Roulette Probability Question

[shummie]

Say you sit down at the wheel and agree to bet $1 38 times on the same number.

Given 0 and 00, Roulette pays 35-1 on a single number bet and you have a 1/38 shot of hitting your number each spin. So after 38 spins, you will on average have $36.

After 38 spins, which is more likely?

(a) You have more than $36.
(b) You have less than $36 (0 dollars).
(c) You are as likely to have 0 dollars as more than $36.

My gut says c. Can anyone confirm?

- Jason

[parachute]

How much money did you start with?

After 38 spins you will have spent $38 and gotten, on average, $35 back. So on average you'll have lost $3, right? I'm not sure how this fits with your $36 question.

[gaming_mouse]

It is pretty unlikely that you will have exactly 36 dollars (ie, that you will break exactly even, minus the house edge). A little less than 1/2 the time you have less than 36, a little less than half the time you have more than 36, and the small remaining portion of the time you have exactly 36.

gm

[pzhon]

[ QUOTE ]
Say you sit down at the wheel and agree to bet $1 38 times on the same number.

After 38 spins, which is more likely?

(a) You have more than $36.
(b) You have less than $36 (0 dollars).
(c) You are as likely to have 0 dollars as more than $36.


[/ QUOTE ]
On average, you have $36. When you are below average, you are exactly $36 below average. When you are above average, you are at least $36 above average, perhaps much more. In fact, if you win more than $36, your average win is about $49.46 above average. That means you are about 49.45/36 as likely to lose as to win, since the times you are below average have to balance the times you are above average.

Situation A happens 36.30% of the time.
Situation B happens 26.42% of the time.

[gaming_mouse]

[ QUOTE ]
On average, you have $36. When you are below average, you are exactly $36 below average.

[/ QUOTE ]

pzhon,

I don't follow this. Your average is $36. Can't you also have any (or at least most) of the numbers between 0 and 36? It seems to me that you are saying if you have less than $36, you have $0. I must be misunderstanding. Can you explain?

[shummie]

Because you only win in $36 increments ($35 payout plus your initial $1 bet).

Assuming a starting bankroll of $36... If you play 36 times and you haven't hit, you have 36 - 36 = $0. If you have hit once, you have $36(start) + $36(win) - $36(bets) = $36.

- Jason

[gaming_mouse]

[ QUOTE ]
Because you only win in $36 increments ($35 payout plus your initial $1 bet).

[/ QUOTE ]

Ahh... I read the question too fast. I thought you were betting on red/black. Please ignore my earlier comments too, as they make no sense given the OP's actual question.

gm

[shummie]

Phzon, thanks for the reply. Your reasoning makes sense to me... all but the $49.45 number. How did you get this? A standard deviation?

- Jason

[pzhon]

[ QUOTE ]
Your reasoning makes sense to me... all but the $49.45 number. How did you get this?

[/ QUOTE ]
I used Mathematica to sum over all of the cases.

Sum[36 (i - 1) Binomial[38, i](37/38)^(38 - i) (1/38)^i, {i, 2, 38}]/Sum[Binomial[38, i](37/38)^(38 - i)(1/38)^i, {i, 2, 38}] //N

49.4569

(I guess I made a mistake and rounded down.)

If you want a good approximation, use a Poisson distribution. For a Poisson distribution with mean 1, the probability of n wins is 1/n! * 1/e.

Sum[(36(n - 1)) 1/n! 1/E, {n, 2, Infinity}]/Sum[1/n! 1/E, {n, 2, Infinity}] = 36/(E-2) ~ 50.12.

Instead of summing over the cases (which is a bit tricky to do by hand) you can use that the above average cases balance the below average case, and the probability of 0 hits is 1/e while the probability of 2 or more hits is 1-(2/e). That also gets you to 36/(e-2).

[college kid]

I'm just not following where you are getting your numbers from. If you do not hit your number, you will be in solution B, if you do hit it exactly once, you will be in solution C, and if you hit it more than once, you will be in solution A. Where on earth did you get your numbers from??? I can't tell just from looking how likely they seem, but I certainly did not follow you in your post. Anybody else care to put in some actual stats equations???

The answers are of course:

Solution A: the probability of hitting your number two or more times

Solution B: the probability of you not hitting your number

Solution C: the probability of you hitting your number exactly once

I don't know the exact figures, but from reading the posts you all seem to not understand the basics of what is going on. All we need is a normal bell curve and the equations necessarily to calculate the likely hood of the number not hitting, hitting once, and hitting more than once. Period.