probability of 18 reds(roulette) in a row explained by a layman

I wanted to explain this to the Australian guy but that thread was closed. Hopefully he'll read this and hopefully it won't be closed. As a layman and not a math guy, I hope I can explain it better as I get his confusion.

So, here goes.

The problem is that each spin being an independent event, the only (random)set of spins that can be calculated is a single spin which at double zero roulette would be 18/38. Once you try to calculate the outcome of 2 or more random spins(a sequence) they are NO LONGER RANDOM. Let us say that instead of 18 reds straight, the odds were calculated for rrbbrbrrbrbbbrbbbr. Now, that may seem like a random set(it is, I just hit the r and b keys randomly till I got to 18) but IT IS NO LONGER RANDOM as the above set has been committed to paper(or computer screen in this case) and if the Wizard tried to calculate the odds of that EXACT outcome it would be the same as 18 straight reds.

Sit at a roulette table and wait for either outcome, 18 straight reds or rrbbrbrrbrbbbrbbbr and you will find they both occur with the same regularity. So if you placed your bets for 18 spins on the pattern rrbbrbrrbrbbbrbbbr you have a 1 in half million chance of winning all 18 spins, the same as winning 18 straight reds in a row.

Now, why can the Wizard predict the future, but the past doesn't count? Let's look at 2 spins instead of 18.

What are the odds of the first spin being red, followed by a black spin? If you sat at the roulette table and the first spin was black, it would not be possible to count that when waiting for a red, then black. So, you would start over with the next spin. If that was black again, you would again have to start over. Now, if the next two spins were r, then b, that happened after 4 spins, not two, right? The odds of a red, followed by a black didn't change but the first two spins weren't red, then black, they were black, black. But you wouldn't be silly enough to count those first two black spins because you were only trying for first red, then black.

Likewise, for 18 reds, if 12 have already occurred you cannot "jump in" because those 12 spins don't change the odds of the future six either.

Well, All that may be more confusing of an answer than I wanted but I hope it helps. Thanks.

Wow, that's good.

I had my own thought. Here goes.

Assume you spin 17 reds in a row, so you have rrrrrrrrrrrrrrrrr -- according to the poster, the probability of the 18-th red is (18/38)^18 rather than 18/38.

The post you made is a point I constantly try and make when someone says "x happened, what are the probability of that?" The answer is always 1, because it happened. And what is the probability that it will happen again?

Another way of seeing the flaw is by means of an old joke about carrying bombs onto a plane. A mathemician is worried about flying on planes so he asks the flight attendant, "what is the probability of there being a bomb on the plane?" She says, "1 in a million!" The next time he boards a plane he gets arrested for carrying a bomb on board. When asked why he had the bomb, he says "the flight attendant told me the probability was 1 in a million of there being a bomb on board, so I figured the probability of there being two bombs on board must be so incredibly small that I would be safe for sure."

There you go.

By the way, your name "darkoz" reminds me of the movie "Zardoz," which surely Mr. Wizard knows well.

In a game like roulette, the law of independent trials applies. 100 reds in a row will occur with the same frequency that any other set of 100 possible outcomes occurs. If there were no 0, the odds of red or black coming up would be exactly 50%, regardless of what had happened before.

That said, for someone pursuing a martingale (doubling up after loss) system, the odds of losing 6 in a row on roulette are slightly less than 1/2^6, or 1/64 6 roll sets. In other words, on average you would only lose 6 times in a row once every 384 rolls (it's actually slightly fewer rolls because of the 0 and/or 00). Of course, over time you would still lose exactly the percentage you were supposed to, but you will walk away a winner much more often than you will lose (your losses when you do lose will take away all winnings and then some

For whatever reason, humans have a very hard time grasping the notion of independent trials - even trained casino personnel. I was once playing a martingale system on mini-bac to kill some time at Mandalay Bay, starting with $50 bets. After about 12 hours, I hadn't yet lost more than 7 in a row, and therefore was up about $45,000 (at $50/hand!). By this time the shift manager was in the pit, and I overheard the pit boss trying to say to the shift manager "he'll never lose" and that they should back me off of the game. The shift manager actually had to tell the pit boss "I know it doesn't seem possible, but eventually it will work out for us if he keeps playing". I actually quit for the night, and the next morning they had raised all of the mini-bac minimums to $200, so I just walked with my $45K :). Apparently even the shift manager wasn't too sure :).

Quote: thefish2010

In a game like roulette, the law of independent trials applies. 100 reds in a row will occur with the same frequency that any other set of 100 possible outcomes occurs. If there were no 0, the odds of red or black coming up would be exactly 50%, regardless of what had happened before.

That said, for someone pursuing a martingale (doubling up after loss) system, the odds of losing 6 in a row on roulette are slightly less than 1/2^6, or 1/64 6 roll sets. In other words, on average you would only lose 6 times in a row once every 384 rolls (it's actually slightly fewer rolls because of the 0 and/or 00). Of course, over time you would still lose exactly the percentage you were supposed to, but you will walk away a winner much more often than you will lose (your losses when you do lose will take away all winnings and then some

For whatever reason, humans have a very hard time grasping the notion of independent trials - even trained casino personnel. I was once playing a martingale system to kill some time at Mandalay Bay, starting with $50 bets. After about 12 hours, I hadn't yet lost more than 7 in a row, and therefore was up about $45,000 (at $50/hand!). By this time the shift manager was in the pit, and I overheard the pit boss trying to say to the shift manager "he'll never lose" and that they should back me off of the game. The shift manager actually had to tell the pit boss "I know it doesn't seem possible, but eventually it will work out for us if he keeps playing". I actually quit for the night, and the next morning they had raised all of the mini-bac minimums to $200, so I just walked with my $45K :). Apparently even the shift manager wasn't too sure :).

How do you martingale mini bac? Were you betting on player?

Yes, Dark Oz should be similar to Zardoz as that is relevant to the Wizard of Oz. Dark Oz refers to a comic book sequel to the Wizard of Oz which I am involved with.

Since you are Dorothy Gale, you should check it out. I wrote the novelization (Dark Oz: Of Courage And Witchcraft) which is on Amazon. Sorry about the plug, but since you noticed the resemblance...

Quote: thefish2010

After about 12 hours, I hadn't yet lost more than 7 in a row, and therefore was up about $45,000 (at $50/hand!).

Winning $45k at $50 per hand for 12 hours using a martingale isn't possible short of playing in the land of OZ.

In the martingale system, after every win you are up $50. That is, any string of losses followed by a win nets you exactly $50. At a normal game pace of 30 spins per hour, you're going to win on average (18/38)-th of those bets, or about 14 of them. If you're very lucky and the dealer is very fast, I'll give you 20 wins per hour. For 12 hours, you will win at most 20*12 = 240 bets, at $50 each gives you 240x$50 = $12,000. There is no way you can get to $45k playing a martingale at a $50 bet playing for 12 hours. Even if you won EVERY bet at a normal game pace, that's only 30 spins * 12 hours * $50 = $18,000.

Your actual expected winnings GIVEN that you didn't have a string of losses that wiped you out and that you are playing $50 at 30 hands per hour is roughly (14 wins per hour)* (12 hours) * ($50 won per win) = $8,400 won.

Then again, you had the experience, so maybe you can fill in the missing details ...

Quote: darkoz

Yes, Dark Oz should be similar to Zardoz as that is relevant to the Wizard of Oz. Dark Oz refers to a comic book sequel to the Wizard of Oz which I am involved with.

Since you are Dorothy Gale, you should check it out. I wrote the novelization (Dark Oz: Of Courage And Witchcraft) which is on Amazon. Sorry about the plug, but since you noticed the resemblance...

Product Description
Based on the long-running comic book series, Dark Oz follows Dorothy Gale, fifteen years after her first visit to the Wizard in Oz. Now in her mid-twenties, she flies through a tornado and crash-lands in Munchkinland only to discover the Scarecrow, Tin Man and Lion have been turned evil. Can she remove the spells? Follow her adventures as one by one she removes the spells, gathers the strength of her forces and rallies to destroy the real culprit-the Nome King. Book one of a trilogy.


About the Author
After initially writing novels, [AUTHOR] became entranced with making films where his audience could share the experience of his storytelling communally. He optioned the rights to the comic book series, Dark Oz, and spent six years shopping it around Hollywood. After all the major studios rejected him, he chose to adapt the screenplay into a novel, returning to his original roots. [AUTHOR] has a BFA in motion picture production. He has worked on many films including Kill Bill 1 and 2, The Aviator, Oceans 12 and the Exorcist Prequel.

Thanks, Dorothy Gale. Yes, that's me.

And I would like to say that it's much harder breaking into Hollywood than it is winning Roulette. At least for me, it has.

But, I'll keep plugging away at both.

Quote: darkoz

Thanks, Dorothy Gale. Yes, that's me.

And I would like to say that it's much harder breaking into Hollywood than it is winning Roulette. At least for me, it has.

But, I'll keep plugging away at both.

I am in awe of your writing bio;"Kill Bill" is one of my absolute favorites of all time. I just saw Quentin Tarantino in person recently. Too bad he didn't talk about KB. But he did say that it took him 10 years to write "Inglorious Bastards."

Yes, winning at Roulette is easier than making a living as a writer ...

Quote: DorothyGale

Quote: thefish2010

After about 12 hours, I hadn't yet lost more than 7 in a row, and therefore was up about $45,000 (at $50/hand!).

Winning $45k at $50 per hand for 12 hours using a martingale isn't possible short of playing in the land of OZ.

In the martingale system, after every win you are up $50. That is, any string of losses followed by a win nets you exactly $50. At a normal game pace of 30 spins per hour, you're going to win on average (18/38)-th of those bets, or about 14 of them. If you're very lucky and the dealer is very fast, I'll give you 20 wins per hour. For 12 hours, you will win at most 20*12 = 240 bets, at $50 each gives you 240x$50 = $12,000. There is no way you can get to $45k playing a martingale at a $50 bet playing for 12 hours. Even if you won EVERY bet at a normal game pace, that's only 30 spins * 12 hours * $50 = $18,000.

Your actual expected winnings GIVEN that you didn't have a string of losses that wiped you out and that you are playing $50 at 30 hands per hour is roughly (14 wins per hour)* (12 hours) * ($50 won per win) = $8,400 won.

Then again, you had the experience, so maybe you can fill in the missing details ...

He seemed to imply it was mini bac, are those hands played significantly more quickly?

I wasn't playing roulette, I was playing mini-bac (on the player side). Sorry I didn't make this more clear in the original post. Mini-bac is very fast - if you switch tables during shuffling, you can do about 120 hands per hour. I was doing 50/100/200/500/1000/2000/5000/10,000, so I was picking up an extra $100-$1,000 here and there. I had to go up to $10,000 once (and won - 1 over 0), and $5,000 several times.

At the end of the day I left with 9 chocolates ($45,000). I had a total of $2,000 of my own cash dropped, so my actual profit was $43,000. It may have been 13 or 14 hours....I started a little after noon and went to bed at 1 or 2. I was pretty drunk by then too.

Quote: DorothyGale

Quote: thefish2010

After about 12 hours, I hadn't yet lost more than 7 in a row, and therefore was up about $45,000 (at $50/hand!).

Winning $45k at $50 per hand for 12 hours using a martingale isn't possible short of playing in the land of OZ.

In the martingale system, after every win you are up $50. That is, any string of losses followed by a win nets you exactly $50. At a normal game pace of 30 spins per hour, you're going to win on average (18/38)-th of those bets, or about 14 of them. If you're very lucky and the dealer is very fast, I'll give you 20 wins per hour. For 12 hours, you will win at most 20*12 = 240 bets, at $50 each gives you 240x$50 = $12,000. There is no way you can get to $45k playing a martingale at a $50 bet playing for 12 hours. Even if you won EVERY bet at a normal game pace, that's only 30 spins * 12 hours * $50 = $18,000.

Your actual expected winnings GIVEN that you didn't have a string of losses that wiped you out and that you are playing $50 at 30 hands per hour is roughly (14 wins per hour)* (12 hours) * ($50 won per win) = $8,400 won.

Then again, you had the experience, so maybe you can fill in the missing details ...

Quote: thefish2010

. I was doing 50/100/200/500/1000/2000/5000/10,000, so I was picking up an extra $100-$1,000 here and there.

OK, mini-bac makes it possible. And you are accelerating the martingale, so that also makes it possible.

You don't win $100 to $1000 here and there. The only way to get those bets out is to have a sequence of losses. For example, to win $1000 means you lost $50, $100, $200 and $500. So you had a net loss of $850 followed by a win of $1000 for a net of $150.

The only time you are ahead in a true martinagale is after a win. For example, if you were betting $50, $100, $200, $400, $800, $1600 etc, then each of the following puts you ahead $50:

W
LW
LLW
LLLW
LLLLW
LLLLLW
LLLLLLW
etc...

The only way NOT to win $50 is to leave the table on anything other than a win (assuming no max bet on the table).

Actually, here and there I was picking up an extra $1K. As I said, I had to go up to $5,000 several times. Each time that happened, I won $5,000, and lost $2000+$1000+$500+$200+$100+$50, which is $3850. 5,000-3850=+$1,150. Anyway, over time, this system is a loser. But it was certainly fun :).

Quote: DorothyGale

Quote: thefish2010

. I was doing 50/100/200/500/1000/2000/5000/10,000, so I was picking up an extra $100-$1,000 here and there.

OK, mini-bac makes it possible. And you are accelerating the martingale, so that also makes it possible.

You don't win $100 to $1000 here and there. The only way to get those bets out is to have a sequence of losses. For example, to win $1000 means you lost $50, $100, $200 and $500. So you had a net loss of $850 followed by a win of $1000 for a net of $150.

The only time you are ahead in a true martinagale is after a win. For example, if you were betting $50, $100, $200, $400, $800, $1600 etc, then each of the following puts you ahead $50:

W
LW
LLW
LLLW
LLLLW
LLLLLW
LLLLLLW
etc...

The only way NOT to win $50 is to leave the table on anything other than a win (assuming no max bet on the table).

Quote: thefish2010

Actually, here and there I was picking up an extra $1K. As I said, I had to go up to $5,000 several times. Each time that happened, I won $5,000, and lost $2000+$1000+$500+$200+$100+$50, which is $3850. 5,000-3850=+$1,150. Anyway, over time, this system is a loser. But it was certainly fun :).

Quote: DorothyGale

Quote: thefish2010

. I was doing 50/100/200/500/1000/2000/5000/10,000, so I was picking up an extra $100-$1,000 here and there.

OK, mini-bac makes it possible. And you are accelerating the martingale, so that also makes it possible.

You don't win $100 to $1000 here and there. The only way to get those bets out is to have a sequence of losses. For example, to win $1000 means you lost $50, $100, $200 and $500. So you had a net loss of $850 followed by a win of $1000 for a net of $150.

The only time you are ahead in a true martinagale is after a win. For example, if you were betting $50, $100, $200, $400, $800, $1600 etc, then each of the following puts you ahead $50:

W
LW
LLW
LLLW
LLLLW
LLLLLW
LLLLLLW
etc...

The only way NOT to win $50 is to leave the table on anything other than a win (assuming no max bet on the table).

Got it, thanks for the extra details.

So long, and thanks for all the fish!