odds versus number of draws to get 6 of 72 items


by rcgldr
Tags: draws, items, number, odds, versus
rcgldr
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#1
Jan5-14, 08:47 AM
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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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lendav_rott
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#2
Jan5-14, 10:09 AM
P: 221
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.
rcgldr
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#3
Jan5-14, 01:18 PM
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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.

davidmoore63@y
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#4
Jan5-14, 03:39 PM
P: 10

odds versus number of draws to get 6 of 72 items


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
rcgldr
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#5
Jan5-14, 05:53 PM
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Quote Quote by davidmoore63@y View Post
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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#6
Jan6-14, 02:47 PM
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The same technique gets you the distribution quite easily (modulo doing actual mathematics). If you can calculate the random variable Xk which is the number of draws required to find one of k unique objects out of 66 of them, then the full distribution to find all six balls is simply
X1+X2+X3+X4+X5+X6

with all six of the random variables being independent.


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