Rounding the 22% loss rate up to 25% for simplicity would imply that in 4 sets of 50 spins, I would average 1 complete loss of X (could be less depending on when the streak started) and 3 wins of .6X yielding a profit of 80% over 4 sets.

I KNOW this has to be wrong or else people smarter than me would have figured it out by now. Please correct me.

You make it sound like, in a set of 50 spins, either you will lose your entire bankroll at some point, or make a profit of at least 60% of your bankroll.

Without a specific method to scrutinize, I think it boils down to this:

On every turn, you are more likely to lose than win.

There is no way to combine a series of losing turns to make a win.

Best of luck.

Quote:LADannyThat's exactly what happens. I will either profit 60% of my bankroll (for the set) or I will lose my bankroll. I'm not asking for validation for that part. I know that part works. I'm really close to even (< 1% so far), but I just had two loosing sets in a row which scared me a little. I created this thread in the hopes of validating the logic in my original post before I commit any more funds to it.

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Well, if you have some system where you are guaranteed to profit by at least 60% of your bankroll after 50 spins if there is not a run of 7 consecutive losing spins, then your logic is correct.

The key word is, "if".

Pardon me, if not others here, for being skeptical about this.

The fact that you mention that you only lose with seven losing spins in a row sounds like it's a Martingale - but a 7-step Martingale requires a bankroll of 127, where either you lose all 127 with 7 losing bets in a row, or win 1 with each winning bet. Even if all 50 bets win, you only profit 50/127, or about 39.37%, of your bankroll.

More encompassing would be “on every win, you’ll be paid less than fair…”Quote:DieterWelcome to the forum.

Without a specific method to scrutinize, I think it boils down to this:

On every turn, you are more likely to lose than win.

There is no way to combine a series of losing turns to make a win.

Best of luck.

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When you make lay bets in craps, for example, you are more likely to win than lose. But you’re still a guaranteed loser given enough time

Quote:Ace2More encompassing would be “on every win, you’ll be paid less than fair…”Quote:DieterWelcome to the forum.

Without a specific method to scrutinize, I think it boils down to this:

On every turn, you are more likely to lose than win.

There is no way to combine a series of losing turns to make a win.

Best of luck.

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When you make lay bets in craps, for example, you are more likely to win than lose. But you’re still a guaranteed loser given enough time

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I don't like the word "fair" in this context, but "even" doesn't fit, either. "True odds", maybe.

I also don't like craps, so it's unlikely to be one of the top 17 examples in my bag of canned responses. (I have mostly come to terms with this personal shortcoming.)

I think banking the game has been shown to be reliably profitable over the last few centuries, unless the croupier wasn't looking.

Speaking of semantics, I’ve always found the term “odds” to be rather odd. What’s the connection between probability and odd ?

Quote:Ace2I’ll assume you like the craps term “free odds” even less

Speaking of semantics, I’ve always found the term “odds” to be rather odd. What’s the connection between probability and odd ?

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Shucks if I know. Best I can manage is to speculate it's not an even money payout.