Q. There's a hot little betting system that'll let you go to a casino roulette table with $1000 and come home a $60 winner nine times out of 10. Interested? Just don't hold us responsible for that 10th visit.
A. You mark down 1-2-3, because you'll be winning your money $6 at a time. Start by betting $4--always the lowest + highest numbers--on either red or black. There are 18 reds on the wheel, 18 blacks and two for the house, giving you 47% chance (18/38) of winning your $4. If you win, cross off the 1 and 3 and bet $2 (the remaining number). If you win again, you've crossed off all numbers and netted your first $6, so you write down 1-2-3 and start over.
If you lose your $4 bet, write down 4--always write down amounts lost--add the 1 and 4, and bet $5. If you win the $5, cross off the 1 and 4, then bet $5 (the 2 + 3) again. If you win on this, all the numbers are crossed off and you're on to a fresh 1-2-3 set. Each loss pushes your bets a little higher until finally when you win, you cover your losses, while netting $6.
There's a catch, of course: You could run out of money before crossing off all the numbers!
A computer simulation by John McGervey ("Probabilities in Everyday Life") beat the house an average of 99 times out of 100 on these 1-2-3 sequences, each winning $6. Winning 10 sequences got to the goal of $60 on a given "visit," and on 90 sequences (9 visits) the computer won $540. But on every 10th visit, on average, the 1-in-100 loss caught up with the computer, scattering its $1,000 to the winds.
There are scores of similar schemes--all useless. Even after winning on 9/10 visits, the computer ended up $460 in the hole. Any random scheme could have done just "as well"!
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