12 thoughts on “Puzzles – 2

  1. regarding the gang of pirate problem if in the end he needs only 50% support
    he could give money to only four of them(oldest 4 other than him) and garner 50% support
    2nd oldest gets =125
    3rd oldest gets =143
    4th oldest gets=167
    5th oldest gets =200
    thus finally our first pirate can make 364

    • You’re right about the four people, but he can be more conservative in the distribution of amounts and keep even more for himself 🙂 Try the case of 2 pirates, 3 pirates and so on…


      • sir if first one dies second one can get a max for himself of 999/5 , if second also dies then third will keep 999/4 and so on.. so after rechecking it seems he will end up with even lesser amount…coz to get their vote he will have to offer them a coin more than what they will get if they get to distribute

      • You will get to see the solution on Tuesday. It is extremely counter-intuitive at first, but if you think about it carefully it makes sense. Remember that I don’t need ALL the others to support me. And also remember that if I die, the ones who I am giving coins to might not be useful to the next guy.


    • They work for any two consecutive numbers….because that is the only case where you can guarantee “either more heads or more tails but not both” In the case of 100 and 98 for example, we would see thatit is possible to have more heads and more tails simultaneously (for example 50 H 50 T v/s 49H 49T)


      • yup got this one… it means if suppose the problem was about 98 and 100 we will hav to list all possible cases( which i itried but found quite taxing) ..but in consecutive there are only two cases so 1/2

  2. Sir, in question of jai and veeru, we are just concerned with more head and not taking tail into account Then Why we are considering either more head or more tail.Why not the cases when H(jai)=H(veeru), like jai getting 50 head and veeru too. This is a case for draw. then how a winner will be decided??

  3. In that case neither will win. There can be 3 cases – J wins, V wins, or tie.The probability of either of them winning is slightly less than 1/2, and that of tie is very very small but non-zero.


  4. hi J, in the coin question

    when I consider the number of coins to be 24, then there is this case P(12 heads)= P(12 Tails) as you have mentioned. So, since this puzzle asks for a winner tie case is ruled out.

    So, the total probability when the no. of coins is even is 1-(1/24)? I am subtracting the probability in which we get a tie.

    Is this correct?

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