Roulette Expected Value

Im trying to make sense of a program that I wrote to simulate different roulette bets.

Lets take a straight up (SU) bet in European Roulette

100$ wager

Expected Value = -2.7$
Variance = ?
Standard Deviation = sqrt(variance)

If I can get this answer, what about for 3 numbers SU, 6 numbers SU?

I'm rusty on this sort of thing, but here's my stab at the first part of your question.

Edit: Yes, I screwed it up. Here are my revised calculations.

EV = {$3500*1 + (-$100)*36}/37 = -$2.7027 as you said. That's 1 roll winning $3500 for each 36 rolls losing $100.
Var = {($3500 - (-$2.7027) )^2 + 36*(-$100 - (-$2.7027) )^2}/37 = 340,803.5062 (units of dollars squared)
Std = sqrt(Var) =$583.7838

For additional numbers bet straight up on the same roll, you should be able to use the same approach as above, except I don't know whether you mean to bet $100 on each of the 3 or 6 numbers or split your $100 over 3 or 6 numbers.

Im betting same amount on each number.

If I do this, can someone explain why variance, stdev will decrease if you bet more numbers?

Quote: beenitty

Expected Value = -2.7$
Variance = ?
Standard Deviation = sqrt(variance)

EV = $100 * (1/37 * 35 - 36/37 * 1) = -$2.70

Variance = $100^2 * (1/37 * 35^2 - 36/37 * 1^2) - EV^2 = $$ 321,351.35 - $$ 7.29 = $$ 321,344.05

Standard deviation = sqrt(Variance) = $566.87


If you bet 3 numbers on 3 different spins, then increase EV by factor 3 ($8.10), and increase standard deviation by factor sqrt(3) to ($981.85). Same with 6.

If you bet 3 numbers on the same spin, then EV is still increased by factor 3, but standard deviation will be a lower (since all bet's wins are mutual exclusive, i.e. anti-correlated):

Variance = $100^2 * (3/37 * 33^2 - 34/37 * 3^2) - EV^2 = $$ 800,204
StdDev = $894.54

Homework is for 6 numbers :)

Quote: MangoJ

Variance = $100^2 * (1/37 * 35^2 - 36/37 * 1^2) - EV^2 = $$ 321,351.35 - $$ 7.29 = $$ 321,344.05

Are you sure about this? I thought that:

Var=E{(X-µ)^2} = (1/37)*(W-µ)^2 + (36/37)*(L-µ)^2

Where:
W = win outcome = +$3,500   (1 time out of 37)
L = loss outcome = -$100   (36 times out of 37)
µ = Expected outcome = -$2.7027

Var = (1/37)*{+$3,500 – (-$2.7027)}^2 + (36/37)*{( -$100) – (-$2.7027)}^2
= (1/37)*(12,268,926) + (36/37)*(9,466.76)
= 331,592.60 + 9,210.90 = 340,803.50 (dollars squared)

What is my mistake?

Yes you are right, I got a wrong sign in there:

Variance = $100^2 * (1/37 * 35^2 + 36/37 * (-1)^2) - EV^2

One can use both formulas, for the identity E{(X-µ)^2} = E{X^2} - 2 E{X} µ + µ^2 = E{X^2} - µ^2.

A slightly different approach.

Roulette simple binomial wagers

For a wager that pays x:1
with a probability of winning p,
the house edge and standard deviation (per unit bet and per 'root decision') are:

ev = $bet * house edge (he)
he = (x+1)*p - 1 
std = (x+1)*Sqrt[p*(1-p)]
p = probability of success

looking at std (standard deviation)
Sqrt[p*(1-p)]
you should know what this is
N*P*Q
variance of 1 bet

but with binomial that treats success as 1 and failure as 0
we want the bankroll movement of +1 and -1 (for even money) thus (x+1) = 1+1 = 2
-1 and +35 for su thus (x+1) = 35+1 = 36
that is the (x+1) term

your su bet
std = (35+1)*Sqrt[(1/37)*(1-(1/37))]

Multiply the SD value by the actual $ wagered.
Done
something like this per 1 unit bet
see if our tables match (I rounded mine)