Yes I do not gamble, I class poker and roulette as games of skill and numbers.

Crux

all the decks are dependent to every bodies game. All the decks are in the same ''box''

Then you should use your beliefs to make quite a bit as a professional roulette player, and after that you can afford to do some prop bets when offered.

All the best.

Not so sure about roulette, but ok.


But you should be able to take this bet then though because under your definition it probably doesn't count as gambling since my skill at calculating the numbers I used in the bet result in my having basically a 100% chance of winning while being able to give you roughly 7.3 times more winning numbers than me. It is an eloquent proposal, don't you think?





If you really don't want to bet on it, still just do it yourself a dozen times or so. Then, assuming you do it honestly, when the number of occurrences over 1,000 trials keeps landing between 640 and 760 you can wonder how I predicted that something that you claim has a chance of occurring of ?/3 would basically certainly occur between 640 and 760 times over 1,000 trials.

Try to understand this then we might get somewhere in this circular discussion.


The only thing that divides the aces you get while playing a game of holdem is time.


Imagine if you could skip time ?


So imagine if you could have a deck from further up the time line, exactly what happens online

Nope

I could be a professional roulette player, but like poker there is a lot of time sitting there playing. I tend get a bit bored even though it is money.

I am not that bothered about money.

No. Perhaps I wasn't clear enough. What I was saying was that, in spite of the fact that there could be no prizes in those two boxes, or prizes in both of them, the odds of any one box containing a prize is still 50%.

Look at it another way. Instead of taking two boxes away, what if I make those my picks? Each box I choose has a 50% chance of containing a prize. But 50% doesn't mean I will have 1 prize every time I choose 2 boxes. Sometimes I will have 2, sometimes I will have none. In fact, the four outcomes are: Both have a prize, the first one I choose has a prize, the second I choose has a prize, or neither has a prize. 2, 1, 1, 0. Four draws, on average, would net me 4 prizes, or one prize per draw. Exactly what we'd expect when making 2 draws at a 50/50 opportunity.

Yes, I do. I know there is still a 50/50 chance of each and every box having a prize, including the 2 that you removed.

It's not a different situation - see above.

hmm, yes it is aces/time you get dealt one hand a time. 52 cards are set variance rather than variance.

Understand that I understand your angle, but! still no. Let me explain another way,


I have 1 box, no other boxes in the scenario, I tell you that in the box there may be prize. What is you chance of a prize?

Why is using a random-ordered deck at random times such a mind numbing event for you?


Are you suggesting that when The System randomly chose Deck 43.... it is not random selection?

Or are you saying because The System randomly chose Deck 43 and because it is random, the top card you get is different from what you "are supposed to get"?

Do you realize in either scenario, the card is random?

Then pay others to do it for you or sell your methods to others. In the casino bonus whoring days I eventually made more from recruiting people to play and clear bonuses (getting a cut of their action) then I did from my own play.

Most people who say they are not bothered by money say it because they have never been able to figure out a way to make money. Pretending your theories and roulette super powers are real - you should be able to easily monetize it, whether you sit for a single spin or not. Perhaps this is too ostensible for you to understand.

All the best.

How many choices are there?

We know with a poker deck, the one box could contain any 1 of the 52 cards

What about in your new question?

Or is it infinite and not remotely related to poker? So it either has or does not have a prize?

Nothing to do with the randomness, it is about the repeat values , for example say you had deck 2 it gave you a top card of the ace of diamonds, then the next hand you was getting the second card deck 45, which was the same card u just had, now this could happen over and over.

There is one box, in the box there is either a prize or there is not , what is the chance of a prize from the box?

lol....sure, in the short term.......but

You ARE aware that over millions of Stars hands, the RNG had pretty spot on distribution when examined by an independent group, right?

At the risk of actually making another post here, that is just as likely to happen when using only one deck.

Deal the top card from Deck One and say it gave you the Ad. At the end of the hand shuffle Deck One. Start the next hand and you get dealt the second card from Deck One, which card is the Ad.

You're just as likely to get the same card in the back to back hands using Deck One for both hands as you are using Deck One for Hand One and Deck 45 for Hand Two.

x = xo + vot + 1/2at2 | (Vƒ)2 = (Vo)2 + 2ax

what's your point?

I am independent of stars and I have examined their setup and determined it is flawed.

I cannot read that , you suppose to have answered 1/2 so i could explain


1/2 = (2=0/1+1/1)

Do you know how to computer program?

If so, it should be very easy to set up a test of your "non-randomness hypothesis".

Do a million trials of the 3x3 array of AKQAKQAKQ.

Shuffle the cards however you want.

Keep track of whatever you want.

Report back "proof" of non-randomness.

Like Lego05 has posted above, it should be easy for you to state what you expect to find (or expect not to find) in such an experiment.

If you are not able to program such a simple simulation, I'd be happy to do so for you.

Just tell me what you are proposing.

Thank you very much.

I don't know what that sequence of 1's, 2's and other symbols is, but the correct answer to this question is that there isn't enough information to answer the question. Or I guess you could answer it that the chance is somewhere in the range of 0% to 100%, inclusive.

pkdk

In your posts where you have said the answer is ?/3, what is your best guess for what ? is?

In poker you have to make your best guess, and it should be 1 in this case, as proved by everyone else except you.

This doesn't mean you will be right exactly 333 times out of 1000 events, but, you will be right between about 310 to 350 times.

When making your best guess, (even though it's not actually a guess before the event), you are assuming you will play infinity events in the future, whether you do or not. Having played a large number of events and then looking back at the actual results, the more events you have played the closer to your having won 1/3 of them will become obvious, although rarely being exactly 1/3.

You seem to be fixated on the specific times that there isn't an Ace, or whatever, on the top of the deck. Those times are irrelevant in the grand scheme of things, as there will be one on the top of the next deck, or the next, or the next, and you will get one, on average, 1/3 of the time, in the future infinity number of events.

Good. Then the existence of the other decks has zero impact on the deck currently in use.