Results 1 to 3 of 3

Math Help - die rolling problem

  1. #1
    Junior Member
    Joined
    Mar 2009
    Posts
    36

    die rolling problem

    hey guys,
    just having trouble with this question.

    if 2 people play a dice game and each rolls a dice, if A rolls a higher dice then B then A wins and vice versa. if they roll the same number its a draw.

    What is the probability that A wins if they play a single game?
    it wouldnt be 1/2 i dont think because they can draw, but both A and B's chance of winning are the same?

    Also if they play 100 games what is the probability A wins 50 or more games?
    would i just use the probability from the question above and do like the
    P(50<X<100) binomial distribution thing.

    thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Mar 2010
    Posts
    18
    Hmmm... I'd be inclined to say that the probability player A wins in a single game is [tex]\dfrac{1}{3}[tex].

    My reasoning, consider it like a game of "Rock, Paper, Scissors". There's only three possible outcomes: Winning, Losing and Drawing and each of these has an equal chances of occuring. Say A rolls a 2, B has to roll a 3 or greater to win but adversely, B could end up in the same situation.

    But this is only speculation. I'm gonna look into a bit more.

    As for the second part, treat the game as if it only had two options: Winning and lossing, where winning has \dfrac{1}{3} chance of occuring and losing has \dfrac{2}{3} chance.

    It's been a long time since I've done statistics, I'm afraid so sorry I can't be of more use.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by b0mb3rz View Post
    hey guys,
    just having trouble with this question.

    if 2 people play a dice game and each rolls a dice, if A rolls a higher dice then B then A wins and vice versa. if they roll the same number its a draw.

    What is the probability that A wins if they play a single game?
    it wouldnt be 1/2 i dont think because they can draw, but both A and B's chance of winning are the same?

    Also if they play 100 games what is the probability A wins 50 or more games?
    would i just use the probability from the question above and do like the
    P(50<X<100) binomial distribution thing.

    thanks
    There are a few ways to think about this here are two different approches

    First one: Conditonal probabilites
    Let W be the event that A wins and the notation that B_1,B_2...B_6 is the event that player B rolls a 1,2,3..6

    Then
    P(W)=P(W|B_1)P(B_1)+P(W|B_2)P(B_2)+...P(W|B_6)P(B_  6)

    Now we need to compute the above probabilities.

    first P(B_i)=\frac{1}{6} for 1 \le i \le 6

    Now P(W|B_1) is the event that A wins given B rolled a 1. This event has prob =\frac{5}{6}.

    P(W|B_2) is the event that A wins given B rolled a 2. This event has prob =\frac{4}{6}. The pattern continues this gives

    P(W)=\frac{5}{6}\frac{1}{6} +\frac{4}{6}\frac{1}{6}+\frac{3}{6}\frac{1}{6}+\fr  ac{2}{6}\frac{1}{6}+\frac{1}{6}\frac{1}{6}+\frac{0  }{6}\frac{1}{6}=\frac{5+4+3+2+1}{36}=\frac{15}{36}

    The 2nd way to get this result is to make a table of all possible outcomes.
    Let the ordered pair (A,B) be the number A and B rolled respectivly then

    (1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
    (2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
    (3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
    (4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
    (5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
    (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

    Now we can just count the out comes there are 36 possible events and
    A > B only on the bottom triangle below the diagonal for a total of 15
    giving the same as the probability above
    \frac{15}{36}
    Last edited by TheEmptySet; March 5th 2010 at 09:23 PM. Reason: fix typo
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Snowball Rolling down a hill problem.
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: September 19th 2010, 10:44 AM
  2. Dice rolling problem
    Posted in the Statistics Forum
    Replies: 2
    Last Post: January 13th 2010, 06:55 AM
  3. rolling dices and coins problem
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: September 4th 2007, 10:37 AM
  4. rolling two dices probability problem
    Posted in the Statistics Forum
    Replies: 2
    Last Post: August 29th 2007, 07:15 PM
  5. rolling two dices probability problem
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: August 28th 2007, 11:27 PM

Search Tags


/mathhelpforum @mathhelpforum