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Need to find the probability of receiving 8 pairs of identical boxes choosing from 8 items 16 times?

ZooJersey

New member
Jun 23, 2019
1
A friend stated they bought 16 crates all which could contain a random C1-C8 item. He then opened the crates and received exactly 2 each of every C1-C8 items.

So, (C1,C1) (C2,C2) (C3,C3) (C4,C4) (C5,C5) (C6,C6) (C7,C7) (C8,C8) is what he ended up with.

He stated this was good because there was an equal chance of getting them. I thought that this was highly unusual, and suggested there was no randomness since he received two of every possible item.

That brings me to the title question. What is the probability of receiving 8 pairs of identical boxes choosing from 8 items 16 times?

I thought it would be 12.5% for drawing a single item and 1.5% of matching a box then [(1.5%)(1.5%)]16 for getting two of all 8.

Any help would be appreciated.

Thank you
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
The probability of getting an item is 1. The probability of getting that same item is 1/8. The probability of then getting another item is 7/8. The probability of getting that same item is 1/8.

Continuing like that the probability of getting 8 items, each twice, is 1(1/8)(7/8)(1/8)(6/8)(1/8)(5/8)(1/8)(4/8)(1/8)(3/8)(1/8)(2/8)(1/8)(1/8)(1/8)= 7!/8^{15}. That is approximately [FONT=Verdana,Arial,Tahoma,Calibri,Geneva,sans-serif]1.432454 x 10^{-10}[/FONT][FONT=Verdana,Arial,Tahoma,Calibri,Geneva,sans-serif].

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