# Thread: Permutation/Probability

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)?

Please advise. Thank you very much.

2. ## Re: Permutation/Probability

for questions like the Q1, what I do is try to keep people fixed (stationary) and then assign places to them. For example if those places mentioned here are chairs, what I do is keep 10 people fixed and then try to find the number of possibilities of assigning chairs to them so eventually end up with P(11,10)

when your routine is followed you get an impossible answer, because you cannot assign 10 people to 11 places (because 1 place is always vacant)

3. ## Re: Permutation/Probability

Okay. So if there are now 11 places but only 9 people, the answer would be P(11,9) right?

Can anybody advise me on question 2 and 3 as well? Thanks.