Some musings on the math inspired by JAW OF THE THIRDS by Dane

[TwoCatSam]

Let's admit the first dozen to hit in a series of THREE spins has less probability of hitting on the second spin and even lesser on the third.

psimones

How do you calculate that? 

Sam

[psimoes]

Let's admit the first dozen to hit in a series of THREE spins has less probability of hitting on the second spin and even lesser on the third.

psimones

How do you calculate that? 

Sam

Sam, I didn't. [...]. From what I wrote back it just seems right.

Here's a comparison. I added a third experiment betting on a single dozen different than last single dozen.
They all have similar LW registries, but if bet for all spins applying different criteria per each spin, even if in a strict pattern, we might capitalize on it.

"math" deleted

[psimoes]

I know it's based on some subverted logic and I just came across an expression that summarizes it: maturity of chances LOL. The I Ching is all about it. I'm trying to build a bridge between two different schools of thought to better understand Randomness, and possibly beat the game.

[psimoes]

I spot an error so far. Second bet loss added to the MM. Total profit +5.

[psimoes]

Well, if you want to put in numbers, we could try like this: We have spin1, spin2 and spin3. A dozen that just hit on spin1 (100%) already happened. It's a given point. Now it has a chance of 10.5% to reappear on spin2. Now we could say it's got 100% since it really happened, but we're looking at past history so this chance is still carrying a state of 10.5%. For spin2 to spin3 we apply the relation from spin1 to spin2.
So, a dozen hits on spin1. The chance of hitting again on spin2 is 10.5. The chance of hitting again on spin3 is 1.1. We analyse spin by spin and make a final judgement from first event to last.

[psimoes]

Sorry about all the editing. I'm making this up as I write along. :-)

[TwoCatSam]

oops

[TwoCatSam]

Quote

    If the last spin was red, the chances of the next spin being red are still 48.6% -- they do not drop to 23.7%. On the other hand, if you hadn't spun the wheel to see the first red result and wanted to know the probability of seeing red over the next 2 spins (and not just on the next 1 spin), the probability would be 23.7%.


I have some opinions on the above, but for the sake of simplicity let's go on.

psimoes

I do not know who the quote was from, but your response to it is in blue.  Would you elaborate on what your opinions are on this quote?

As I see it, a person must believe that he will see red 23.7% of the time in the next two spins OR he must discount one of the firm tenants in math.  If I flip and coin and say to you, "I will get heads.", I have a 50/50 chance.  If I say to you, "I'll get two heads in a row.", then I have only a 25% chance of being right. 

Do you agree that the chance of hitting heads--in a row--decreases by 50% each spin?  50, 25, 12.5, 6.25 and on to the edge of the universe.

Sam

[psimoes]

Sam, the conjectures in my last six or seven posts were like the prosecutor that lost the case after a loooooooong trial. Dumb thinking and I wrote a ton of crap! Sorry for wasting the time of those who read it.

Let's say I was misguided by insufficient data stated by the following:

If you hadn't spun the wheel to see the first red result and wanted to know the probability of seeing red over the next 2 spins (and not just on the next 1 spin), the probability would be 25%.

Because:

ANY two outcomes over two spins have 25% probabilities of hitting, not just repeaters.

RB face the odds of RR, BB or BR hitting as well. RR, BB, RB, BR all have the same 25% probablities. This is useless in terms of bet selection and even more at "trying to decode random"...

So forget all the crap about repeating dozens. It seemed right at the time because they were based on false premises and results proved it was wrong.

Odds don't change. Bankroll does.

Here's my opinions saved for later:

We arrive at the table with 1 unit and decide to bet on two reds in a row. The odds are 3:1. In case of winning, we should expect a return of 3 units.
But we have to make one bet per spin. Seeing the odds are 1:1, we bet 1 unit on Red and expect a return of 1 in case of winning the first trial.

If the last spin was red, the chances of the next spin being red are still 50% -- they do not drop to 25%.

If every trial is independent, the odds per spin are always 1:1 so we bet another unit on Red.

Red hits a second time, we're paid 1 unit. Final profit however is 2:1. What happened?

We should have increased the second stake to two units. We'd win 2 units and net result would be +3.

To recap, the outcomes are what they are. Past results reflect only in our money management. The second bet of 2 units only seems riskier if we think of it as an independent "event". Looking at the bankroll history, it still carries the weight of 1 unit.

Now I ask: do we have an edge by using positive progressions?

[TwoCatSam]

Below is what troubles me.  Do you think the quote is wrong?



Let's say I was misguided by insufficient data stated by the following:

Quote

    If you hadn't spun the wheel to see the first red result and wanted to know the probability of seeing red over the next 2 spins (and not just on the next 1 spin), the probability would be 25%.

[psimoes]

Do you agree that the chance of hitting heads--in a row--decreases by 50% each spin?  50, 25, 12.5, 6.25 and on to the edge of the universe.

Sam

Sorry, forgot to answer that question (although it might already been answered):

For now I believe the chance of winning all four trials in a row is 6.25%. 15:1 odds against it.

[psimoes]

Below is what troubles me.  Do you think the quote is wrong?



Let's say I was misguided by insufficient data stated by the following:

Quote

    If you hadn't spun the wheel to see the first red result and wanted to know the probability of seeing red over the next 2 spins (and not just on the next 1 spin), the probability would be 25%.

No, sorry I didn't express it well. By "insufficient data" I meant they could have added something like "the above statement is also true for BB and RB or BR". I erroneously took the single example posted (RR) and took it for absolute.

But it does still raise a few questions, such as:

What happens AFTER someone wins four ECs in a row?

[TwoCatSam]

What happens AFTER someone wins four ECs in a row?

They have the exact same chance of winning number five.

Unless you delve into voodoo or magic, the past numbers of the wheel do not exist.  The next spin is the first one that wheel has ever spun.

Yep, that's the truth.

Sam

[psimoes]

Probabilities do not explain why RRRRRRRRRR happen, they just tell us they have the same 0.09765625% chances of happening as RRRRRBBBBB, BRBBRBRRRB, or whatever combination of outcomes we can think of, in any given context. RRRRRRRRRR is as "rare" as BRBRBBRRBR. We just notice the former because it's so easily noticeable and classify it as a "pattern", but the latter is also a "pattern" if we look for it; if you see my point.
We learn something new everyday.

[psimoes]

What happens AFTER someone wins four ECs in a row?

They have the exact same chance of winning number five.

Unless you delve into voodoo or magic, the past numbers of the wheel do not exist.  The next spin is the first one that wheel has ever spun.

Yep, that's the truth.

Sam

Sam, the next spin being always the first one is right. But the person who keeps betting should be concerned about the past history of winnings, I think. We have always 50% chance of winning each bet, but not every bet. Otherwise the probability would say so. When a person has just won four ECs in a row and goes for a fifth, don't you think that person might be pushing their luck?