Throwing dice

This was an interesting question posted somewhere on the Internet. I rephrased it a bit as I understood it.

Suppose you're playing the following game: Each time you throw a dice you get the result payed out. If you throw 4 or more, you can throw again; otherwise, the game stops. How much are you willing to pay to play the game? 

The solution I came up with:

 If you write down a tree, you can see you're calculating the limit of: 


(1/6+...+6/6) + (3/6)( (1/6+...+6/6) + (3/6)( (1/6+...+6/6) + ... )) 


of which the fixed point is 


ϕ = (1/6+...+6/6) + (3/6)(ϕ) 


is ϕ = (7/2) + (1/2)(ϕ) 


is (1/2)(ϕ) = (7/2) 


is ϕ = 7


At most 7.


Corollary: if you take ϕ = (2/2) + (5/2) + (1/2)(ϕ) it is trivial to expand ϕ to (2/2) + (7/4) + (12/8) + (17/16) + ...