Possibly dumb too.

Suppose an American Roulette player decides to bet $1 on a single number each spin until they either win a spin or lose 26.

The chance of profiting is 1 - (37/38)^26 = 50.01136 %

The EV is Σ[(36-k)(37/38)^k-1]/38 from k=1 to 26
- 26(37/38)^26

= -1.000227222

The EV per dollar risked is (above)/26 = -0.0384702278

So has this player reduced the house edge from 1/19 (5.263%) to 3.847% ?


Almost half the time, the player will win before the 26th spin and thus will not be placing $26 worth of wagers.

Their average amount wagered (also the avg duration of the session) is

Σ[k(37/38)^k-1]/38 from k=1 to 26
= 38 - Σ[k(37/38)^k-1]/38 from k=27 to ∞
= 38 - 27(37/38)²⁶ - 38(37/38)²⁷
= 6.007271

But 6.007271 • -1/19 is only -0.31617, so the EV is not (edge)•(avg total wager)

So if 3.847% isn't due to the avg session length being <26 then what's causing it, magic?

I don't understand why you've calculated EV this way? If you want to enumerate all possible outcomes multiplied by the probability of each outcome, use the binomial expansion. I've formatted it so it can be inputted into the wolfram alpha online solve to easily check for yourself.

EV = sum((36k-26)*(37/38)^(26-k)*(1/38)^k*binomialCoefficient[26, k], 0, 26)
= -1.3684
Divide by 26 and you get -.05263, exactly the EV of a single wager.

Bettor stops after he wins.

Right that's why it isn't a binomial distribution.

Clearly the fact that the EV isn't -26/19 has something to do with there being fewer than 26 guaranteed trials, but I thought it would be -(avg amount wagered)/19.

If we change the strategy, it is that. Let's say the player will bet red/black until they either double up or go bust.

If they go all-in on the first spin then the EV is -26/19.

If the bet $1 each spin:

P(double up) = (9/19)²⁶ / [(9/19)²⁶ + (10/19)²⁶] = .0606896133 = p

EV = 26(2p-1) = -22.844

avg amount wagered = avg duration = 19(26 - 52p) = 434.0386621

-434.0386621/19 = -22.844

So for the red/black strategy, the EV a multiple of the avg # of spins. Why isn't it for the number strategy too?

Mystery solved: I had miscalculated the average amount wagered.

k=26 is different because the game will last 26 spins whether the 26th spin is won or lost.

So the avg duration is that sum from k=1 to 25, plus 26(37/38)²⁵

= 38 - (26*37 + 37²)(37/38)²⁴/38 + 26(37/38)²⁵

= 19.00431714

Multiply that by -1/19 and you get -1.000227218

praise the lord!

good job

The important point to understand is that the fact that EV is the disadvantage times the average total wager is all you need to PROVE that there was an error in the calculation. It is a RIGOROUS proof that does not require you to find the error.