I'm stuck on a mathematical riddle from Dudeney's Amusement in Mathematics.

I will write here the text of the riddle and my attempt. I just don't get how to proceed further.

Seven men, whose names were Adams, Baker, Carter, Dobson, Edwards, Francis, and Gudgeon, were recently engaged in play. The name of the particular game is of no consequence. They had agreed that whenever a player won a game he should double the money of each of the other players—that is, he was to give the players just as much money as they had already in their pockets. They played seven games, and, strange to say, each won a game in turn, in the order in which their names are given. But a more curious coincidence is this—that when they had finished play each of the seven men had exactly the same amount—two shillings and eightpence—in his pocket. The puzzle is to find out how much money each man had with him before he sat down to play.


I started to mind like this: first of all $2$ shillings and $8$ pence are $32$ pence ($1$s = $12$p).

Hence at the end they all have $32$ p, which means in total there are $224$ p since there are seven players.

Now, Assume the players are called $A B C D E F G$.

$A$ is the first and he possesses $a$ pence, hence the other ones, in total, have $R$ pence.

The first useful equation is trivial and its

$$a + R = 224$$

Now: $A$ wins first, hence he has to pay everyone else twice what they have.

But $R$ is what they all have, hence he pays $R$: after the first turn $A$ remains with $a - R$ pence.

But then he always loses (other wins), and each turn until the end $A$ gets his sum doubled. In this way the turn are described by:

  1. $a - R$

  2. $2(a-R)$

  3. $4(a-R)$


  1. $64(a.R)$

After seven rounds, $A$ has $64(a-R)$ pence, which must be equal to $32$ since the text says it's what they all have at the end.

Trivially we conclude

$$ \begin{cases} a + R = 224 \\ a - R = 1/2 \end{cases} $$

Which means $a = 112.25$ which in farthings is $449$ f.

Now my problem is to understand how to get the sum of the other players. I tried with a similar method but I failed.

Any help?

  • 6
    $\begingroup$ Why don't you analyze it in reverse? How much did each have when G won? How much when F won? Etc. $\endgroup$ – Aravind Mar 16 '17 at 16:22
  • $\begingroup$ What does "a queer coincidence" have to do with this problem? Is that the name of the riddle in the book? $\endgroup$ – Stella Biderman Mar 16 '17 at 16:30
  • $\begingroup$ @StellaBiderman yes it's the name of the riddle! $\endgroup$ – Turing Mar 16 '17 at 16:33
  • 1
    $\begingroup$ @StellaBiderman: Yes, that is in fact the name of this riddle, which appears on page 2 of the book (1917). $\endgroup$ – hardmath Mar 16 '17 at 16:34
  • $\begingroup$ @Aravind Smooth! Seems like I have to take loopholes instead of creating equations :D $\endgroup$ – Turing Mar 16 '17 at 16:40

It is essentially bookkeepping (or literally bookkeeping, since they are playing for money). Working backwards, each player has half as much as the subsequent round, except the "winner" of the next round additionally has half the total of $7\times 32=224$, so $112$ more than half.

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  • $\begingroup$ Thanks to this table I realised that the riddle was incredibly easy -.- Thank you!! $\endgroup$ – Turing Mar 16 '17 at 16:59

Hint: After $G$ wins they each have $32$, so before the last game the first six had $16$ and $G$ had $128$. Then before $F$ wins they had ????

  • $\begingroup$ I don't know.. Using that method I get a wrong answer for $A$. $\endgroup$ – Turing Mar 16 '17 at 16:50
  • $\begingroup$ Without seeing your work I can't tell what is wrong. What do you have before F wins? $\endgroup$ – Ross Millikan Mar 16 '17 at 16:52
  • $\begingroup$ My bad! All ok :D $\endgroup$ – Turing Mar 16 '17 at 16:59

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