Sunday, August 16, 2009 – Answer to the 3 Prisoner Puzzle, Plus Two New Puzzles
Last Sunday, August 9, 2009, I gave readers the great “3 Prisoner Puzzle”. In a day or two I put the same puzzle up on Youtube.com and direct people to this page to see the answer. However you saw the puzzle, you can see the answer, and two new puzzles, by clicking on the file name below to see the pdf file.
Answer to the “3 Prisoner Puzzle”, plus Two New Puzzles
20 Responses to “Sunday, August 16, 2009 – Answer to the 3 Prisoner Puzzle, Plus Two New Puzzles”
Leave a Comment
Lessons
Pages
August 29th, 2009 at 5:32 pm
I GOT THE SISTER BROTHER THING.
HIS FATHERS SON IS HIM
Its the person who has no brothers/sisters
September 6th, 2009 at 4:28 pm
The three prisoners answer seems paradoxical to me. If Igor’s disk is white, each prisoner will see two white disks. They each have the same information, and we’re told they’re are equally intelligent, so they should all come to the same conclusion at the same time, shouldn’t they? The fact that Igor takes this line of reasoning “one step further” suggests that he’s more intelligent; but if the prisoners aren’t equally intelligent, Igor can’t base his arguent on the assumption that they’ll have reasoned along the same lines as he would or come to any conclusions beyond the obvious (supposing that they don’t see two black disks).
In fact, couldn’t we reason in the opposite way? We’re told there’s one correct answer. We know that Igor alone gets it. The other prisoners are as clever as him but haven’t worked it out. Therefore Igor must have different information from the other prisoners. Since Igor sees two white disks, this means the other prisoners must each see one white and one black disk. Therefor Igor must have a black disk!
September 24th, 2009 at 3:38 pm
dear math hiker, my cousin presented me with an interesting math problem yesterday, and I’ve been trying to solve it ever since: A tree has one-million leaves, if one leaf falls on october 29, 2 leaves fall the next day, three the next and so on… what will the date be when all the final leaves fall, I like math alot, but I’ve never been so challenged before.
October 3rd, 2009 at 10:08 am
I agree with Gail, upon looking at the answer I was faced with the same conclusion that she has reached, therefore the question is flawed. Either they would have all reached the same solution at the same time, or none would have- had their intelligence been equal. The fact that they should have each been of equal intellect therefore means we could have taken the perspective from any of the 3 prisoners had the same information been provided, and thus drawn the same conclusions, though as Igor had the intelligence to adopt the perspective of the others meant that he was above their intelligence and could not have therefore assumed the thinking processes of the others.
October 15th, 2009 at 3:47 pm
Dear MathHiker,
I just want to say thank you so much for your youtube videos, you make so much sense, my math teacher is a bore, but you made so much sense in less then 5 minutes.
Thank you so much.
December 14th, 2009 at 9:59 am
i couldnt find the answer but i dont think the president asked a question, did he? so igor guessed but he was let free.. so unless im missing something i dont really get it??
but i think the answer is white and as i was thnking about my logic i realized his color could also be black.
the story didnt even really make sense to me because the president didnt ask a question. ???
January 21st, 2010 at 1:33 pm
Answering Gail’s question, they will all have fallen off in 1414 days using a similar formula of mine.
March 23rd, 2010 at 2:32 am
In my opinion (I’m not a high school (Grade 6 Filipino student) student so… I’ll just be opening this out (and I can’t open the file ) ) I was thinking… If they each prisoner’s backs then the prisoners could ask eash other… duhh…
March 24th, 2010 at 3:49 am
I think, the answer is black, since the three prisoners were placed in a dark dungeon, they should not see any black colors. They should see only white colors.
March 27th, 2010 at 1:15 pm
Each prisoner is trying to win. If you are NOT the winner, then you get two more years added to your sentence, so they would not want to tell each other what they are seeing.
March 27th, 2010 at 1:16 pm
But one of the prisoners is already seeing two whites, so there are two blacks and a white left. There was a light, so they could observe a black on someones back it that was the case.
May 31st, 2010 at 3:56 pm
the brother/sister riddle – THAT MAN is his son
June 27th, 2010 at 11:41 am
Gail:
Xatnu is right, it’s 1414 days. The formula is:
1,000,000= (x+1)*(x/2) .
which is (rounded to nearest hundredth):
0=(x+1414.72)(x-1413.72)
So x is either 1413.72 or -1414.72. Since days can’t be negative, and any portion of a day constitutes that day, we round the positive answer up to 1414 days. Now that is just less than 4 years, so we can assume that at least one of those years will be leap. For simplicity let’s assume the forth year is leap, so let’s subtract 365 three times. That leaves us with 319 days. Now jump forward a month at a time subtracting 31 days (Oct), 30 days (Nov), 31 (Dec), 31 (Jan), 29 (Feb), 31 (Mar), 30 (Apr), 31 (May), 30 (Jun), and 31 (Jul), leaving us on August 28th (since Oct 29th was day #1, Oct 28th was day #0, and we are adding months to day #0 not day #1) and we now have 14 days to go. Now subtract 3 days for the rest of August, and our answer is September 11th.
July 16th, 2010 at 5:05 pm
“But THAT MAN’S father is my father’s son”
My father’s son would be ME, because I am an only child, it could not be a brother or sister.
Since my father’s son (me), is THAT MAN’S father…I am therefore THAT MAN’S father…thus, THAT MAN is my son.
“Who is THAT MAN?”
THAT MAN is my son.
July 30th, 2010 at 9:52 am
RIght you are !!!
September 4th, 2010 at 3:05 pm
that man could also be your nephew
September 4th, 2010 at 3:08 pm
and for the prisoners,
igor has a white disk, because he would know that if either of the other guys saw a black and white they would know that the other guy would see two blacks if they were black and leave, but they didnt leave. so igor knows none of them see any blacks
September 11th, 2010 at 5:32 pm
3 prisoner problem.
The “deductive” reasoning employed by the author considers only one of each disjuncts of the disjunction…
For those of you with no experience in symbolic logic…
When Igor considers the position of Qwerty, he is faced with an either or dilemma. From the perspective of Igor, “Qwerty either sees a white disk on my back or a black one.”
The author of the solution only considers one part of the either/or statement. Igor is still faced with the either/or dilemma. Either Qwerty and Uiop each see two white disks, or they see one white and one black. All that can be confirmed is that Uiop would infer that he was wearing a white disk.
All of the conditionals that were used to deductively reason your way through the problem to the solution are unsound. Hence, the “logic” employed to solve the problem fails. Your argument may be valid in form, but it is unsound!
February 4th, 2011 at 5:40 am
Sir, i think the answer to 3 prisoner puzzle would be THE BLACK DISK…. as the logic would be… supposedly, A was the person to deduce first. he saw 2 white disks but none of the other two were able to ans first because they each saw one white and one black… if they would have seen 2 whites they would have guessed there to be white itself…..
September 25th, 2011 at 2:20 pm
IGOR ( lets call him prisoner C) has a white card at his back. Now let us consider C had BLACK, then prisoner A would have seen a BLACK on C and a WHITE on A (puzzle says A and B has WHITE). Now A cannot make any decision as he sees BLACK on C ( assumed) and WHITE on B. BUT B would have made a IMMEDIATE decision as follows ” A can see a BLACK on C and had he seen a BLACK on me, he would have gone way, BUT he did not, hence mine has to be WHITE” and with logic B should have gone out. BUT B DID NOT DO IT. Thus C is clear that he has WHITE on his back.