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Odds versus number of draws to get 6 of 72 items 
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#1
Jan514, 08:47 AM

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P: 7,128

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. 


#2
Jan514, 10:09 AM

P: 223

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  (A1)^n] / A^n = 0.5 , solve for N somehow and I imagine that would be the answer. I think n = ln(0.5) / ln[(A1)/(A)] , which is like 9.9021 * 10^7 rolls, sounds kind of tedious. 


#3
Jan514, 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.



#4
Jan514, 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 


#5
Jan514, 05:53 PM

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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. 


#6
Jan614, 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 X_{k} 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
X_{1}+X_{2}+X_{3}+X_{4}+X_{5}+X_{6} with all six of the random variables being independent. 


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