# On Lucky Sevens

The Baron's most recent game consisted of a race to complete a trick of four sevens, with the Baron dealing cards from a pristine deck, running from Ace to King once in each suit, and Sir R----- dealing from a well shuffled deck. As soon as either player held such a trick the game concluded and a prize was taken, eleven coins for the Baron if he should have four sevens and nine for Sir R----- otherwise.
The key to reckoning the equity of the wager is to note that it is unchanged should the Baron and Sir R----- take turns dealing out the rest of their cards one by one after the prize has been taken.
If Sir R----- had lost the game then he would most assuredly deal a seven after the Baron and, since the Baron's last seven is seven cards from the end of the deck, there consequently should have been at least one seven in the last seven cards of Sir R------'s deck.
The fairness of the wager is therefore identical to one in which Sir R----- draws first and in which the first player to deal a seven from the bottom of the deck loses. Indeed, I made this observation to the Baron, but am not sure that he grasped its significance.

Now, we can say with full confidence that there are either one or more sevens in Sir R-----'s last seven cards or that there are none and that the chance of the former is therefore one minus the chance of the latter. That probability is the product of the chance that each card drawn is any but one of the four sevens which, given that each draw reduces the number of cards in the deck by one, is given by
$p = \frac{48}{52} \times \frac{47}{51} \times \frac{46}{50} \times \frac{45}{49} \times \frac{44}{48} \times \frac{43}{47} \times \frac{42}{46}$
We can cancel out the 48, 47 and 46 that appear in the top and bottom of the fractions, but the calculation still appears somewhat daunting
$p = \frac{1}{52} \times \frac{1}{51} \times \frac{1}{50} \times \frac{45}{49} \times 44 \times 43 \times 42$
Fortunately, we can simplify our work still further if we factorise each term into a product of primes
\begin{align*} p &= \frac{1}{2^2 \times 13} \times \frac{1}{3 \times 17} \times \frac{1}{2 \times 5^2} \times \frac{3^2 \times 5}{7^2} \times \left(2^2 \times 11\right) \times 43 \times (2 \times 3 \times 7)\\ &= \frac{2^3 \times 3^3 \times 5 \times 7 \times 11 \times 43}{2^3 \times 3 \times 5^2 \times 7^2 \times 13 \times 17}\\ &= \frac{3^2 \times 11 \times 43}{5 \times 7 \times 13 \times 17}\\ &= \frac{4,257}{7,735} \end{align*}
Sir R-----'s expected winnings are therefore
\begin{align*} p \times 9 - (1-p) \times 11 &= \frac{4,257}{7,735} \times 9 - \left(1 - \frac{4,257}{7,735}\right) \times 11\\ &= \frac{4,257}{7,735} \times 20 - 11\\ &= \frac{4,257 \times 4}{7,735 \div 5} - 11\\ &= \frac{17,028}{1,547} - 11\\ &= \frac{17,028 - 1,547 \times 11}{1,547}\\ &= \frac{17,028 - 17,017}{1,547}\\ &= \frac{11}{1,547} \end{align*}
and, given that they are slightly biased in his favour, I should have happily suggested that he take up the Baron's wager.
$$\Box$$

Based upon an article I wrote for ACCU's CVu magazine.

### Gallimaufry  AKCalc
ECMA  Endarkenment
Turning Sixteen

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