That one I'm pretty low on ideas of how to approach it.

Five pirates of different ages have a treasure of 50 gold coins. On their ship, they decide to split the coins using this scheme: The oldest pirate proposes how to share the coins, all of them will then vote for or against it. If more than a half of pirates vote for it, then the coins will be shared that way. Otherwise, the pirate proposing the scheme will be thrown overboard, and the process is repeated with the pirates that remain. Assuming that all 5 pirates are intelligent, rational, greedy and do not wish to die, how much will the 2nd youngest pirate get?

  • 97
  • 98
  • 0
  • 20

An important detail in the problem is missing: you need to specify how a pirate will vote if he gets the same amount of money either way, and does not die either way. I will assume that in this case the pirate is malicious and will automatically vote against.

Now, we start with the case of $1$ pirate, and slowly increase the number of pirates.

  • If there is ever one pirate remaining, she will just take all the 50 coins for herself and vote for her own proposal.

  • If there are two pirates remaining, the younger pirate knows that she can get $50$ coins herself if the older pirate's proposal fails. Thus she votes against the proposal no matter what. The older pirate dies, and the younger pirate keeps $50$.

  • If there are three pirates remaining, the second-to-oldest pirate does not want to die, and he knows he will die if the proposal does not go through. Thus he votes for the proposal no matter what. Given this, the oldest pirate takes all the money ($50$) for herself, and gets $2$ out of $3$ votes (including her own), and the proposal passes.

Here is where it gets interesting.

  • If there are four pirates remaining, the oldest needs to win $3$ votes for his proposal to pass (including his own vote). The second-to-oldest will vote against no matter what, because she knows if the proposal fails she will get all of the gold. So the oldest cannot win her vote. But the two youngest pirates know they will get nothing if the proposal fails. Thus the oldest pirate need only bribe them with $1$ gold each, and they vote for him. So, the oldest pirate proposes $48$ for himself and $1$ for each of the two youngest, and the proposal passes.

  • Finally, if there are five pirates remaining, the oldest also needs to win $3$ votes. In order to win the second-to-oldest's vote, she needs to pay him $49$ gold. In order to win the third-to-oldest's vote, she needs to pay her $1$. And in order to win the two youngest's votes, she needs to pay them $2$ each. So she pays $2$ to one of the two youngest, and $1$ to the third-oldest, and keeps $47$ for herself.

The 2nd to youngest pirate will either get $0$ or $2$, depending on how the oldest pirate chooses to propose. Thus this problem is ill-posed. (I also am confused why $97$ and $98$ are listed as answers with only $50$ gold total...)