It's a poker discussion until you start asking people if they know what ostensible means, and incorrectly accuse people of lying and not being objective. Then it becomes an English lesson. But of course, just like the poker discussion, you are convinced that you are never wrong, even when it is clearly demonstrated that you are.

I'm close to done with this nonsense, but I suppose I can play along a little longer.

As usual, you have made your question rather unclear. I *think* you're asking if I choose from a group (y) that consists of only a 1, what the odds are of getting a 1, which is of course 100%, and if I choose from a group (x) that consists of a 1 and a 2, what the odds are of getting a 1 with a single choice, and of course that's 50%. So, what's your point?

Unlike the others, I never bothered trying to debate your obsession, because I knew there was no point. You are not a standard throwaway paranoid riggie, rather you are someone with a genuine mental health issue, so while I might toss in a little trolly post (cats melting steel etc), I take it easy on people like you, because I believe in the end you should get some treatment for your issues, even though I know that you never will.

Your personalized obsession is a strange one, and you do remind me of a staking app long ago who had a 10,000 word application with detailed beliefs of how the rooms worked with coaching sites and since we had people in coaching sites that by osmosis the cards would be better (he had all sorts of very creative math to go with his beliefs).

That guy was quite a bit smarter than you in general (you have the profile of a very narrow person that overdeveloped a specific thinking process), but his mental health condition really crippled him, and he knew it. I actually worked with him a couple months, and it was a very memorable experience, and he made some progress, but the demons he had were too much for him in the end.

Your demons seem more limited and semi contained, so while you should likely get some help, as long as they are not having too much of an impact on your actual life - I suppose get the most value you can out of them. If that comes in the form of debating a half dozen guys about what 4 divided by 2 means, then so be it.

Better luck with your next ten years.

All the best.

It must have been clear because you have answered it correctly.


Let us start on a nicer note, Hi my name is Steve.

Do you agree your answers are not equal

x ≠ y ?

You could be that one person smart enough to understand, you should never judge a person by other comments.

I have been on science for years, I have learnt a lot about vectors etc. Try a discussion you might understand.

At this rate, it'll be April before we get anywhere. No, 50 does not equal 100.

Right good, and if you ''play'' along it will be much faster I assure you


We are in agreement it is not equal


y
12
12
12
12
12x


OK the same scenario as just,

All X = 1/2

Y1 = 5/5

You agree with this?

Confirmation we understand so far.

I barely read any details of your posts after the first few, because I did not need to to see your mental condition at work. The only thing that surprised me was the effort others made trying to convince you of the errors in your logic.

I don't find your condescension offensive, because I see it for what it is - your personal defense mechanism to protect your beliefs, and your beliefs are the major component of who you are as a human. It is about you, not others.


As I said, you present yourself as extremely narrow. If you quantify that as "vectors" then so be it, the label or topic does not matter too much.

No doubt others who actually know you in real life have encouraged you to get professional help for your mental condition, and they would have more invested in you than me (as my investment is 0), so perhaps one day you will listen to them before you consume another decade or more with your personal obsession.

All the best.

Ok Mont, but I can assure it is not a belief , and in a few minutes BOBO is going to be the first person to understand with me. You only need to take up the conversation from a few posts ago and you will get it.

Yes, pkdk, yes. If I choose from Y (column 1), there is 100% chance of me getting a 1, and if I choose from x (row 5), there is a 50% chance of me getting a 1.

I know to you it is not a belief - it is a lot more. A good portion of your life is dedicated to it.

You have been thinking about this and preaching this this for over a decade, and if that is the closest you have come mentally to finally believe you have an ally, then I would qualify that as a definite "cry for help" on your part.

I would suggest you speak with people you know in real life who care about your well being. No doubt you likely alienated many of them with your obsession, but perhaps this would be the time for you to start accepting their help, and if you do so then good luck getting the treatments that would be best suited for your conditions.

All the best.

Right cool, we are so close. The next part explains the problem, let's just hope I can ask the question correctly.


Same scenario,


y
a12
b12
c12
d12
e12x


I am going to ask you to pick any two x rows. WE will have the left aligned value we have been discussing. I have labelled them a-e to make it simpler.


I will do the same and explain the ''process'' of my choice.


I choose first d : I have a 1/2 chance of the left aligned value being a 1


I choose second a: I have a 1/2 chance of the left aligned value being a 1


problem - BUT , I have a 100% 5/5 chance it is a repeat value of the first choice.


Understand the red part, that is the problem in essence.


Do you agree with what I have explained to this particular scenario?

Actually, we need to back this up.

The only way this works is if you've placed the numbers exactly as you laid out, but I don't know what they are, and I know they aren't randomly placed. Is that what you mean?

I suppose it would also work if we both know what the numbers are, but I randomly choose, perhaps by rolling a die, or using a coin.

You have jumped the gun slightly, I was going to explain the random scenario in the next few posts keeping to the same scenario.


Could you please answer the last question, then we can move on to the next demonstration before we do the random one.

p.s a dice or coin cannot explain this.

In the final scenario of known positions,

a1
b1
c2
d1
e1


You have just received a 1, what is your chance of a repeat value by randomly picking one of the above?


p.s Welcome to OZ ''Dorothy'' because I think you understand.


added- quite clearly the answer is 4/5 and no longer 1/2 although each first value still has a 1/2 chance of being a 1, it is a paradox at the least.

So what's your rig theory pkdk ?

My theory is that it is broken by the distribution system, they do not know it is broke because the numbers work out in general. However on really close inspection, the numbers do not quite work out .
It is something that ''they'' would have never thought of, it is not rigged .

and the easy fix formula

x Δ y Δ x

If when the new deck comes to the table, then the cards were randomly dealt from the deck to all players, meaning not the top 18 cards in a 9 seat, then it would be fixed. Maybe....

No, I can't answer the last question properly without understanding what I am or am not able to see, and whether the numbers are placed randomly. That's why I asked.

Actually, perhaps I can answer your question in spite of the incomplete information.

This:

And this:

Do not align.

If you have a 50% chance of choosing a number in a particular row, that implies that either the number placement or your choice is random. If it wasn't, you'd have a 100% chance of choosing the number you want.

But if that's the case, then you would have a 50% chance of it being a repeat value.

In technical terms

x is continuous over time where y is discrete over time.

You have a 50% chance of a repeat value for all x 1/2 , but if your new event is a choice from y , you have a 4/5 chance of a repeat value in the known value scenario.


12
12
21
12
12

Once you change from x to y, the right values are removed from the equation , they no longer have any affect on the new chance.

1
1
2
1
1

4/5 , there is no 1/2 once the x to y change is made.


Start off with one deck of cards, you are getting the top card, 1/52 chance of any variable


Then swap to using 52 decks, you are still getting the top card, you have a 1/52 chance of getting any particular variable, but you now have ?/52 chance of getting a repeat value, the chance of a repeat value suppose to be 1/52.
You have switched from x to y .



a{win}{loss}
b{loss}{win}
c{loss}{win}
d{win}{loss}
e{loss}{win}


I randomly get a win, what are my chances of a repeat win with a random choice? (left column as always)

Your choice is independent of x, dependent to y

With the incomplete information provided about the scenario, this makes no sense to me. If I'm randomly choosing each value, my chance of a repeat value is 50% - that is quite obvious. For the answer to be different, there must be a condition you haven't made clear.

And with that, I'm off to bed.

Your choice is independent of x, dependent to y


goodnight , thanks for the discussion

Your theory is just plain wrong. None of your set theory gibberish is relevant at all.

Your theory boils down to the idea that there will be a different distribution of cards when dealt from multiple random decks (or from simulated decks) vs being dealt from a single fully reshuffled deck. There won't be. This is very easily proven with simulations. Do them and prove it to yourself empirically, and stop trying to prove it on paper. It's impossible for a player to tell the difference between the two methods even if a billion hands are dealt.

I can see how you may have convinced yourself of this with your fuzzy logic, but it's wrong.

Ostensible is not a noun.

You cannot see the ostensible. It is a descriptor not a thing. It is an adjective.