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Pigeohole principle  rolling die 
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#1
Apr2510, 07:35 PM

P: 283

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, 


#2
Apr2510, 07:53 PM

P: 403

Well, the die has six different faces. Do you really need to throw it 6^{2} times to get the same result twice? Think of the different faces as the "holes" and the each throw's score as the "pigeons".



#3
Apr2510, 08:01 PM

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 


#4
Apr2510, 08:16 PM

P: 403

Pigeohole principle  rolling die
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? 


#5
Apr2510, 09:15 PM

P: 283

so the whole process grows by 6(n1) +1 


#6
Apr2510, 09:24 PM

P: 403

Yes.



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