# On The Fours Reawakens

Recall that the Baron's latest game involved he and Sir R----- each casting four four sided dice. For each pair of dice formed from one of the Baron's and one of Sir R-----'s with matching faces, Sir R----- should have received thirteen coins having given twenty eight to play.

To reckon the outcome of this wager we can proceed backwards from Sir R-----'s fourth cast to his first, calculating the expected number of matches as we do so. There are but four arrangements of the Baron's dice that might remain had Sir R----- made no matches with his first three throws; one pair and two singles, two pairs, a triplet and a single, and four of a kind. Using different letters to represent different, albeit unspecified, faces the expected number of matches for these are
\begin{align*} \mathrm{E}_1[\{a,a,b,c\}] &= \tfrac34\\ \mathrm{E}_1[\{a,a,b,b\}] &= \tfrac12\\ \mathrm{E}_1[\{a,a,a,b\}] &= \tfrac12\\ \mathrm{E}_1[\{a,a,a,a\}] &= \tfrac14 \end{align*}
Similarly, there are three arrangements that might remain had Sir R----- previously made one match, having expectations of a final match
\begin{align*} \mathrm{E}_1[\{a,b,c\}] &= \tfrac34\\ \mathrm{E}_1[\{a,a,b\}] &= \tfrac12\\ \mathrm{E}_1[\{a,a,a\}] &= \tfrac14 \end{align*}
Had he made two matches then the possible states of the Baron's remaining dice and Sir R-----'s expected matches for them are
\begin{align*} \mathrm{E}_1[\{a,b\}] &= \tfrac12\\ \mathrm{E}_1[\{a,a\}] &= \tfrac14 \end{align*}
Finally, if but one of the Baron's dice remains unpaired, we have
$\mathrm{E}_1[\{a\}] = \tfrac14$
To figure Sir R-----'s expectation when he has two dice to cast we must include his expectation for his remaining die as well as the pair, or lack thereof, he should make with the first of them. If the Baron has a pair and two singles this comes out to
\begin{align*} \mathrm{E}_2[\{a,a,b,c\}] &= \tfrac14 \times \left(1+\mathrm{E}_1[\{a,a,b\}]\right) + \tfrac14 \times \left(1+\mathrm{E}_1[\{a,a,c\}]\right)\\ &+ \tfrac14 \times \left(1+\mathrm{E}_1[\{a,b,c\}]\right) + \tfrac14 \times \left(0+\mathrm{E}_1[\{a,a,b,c\}]\right)\\ &= \tfrac12 \times \left(1+\mathrm{E}_1[\{a,a,b\}]\right) + \tfrac14 \times \left(1+\mathrm{E}_1[\{a,b,c\}]\right) + \tfrac14 \times \left(0+\mathrm{E}_1[\{a,a,b,c\}]\right)\\ &= \tfrac12 \times \tfrac32 + \tfrac14 \times \tfrac74 + \tfrac14 \times \tfrac34 = \tfrac{12+7+3}{16} = \tfrac{11}{8} \end{align*}
For the remaining sets of four dice we have
\begin{align*} \mathrm{E}_2[\{a,a,b,b\}] &= \tfrac12 \times \left(1+\mathrm{E}_1[\{a,a,b\}]\right) +\tfrac12 \times \left(0+\mathrm{E}_1[\{a,a,b,b\}]\right) = \tfrac12 \times \tfrac32 + \tfrac12 \times \tfrac12 = 1\\ \mathrm{E}_2[\{a,a,a,b\}] &= \tfrac14 \times \left(1+\mathrm{E}_1[\{a,a,a\}]\right) + \tfrac14 \times \left(1+\mathrm{E}_1[\{a,a,b\}]\right) + \tfrac12 \times \left(0+\mathrm{E}_1[\{a,a,a,b\}]\right)\\ &= \tfrac14 \times \tfrac54 + \tfrac14 \times \tfrac32 + \tfrac12 \times \tfrac12 = \tfrac{5+6+4}{16} = \tfrac{15}{16}\\ \mathrm{E}_2[\{a,a,a,a\}] &= \tfrac14 \times \left(1+\mathrm{E}_1[\{a,a,a\}]\right) + \tfrac34 \times \left(0+\mathrm{E}_1[\{a,a,a,a\}]\right) = \tfrac14 \times \tfrac54 + \tfrac34 \times \tfrac14 = \tfrac{5+3}{16} = \tfrac12 \end{align*}
Similarly, for those of three
\begin{align*} \mathrm{E}_2[\{a,b,c\}] &= \tfrac34 \times \left(1+\mathrm{E}_1[\{a,b\}]\right) + \tfrac14 \times \left(0+\mathrm{E}_1[\{a,b,c\}]\right) = \tfrac34 \times \tfrac32 + \tfrac14 \times \tfrac34 = \tfrac{18+3}{16} = \tfrac{21}{16}\\ \mathrm{E}_2[\{a,a,b\}] &= \tfrac14 \times \left(1+\mathrm{E}_1[\{a,a\}]\right) + \tfrac14 \times \left(1+\mathrm{E}_1[\{a,b\}]\right) + \tfrac12 \times \left(0+\mathrm{E}_1[\{a,a,b\}]\right)\\ &= \tfrac14 \times \tfrac54 + \tfrac14 \times \tfrac32 + \tfrac12 \times \tfrac12 = \tfrac{5+6+4}{16} = \tfrac{15}{16}\\ \mathrm{E}_2[\{a,a,a\}] &= \tfrac14 \times \left(1+\mathrm{E}_1[\{a,a\}]\right) + \tfrac34 \times \left(0+\mathrm{E}_1[\{a,a,a\}]\right) = \tfrac14 \times \tfrac54 + \tfrac34 \times \tfrac14 = \tfrac{5+3}{16} = \tfrac12 \end{align*}
and those of two
\begin{align*} \mathrm{E}_2[\{a,b\}] &= \tfrac12 \times \left(1+\mathrm{E}_1[\{a\}]\right) + \tfrac12 \times \left(0+\mathrm{E}_1[\{a,b\}]\right) = \tfrac12 \times \tfrac54 + \tfrac12 \times \tfrac12 = \tfrac{5+2}{8} = \tfrac78\\ \mathrm{E}_2[\{a,a\}] &= \tfrac14 \times \left(1+\mathrm{E}_1[\{a\}]\right) + \tfrac34 \times \left(0+\mathrm{E}_1[\{a,a\}]\right) = \tfrac14 \times \tfrac54 + \tfrac34 \times \tfrac 14 = \tfrac{5+3}{16} = \tfrac12 \end{align*}
Next, Sir R-----'s expectation when he has three dice to roll and the Baron has four remaining are
\begin{align*} \mathrm{E}_3[\{a,a,b,c\}] &= \tfrac14 \times \left(1+\mathrm{E}_2[\{a,b,c\}]\right) + \tfrac12 \times \left(1+\mathrm{E}_2[\{a,a,b\}]\right) + \tfrac14 \times \left(0+\mathrm{E}_2[\{a,a,b,c\}]\right)\\ &= \tfrac14 \times \tfrac{37}{16} + \tfrac12 \times \tfrac{31}{16} + \tfrac14 \times \tfrac{11}{8} = \tfrac{37+62+22}{64} = \tfrac{121}{64}\\ \mathrm{E}_3[\{a,a,b,b\}] &= \tfrac12 \times \left(1+\mathrm{E}_2[\{a,a,b\}]\right) + \tfrac12 \times \left(0+\mathrm{E}_2[\{a,a,b,b\}]\right) = \tfrac12 \times \tfrac{31}{16} + \tfrac12 \times 1 = \tfrac{31+16}{32} = \tfrac{47}{32}\\ \mathrm{E}_3[\{a,a,a,b\}] &= \tfrac14 \times \left(1+\mathrm{E}_2[\{a,a,a\}]\right) + \tfrac14 \times \left(1+\mathrm{E}_2[\{a,a,b\}]\right) + \tfrac12 \times \left(0+\mathrm{E}_2[\{a,a,a,b\}]\right)\\ &= \tfrac14 \times \tfrac32 + \tfrac14 \times \tfrac{31}{16} + \tfrac12 \times \tfrac{15}{16} = \tfrac{24+31+30}{64} = \tfrac{85}{64}\\ \mathrm{E}_3[\{a,a,a,a\}] &= \tfrac14 \times \left(1+\mathrm{E}_2[\{a,a,a\}]\right) + \tfrac34 \times \left(0+\mathrm{E}_2[\{a,a,a,a\}]\right) = \tfrac14 \times \tfrac32 + \tfrac34 \times \tfrac12 = \tfrac{3+3}8 = \tfrac34 \end{align*}
and those when the Baron has but three
\begin{align*} \mathrm{E}_3[\{a,b,c\}] &= \tfrac34 \times \left(1+\mathrm{E}_2[\{a,b\}]\right) + \tfrac14 \times \left(0+\mathrm{E}_2[\{a,b,c\}]\right) = \tfrac34 \times \tfrac{15}8 + \tfrac14 \times \tfrac{21}{16} = \tfrac{90+21}{64} = \tfrac{111}{64}\\ \mathrm{E}_3[\{a,a,b\}] &= \tfrac14 \times \left(1+\mathrm{E}_2[\{a,a\}]\right) + \tfrac14 \times \left(1+\mathrm{E}_2[\{a,b\}]\right) + \tfrac12 \times \left(0+\mathrm{E}_2[\{a,a,b\}]\right)\\ &= \tfrac14 \times \tfrac32 + \tfrac14 \times \tfrac{15}{8} + \tfrac12 \times \tfrac{15}{16} = \tfrac{12+15+15}{32} = \tfrac{21}{16}\\ \mathrm{E}_3[\{a,a,a\}] &= \tfrac14 \times \left(1+\mathrm{E}_2[\{a,a\}]\right) + \tfrac34 \times \left(0+\mathrm{E}_2[\{a,a,a\}]\right) = \tfrac14 \times \tfrac32 + \tfrac34 \times \tfrac12 = \tfrac{3+3}8 = \tfrac34 \end{align*}
Finally, we can deduce that Sir R-----'s expected number of matches at the outset are given by
\begin{align*} \mathrm{E}_4[\{a,b,c,d\}] &= 1 + \mathrm{E}_3[\{a,b,c\}] = 1 + \tfrac{111}{64} = \tfrac{175}{64}\\ \mathrm{E}_4[\{a,a,b,c\}] &= \tfrac12 \times \left(1+\mathrm{E}_3[\{a,a,b\}]\right) + \tfrac14 \times \left(1+\mathrm{E}_3[\{a,b,c\}]\right) + \tfrac14 \times \left(0+\mathrm{E}_3[\{a,a,b,c\}]\right)\\ &= \tfrac12 \times \tfrac{37}{16} + \tfrac14 \times \tfrac{175}{64} + \tfrac14 \times \tfrac{121}{64} = \tfrac{296+175+121}{256} = \tfrac{37}{16}\\ \mathrm{E}_4[\{a,a,b,b\}] &= \tfrac12 \times \left(1+\mathrm{E}_3[\{a,a,b\}]\right) + \tfrac12 \times \left(0+\mathrm{E}_3[\{a,a,b,b\}]\right) = \tfrac12 \times \tfrac{37}{16} + \tfrac12 \times \tfrac{47}{32} = \tfrac{74+47}{64} = \tfrac{121}{64}\\ \mathrm{E}_4[\{a,a,a,b\}] &= \tfrac14 \times \left(1+\mathrm{E}_3[\{a,a,a\}]\right) + \tfrac14 \times \left(1+\mathrm{E}_3[\{a,a,b\}]\right) + \tfrac12 \times \left(0+\mathrm{E}_3[\{a,a,a,b\}]\right)\\ &= \tfrac14 \times \tfrac74 + \tfrac14 \times \tfrac{37}{16} + \tfrac12 \times \tfrac{85}{64} = \tfrac{56+74+85}{128} = \tfrac{215}{128}\\ \mathrm{E}_4[\{a,a,a,a\}] &= \tfrac14 \times \left(1+\mathrm{E}_3[\{a,a,a\}]\right) + \tfrac34 \times \left(0+\mathrm{E}_3[\{a,a,a,a\}]\right) = \tfrac14 \times \tfrac74 + \tfrac34 \times \tfrac34 = \tfrac{7+9}{16} = 1 \end{align*}
To reckon Sir R-----'s overall expectation we must take the probabilities that these arrangements of the Baron's dice should be observed
\begin{align*} \Pr(\{a,b,c,d\}) &= \left(1 \times \tfrac34 \times \tfrac12 \times \tfrac14\right) \times 1 = \tfrac{3}{32}\\ \Pr(\{a,a,b,c\}) &= \left(1 \times \tfrac14 \times \tfrac34 \times \tfrac12\right) \times 6 = \tfrac{9}{16}\\ \Pr(\{a,a,b,b\}) &= \left(1 \times \tfrac14 \times \tfrac34 \times \tfrac14\right) \times 3 = \tfrac{9}{64}\\ \Pr(\{a,a,a,b\}) &= \left(1 \times \tfrac14 \times \tfrac14 \times \tfrac34\right) \times 4 = \tfrac{3}{16}\\ \Pr(\{a,a,a,a\}) &= \left(1 \times \tfrac14 \times \tfrac14 \times \tfrac14\right) \times 1 = \tfrac{1}{64} \end{align*}
which when multiplied by their respective expectations yield
\begin{align*} \Pr(\{a,b,c,d\}) \times \mathrm{E}_4[\{a,b,c,d\}] &= \frac{3}{32} \times \frac{175}{64} = \frac{525}{2,048}\\ \Pr(\{a,a,b,c\}) \times \mathrm{E}_4[\{a,a,b,c\}] &= \frac{9}{16} \times \frac{37}{16} = \frac{333}{256}\\ \Pr(\{a,a,b,b\}) \times \mathrm{E}_4[\{a,a,b,b\}] &= \frac{9}{64} \times \frac{121}{64} = \frac{1,089}{4,096}\\ \Pr(\{a,a,a,b\}) \times \mathrm{E}_4[\{a,a,a,b\}] &= \frac{3}{16} \times \frac{215}{128} = \frac{645}{2,048}\\ \Pr(\{a,a,a,a\}) \times \mathrm{E}_4[\{a,a,a,a\}] &= \frac{1}{64} \times 1 = \frac{1}{64} \end{align*}
the sum of which is the value that we require
$\frac{525}{2,048} + \frac{333}{256} + \frac{1,089}{4,096} + \frac{645}{2,048} + \frac{1}{64} = \frac{1,050+5,328+1,089+1,290+64}{4,096} = \frac{8,821}{4,096}$
Sir R-----'s expected winnings are consequently
$\frac{8,821}{4,096} \times 13 - 28 = \frac{8,821 \times 13 - 28 \times 4,096}{4,096} = \frac{114,673-114,688}{4,096} = -\frac{15}{4,096}$
and I should have recommended that he declined the Baron's invitation!
$$\Box$$