# Conditioned probability of two sequential experiences

How can I calculate the probability of getting a number smaller than a previous random picked number?

Details:

I have a first independent experience where a set of n random numbers are picked from 1 to M.

(n <<< M)

From the picked numbers, I choose the smaller number.

Then, I repeat this experience again (independently).

How can I calculate the probability of this second number being smaller or equal than the first one?

• what do you mean by a minor number? – avz2611 Oct 24 '17 at 10:39
• Sorry, English isn't my mother language. I meant to say: smaller number. – Pedro Simões Oct 24 '17 at 10:45
• Not entirely sure I am following the procedure. If I have understood it, the two numbers are chosen in exactly the same way. If so, then the probability that the second is smaller is $.5$, by symmetry, if the process is continuous. otherwise it is $.5\times (1-p_{tie})$ where $p_{tie}$ denotes the probability that the two values coincide. – lulu Oct 24 '17 at 11:19
• To illustrate, suppose my numbers are obtained by tossing a fair die. Then the probability of a tie is $\frac 16$. The probability that the second toss is strictly lower than the first is $\frac 12\times \left(1-\frac 16\right)=\frac 5{12}$. You can check this by listing all possible rolls. – lulu Oct 24 '17 at 12:02
• Then the exact same computation goes through. Now $p_{tie}=2^{-256}$ so the probability you want is $\frac 12\times \left(1-2^{-256}\right)$ which is certainly very close to $.5$ (as $p_{tie}$ is very close to $0$). – lulu Oct 24 '17 at 12:29

If the numbers are drawn so there is zero probability of ties and all orders are equally likely, then the probability that the $n+1^{\text{th}}$ number drawn will be smaller than all of the previous $n$ numbers drawn is $\frac{1}{n+1}$, by symmetry
An equivalent approach counting the possible orders could calculate $\frac{1 \times n!}{(n+1)!}$ to give the same answer