Dice game

  • Thread starter Fiction123
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  • #1

Main Question or Discussion Point

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?
 

Answers and Replies

  • #2
315
1
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.
 
  • #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?
 
  • #4
315
1
your b_{n+1} is not correct
 
  • #5
Oh yes, it has to increase with n.
 
  • #6
c_{n+1} = (somethin with) 4/6*p_2_3_4_5 + 1/6*b_n + 1/3*p_1_6
 
Last edited:
  • #7
Nobody with an answer?
 
  • #8
315
1
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.
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.
 
  • #9
Thank you very much! I'll try!
 

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