Let’s denote participants “A” (Johnny), “B” (John), and “L” (Lola). x, y – discount factors for A and B, correspondingly. Analysis.

• There are several factors (assumptions) to be taken into consideration: (A1) Envy vs. ‘no envy’: whether ‘A’ and ‘B’ indifferent towards the utility increase of the other player. (A2) Informational transparency of the preferences of A,B: information can be known or unknown to players; (A3) Evaluation rule of “L”: we have to make assumptions about ‘love’ and ‘friendship’. (A3.1)‘Love’ may mean abnegation (desire to leave as much as possible of the pie to ‘L’). In this case, ‘ differential penalty function’ should be added to the model ( if someone eats more than fair half – he gets penalized, the more he eats the more social disdain he gets)
• (A3.2)‘Friendship’ may mean a) Equality (without taking into account utilities) b) desire to take into account the demand (utility function) of the partner c) Generosity – ultimate intention to benefit a friend
• (A4) Information transparency of the evaluation rules of ‘L’ for players
• (A5) stationarity of the utility functions (utility does not depend on time
• (A6) players are strategically rational (able to calculate long term consequences)

Scenario 1 (social choice – A3.1 and A3.2 hold).

If ‘L’ evaluates love and friendship of contenders, the ‘most social’ and generous solution could be to leave the entire pie untouched to ‘L’, presuming that “L” likes pies. Another social option is to leave one third to ‘L’ and divide the remaining part between A and B equally. This approach demonstrates both love and friendship. The third best choice is to divide the pie equally between A and B, leaving nothing to L. This would prove the friendship only.

Scenario 2 External observation does not matter, self-interests only are considered by the players, full information is available

Solve the problem for n=6 starting from the end. At the last step, counter offer goes to A, and B proposes (0,1). If (A1) holds, A accepts the offer. Now consider the step 5. Moving from the step 5 to the step 6 , B suffers loss of y. This means that A can say: listen, you are indifferent towards my gain, so just give me the residual (1-y) and get at the step 5 your ‘y’, which you obtain anyway at the step 6.

So, at the step 5 we get a payout of {(1-y),y}. Similarly, B, when he makes his offer from the step 4, understands that A will be indifferent between getting his share ‘x’ times less at the step 5 or at step 6. (if A5 holds).

For that reason, the proposal of B at the step 4 is {x(1-y); 1-x(1-y)}.

B gets this proposal from the step3 from A, who is aware of the discount y. Thus, his offer becomes { 1-y(1-x(1-y)); y(1-x(1-y)) }.

Again, A gets input from the step 2: a rational friend B takes into account the discount x. {x(1-y(1-x(1-y))) ; 1-x(1-y(1-x(1-y)))}. Finally, A’s rational offer at the step one, which should be accepted immediately, is { 1-y(1-x(1-y(1-x(1-y)))); y(1-x(1-y(1-x(1-y))))}.

Opening brackets, we get the solution for n=6: { 1- (y – yx + xy 2 - x 2 y 2 + x 2 y 3 ) ; y – yx + xy 2 - x 2 y 2 + x 2 y 3 }

If “L” doesn’t come, the game becomes infinite. Denote: A’s offer (a1,a2) and B’s (b1,b2)). We can note that in the equilibrium a2=y b2, and b1=x a1. Taking into account that a2=1-a1, and b2=-b1, we get A’s proposal {(1-y)/(1-xy); y(1-x)/(1-xy)} the equilibrium solution that should be accepted by B.