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).

[elitegimp]

[ QUOTE ]
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

[/ QUOTE ]

Here's my way of doing it -- the probability that you win on any given roll is 1/38, or 2.632% of the time. In order to end with $0, you need to be wrong all 38 times -- this happens (1-1/38)^38 = (37/38)^38 = 0.3630. So there is a 36.3% chance you lose all 38 rolls.

In order to end with $36, you need to be right exactly once, and wrong the other 37 times. This happens 38*(1/38)*(37/38)^37 = 0.3728 or 37.28% of the time.

Recap:
You go broke 36.3% of the time
You end with exactly $36 37.28% of the time.
Therefore you end with more than $36 1-.363-.3728=0.2642 or 26.42% of the time.

So you are most likely to have exactly $36, but you are more likely to have $0 than more than $36. This makes sense -- you are more likely to lose 38 times than you are to win at least twice.

Note: the 38 in
[ QUOTE ]
38*(1/38)*(37/38)^37 = 0.3728 or 37.28% of the time.

[/ QUOTE ]
is the number of possible rolls for you to be right on. There is a (1/38)*(37/38)^37 chance that you are right on a specific roll (i.e. the third roll), but you just want the chance that any of the 38 are correct.