# On May The Fours Be With You

In their most recent wager Sir R-----'s goal was to guess the outcome of the Baron's roll of four four sided dice at a cost of four coins and a prize, if successful, of forty four. On the face of it this seems a rather meagre prize since there are two hundred and fifty six possible outcomes of the Baron's throw. Crucially, however, the fact that the order of the matching dice was not a matter of consequence meant that Sir R-----'s chances were significantly improved.

Specifically, if Sir R----- were to have chosen four different faces for his dice then he would certainly have had a match for the first die rolled by the Baron. Moreover he should then have had three chances out of four of matching the second, two chances out of four of matching the third and one chance out of four of matching the last. His expected winnings would consequently have been
$1 \times \frac34 \times \frac24 \times \frac14 \times 44 - 4 = \frac{6}{64} \times 44 - 4 = 4 \times \left(\frac{3}{32} \times 11 - 1\right) = 4 \times \frac{33-32}{32} = 4 \times \frac{1}{32} = \frac18$
By a pleasing coincidence, if we round $$\pi$$ to three decimal places, which is to say $$3.142$$, the digits are a winning choice for Sir R-----!

Now, if he were to have made any other choice then there would have been at least one face that would have represented a loss for him had the Baron's thrown it with any of his dice. His chance of victory would hence have been no greater than
$\frac34 \times \frac34 \times \frac24 \times \frac14 = \frac{18}{256} = \frac{9}{128}$
yielding an expected prize less than or equal to
$4 \times \left(\frac{9}{128} \times 11 - 1\right) = 4 \times \frac{99-128}{128} = 4 \times -\frac{29}{128} = -\frac{29}{32}$
It would therefore have been in Sir R-----'s interest to accept the Baron's wager, provided that he chose his dice wisely.
$$\Box$$

### Gallimaufry  AKCalc
ECMA  Endarkenment
Turning Sixteen

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