Given 200 cards where each card has a unique number from 1 to 200.

We randomly pick 30 cards (the order we pick them matters). What is the probability the unique numbers of the cards we pick are in ascending order?

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    $\begingroup$ There are $30!$ possible orders and all are equally likely....so $\frac 1{30!}$ (not a very large number). $\endgroup$ – lulu Apr 6 '16 at 21:22
  • $\begingroup$ @lulu: can you please explain why? Why we don't use the 200 anywhere? $\endgroup$ – MATH000 Apr 6 '16 at 21:24
  • $\begingroup$ The $200$ is a red herring. When you say you randomly choose cards I interpret that to mean that every (ordered) set is equally likely. That is to say $\{1,2,3\}$ is exactly as likely as $\{3,2,1\}$. $\endgroup$ – lulu Apr 6 '16 at 21:29
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    $\begingroup$ If you want to proceed by force: the probability that the least card is in slot $1$ is $\frac 1{30}$. The probability that the second smallest card is in slot $2$ (given that we know it isn't in slot $1$) is $\frac 1{29}$. And so on. $\endgroup$ – lulu Apr 6 '16 at 21:31
  • $\begingroup$ @lulu: thank you very much of your help! {1,2,3} is not the same as {3,2,1} that's why i said order matters. $\endgroup$ – MATH000 Apr 6 '16 at 21:33

Just to elaborate on the comments, I'll give two different arguments. The first, based purely on symmetry, and the second based on counting.

I. One way to get a uniform distribution is to select an unordered subset (with $30$ elements) and then choose a random permutation of it. Here, we don't care which unordered subset we choose and we only want one of the $30!$ permutations. As the permutations are equally probable, the answer is $ {\frac 1{30!}}$.

II. First we count the number of ordered subsets. As noted in the posted solution of @Hamid there are $$\frac {200!}{(200-30)!}$$

How many are in ascending order? Well as any unordered subset can be put in ascending order in exactly one way there are $$\binom {200}{30}=\frac {200!}{(30!)\times (200-30)!}$$ Hence the probability is the ratio $$\frac {200!}{(30!)\times (200-30)!}\times \frac {(200-30)!}{200!}=\frac 1{30!}$$


This answer is for the case when cards are required to be "consecutive", not in "ascending order"

The total number of possibilities is $$ \frac{200!}{(200-30)!} $$ The possible number of ordered choices is $200-30+1 = 171$$^*$. So the probability should be $$ \frac{200-30+1}{\frac{200!}{(200-30)!}} $$

$^*$ This comes from $1\to 30$, $2\to 31$, $\cdots$, $171\to200$.

Example: let it $4$ and $2$, your choices are $12,13,14,21,23,24,31,32, 34, 41,42,43$. What you are looking for are $12,23,34$. So the probability is $$ \frac{4-1+2}{\frac{4!}{(4-2)!}}=\frac{3}{12} $$

  • $\begingroup$ The OP specified "ascending order" not "consecutive". $\endgroup$ – lulu Apr 6 '16 at 22:36
  • $\begingroup$ Oh, I see! Thanks for the comment, I will just leave it and make a note about this. $\endgroup$ – user164550 Apr 6 '16 at 22:42
  • $\begingroup$ So you are right, it is $\frac{1}{30!}$ $\endgroup$ – user164550 Apr 6 '16 at 23:05

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