deeper probability test (1 Viewer)

[OLDMAN]

A test on recursive probability tree.

Di and Dot bet on the total roll of two standard dice. Di bets that a 12 will be rolled first. Dot bets that two consecutive 7's will be rolled first. They keep rolling until one wins. What is the probability that Di will win?

Sorry about the silly names, can't help myself.

 

[Affinity]

Is this the 'finite state machine' method?

define 2 'states'

(1) The initial state or when the last toss was not a 7
(2) Last toss was a 7.

Let A and B be the probability of Dot winning at state 1 and 2 respectively.

A = (29/36)A + (1/6)B ....................(1)
B = 1/6 + (29/36)A ........................(2)

solving, gives A=6/13, Dot's chance of winning
so Di's chance = 7/13

 

[OLDMAN]

Check Affinity's answer using "4-unit" methodology.

The key to problems of this type :mutual exclusivity, and geometric series.

Consider all possible outcomes.

Di wins 1st toss.............................1/36.............Dot losses
Neither 7 nor 12 1st toss..............29/36...........Back to "Start"
7 1st toss, Di wins 2nd toss..........6/36*1/36....Dot loses
7 1st toss, neither 7/12 2nd toss..6/36*29/36..Back to "Start"
7 1st and 2nd toss............................1/36..........Dot wins.

P(Di wins)=1/36+1/36*P(Back to "Start")+1/36*P(Back to "Start")^2+1/36*P(Back to "Start")^3+....
Now P(Back to "Start")=29/36+6/36*29/36=203/216

P(Dot wins)=6/13

 

[Affinity]

let me explain the other solution in a bit more detail:

The chance of Dot winning is constant in each state (1 and 2), it doesn't depend on the number of throws prior to it.

Examining state 1,

1/6 of the tosses will sum to 7, which changes the state to (2)

1/36 of the tosses will sum to 12, in this case Di wins

29/36 of the tosses will not sum to 7 nor 12, and the state remains (1).

therefore A= (1/6)B + (29/36)A

examining state (2)

1/6 of the throws will result in a 7 & Dot wins.

1/36 of the throws will be a 12 and Di wins

29/36 of the throws will turn out to be neither, which changes the state back to (1)

hence B= 1/6 + (29/36)A


Di's chance of winning could be found by this method directly too (without finding dot's first)

let Di's chance of winning in states 1 and 2 be C and D respectively

C = 1/36 + (1/6)D +(29/36)C
D = 1/36 + (29/36)C

solving,

C = 1/36 + (1/6)*[1/36 + (29/36)C] +(29/36)C

C = 7/216 + [(203)/216]C

(13/216)C = 7/216

C = 7/13, Di's Chance of winning initially