Yesterday I went and played roulette. Started off with a bankroll of 600 dollars. My earnings and losses fluctuated for about forty minutes...and then I began losing mostly, until, about an hour after I first began, I had one five-dollar chip. I exchanged it for five one-dollar chips. I placed four dollars on 0 and one dollar on 35.

0 hit!

From there, I started betting about twenty dollars per spin - mostly filling sections (which is my favorite playing strategy), so in other words I'd put about three chips on every number from 25 to 36 and then a few splits.

About an hour later, I had a total of 600 dollars. I didn't want to risk my luck and stopped then.

I was told by a friend that the changes of this happening, to be down to the last 5 dollars, and then win 600 dollars (actually, that was the original amount I started with) are one in a million. I don't think it's one in a million, or not even one in a thousand, but how lucky would this be considered?

No straight-up bet in roulette is outside 1 out of 38.

roulette, strategy, one in a million... what?

I think any degen gambler has done something similar before. I once was playing at a $5 black jack game and ended up splitting three times and doubling twice and lost it all. It left me with $2.50 and the dealer told me it was ok to play under the limit since it was my last chip. I ran that 2.50 into $800.

I am pretty sure we all have a story so I would say its not 1 in a mil, maybe 1 in 3

Odds of turning $5 into $600??? Well you are 1 in 38 to win $175 if you bet $5 on 1 number. Then let's say you put all of that on black twice... You arr looking at around 1 in 156 as a worst case scenario so not even close to one in a million

Thank you for the clarification. My goodness - 800 after being down to the last 2.50. That's better luck than I had!

Sweet run! Congrats on not going $600 to $5 to $600 to $0 as well. Sounds like a pretty good day.

yeah but it sure wsn't a quick run. I won my 2.50, let the 5 ride, let the 10 ride, let the 20 ride and played normal with the 40 and approx 6 hours later I went home.

So OP broke even on a session and wants to know the odds?

Roulette tactics = Banzai Charge tactics! Have a good time and good luck to you sir!

Pretty much this. Every random walk that gets you from point A to point B in the same number of steps, happens with the exact same probability.

The excursion distances are irrelevant, only the endpoints and number of steps matter when calculating the probability. Believing anything different is gambler's fallacy (that one spin affects the next one).

That means being down $595 at one point, is irrelevant to answering the question. The question is simply, what is the chance I bet N times (assuming equal bets) in a 5% house edge game and end up even.

Are you sure?

In the case of a binary choice (win/lose each bet), yes.

Take 10 bets and these two ways to break even (of many).
+++++-----
and
+-+-+-+-+-

These have the same probability of occurring, but one has a a maximum excursion of 5, and the other a maximum excursion of 1.

Let's look at why. We start with + in both cases.
Then 50% +/- for 2 possible 2-bet sequences with equal probability
++
+-
Then 50% +/- for 4 subsequent 3-bet sequences all equal probability
+++
++-
+-+
+--
Then 50% +/- for 8 subsequent 4-bet sequences all equal probability
++++
+++-
++-+
++--
+-++
+-+-
+--+
+---

If we continue this we get two paths to the two 10-bet sequences I showed at the beginning, and both of them will be a summation of identical probabilities.

It should be obvious that all binary sequences we can construct of a given length and common endpoint always have the same probability to occur in a random walk. Substitute 1s and 0s if it helps conceptualize it.



Edit: This thread might be helpful too. They discuss why the dice sequences (a) 11111111111111111111; or (b) 63335643331622221214.
have the exact same probability to occur.

Continuing the above, Wiki confirms what I said too:

"The number of different walks of n steps where each step is +1 or −1 is clearly 2n. For the simple random walk, each of these walks are equally likely."

http://en.wikipedia.org/wiki/Random_walk

This means my statement about walks of n steps with a common endpoint is just a subset. ALL walks with n steps have the same probability regardless of endpoint.

Yes, but that's trivial. The final results will represent a distribution centered around the expectation. The two endpoints have the same frequency as any other, but only represent a result that you will experience once.