A forum for good logic/math puzzles.

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Postby Jayraj » Sat Nov 24, 2018 3:37 am UTC

Jack invited two of his friends, Paul and John, to play poker at his place with gold coins.

After the game, they piled up all the winnings on the table.

Jack claims to have won half the pot and took some coins.

Paul claims to have won one-third and took some coins.

Finally, John took the rest.

Jack realised that he took too much, thus returning half on the table.

Paul decided to follow and returned one-third.

And John at his turn returned one-sixth.

The coins left on the table was then shared out equally and they had 42 pieces each.

Jack has now half of the original pot.

Paul has one-third and John has one-sixth.

What is the pot value originally and how much did each of them take at first?

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Joined: Tue Feb 03, 2009 2:05 pm UTC

Re: Poker

Postby Yat » Mon Nov 26, 2018 1:15 pm UTC

Brutally arithmetical solution:
Call A, B and C the number of coins each initially takes, translate what we know into three equations (for example, 2B/3+42 = (A+B+C)/3, because after Paul removed 1/3 of his initial stock and getting 42 coins back, he has 1/3 of the total), then solve the system. 282 coins are on the table, Jack takes 198, puts 99 back and gets 42, he is left with 141 which is half of the pot. Paul takes 78, puts 26 back and takes 42 again, to end up with 94, John takes the remaining 6, puts 1 back and takes 42, to end up with 47.
The first division is 198+78+6 = 282.
The final division is 141+94+47 = 282.
What's on the table between the two is 99+26+1 = 126 = 3*42.
There probably is a smarter way to do this.

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