# Pigeohole principle - rolling die

by jwxie
Tags: pigeohole, principle, rolling
P: 283
 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,
 P: 402 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".
 P: 283 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
P: 402
Pigeohole principle - rolling die

 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?
P: 283
 Quote by JSuarez 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
 P: 402 Yes.

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