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If I draw at random from a set of 50 numbers( say 1:50), five numbers without replacing the drawn number, and then repeat this process ten times starting each 'draw' from the full set, how likely is it that I will draw every number of the set? How do I calculate this? Would it be 50( the number of 'draws') divided by 500 ( the total number from which the draws were taken) i.e.10% or is it more complicated than this?

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  • $\begingroup$ Is this for some subject? $\endgroup$
    – Glen_b
    Commented Jan 12, 2015 at 12:36
  • $\begingroup$ The answer to the question as written is 100%, because you never replace any of the numbers, whence after ten draws all 50 numbers have been taken. Because this is a trivial observation, leading to an uninteresting question, presumably "repeat this process" begins by replacing the five numbers in the set. That makes this a generalization of the Coupon Collector's Problem. $\endgroup$
    – whuber
    Commented Jan 12, 2015 at 15:04
  • $\begingroup$ Out of pure curiosity. whuber read the question incorrectly, by the way $\endgroup$ Commented Jan 12, 2015 at 15:14
  • $\begingroup$ Please explain: after all, a formula in an answer to the duplicate agrees exactly with the answer you accepted here! $\endgroup$
    – whuber
    Commented Jan 13, 2015 at 9:28

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I define trial as the process of five draws. In the first trial, your chance to draw five different numbers is 100%. So, your probability will depend only on the last nine trials. In the second trial, you calculate first the probability to draw any number except the ones from the first trial, i.e 45/50. Now, as you have already six different numbers, the probability for drawing again a different number is 44/49, and so on. Since you replace after every trial all numbers, your probability of drawing in the first draw of the third trial any number that wasn't drawn before is 40/50. And then you proceed as before. At the end, you need to multiply all probabilities for each draw, because you have a logical And, i.e. you draw a number And all others before must be different from it. I calculated with R and it resulted in 2.694293e-20. The probability is almost zero! Here is the code:

n <- 45:1 #numerator with desired outcomes
d <- rep(50:46,9) #denumerator with possible outcomes
p <- prod(n/d) # logical 'and': multiplication
p #print the probability
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  • $\begingroup$ Thank you. I thought that I was being naive and that the probability would be low - didn't 'expect' it to be so vanishingly small though! $\endgroup$ Commented Jan 12, 2015 at 13:19
  • $\begingroup$ Is it possible to estimate how many trials would be necessary to draw all fifty numbers, as a matter of interest? $\endgroup$ Commented Jan 12, 2015 at 15:05
  • $\begingroup$ The duplicate thread answers that question, too. $\endgroup$
    – whuber
    Commented Jan 12, 2015 at 15:09
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    $\begingroup$ (+1) Using the formula in the duplicate for $P(X_{10}(A)=50)$ gives the answer as $$\frac{104586258561652151699128857}{3881770614586446846243219560109206166400000000}\approx 2.6942926\times 10^{-20}.$$ It also shows that one must conduct $41$ draws in order to have a better than $1/2$ chance to collect all $50$ numbers (the chance then is approximately $0.501362$). It will require $81$ draws to have better than a $99\%$ chance of collecting them all, etc. $\endgroup$
    – whuber
    Commented Jan 12, 2015 at 15:11

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