Quote:
Originally posted by Kevsh
There is a roulette theory about betting black (or red) that sounds great on paper:
You start by betting $5 on black.
If you win, fine, you pocket $5 profit.
If you lose, you double the bet ($10).
Since it pays out 100% of the original bet, a win on the 2nd try will net you $5
(You bet $5, lost, then $10, won. $15 spend but you won $20 = $5 net)
If you lose the 2nd (and so on) you keep doubling your bet. No matter how long it goes red it will eventually go black and you will $5 no matter what the bet ($5,$10,$20,$40 ... $512, $1024)
A friend had this idea our first trip to Vegas. The thinking was, what are the odds it will go black 9 times in a row? (Which would have meant you gambled $2555 total so far)
Well, mathematically, 1/512 chance. Fairly small right? Wrong. As the above post points out, past history has no effect so really its 50/50 each time (1/2).
As my friend found out $2555 later - and imagine having that much on a roulette wheel? If he last the next bet would have been $2560 (and over $5k total) which he didn't have.
We won, luckily, but sh1t his pants along the way. Lesson learned.
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Your friend's theory was correct (if there were no 0 and 00). That is the reason they have 0 and 00 on the wheel. The house always wins. Black and Red are not 50% bets. They may pay 2:1 but its not a 1:2 bet. Its a 9:19 bet.
Over enough spins (a million+) the # of red hit equals # of black hits, but # red hits + # of black hits != total # of hits.