probability (expected value) help pls...?

Question

the probability that a roulette wheel stops on a red number is 18/37. For each bet on "red, you are returned twice your bet (including your bet) if the wheel stops on the red number, and lose your money if it does not .

if you bet $1 on each of 10 consecutive plays, what is your expected winnings?

Answer 1

you can't figure that out because in those 10 consecutrive plays ur propablility is losed 5 times win 5 times but u do not know in what order that will occur

Answer 2

On one bet, potential gain is +1 with probability 18/37.
On one bet, potential loss is -1 with probability 19/37.
Expectation is +1(18/37) - 1(19/37) = -.02703
In 10 bets, expectation is 10 times that, or -0.2703

Answer 3

p=probability of winning $1 on a red bet 18/37=0.4864865
q=probability of losing $1 on a red bet 19/37=0.5135135
on each bet your winning expectations are $1x.4864865
and your losing expectations are $1x0.5135135
on each bet you expectation is -$0.027027.
If you bet an inifite number of times, your average loss per 10 spins would be 0.27027 times a single bet amount.

Answer 4

p = 0.4865
q = 1 - p = 0.5135
wins 10C(wins) P(nwins) nP
0 . . . . . . . . . 1 . . 0.0013 0.0000
. 1 . . . .. . . . 10 . . 0.0121 0.0121
. 2 . . .. . . . . 45 . . 0.0515 0.1030
. 3 . . .. . . . 120 . . 0.1301 0.3903
. 4 . . .. . . . 210 . . 0.2157 0.8627
. 5 . . .. . . . 252 . . 0.2452 1.2260
. 6 . . .. . . . 210 . . 0.1936 1.1615
. 7 . . .. . . . 120 . . 0.1048 0.7336
. 8 . . .. . . . . 45 . . 0.0372 0.2978
. 9 . . .. . . . . 10 . . 0.0078 0.0705
10 . . .. . . . . . 1 . . 0.0007 0.0074
55 . . . . . 1024 . . 1.0000 4.8649
4.8649 / 55 = 0.0885
nP is numerically the same as the winnings for n wins in dollars. Therefore, the expected winnings, rounded to the nearest penny is $0.09