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Topic: Furthest number picked to a random number picked. (Read 575 times) |
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Lightboxes
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Furthest number picked to a random number picked.
« on: Sep 20th, 2003, 1:38am » |
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You are in a communications class. Today is the day where the students have to do speeches in front of the class. You dread public speaking and would prefer to prepare yourself more so that you won't be so nervous. So rather than having to speak today, you hope to speak the next class time. However, the teacher has run out of volunteers and is now forced to randomly pick students. For the teacher to do this, he decides to randomly pick a number from 1 - 100 including 1 and 100. The student closest to that number will have to do the speech.
There are only 5 of you left who have not spoken in front of the class yet. There is only time for one more speech before class ends. You get to pick the first number.
What number do you pick to lower the probablity of you being the closest/having to speak today?
_.:=-=:._ Assume they can't pick the same number.
_.:=-=:._ Assume 2 hates only you and will increase your probability by as much as possible when convenient. In other words, if 2 has the opportunity to pick two or more numbers such that either one will give 2 the lowest possible probability then 2 will pick the number that increases your probability. 3 hates only 2. 4 hates 3 and 5 hates 4.
_.:=-=:._ Assume that the students that pick numbers know what the previous students picked. In other words, student 2 knows what you picked...student 3 knows what you and student 2 picked...etc.
_.:=-=:._ Assume that in the case of a tie, then only those who tie will have to duel to a fight to the death! And the one that dies last in the duel, will have to do the speech. But if your dead, you won't have to go! Then the class ends.
_.:=-=:._ Assume the other students are playing by an optimal strategy. In other words, they are trying to get the lowest probability possible of not being the closest.
_.:=-=:._ Assume the number the teacher picks is perfectly random; as in not picked by a human but rather a computer that has the ability to pick a number that gives equal chance to all the numbers of being picked.
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| « Last Edit: Sep 21st, 2003, 8:11pm by Lightboxes » |
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Sir Col
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Re: Closest number picked to a random number picke
« Reply #1 on: Sep 20th, 2003, 4:18pm » |
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I'm not much good at these problems, so I've probably missed some clever trick...
::
If we started by selecting the middle, the others would all select one side of 50, and fill up that section; that would leave a large range for which we would be selected.
In fact if we pick anything but an extreme, the other players will all occupy the smallest section.
So I would pick one of the extremes, say 1. This way, a random number cannot appear below me, and I only need concern myself with numbers equal or greater.
By the same logic, the second person would pick the other extreme, 100.
Now the 3rd person is not going to select the middle, otherwise the 4th and 5th will select one side, leaving a large range which is closer to 50. So the 3rd person will select nearer to one extreme, say 25.
However, the 4th person will now select somewhere inside the smaller gap, say 12 or 13. The 5th person will then pick somewhere around 6.
In this case I have done well, but if the 3rd person selected the other extreme, say 75. The 4th would select 87 or 88, and the 5th would pick abou 94.
It seems that there is no guaranteed optimal stragegy for the 1st person, and it is very much dependent on luck.
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Icarus
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Re: Closest number picked to a random number picke
« Reply #2 on: Sep 20th, 2003, 5:34pm » |
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It sounds to me like 1 picks "1", 2 picks "100" to maximize 1's chance. 3 picks "2", to maximize 2's chance. 4 picks "99" to maximize 3's chance, 5 picks "3" to maximize 4's chance.
Except... 2 knows this, and he wants you (1) to have the max chance, and following through with this course of action leaves you with a chance of only 1/100. So he wants to pick something so that the actions of 3, 4, 5 will still leave you with as big a chance as possible. Similarly, 3, 4, 5 understand things as well. Only 5's response is so easily predictable.
Actually, I think this puzzle needs some clarification before a definite answer is possible. Are the hatreds of 2, 3, 4, and 5 so great that they would increase their own chances in order to increase the chance of the object of their malice? If so, by how much?
This puzzle has some similarities to the "Greedy Pirates" puzzle in the Hard forum. And I think will require some clarification before a definitive answer is possible.
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Lightboxes
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Re: Closest number picked to a random number picke
« Reply #3 on: Sep 20th, 2003, 7:55pm » |
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I actually had this happen to 5 students in my Comm. class but the students didn't hate eachother. So basically, I made this up.
The reason I had the hatred in there, is because there are too many possibilities where almost everyone gets the same probabilitiy. I wanted there to be one solution in other words.
That's what I was thinking Icarus...
So ::Since student 3 picks 2 then student 2 should pick 2 since student 1 will have his chances decreased to 1/100 ANYWAY!
3 will do the same (3 will pick 3)...I think...etc.
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Icarus
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Re: Closest number picked to a random number picke
« Reply #4 on: Sep 20th, 2003, 8:21pm » |
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There is no advantage whatsoever to student 2 in picking "2". But depending on the behavior of other students in a way which is not definite from your statements, it might be to 2's advantage to pick 99 or 98.
To make this a little clearer, I am renaming the students A, B, C, D, E (in order). That way I don't have numbers picking numbers. Here is an example of what I am refering to: A picks 1. B picks 96. C sees that B's chances are equally maximized by picks of 2 or 100, picks 100 to minimize his own outcome - leaving A exposed! D sees that there is no way he can increase C's probability, picks 97, as this does not decrease C's probability, but drastically decreases his own. E similarly picks 99, since it greatly decreases his own chances without reducing D's.
The question is, will B consider this a better solution than the original, since the same increase in chance that occurs for A also occurs for himself? Which does he prefer - to stick it to A even at the cost of his own interests, or to save his own bacon at the cost of letting A off the hook?
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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Lightboxes
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Re: Closest number picked to a random number picke
« Reply #5 on: Sep 21st, 2003, 12:43am » |
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P=Probability
I think I just figured it out. It's actually quite simple.
You have to remember assumption #2..."when convenient...".
::A will pick 1 or 100. Obviously he should do that as Sir Col has said before:
Quote:| a random number cannot appear below me |
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E is screwed no matter what happens because he will be last and only have ONE AREA (range) of choice.
Explanation, working backwards:
Assume E does have a choice between 2 areas to pick from. Obviously he will pick the smaller area. But how did that area appear there? There can't be two areas because that would mean the last person (D) picked a spot that would not make a difference to (C). But how can that happen? C not being affected at all? That would mean B picked a number such that C had the option of two areas. But how did that happen? That means B picked a number such that two areas gave C that option...which would mean that B did not
follow the assumption #2. He didn't pick a number that maximized A's probability WHEN CONVENIENT and minimizing his own (b's own).
In other words, the first thing these students will do is to minimize their probability according to assump. #2. When convenient they will increase the previous student's probability.
(Because of the above) Therefore, E should have the worst possible spots to pick. If this is true, then D should almost be equally screwed or basically, the second to first screwed person. Using this information A,B, and C will pick numbers such that E will be forced to pick a certain number and D will be forced as well. What number is that?
A will pick 1 knowing full well that if B picks 100 (the best spot because of assumption #2) then C will pick 2 forcing a probability of 1/100. So A is picking that spot to actually lower is probability as low as possible. B will pick 100 and not any other number because of assumption #2, but B is not picking that number ONLY because he hates 1 but rather because C and D will help him get the lowest probability AND because he hates 1! Why? Because C will pick follow the same rules and pick 2 forcing D to pick 99. And E will be forced to pick 3.
In other words...as the students will say under their breath...
A: "I will minimize my probability"
B: "I will minimize my probability...wait a minute...my probability is the same (excluding C,D, and E for now) no matter what number I pick (2-100) but I do hate A so I'll pick 100 and in the process, if someone goes directly in front of A then obviously someone will do the same to me (Since this is optimal) and I will have a 1/100 chance! So who cares about A if I get 1/100!"
C: Same as B.
D: "I'm screwed."
E: "I'm screwed."
RESULT:
D has a P( 48/100 ) because the number right in between them makes it a tie. So that doesn’t count.
E has a P( 48/100 )
Now, what happens from here?
D knows that E will position himself right in front of C. So the question for him is…”Am I really screwed?” What if I (D) pick a number such that my P is the same or lower, and gives C a higher P while at the same time forcing E to pick a number such that it does not lower C’s P. If A=1, B=100, and C=2 then D would HAVE to pick 52 in a feeble attempt to force E to pick 99 but that makes D’s P way to high and not worth it. Basically, this just reinforces the fact that D and E are screwed because of A, B, and C.
In other words, it’s kinda similar to the greedy pirates as someone said earlier.
In other words, A B C are playing the optimal way…getting P(1/100).
SUMMARY:
A picks 1. B picks 100. C realizes what's going on and picks 2. D realizes what's going on but is screwed. E is screwed.
To understand and remember why they pick the numbers they do...you have to keep in mind the WHOLE explanation. Not just part of it.
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| « Last Edit: Sep 21st, 2003, 1:24am by Lightboxes » |
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Re: Closest number picked to a random number picke
« Reply #6 on: Sep 21st, 2003, 2:03pm » |
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I think the hatred aspect makes the problem overly simple. Suppose you eliminate that condition and make their thought process as follows: each will minimize his own probability; but given equal choices, they will spite the previous player if that player could have helped them without hurting his own chances. If he did the best he could for them, then they will do the best they can for him. That way number 4 knows that if he picks 99, 5 will spite him and pick 3, (because 4 could have picked 98, allowing 5 to get a 1 percent number, without hurting 4 at all). Then each will be sitting with a miserable 48.5% chance. But if 4 picks a middle number, 5 will try to help 4 out if he can do so at no cost to himself.
SO 4 picks 51 and 5 picks 52. And 4 of the 5 students end up with equal odds. Number 3 would like to use the same principle to make 1 carry his fair share of the odds, but if 3 picked any middle number, 4 and 5 would both pick on the same smaller side of 3. Greed always trumps altruism.
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