Does the “La Partage” rule in Roulette change the value of “q” when calculating Standard Deviation.

Standard Deviation for Even Chance bet,

EV ± ( Bet * 2 * √(n* p * q) )

Example WITHOUT La Partage,
18 ways I Win, 19 ways I Lose.
n = 37 spins
p = (18/37)
q = (1-p) = (19/37)
Bet = $5.00
EV = -$5.00 after 37 spins.

1 STD =
-$5.00 ± ( $5.00 * 2 * √(37*(18/37)*(19/37)) ) =
-$5.00 ± $30.4027025826
Between ≈ -$35.40>...<+$25.40

Example WITH La Partage,
18 ways I Win, 18 ways I Lose and 1 way to lose 0.5.
n = 37 spins
p = (18/37)
Probability of Loss = (18/37) + (0.5/37) = (18.5/37)????
OR is q still (1-p) = (19/37)
Bet = $5.00
EV = -$2.50 after 37 spins.

1 STD =
-$2.50 ± ( $5.00 * 2 * √(37*(18/37)*(18.5/37)) ) =
-$2.50 ± $30.00
Between = -$32.50>...<+$27.50

I know q = (1 - p) but does that change with the La Partage rule? If I use q = (1 - p) i get (19/37) for both examples above. But I also always assumed q = “probability of loss” so then I get (18.5/37) with La Partage. Maybe I shouldn’t assume. Haha!

I know the difference is very minor but I was just curious. Thanks!

I don't think it's as easy as that. It's no longer an "even money bet" as there are three possible results.
There are now three probabilities involved:
p = the probability of a win of 1
q = the probability of a loss of 1
r = the probability of a loss of 1/2
p + q + r = 1
The mean M = p - q - r/2
The variance, which is the square of the SD, is:
p (1 - M)² + q (-1 - M)² + r (-1/2 - M)²
Over N bets, the mean is nM, and the SD is √(N (p (1 - M)² + q (-1 - M)² + r (-1/2 - M)²))

Thank you ThatDonGuy! I appreciate you taking the time to explain it.

p is probability of success and q = 1 - p is the probability of failure. Q is applicable only if there are two possible outcomes: success or failure. La partage roulette has 3 possible outcomes.

However, we can reformat the bet into binary outcome bet and get a q value. The variance V of the game is .9795 and it’s return R is 73/74. 1 / (V/R^2 +1) = .4984 which is our new p value. The payout is R/P - 1 = .979 to 1. The q value is 1 - p = .5016. For evaluating your expected result and variance, this bet is statistically equivalent to the original bet.

Using round numbers, the new bet would be like having a wheel with 153 black, 153 red and 1 green pockets and you get paid .98 to 1 for a win on red or black. No partage

Thanks Ace2! I appreciate the great explanation of changing it to a binary outcome.