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.

Ah, I admit to only skimming there.

I think NewOldGuy is missing the actual question.

OP's question can be reinterpretted as....

What are the odds of starting with $5 and running it up to $600?

Surely, this is a completely different question than....

What are the odds of starting with $600, playing roulette, and then ending with $600?

I think the argument boils down to how you interpret OP's question.

I interpret as question #1, I think NewOldGuy interprets it as question #2.

Let's do that -- on the assumption that losing $595 is within everyone's grasp

I'm guessing that the only way OP was allowed to make the wager is because it was the end of a color or buy-in -- so he had to bet it all on the first wager. That makes it a lot simpler -- so if we know what the minimum wager inside was, and what the minimum unit size was, we could answer that.