**4=5: July 21, 2010**

(4-9/2)^2=(5-9/2)^2

(-1/2)^2=(1/2)^2

4-9/2=-1/2, while 5-9/2=+1/2

-1/2 squared equals 1/4, as does +1/2 squared. But just because (-1/2)^2=(+1/2)^2,

it doesn't mean that -1/2=+1/2.

**Pirates, Part 2: July 14, 2010**

For this we again work backwards, assuming that pirates #1-4 have been executed and we're just left with #5 and #6. #5 gets to choose the plan and proposes that he keep the coin for himself. #6 will vote against it, but #5's proposal wins in a tie (he votes for the plan) and he ends up keeping the coin.

#5: 1

#6: 0

Move to the 3-pirate case, with #4, 5, and 6. #4 proposes the plan. He needs one other's support, so gives the coin to #6. #4 and #6 vote for the plan and it passes.

#4: 0

#5: 0

#6: 1

In the 4-pirate case, #3 is proposing the plan and needs one other pirate to support it. He can give the coin to either #4 or #5 and the plan will pass.

#3: 0

#4: 1

#5: 0

#6: 0

The 5-pirate case is interesting, because no matter who #2 gives the coin to, there are going to be three others who oppose the plan and would rather see him die. So #2 is doomed to be executed.

#2: X

Back to the original scenario posed in the question. #1 is proposing a plan. If #1 dies and #2 is put in charge, #2 will be doomed to be executed, so #2 is inclined to support whatever plan #1 comes up with. Therefore to get the support of half the pirates, #1 needs to give the coin to either #3, 4, 5, or 6. Then he, #2, and the recipient of the coin will vote for the plan.

**The Blind Date Bachelor: July 7, 2010**

I found out soon after posting this that I should have named it after MTV's "Next", because the two games share more similarities.

The best solution that I know is to not choose the first two dates, but remember the best of the two. If he likes his third date better than either of the first two, choose her. If he doesn't like her as much, skip and marry the fourth date. 10 our of 24 times, this will result in the bachelor marrying his best match. This is better than choosing at random, which would be right 6 out of 24 times.

**Pirates: June 30, 2010**

I liked Ned's solution best:

I worked it backwards. Let's say the top 3 pirates have walked the plank. Now the deal is easy:

#4: 100

#5: 0

Ok, back up one step and there are three pirates left. #5 knows if he votes down the offer, he will get 0 as above so he has to accept whatever is offered. If he votes with #3, then it doesn't matter what #4 says, so #4 gets nothing:

#3: 99

#4: 0

#5: 1

Back up to 4 pirates left. #2 offers 1 coin to #4 and keeps the rest for himself. #4 knows that if he rejects the deal, #3 will be in charge and he will get 0 coins, so votes in favor of #2's deal.

#2: 99

#3: 0

#4: 1

#5: 0

Finally, back to the original situation. The 4 pirate scenario is a stable situation, and all the pirates know that it will never get past that. Therefore, #1 only has to offer #3 and #5 1 coin to have them go along with the plan. Therefore, the final situation is like this:

#1: 98

#2: 0

#3: 1

#4: 0

#5: 1

#2 and #4 get totally screwed, but what are they going to do about it?