Puzzle #97: Pirates and Gold Coins – Part II

I had posted a puzzle earlier (Puzzle #3) – https://alokgoyal1971.com/2013/02/10/puzzle-3-five-pirates-and-looted-gold-coins/. This puzzle is an extension of the same puzzle, and also attributed to the same family – this time to Alok Mittal from his Mathematical Circles group.

Original puzzles goes as follows:

Five pirates have looted 100 gold coins. Now they have to divide their loot.These pirates are very greedy and cruel,so they want to maximize their profit and they don’t shy to kill their pirates. For distribution they decide that senior most pirate will proposes a distribution of the loot. All the pirates will vote, and if at least half accept the proposal, the loot is divided as proposed. If not, the most senior pirate will be killed, and they start over again with the next senior pirate. What solution does the most senior pirate propose? So that he can get the maximum profit, and of course live to have it!

Variation of this puzzle:

Assume there are 6 pirates now – A, B, C, D, E and F. A is the senior most, F being the most junior in that order. They have only 1 gold coin. Like before, for a proposal to be accepted, at least half need to vote in favor. What should be A’s strategy?

To avoid any doubt about the pirates’ behavior, following defines the order of prioritization for them:

– Desire to live

– Greed (have as many gold coins as possible)

– Cruelty – all else being equal, they would rather see someone else being killed?

What is A’s strategy?

As always, please send your answers directly to me at alokgoyal_2001@yahoo.com. If you like the puzzle, please share it with others. If you have interesting puzzles to share, please send them to me at my e-mail given above.


Happy strategizing!


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4 Responses to Puzzle #97: Pirates and Gold Coins – Part II

  1. P R KUMAR says:


    A new puzzle: Two persons relative to a teacher are talking to each other. One is father of the teacher’s son and another is son of the teacher’s father. But the teacher is not there in the conversation. Who is talking to whom?

  2. P R KUMAR says:

    Dear sir,

    Following is not only a puzzle but my doubt also. So please clarify:

    A company has 12 salespersons, of which 6 are men and 6 are women. The company wishs to send them to 3 cities, 2 men and 2 women to each city. In how many ways can this be done ?

  3. Karan says:

    Ways to choose 2 men and 2 women out of 12 people will be 6c2 x 6c2 which equals 225.Out of the remaining 4 men and 4 women you can choose 2 men and 2 women using 4c2 x 4c2 which equals 36. Only 2 men and 2 women being left now there is only one way to pick them up . Hence total cases are 225x36x1x3! Factorial three because the cities can be different for a different group of people. 🙂

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