it with pre-determined weighings.
http://www.iwriteiam.nl/Ha12coins.html
@markr,
yeah, but you used a website -_-
@youlikeit,
There was no such thing as a web site when I solved it. As I stated, my method was different then the one outlined at the link. It would be difficult to document clearly. The link is for others to check out if they are interested because of the neatness of the solution.
k heres the next one, im running out though, so if anyone would like to post their own, thatd be fun.
a couple has 2 children. the first is a boy, what is the chance that the other child is a girl? (its not 50%)
@markr,
you mean than but it should be to
i'm glad it wasn't a boy born on wednesday that'd be too tricky
and you've butchered your question likeit
@laughoutlood,
When you toss a coin why is it more likely to show tails than heads?
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