On Flannel


The Baron's game of Flannel is one that my fellow students and I are familiar with, it being curiously identical to the game of Sweat-Cloth that is much played in the taverns near our college.

On the face of it, it seems to be an equitable wager, but in truth Sir R----- is at a distinct disadvantage. The simplest explanation[1] as to why is to consider the possible outcomes if he were to bet one coin on each value.
Now if the Baron were to roll three different numbers then Sir R----- would profit by the three coins he had staked upon them and lose the three he had staked upon others; a perfectly equitable outcome. However, if the Baron rolled a double then Sir R----- would have recieved two coins for it, one for the remaining single and lost four for those that didn't show. The situation would have been even worse for Sir R----- if the Baron had rolled a triple; in this case he should have won three coins but lost five!

To figure by just how much the game favours the Baron let us assume that Sir R----- staked a single coin upon the number six.
To calculate his expected winnings we need the probabilities of the Baron rolling zero, one, two or three sixes.
The probability of no sixes is simply
\[ \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{125}{216} \]
The probability of one six is given by
\[ \left(\frac{1}{6} \times \frac{5}{6} \times \frac{5}{6}\right) + \left(\frac{5}{6} \times \frac{1}{6} \times \frac{5}{6}\right) + \left(\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}\right) = \frac{75}{216} \]
Note that there are three terms to sum because the six might turn up on any of the dice.
Similarly, the probability of two sixes is
\[ \left(\frac{5}{6} \times \frac{1}{6} \times \frac{1}{6}\right) + \left(\frac{1}{6} \times \frac{5}{6} \times \frac{1}{6}\right) + \left(\frac{1}{6} \times \frac{1}{6} \times \frac{5}{6}\right) = \frac{15}{216} \]
where each term corresponds to the non-six turning up on each die.
Finally, the probability of three sixes is
\[ \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{216} \]
Now if we multiply Sir R-----'s payoff each outcome by their respective probabilities we recover his average winnings
\[ -1 \times \frac{125}{216} + 1 \times \frac{75}{216} + 2 \times \frac{15}{216} + 3 \times \frac{1}{216} = -\frac{17}{216} \approx -0.08; \]
This confirms our conclusion that Sir R----- should expect a loss and should therefore have declined the Baron's invitation.


[1] Knizia, R. Dice Games Properly Explained, Blue Terrier Press, 2010.

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