My solution:
Bill Earl
This is a real cut-throat problem that ends up with a very lopsided solution. The basic strategy for all participants is to buy just enough votes to win, and not pay anymore than you have to.
In the perfect solution only one participant actually has any say in the result. However, that solution requires an understanding of what each participant would do if they did have an opportunity to choose a strategy.
To simplify things, let's call the 5 participants A, B, C, D and E. Since participants will be eliminated from left to right, each participant must only be concerned with winning enough votes from those to his/her right. To win their votes, you need to understand what their situation would be if you are eliminated. So we start with the rightmost participant (E) and work backwards:
E is simple. If he is the only one left, he gets all the loot.
D has no hope at all if it comes down to D & E. He needs to get a majority, but E has no reason to vote for him. If it gets this far, D gets nothing.
C is in a good spot. He needs just 1 more vote to stay in the game. D will lose everything if C is eliminated, so his vote can be bought for just 1 bar. C takes 99 and now E gets nothing.
B needs two more votes to stay in. He'd have to give C all 100 bars to buy his vote, so he needs to convince D and E. D already has a guaranteed 1 bar, but he could be bought for 2. E stands to get nothing, so his vote can be bought for just 1 bar. Now C gets nothing and B walks away with 97 bars.
A also needs only 2 more votes to stay in. If A is eliminated, C will get nothing and E will get one bar, so these are the two cheapest votes. C's vote can be bought for 1 bar and E can be bought for 2. A takes home 97. B and D go away empty handed.