Separate profit from cash in a dice game

The source uses the usual independent fair-dice model, giving 36 equally likely ordered outcomes. The complete sum table is

There are six cells with a sum at most 4, and thirty with a sum at least 5. Thus the payment probabilities are $1/6$ and $5/6$.

Let A be Alfred’s profit (gain or winnings measured as net change) in one game. This distinction follows the Teacher guidance: his cash balance is $100+A$, not A.

$E(A)=-x(5/6)+(x+8)(1/6)$. Hence expected cash is $100+E(A)=100-x(5/6)+(x+8)(1/6)=304/3-(2/3)x=(304-2x)/3$ pounds, as required.

Writing $A_i$ for the profit in game i, the Teacher completion is $100+E(A_1)+E(A_2)+E(A_3)+E(A_4)+E(A_5)+E(A_6)=100+6E(A)=100+6[-5x/6+(x+8)/6]=100-5x+x+8=108-4x$ pounds. Linearity of expectation only needs each game to have the stated marginal law.

Let B be Bertie’s one-game profit.

$E(B)=x(5/6)-(x+8)(1/6)$. The Teacher completion gives $100+E(B)=100+5x/6-(x+8)/6=296/3+2x/3$ pounds.

As the Teacher note asks, explain fairness: this is a zero-sum transfer game, so equal expected net winnings means each player’s expected profit is zero. Since both start with the same £100, it is equivalent to equating their expected cash balances. The source algebra is $(304-2x)/3=296/3+2x/3$, then $304-2x=296+2x$, $4x=8$, giving $x=2$.

For $x=3$, the profit law is

$E(A)=-3(5/6)+11(1/6)=-2/3$, and $E(A^2)=(-3)^2(5/6)+11^2(1/6)=83/3$. Thus $\operatorname{Var}(A)=E(A^2)-[E(A)]^2=83/3-(2/3)^2=245/9$. Finally $\operatorname{Var}(100+A)=\operatorname{Var}(100)+\operatorname{Var}(A)=0+245/9=245/9$ pounds squared. The zero variance of the constant is explicit in the Teacher completion; this is the constant-shift rule.