« on: 29/06/2015 19:15:50 »
You should really post this in New Theories. You can either delete this post and repost there or ask one of the moderators to move it.
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....we do not select the decks, when our table finishes a hand, we get a random new deck from the already shuffled decks.It doesn't matter who selects the new deck or in what order, whether it is the player, the dealer, an onlooker or a computer.
So does every other table.
and yesand no
''should take account of this 'infinite vertical y axis' and so the probability for a particular deck is not 4/52 but a complex combination of infinity, time and the x, y axes.''
table one table 2 table 3 table 4 table 5Doesn't matter.
deck one deck 2 deck 3 deck 4 deck 5
table 3 finishes their hand first so get the next deck.
what are the odds of throwing (a) at least and (b) exactly one 1 in 6 throws?I don't think you are going to get an answer to this, given his poor understanding of probability, unless someone tells him.
I somehow think this is the problem you are trying to solve.At times it is very difficult to understand what he is trying to solve, as some examples appear to be related to selecting from rows and columns.
Table one player 2 receives deck 1,8,11,56,72, luckily by timing receiving good starting hands.
Table two player 2 receives deck 2, 9,12,55,70, unluckily receiving poor starting hands.
The third sequence contains an excess of tails so might draw the attention of an amateur statistician but a professional would tell him that the excess is not statistically significant - yet.I was wondering whether he knew enough to avoid that obvious trap. I suspected from his previous posts that he didn't.
4/52 to get an ace is not the same as x/x, if we do not know how many decks there are, and we do not know how many aces have fell in the position of the top card, we can not say 4/52.No it wouldn't. Mathematical gibberish.
it would be would it not?
I will show you I understandThis is wrong, if you have already thrown a 1, the odds of your second throw being a 1 is 1/6.
if I rolled a number 1 with a dice, the next go my odds are 1/6^2 to roll another number 1.
OK imagine your game, you have a single deck and I have 100000000 decks and can randomly choose any deck, how many of those decks have an ace as the top card?To correctly work out the probability you also have to consider the decks that don't have an ace as the top card and there are far more of those.
You have a 4/52 per every shuffleThere you have it, we both have the same chance of winning for any particular game.
A 1/52 chance of any particular card
I have a 1/52 chance of any card and also a 4/52 chance of an ace being in my seat alignment from any individual deck,
But what is my cross odds,There is no such thing as cross odds
what is the odds that a pick a deck that as already been shuffled that as an ace aligned to my seat?You have chosen a particular set of 5 packs, but there are many more sets where you don't get an ace. In order to understand probability you have to consider not only the favourable sets, but also the unfavourable ones.
I am not relying on the shuffle, I am relying on deck choice.in this situation my cross odds are 2/5
I will try science , ...................I don't know what you think science is, but this is not it.
.............., x is short term and y is long term I do not have two lifetimes,
Life would just get so incredibly boring after a while that it wouldn't be worth living for.You would just have to hope that there were lots of exciting new discoveries in science, .......or lots more pseudo scientists to do battle with.
Hi Colin you are getting colder and away from the thinkingOh no I'm not, I'm getting warmer if not hot, because I think I can see where the problem is.
I use the word theory in it's very loosest sense. The same sense that most of the new theorists use it. (jefferyH, I don't include you in that, you know what a theory is)Quote from: Colin....but used as a new theory platform.
I'm glad you identified it. I would never have suspected it of being a theory!
the first card, the odds of an ace are 4/52.These probabilities are only valid if the first 2 players do not receive an ace. Ok with that?
the second card. Again 4/51odds of receiving a ace.
the 3 rd card my odds are also 4/50
You are then asked to choose a random deck from several decks that are all ready shuffledSo how does play proceed? I need to see the full scenario. Do I receive cards from this new deck and you and Alan from the original deck? If so, then yes unequal game. If we all play from the new deck in the same way as described at the beginning, then it is an equal game.
X is 52What you are describing here does not change the probability of receiving an ace from the 1st or 2nd cards in the pack chosen. It doesn't matter which pack you choose.
Y is repeats etc
X is not equal to y
One deck is always 52 different cards , so imagine card two is an ace, then imagine several decks that card 2 was also an ace, partitioning these decks is several other decks that card 2 is not an ace, you then pick random decks , by luck you pick every deck that gives you an ace, this is not normal to a game of poker.
If we labelled the winning decks red and the losing decks blue and randomly shuffled the decks ,a sort of roulette would happen if distribution was based on random timing of the wheel
Isn't it great. You start a thread to gather opinion and hopefully learn something. Then it is hijacked and all the opinion is lost in the dross. We might as well just all PM each other.I'm convinced these random dross comments put newcomers off posting, it's almost as if the question isnt taken seriously but used as a new theory platform..
what you have shown here is not the same as the scenario I gave above. Let me give some examples to show how different - the scenario is everything.
play the x axis has 3 decks, or play the Y axis as 3 decks, see the difference?
Say player 1 draws 5 cards from his pack, player 2 then draws 5 from his own pack. On the next hand player 1 draws 5 cards from his pack without replacing the 1st 5. But player 2 gets a new pack. In this case the probabilities will be different for each player.
Can you confer my scenario is the correct logic and maths?I don't know about the logic because you haven't fully described the scenario.
Would you agree that the player could intercept values by chance and that the maths is player (a) intercepts x,y over time
No, there was a docmentary about his life many years ago and some facts really stood out. Complex beliefs, alchemy etc.
That's quite true. Wikipedia confirms what you just said. Is that where you learned it?
I think that the number 7 is special for some reason. I.e. I think most people think of it as their favorite number or their lucky number if they have such views. I always liked the number 7 but I think that's because channel 7 was the one that came through the clearest on our TV when I was a kid.It is special in some Jewish/Biblical traditions. 7 days, 7 tribes, 7 candlesticks (perhaps representing tribes??), 7 years to work for hand of lady, 7 dreams I seem to remember or was it 7 years famine? Hence the numeroligists pick it up, don't know much about it, just the odd opinion here & there.
all x axis would be 1/3 where y axis you can clearly observe is different.
The sequence of the first ten throws are
This is all good, based on only you playing, now lets consider that there is 2 players, I and you, except by randomness, the already results, are distributed to us both.
you get the already tossed 3rd toss, and the 5th toss, and the ninth toss.
You receive 3 tails in a row.
can you understand that?
Well, without explanation we are confused because in post #1 you say that X=player seat, but above you provide 2 different definitions of X. Confused we are.
I assume you know what it means without explanation. X is any one of the 52 variants of x axis. X in the y axis is any one of the 52 variants of x*∞. There is an infinite amount of rows of 52, y axis.
So lets say row 10 , column 1, there is an X with the value of being an ace.No, because the phrase "By random timing this could be intercepted of the distribution" is incomprehensible.
By random timing this could be intercepted of the distribution. Do you agree?
In the above I have a 2/4 chance of receiving a 1 if my go is first of the distribution.