What is the odds of picking numbers 1 through 10 IN ORDER unseen from a pot? (only numbers are 1,2,3,4,5,6,7,8,9,10).

Question

What is the odds of picking numbers 1 through 10 IN ORDER unseen from a pot? (only numbers are 1,2,3,4,5,6,7,8,9,10).

Answer 1

The answer depends on whether the selection happens with replacement or without.

With replacement, there are 10¹⁰ possibilities for picking ten numbers, so the probability of a particular sequence is p=1/10¹⁰. To obtain the odds, we use the formula:

odds = p/(1-p) = (1/10¹⁰)/(1 - 1/10¹⁰) = 1/(10¹⁰ -1)

Without replacement, there are 10! possibilities, so the probability of a particular sequence is 1/10!. To obtain the odds, we again use the formula:

odds = p/(1-p) = (1/10!)/(1 - 1/10!) = 1/(10! -1)

Hope this helps and let me know if you need more clarification on any step!

Answer 2

It's not clear whether the numbers would be replaced or not once selected. However, assuming they were just removed from the pot one at a time, the number of possibilities would be:

Selecting 1 - 10! / (10 - 1)! = 10 (or 1 in 10 chances)

Selecting 2 - 9! / (9 - 1)! = 9 (or 1 in 9 chances), etc.

The chance of selecting all 10 in oorder is therefore:

1/10 * 1/9 * 1/8 * 1/7 * 1/6 * 1/5 * 1/4 * 1/3 * 1/2 * 1/1 = 1/10! = 1/3,628,800

Therefore, the odds would be 1 in 3,628,800 (or roughly 0.0000276%).

Answer 3

1/10 x 1/9 x1/8 x1/7 x1/6 x1/5 x1/4 x1/3 x1/2

= 1/262880

= 3.80401704 x 10^-6

=0.00000380401704

=.000380401704%

= about .0004%

= about 1/40 of 1% chance or probability of selecting randomly and getting all 10 in order 1 to 10

odds are 262,879:1 against it happening

odds of it happening are 1:262,879

on average for every one time it happens, 262,879 times it won't happen