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.

Yes. But the main point in response to the OP is that being down $X at some point and ending up even, has a probability that is totally unrelated to the maximum amount he was down during the session. His question (and his friend's response) strongly implied that they thought it mattered. It doesn't.

I think it does. Even though each unique sequence of events has the same probability -- when you look at final results, many outcomes can be arrived at in various ways -- the outcome will have a higher frequency.

Let's say a player has 38 wagers, plays one a time, loses 37 and hits 1, for a loss of two . Expected. There are a lot of ways to get there -- you simply need to hit a number on any one of the 38 spins. If, on the other hand, the player was at one point down 37 bets, there's only one way to get there.

The combination of those conditionals will add up to the same number either way you do. Try it.

Why do you think it matters that the single "hit" is the last one or some other one out of 38 tries? Only if it is the last one do we hit the maximum excursion of being down 37.

It's a lot simpler to analyse it using even money bets, but the principle is the same.

Because it is the only combination that fits the added information that the players was at one point down 37 -- even though the results are the same. I've forgotten OP's question, but it was something along the line of starting with 600, losing 595 and then getting even. While getting even isn't that remarkable, having been down 595 and then getting even combined is a lot more remarkable.

Sure there are multiple ways to stay within some number of even, but every one of those ways is different, and they all have the same chance to happen as the maximum excursion path. I think in your thoughts you are combining multiple similar paths and treating them as one. Sure they are more alike than the one that goes farthest away and comes back, but they are not exactly alike.

EVERY combination of wins/losses that leads to break even is unique and has only one way to follow that path. Including that one. Again, the maximum excursion has nothing to do with the probability of that path having been taken to the result. That's the point I've been making.

The original question was something along the lines of "zomg what are the odds of being down to one bet left and then winning it all back and breaking even?" Well, exactly the same chance as all other sequences of N bets that end up breaking even, including ones where you were never more than a couple bets down. We could just as well ask, "zomg what is the chance we go from 5 bets down to 10 bets up to 8 bets down to 20 bets up to breakeven?" Well, if the step count is the same and we've described all the excursions, it's the same probability as what happened to the OP.

I can't explain it to you any better than I have. You can easily prove I'm right by doing some examples with small N.

I really think this just boils down to different interpretations of the question asked, or the question that should have been asked had it been less ambiguous. Probability questions need to be specific.

But the OP was not looking at any specific path -- only that it took him to $5 and returned him to even. While the specific path taken has the same probability as any other specific path -- there are many fewer specific paths that would include both $5 and even than there are that simply end at even.

There are also many paths that lead to being down to one bet left, it isn't just a straight line. Even OP said he fluctuated around a lot before being down to his last bet. I'm saying there are just as many possible paths that include being down to one bet and getting even, as there are being only down only one bet from even and getting even. Both actually have infinite sequences because we can fluctuate around any level.

I think the disagreement here is that you are answering the question you think the OP is asking, but not what he actually wrote. He asked for the chance to be down to his last bet and get back to even. He didn't give any specifics about the path he took or how many bets he used, other than saying it wasn't a straight line down and back. My answer was that it is the same chance no matter how far away he got from even before coming back, because all we know is the maximum excursion and the end result, we know nothing about the path he took. I still say my answer is correct for the question asked with the information given.

I think I then got off track by trying to say if the number of steps N were the same then it didn't matter what the maximum excursion was, and I believe now that was wrong on that part specifically, and that's where you corrected me. But for the OP question, we don't know N.

And actually I just realized that without knowing N or some other information about the path, the OPs question is pointless anyway, because the chance of any 2D random walk returning to the origin eventually is always 100%. The question needs to be more like, "what is the chance of an N step random walk (N bets played) with a maximum excursion of X (most I was down), ending at the origin (break-even)?" And then we could put in two values of X to compare. There are some formulas for computing that probability.

/all for me

Doesn't a random walk of a -EV bet diverge to infinity?

Yes, but I was treating short sessions as if they were 50/50 bets just to make the discussion easier.

It was actually your post that started my long thread hijack.

Basically I was trying to make the point that the answer to the OP's question didn't change based on how much he was down during the session. It turns out the question and the given information needs to be more specific to say that. I think my last two posts above sum that up.