1. ## Permutation/Probability

Question 1
There are 11 places... 1st place, 2nd place ... 11th place. There are only 10 people.

P(n,r) = n! / (n-r)!
P(10,11)= 10! / (10-11)!
= 10! / (-1)! (Impossible)

What will happen if r is bigger than n?

Question 2
Roulette - Wikipedia, the free encyclopedia

The expectation for getting '0' = - 2/38 (per 1 unit)
The expectation for getting 'even' = - 2/38 (per 1 unit)
The expectation for getting '1-6' = - 2/38 (per 1 unit)

However, if I combined and placed 3 units on getting '0', 'even' and '1-6',
The expectation will be = - 2/38 (per 1 unit)
Since combination will not gain, should i divide the money to buy few things or play one bet for a while then play another bet?

Question 3
This is related to a game called craps.
There are two dices.

If the dices rolled 4, the games continues. After that, the game will end if either 4 or 7 appears first. For example, if the 2nd rolls are 5, the game will continue until either 4 or 7 appears.

Probability = 3/36 * 3/(6+3)

Why is it 3/(6+3)?