Pigeohole principle - rolling die

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Discussion Overview

The discussion revolves around the application of the pigeonhole principle to determine the number of times a single die must be rolled to achieve repeated scores. Participants explore the conditions for obtaining at least two, three, and n times the same score, where n is greater than or equal to 4.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant suggests that to get the same score at least twice when rolling a die, one might think to roll it 6^2 times, but expresses confusion about the application of the pigeonhole principle.
  • Another participant clarifies that with 6 faces on the die (the "holes"), rolling it 62 times is not necessary to achieve a repeated score.
  • A participant calculates that to get at least two scores in one hole, 7 rolls are needed, but questions the reasoning behind needing 13 rolls to get three scores in one hole.
  • There is a discussion about the number of "pigeons" (scores from rolls) increasing while the number of "holes" (die faces) remains constant at 6, leading to confusion about the required number of rolls.
  • One participant suggests that to achieve three scores in one hole, the reasoning applied to the first question should be extended, indicating that 8 rolls may not suffice.
  • Another participant agrees and elaborates that to get a third score for a number that already has two, additional rolls are necessary, leading to a formulaic approach of 6(n-1) + 1.

Areas of Agreement / Disagreement

Participants express differing views on the number of rolls required for achieving repeated scores, with some calculations leading to confusion and disagreement about the correct application of the pigeonhole principle.

Contextual Notes

Participants are working through the mathematical reasoning behind the pigeonhole principle as applied to rolling a die, with some assumptions and calculations remaining unresolved.

jwxie
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How many times must we roll a single die in order to get the same score:
a) at least twice?
b) at least three times?
c) at least n times? for n >= 4

Well I know that P.P says that for m objects, and n boxes, where m > n, there must be one box of n that contains at least two or more objects.

It seems simple, but the application is not.

So for the given problem, like a
There are 6 sides for a single die. And I thought I should do 6^2, since we want to the get the same number at least twice.

I get really stuck at solving this. Can you guys kindly guide me through?

Thanks,
 
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Well, the die has six different faces. Do you really need to throw it 62 times to get the same result twice? Think of the different faces as the "holes" and the each throw's score as the "pigeons".
 
Okay, in order to get one hole get two pigeons, we need n holes, and n+1 pigeons, so for (a) we need 7, where n = 6.

to get three pigeons in one hole, and we still have 6 (n) holes, i thought we just need another pigeons, total of 8 pigeons, but the answer key said 13. how come?

thanks jsuarez
 
we still have 6 (n) holes, i thought we just need another pigeons, total of 8 pigeons

No, for all questions you have just six "holes" (the faces), but the number of "pigeons" (each pigeon is the score at each throw) increases.

So for the second (and third) questions, just apply the same reasoning that you applied on the first. You should be able to see that 8 doesn't work; just look at the following eight (possible) scores: 1 4 3 2 4 2 1 6

So, how many throws do you need for the same score to appear at least three times?
 
JSuarez said:
No, for all questions you have just six "holes" (the faces), but the number of "pigeons" (each pigeon is the score at each throw) increases.

So for the second (and third) questions, just apply the same reasoning that you applied on the first. You should be able to see that 8 doesn't work; just look at the following eight (possible) scores: 1 4 3 2 4 2 1 6

So, how many throws do you need for the same score to appear at least three times?

oh right. in order to get another one for the one that has two already, we need another round (which means 6 more)

so the whole process grows by 6(n-1) +1
 
Yes.
 

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