Mensa Math Brain Teasers (Solution 2)

 In this  simple dice game, let p be the probability of winning on a single turn. 

Let q= 1- p be the probability of not winning.  Jeff wins with probability p on his first turn.  The probability that neither Jeff nor George win on their first turn is q², in which case the game essentially starts over.  Therefore p _Jeff , the probability that Jeff wins =  p + q² p _Jeff .  

Further:

P _Jeff   =  p/(1 -  q²) =  p/ {(1 + q) (1- q)} = p/ (1 + q)p = 1/ (1 + q)  

p =   1/6 (The probability of throwing a 7 with two fair dice)

Then:  q = 5/6

P _Jeff   =   1/ (1 + q)  =  1/ (1 + 5/6) =  1/ (11/6)  = 6/ 11  

Now, let G be the average length of the game.

Then G = 1 with probability p.

G = G + 1 with probability q.

G = 1 (1/6) + (G + 1) (5/6) =  (5/6) G + 1

Subtract (5/6) G from both sides:

(1/6) G =1  ® G =6

The average length of the game is 6 rolls of the dice.