I’m no logician, lateral thinker, or mathematician, so I guess that makes me bored. Not sure if this makes sense or would even work, but for what it’s worth, here goes.
Let’s play some roulette. We could try this at the crap table with Pass and Don’t Pass, but roulette offers a much simpler result.
We’re going to pair off our gamblers. For every spin of the wheel, one player in each pair will bet one $5 chip on Red, the other will bet one $5 chip on Black. (Red and Black are even money wagers.) After each spin is done, we should have one winner and one loser (though not in all cases–see below), and it is time for the next pair of gamblers to go to the table. Each pair only plays one spin of the wheel, and no gambler plays again until all others have played at least once.
In the first 38 spins, assuming the numbers work the way they should over time, after each of our 76 (so far) gamblers has played one spin of the wheel, we should have 36 winners, and 40 losers, since neither Red nor Black wins on Zero or Double Zero. Our 36 winners each have $105 now, and we send them home with more money than they came in with.
We still need 15 winners, however, if we are to fulfil the terms of the OP. (36 winners + 15 winners = 51 winners, a majority of the original 100.)
Left are 40 losers, each of whom have have $95; and 24 gamblers, each of whom have not yet played and thus have their original $100 stakes.
We’ll send the 24 who haven’t played yet to the table. They pair off and each wagers one $5 chip on Red or Black for the next 12 spins. For the following 20 spins, we pair off our original 40 losers, each of whom wagers $10 on Red or Black, as they did before. (I know this is basically following the Martingale system, but because we’re using pairs of disciplined players in rotation, and only need to achieve 51 winning players, we won’t need to continue the Martingale to its normally ruinous end.) To complete our series of 38 spins, we need 12 players to play 6 spins–we’ll use the ones who have only played once so far if we can, so they only have to wager $10.
Again, assuming the numbers work in the second set of 38 spins, we have 36 winners. No winner should have more than $105, but that still counts as a profit as far as the OP conditions go. With our original 36 winners, and the new group of 36, we have 72 winners. This is more than enough to fulfil the conditions of the OP (a majority of players being 51), and provides a little leeway in the cases where Zero and Double Zero come up a little more often than they should.
Hm. Using this method, I don’t know what the probability is that a majority would walk out winners, but it’s a thought anyway.