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

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In summary, the conversation discussed the probability of receiving 8 pairs of identical boxes when choosing from 8 items 16 times. The friend had opened 16 crates and received exactly 2 of each C1-C8 item, which was considered highly unusual. The probability was calculated to be approximately 1.432454 x 10^{-10}.
  • #1
ZooJersey
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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
 
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  • #2
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].

[FONT=Verdana,Arial,Tahoma,Calibri,Geneva,sans-serif] [/FONT]
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What is the probability of receiving 8 pairs of identical boxes when choosing from 8 items 16 times?

The probability of receiving 8 pairs of identical boxes when choosing from 8 items 16 times is extremely low. This is because there are a total of 8!/(2!)^8 = 1,134,960 possible combinations, but only one combination will result in 8 pairs of identical boxes.

How can I calculate the probability of receiving 8 pairs of identical boxes?

To calculate the probability, you can use the formula P = (n!/(r!*(n-r)!))*(1/n)^r, where n is the total number of items, r is the number of items chosen, and P is the probability of getting the desired outcome. In this case, n = 8, r = 16, and the desired outcome is 8 pairs of identical boxes.

Is there a way to increase the probability of receiving 8 pairs of identical boxes?

No, the probability cannot be increased as it is based on the total number of possible combinations. However, you can increase your chances by choosing from a larger pool of items or by increasing the number of times you choose.

What is the expected number of times I will receive 8 pairs of identical boxes when choosing from 8 items 16 times?

The expected number of times you will receive 8 pairs of identical boxes can be calculated using the formula E(x) = n*(1/n)^r, where n is the total number of items and r is the number of items chosen. In this case, E(x) = 16*(1/8)^8 = 0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

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