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The gamblers' fanny pack
of tricks includes ways to manipulate the chance of winning on any
round. Players have opportunities not merely to shave the odds,
but to flip them and be favored to win. It's done by using, not
breaking, the laws of probability. But, better than a 50-50 chance
doesn't mean having an edge.
The
approach is especially straightforward in roulette, which therefore
serves as a prime but not the sole example. Other than in casinos
where 0 or 00 either lose only half or "imprison" money
for the next spin on "outside" bets, the house advantage
in roulette is the same on every wager regardless of the likelihood
of winning. The trade-off involves balancing greater or lesser odds
with smaller or larger payoffs per dollar at risk.
The idea is
simple. Say you bet $12 on a single spot; the odds are 37-to-1 against
you but a win brings $420. Bet the $12 as $6 on each of two spots
(or $12 on the line between two numbers); you're fighting odds of
18-to-1 and are up to net $204. At $4 on each of three numbers,
odds are an adverse 11.67-to-1 to snatch $132. Likewise, $3 on each
of four spots (or $12 on a four-number corner) has you in an 8.5-to-1
battle to grab $96. Go to $2 on six different numbers and you're
outweighed 5.33-to-1 to earn $60. At $1 on each of a dozen positions,
it's a 2.17-to-1 ordeal looking for $24. Carry the reasoning further,
to $6 on each of two columns; you're favored to win this bet, having
put odds of 12-to-7 on your side, but you're risking $12 for a $6
profit. Just beyond the limits of logic, or maybe you think it's
a ruse to bamboozle the bigwigs into coddling you with comps for
pretending to play with what a book you bought said is negligible
exposure, you could bet $1 on each of the 12 three-number rows;
you've given yourself odds of 18-to-1 to break even and not lose.
Of course, as
every erudite solid citizen knows from Aesop's Fables, one swallow
does not make a summer. Nor does one bet a session. Unless, that
is, it happens to be something like a winning wager on a giant jackpot,
but that's another fable.
The key question
transcends the projected results of single rounds and involves how
folks can expect to do in sessions of reasonable duration on bets
having various odds. The accompanying list gives an indication of
the answer. The data assume bets of $12, as indicated, consistently
for 100 spins -- about two hours at a moderately-paced table. The
entries then show the range where players would have 90 percent
probability of finishing. Stated differently, an individual would
have under 5 percent chance of ending the session deeper in the
hole than the "worst" figure and also less than 5 percent
probability of quitting after the two hours with more profit than
the "best" value.
Range
over which a player would have 90 percent chance of finishing,
with bets of $12 distributed as shown at double-zero roulette
| Bet
|
Worst |
Best
|
| 1 spot
|
-1,200 |
+1,074
|
| 2
spots |
-856 |
+730 |
| 3 spots |
-702 |
+576 |
| 4 spots |
-608 |
+482 |
| 6 spots |
-495 |
+369 |
| 12 spots
|
-338 |
+212 |
| 2 columns |
-206 |
+80 |
| 12
rows |
-108 |
-24
|
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