# Probability of drawing cards in ascending order

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?

• There are $30!$ possible orders and all are equally likely....so $\frac 1{30!}$ (not a very large number). – lulu Apr 6 '16 at 21:22
• @lulu: can you please explain why? Why we don't use the 200 anywhere? – MATH000 Apr 6 '16 at 21:24
• 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 $\{3,2,1\}$. – lulu Apr 6 '16 at 21:29
• 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. – lulu Apr 6 '16 at 21:31
• @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. – MATH000 Apr 6 '16 at 21:33

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 $$\frac {200!}{(30!)\times (200-30)!}\times \frac {(200-30)!}{200!}=\frac 1{30!}$$
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}$$• The OP specified "ascending order" not "consecutive". – lulu Apr 6 '16 at 22:36 • Oh, I see! Thanks for the comment, I will just leave it and make a note about this. – user164550 Apr 6 '16 at 22:42 • So you are right, it is$\frac{1}{30!}\$ – user164550 Apr 6 '16 at 23:05