Odds versus number of draws to get 6 of 72 items

  • Thread starter rcgldr
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In summary, using a program to test the average number of draws needed to see all 6 unique colored balls at least once in a jar filled with 66 white and 6 colored balls, the result is about 176 draws. This can be calculated by finding the expected number of draws for each unique ball and adding them together. While the original problem statement was about the odds of getting 6 of 72 items, the average number of draws to see all 6 unique balls was deemed sufficient. The same technique can also be used to calculate the distribution curve for finding all 6 balls.
  • #1
rcgldr
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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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  • #2
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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  • #3
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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  • #4
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
davidmoore63@y said:
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.
 
  • #6
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.
 

1. What is the concept of odds versus number of draws in relation to getting 6 of 72 items?

The concept of odds versus number of draws refers to the likelihood of selecting 6 specific items out of a total of 72 items in a random draw. It considers the probability of getting the desired outcome (6 items) over a certain number of attempts (draws).

2. How do you calculate the odds of getting 6 of 72 items in a single draw?

To calculate the odds of getting 6 of 72 items in a single draw, you would use the formula: (number of desired outcomes ÷ total number of possible outcomes) x 100. In this case, the number of desired outcomes is 6 and the total number of possible outcomes is 72. So the calculation would be (6 ÷ 72) x 100 = 8.33%. This means that in a single draw, the odds of getting 6 of 72 items is 8.33%.

3. What is the relationship between the number of draws and the odds of getting 6 of 72 items?

The relationship between the number of draws and the odds of getting 6 of 72 items is that as the number of draws increases, the odds of getting 6 items also increases. This is because with each additional draw, there is a higher chance of selecting the desired items. For example, if you were to draw 10 times instead of just once, your odds of getting 6 of 72 items would be significantly higher.

4. Is it possible to guarantee getting 6 of 72 items in a certain number of draws?

No, it is not possible to guarantee getting 6 of 72 items in a certain number of draws. The odds of getting the desired outcome may increase with more draws, but there is always a chance that the desired items may not be selected. However, the more draws that are done, the higher the odds become.

5. Are there any factors that can affect the odds of getting 6 of 72 items in a certain number of draws?

Yes, there are several factors that can affect the odds of getting 6 of 72 items in a certain number of draws. These may include the randomness of the selection process, the number of items available, and any external factors that may influence the outcome. Additionally, the odds may also be affected by any biases or patterns in the selection process.

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