# Odds versus number of draws to get 6 of 72 items

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number of draws versus odds to get 6 of 72 items

update - changing the problem statement:

You have a jar filled with 66 white, and 6 unique balls: 1 red, 1 blue, 1 green, 1 purple, 1 yellow, and 1 cyan ball. You draw the balls one at a time and return them each time, recording what you've drawn.

What is the average number of draws before you see all 6 unique colored balls at least once?

Using a program to test this, it seems that about 176 draws are needed.

I could modify the program to determine the average success rate versus number of draws to get the answer to the original problem statement, but what I really wanted was the average number of draws for success. I could probably do a standard deviation on this, but that's not really needed either.

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If we get to pick the balls 6 at a time and we would only have 1 time to pick, there is only 1 set out of Combinations of 6 out 72 that will net us the winner.

If we had a 12 sided dice and we want to roll for a specific number an average of 50% of the time it would be
(12^n - 11^n) / 12^n = 0.5 , where n is the total number of dicerolls.
In essence, we have to roll that gazillion sided dice an N times so that we would get that 1 specific result an average of 50% of the time.

Total combinations are
156 238 908 so call that A
[A^n - (A-1)^n] / A^n = 0.5 , solve for N somehow and I imagine that would be the answer.

I think n = ln(0.5) / ln[(A-1)/(A)] , which is like 9.9021 * 10^7 rolls, sounds kind of tedious.

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With the change to the original problem statement, it seems the average number of draws to see at least one of each of 6 unique balls is about 176 draws.

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The probability of getting a unique ball on the first draw is 6/72. Therefore the expected number of draws needed to obtain the first unique ball is 72/6. Now, the probability of getting the second unique ball on the next draw is 5/72, so the expected number of draws to get the second unique ball is 72/5. Continuing this reasoning, the total expected number of draws to get all the unique balls is:

72/6 + 72/5 + 72/4 + 72/3 + 72/2 + 72/1 = 176.4

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72/6 + 72/5 + 72/4 + 72/3 + 72/2 + 72/1 = 176.4
I also thought of this same approach, but I wasn't sure if that approach wasn't missing something. Since the program output matched the results (about 176.36), I assume it's probably ok.

Originally I was trying to figure out the number of draws versus chance of success, but realized that just getting an average number of draws would be good enough. As mentioned, I could modify the program to get a distribution curve, but just knowing the average is good enough.

Office_Shredder
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