Results 1 to 3 of 3

Math Help - Permutation/Probability

  1. #1
    Newbie
    Joined
    Apr 2010
    Posts
    20

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member BAdhi's Avatar
    Joined
    Oct 2010
    From
    Gampaha, Sri Lanka
    Posts
    252
    Thanks
    6

    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)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2010
    Posts
    20

    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Is this a permutation?
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: April 4th 2011, 12:43 AM
  2. Permutation
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: September 3rd 2009, 09:02 AM
  3. permutation/ probability question part-2
    Posted in the Statistics Forum
    Replies: 5
    Last Post: March 10th 2009, 11:34 AM
  4. permutation 3
    Posted in the Math Topics Forum
    Replies: 11
    Last Post: February 10th 2009, 12:02 PM
  5. Permutation
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: August 7th 2008, 11:01 AM

/mathhelpforum @mathhelpforum