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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?

MATH000
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    There are $30!$ possible orders and all are equally likely....so $\frac 1{30!}$ (not a very large number). – lulu Apr 06 '16 at 21:22
  • @lulu: can you please explain why? Why we don't use the 200 anywhere? – MATH000 Apr 06 '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 06 '16 at 21:29
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    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 06 '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 06 '16 at 21:33
  • To expand on @lulu comment. Pick 30 cards from the 200. Now, keeping them in the same order, renumber them 1-30, based on how large the number is. What is the chance that the sequence is 1,2,...,30? Clearly, it's 1/30! – Bonnaduck Apr 07 '16 at 03:11

2 Answers2

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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 $$\frac {200!}{(30!)\times (200-30)!}\times \frac {(200-30)!}{200!}=\frac 1{30!}$$

lulu
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  • could you explain the meaning of the 2nd term? The 1st term is the number of ways to select 30 cards out of 200. What is the 2nd term? – Mining Sep 21 '21 at 13:37
  • @Marco That's the (reciprocal of) the number of ordered subsets with $30$ elements. – lulu Sep 21 '21 at 13:44
  • ok so it is the probability of selecting an ordered set? – Mining Sep 21 '21 at 13:52
  • @Marco. Not following. You asked about the second term. That's the reciprocal of the number of ordered sets. It's not meant to be a probability. – lulu Sep 21 '21 at 13:58
  • @Marco My second method starts with the remark that the desired probability is $\frac {\text {number of increasing subsets with 30 elements}}{\text {number of ordered subsets with 30 elements}}$. That's all that's going on there. – lulu Sep 21 '21 at 13:59
  • yeah sorry I misunderstood one thing. It is clear now. – Mining Sep 21 '21 at 14:23
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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 $$ \frac{4-1+2}{\frac{4!}{(4-2)!}}=\frac{3}{12} $$