*
Then why not go to the source?
http://cartalk.cars.com/Radio/Puzzler/index.html
Here is the latest puzzle:
New Puzzler: Grasshopper Race
RAY: Imagine there is a hat sitting on the table. And there are two contestants. You, Tommy, will be one of the contestants, and we’ll call the other one Vinnie.
You reach into the hat, Tommy, and pull out a number. Then, Vinnie does the same. Now, the reason this hat is magical is that, no matter what number you pull out, Vinnie will always pull out a number that is either one above or one below your number. For example, if you pull out a two, Tommy, you know Vinnie has pulled out either a one or a three.
TOM: I’m with you man! Keep going. I like it.
RAY: To make it simple, we’ll limit the numbers to between one… and infinity.
So, each of you pulls out a number. Let’s say you pick three and Vinnie picks two. I’m the moderator, and I ask Tommy, “Do you know what number Vinnie has?” Tommy looks at his number, which is 3, and says, “No, I don’t.”
I then ask Vinnie, “Do you know what number Tommy has?” He looks at his number two and says, “Yes.” He knows Tommy has to have a 3.
TOM: Whoa, I’m still with you. You bet.
RAY: Now, here’s the tricky part:
Regardless of the numbers that are picked, and assuming that both contestants answer truthfully, if I keep asking the question of both of them, eventually one contestant will know what number the other contestant has.
In other words, if I ask Tommy, then ask Vinny, then Tommy again… then Vinny again… eventually one of them will know the other’s number.
The question is this:
How come is that?
Replies
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Doesn't work if you have any number and believe in a closed universe and infinity closes around into negative infinity.
*And....if ya ever drop that hat into the i correction...b look out!near the stream,aj
*Roger,Eventually, either Tommy or Vinnie will draw the 4.If Tommy draws the 4, he knows that Vinnie must have a 5, since he has previously drawn the 3.On the other hand, if Vinnie draws the 4, he knows that Tommy has the 5, since he knew Tommy previously had drawn the 3.Any milkbones?
*Because eventually one of them will draw the number one. Per the rules the other number must be two since zero is not in the set of numbers 1-infinity?No, it can't be that - you said 'regardless of the numbers picked'??? If there's only one of each number being picked then Ken is right - they can figure it out by elimination.
*Jeff,I think as the problem was presented, Ray, the moderator, changed horses in the middle of the stream. First he says, "For example, if you pull out a two, Tommy, you know Vinnie has pulled out either a one or a three."Then, in a later statement, Ray says,"To make it simple, we'll limit the numbers to between one... and infinity."The number "one" is NOT "between" 1 and infinity. So I eliminated it from the hat at that point. I think the key to solving the problem was in Rays statement"RAY: To make it simple, we'll limit the numbers to between one... and infinity." Therefore, the number 4, was the next number in line for me to look to for a solution.
*So as soon as they choose either a one or a number that is one away from a number the opponent has chosen, they get it right. And with an infinate number of possible numbers, that would most likely take...lets see if I can do the math...an infinate amount of time.
*Ted,Where are you when we need you?
*i Regardless of the numbers that are picked, and assuming that both contestants answer truthfully, if I keep asking the question of both of them, eventually one contestant will know what number the other contestant has.As written, it translates to a set of numbers being picked, and the question repeatedly asked without a new set of numbers picked.Semantics, the future is semantics...
*Splintergroupie's on to it. Another assumption is that the two contestants are pretty bright and also are honest in their answers. Inclusion or exclusion of 1 doesn't matter, in fact, any two different integers (positive, zero or negative) could be chosen...
*I agree. And given the way it is written, I don't see the answer.Rich Beckman
*Vinnie or Tommy can answer truthfully and say the other does NOT have a ___.
*Reading it again, there isn't really an implication that numbers will continue to be picked after the last example. Therefore (as could be implied) Tommy and Vinnie could be each holding one number and the questioning (but not picking) continues, in other words, the guessing continues with only one set of two numbers until the correct answer is reached - eventually?
*Re-readng it myself, only one of two responses is called for, a "yes" or a "no". No guessing allowed, Jeff, and it does away with my theory of negatives as well...unless there's code involved, a sorting program...Hey, Roger, this isn't the Grasshopper Puzzle, this is the Magic Hat Puzzle!
*Assuming that one pair of numbers is drawn, and the questioning continues until one of the participants knows the answer, and assuming the number one is included in the given set of numbers: Tommy has three, Vinnie has two. Question is asked of Tommy first, who knows Vinnie must have a two or a four, but doesn't yet have enough information to know for sure. Question is asked of Vinnie, who knows that Tommy must have a one or a three. But, if Tommy had a one, then Tommy would have known that Vinnie had a two, since zero is not part of the given set of numbers. Since Tommy told the truth, in that he did not know Vinnie's number the first time he was asked, it follows that Tommy must have a three. Simple as that.Now lets try another.Tommy has three, Vinnie has four. Question is asked of Tommy first, who knows Vinnie must have a two or a four, but doesn't yet have enough information to know for sure. Question is asked of Vinnie, who knows that Tommy must have a three or a five, but doesn't yet have enough information to know for sure. Question is asked of Tommy (second time), who now has the answer. If Vinnie had two, he would have known Tommy has three, since Vinnie would have ruled out Tommy having one the first time Tommy answered. Therefore Tommy knows Vinnie has four. Simple as that.I assume this logic process could be carried out for any pair of whole numbers? I tried using six and seven as the pair of numbers, but somehow I got lost in my thought process and gave up!Gregg
*i since Vinnie would have ruled out Tommy having one the first time Tommy answered.Doesn't this only work because number one is a special case in the example, at the end of the number set? But if all numbers are included...hmm...still stumped.
*Good reasoning. I bet you've got it.Rich Beckman
*Looks like Gregg has cracked it. Great job.
*
Then why not go to the source?
http://cartalk.cars.com/Radio/Puzzler/index.html
Here is the latest puzzle:
New Puzzler: Grasshopper Race
RAY: Imagine there is a hat sitting on the table. And there are two contestants. You, Tommy, will be one of the contestants, and we'll call the other one Vinnie.
You reach into the hat, Tommy, and pull out a number. Then, Vinnie does the same. Now, the reason this hat is magical is that, no matter what number you pull out, Vinnie will always pull out a number that is either one above or one below your number. For example, if you pull out a two, Tommy, you know Vinnie has pulled out either a one or a three.
TOM: I'm with you man! Keep going. I like it.
RAY: To make it simple, we'll limit the numbers to between one... and infinity.
So, each of you pulls out a number. Let's say you pick three and Vinnie picks two. I'm the moderator, and I ask Tommy, "Do you know what number Vinnie has?" Tommy looks at his number, which is 3, and says, "No, I don't."
I then ask Vinnie, "Do you know what number Tommy has?" He looks at his number two and says, "Yes." He knows Tommy has to have a 3.
TOM: Whoa, I'm still with you. You bet.
RAY: Now, here's the tricky part:
Regardless of the numbers that are picked, and assuming that both contestants answer truthfully, if I keep asking the question of both of them, eventually one contestant will know what number the other contestant has.
In other words, if I ask Tommy, then ask Vinny, then Tommy again... then Vinny again... eventually one of them will know the other's number.
The question is this:
How come is that?