Since there are 37 pockets on the wheel, and you win on
18 of them, the probability of winning is 18/37 and the
probability of losing is 19/37. So the expected value of a
$1 bet is ($118/37)($119/37) =
$0.027. For each dollar you bet, your expected loss is
2.7 cents, exactly the same as for bets on a single number.
Since the probability of red or black is independent of
the results of previous spins, the expected value of each $1
Fred bets is still a loss of 2.7 cents. Fred has fallen
victim to the gambler's fallacy. The expected
value of each bet at roulette is always the same, whatever
"strategy" you use.
The house wins $1 on 37 outcomes and loses $35 on one.
So for each $1 bet, the expected value for the house is
($137/38)($351/38) = $0.053. The
house edge is 5.3%. You can expect to lose your money twice
as fast, on average, on a wheel with two zeros.
The chance that the first number drawn matches one of
your six numbers is 6/20. The chance that the second number
drawn matches one of your remaining five numbers is 5/19
(since there are now only 19 numbers to draw from).
Continuing in this way, the probability of matching all six
numbers is
6/205/194/183/172/161/
15 = 0.0000258, or about 1/40,000.
If 120,000 people play, the takings are $120,000 and
the winnings are $72,000. If you could ensure that you share
the winnings with two other people, your share would be
$24,000. Taking the chance of winning to be 1/40,000, the
expected value of a ticket is ($24,0001/40,000)
$1 = $0.40. If you could ensure
that you share the winnings with one other person, your share
would be $36,000, and the expected value of a ticket would be
($36,0001/40,000) $1 =
$0.10. If you could ensure that you do not share the
prize, the expected value of a ticket would be
($72,0001/40,000) $1 =
$0.80. If you could somehow ensure that you always pick
numbers that nobody else picks, the expected value of playing
this lottery is positive.