Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Dice game

  1. Apr 14, 2009 #1
    I have got a problem.


    The dice is thrown until 1 has been rolled twice. The two 1´s do not need to be rolled consecutively.


    Whenever the counter of the 1´s is one and if a 6 is rolled, then the counter of the 1´s is reset to zero, i.e. at least two more 1´s have to be rolled in order to end the game.


    Questions:

    What is the average number of dice throws ?
    How often is the counter of the 1´s reset to zero on average in the whole game?
     
  2. jcsd
  3. Apr 14, 2009 #2
    Define
    a_n = probability that the 1's counter is 0 after the nth roll
    b_n = probability that the 1's counter is 1 after the nth roll
    c_n = probability that the game ends after the nth roll

    you know:
    a_1 = 5/6, b_1 = 1/6, c_1 = 0
    a_{n+1} = 5/6 a_n + 1/6 b_n
    Find similar terms for b_{n+1} and c_{n+1}, then solve the recurrence, and find the average of the c's to answer your first question.
     
  4. Apr 14, 2009 #3
    Thank you mXSCNT, but I dont know the term for the c_{n+1}

    a_{n+1} = 5/6 a_n + 1/6 b_n

    b_{n+1} = 1 - a_{n+1}

    c_{n+1} = ?????

    Is it the markow chain?
     
  5. Apr 14, 2009 #4
    your b_{n+1} is not correct
     
  6. Apr 14, 2009 #5
    Oh yes, it has to increase with n.
     
  7. Apr 14, 2009 #6
    c_{n+1} = (somethin with) 4/6*p_2_3_4_5 + 1/6*b_n + 1/3*p_1_6
     
    Last edited: Apr 14, 2009
  8. Apr 17, 2009 #7
    Nobody with an answer?
     
  9. Apr 17, 2009 #8
    Let me explain how I got a_{n+1}.

    a_{n+1} = P(1's counter is 0 after roll n+1)
    = P((1's counter was 0 after roll n, and then rolled something other than 1) or (1's counter was 1 after roll n, and then rolled a 6))

    Here I am simply enumerating the different ways that the 1's counter could become 0 at roll n+1.

    = P(1's counter was 0 after roll n, and then rolled something other than 1) + P(1's counter was 1 after roll n, and then rolled a 6)

    Because the 2 possibilities are mutually exclusive, the probability of either of them is the sum of the probabilities of each.

    = P(n+1st roll is not 1) P(1's counter was 0 after roll n) + P(n+1st roll is 6) P(1's counter was 1 after roll n)

    Because rolling a number is independent of the 1's counter, the probability of both happening is the product of the probabilities of each happening.

    = 5/6 a_n + 1/6 b_n

    Here I am simply substituting in the actual probabilities, and using the definitions of a_n and b_n.

    Perhaps now you can find b_{n+1}, following my example.
     
  10. Apr 18, 2009 #9
    Thank you very much! I'll try!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Dice game
  1. Dice game (Replies: 2)

  2. The Dice Game (Replies: 2)

Loading...