Interesting post found at another website: " We asked the following question of interview candidates, and gave them twenty minutes to formulate their answer. You decide to play poker in a casino. But on your way to the poker pit, you are lucky enough to stumble upon a promotional roulette wheel that has been set up differently from a normal wheel. A standard wheel has eighteen black, eighteen red, and one green space (though some have two green spaces). There are still thirty-seven spaces on the wheel, but - There are four spaces marked Red-16 (the three extra spaces replacing three Black numbers) , - There are three spaces marked Red-14 (the two extra spaces replacing two Black numbers). So there are Twenty-three spaces marked Red = {Sixteen singles, four Red-16, three Red-14} , Thirteen spaces marked Black = {Thirteen singles} , One space marked Green = {One Green-0}. - You have a starting bankroll of $1000, and are allowed to play for ten spins. - You are only allowed to place one of these types of bets: -- Any single number (including R-16, R-14, G-0), which pays off at 35-1 , -- Red or Black, which pays off at 1-1. - You are allowed to place a bet for any amount (up to your current bankroll). - You are allowed to place multiple bets on the same spin. - You can use any winnings for future bets (e.g., if you bet $1000 on Red on spin one, and win, you now can place up to $2000 worth of bets on spin two.) What is your strategy?"
Yeah, my bad. The OP said "you decide to play roulette in a casino". But I changed it because I figured anybody stupid enough to play regular roulette couldn't figure out the odds on the favorable game.
The way I read it, the bet on either R-16 or R-14 wins 1:1 if the outcome is any red, except when the outcome is R-16 and R-14, in which case the bet wins 35:1. Therefore, it doesn't make sense to bet on just Red.
If I were lucky enough to hurt myself when I stumbled on the wheel, I'd sue the bastards then go play poker with NoDoji.