le Monde puzzle [#745]

The puzzle in Le Monde this weekend is not that clear (for a change!), so I may be confused in the following exposition:

Three card players are betting with a certain (and different) number of chips each, between 4 and 9. After each game, the looser doubles the number of chips of the winner (while the second keeps her chips). The game stops if the looser cannot pay the winner. Find the initial configuration such that, at some point in the party, all players have the same number of chips.

So, if (x1,x2,x3) is the chip configuration at time t for the ordered players, (2x1,x2,x3-x1) is the chip configuration at time (t+1). Rather than running an exhaustive search, given the limited number of possibilities, I decided to search at random, ending up with the R function

lemonde=function(chip,chap){

x=rep(-1,3)
while (min(x)<0){

  start=x=sample(chip:chap,3)
  while (length(unique(start))==1)
    start=x=sample(chip:chap,3)

  while ((min(x)>-1)&&(length(unique(x))>1)){
    x=sample(x)  #random winner
    x=c(2*x[1],x[2],x[3]-x[1])
    }}

list(start=sort(start),finish=x)
}

leading to

> lemonde(4,9)
$start
[1] 7 8 9

$finish
[1] 8 8 8

with a unique starting point. More interestingly, other configurations may have several starting points. Of course, a mathematical analysis of the problem would bring more light on the difference. Maybe the issue of Le Monde next weekend (i.e., tonight!) will be enough.

> lemonde(4,10)
$start
[1]  5  9 10

$finish
[1] 8 8 8

> lemonde(4,10)
$start
[1]  6  8 10

$finish
[1] 8 8 8

> lemonde(4,10)
$start
[1] 7 8 9

$finish
[1] 8 8 8

3 Responses to “le Monde puzzle [#745]”

  1. […] a loooong break, here is one Le Monde mathematical puzzle I had time to look at, prior to going to Dauphine for a […]

  2. When the game ends, the player who just won has an even number of chips. Hence the only possible end distributions are (6,6,6) and (8,8,8).

    It is easy to check that the only distributions leading to (6,6,6) are (3,6,9) and (3,3,12), neither of which is admissible. Hence the end distribution is (8,8,8).

    If there are 24 chips in total, the initial distribution must be (7,8,9) or (6,9,9). (6,9,9) cannot work: all these numbers are multiples of 3, hence this will still be true at all subsequent rounds.

    Starting at (7,8,9) there is the path
    (7,8,9)->(14,8,2)->(12,8,4)->(8,8,8).

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.