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I have no idea what you're trying to derive here...I assume it has something to do with cards (1/52) and probability.1/∞ is not defined itself, and so isn't a good part of a deffinition.what does "player seat" mean?what does "∅ = random" mean? is it a random number (generated how/chosen from what set with what probability function) is ∅ a randomizing function?

x={1/52}/ty={1/∞}/tX=player seatt=timeP=probabilities∅=randomP(x)=P(X∩y)/∅t

It's all meaningless. You clearly ignored that i said about stating the meaning of the math you're writing. Without that it's all meaningless.

Quote from: PmbPhy on 19/06/2015 06:30:00It's all meaningless. You clearly ignored that i said about stating the meaning of the math you're writing. Without that it's all meaningless.To be honest even with explanation this is meaningless. He is just stringing together half understood terms in a totally illogical way.I suppose it's easier than learning real maths and logic!

Quote from: Colin2B on 19/06/2015 07:36:49Quote from: PmbPhy on 19/06/2015 06:30:00It's all meaningless. You clearly ignored that i said about stating the meaning of the math you're writing. Without that it's all meaningless.To be honest even with explanation this is meaningless. He is just stringing together half understood terms in a totally illogical way.I suppose it's easier than learning real maths and logic!Quite right my friend. And as ChiralSPO mentioned, 1/∞ is total nonsense. Anybody who knows what ∞ means knows better than to treat it like a number and thus make any attempt to try to divide one by it. Something like is rooted in complete ignorance.

1/x → 0 as x → ∞, so why not just write 0?

y=0 ? that would not make any sense?

probability of X from y is infinite

Quote from: Thebox on 19/06/2015 18:31:43y=0 ? that would not make any sense?Correct, it does not make any sense, but that is what you wrote!Nothing else makes sense either.Quote from: Thebox on 19/06/2015 18:31:43probability of X from y is infiniteProbability can only take values from 0 to 1. Infinite probability is meaningless.Why are you using time as a value. To paraphrase Tina Turner "What's time got to do with it?"??

I am using time, because time decides what point of Y you receive. Intersection of x,y by random timing of a tables hand.

every element of x is an infinite element of y.

But I think I am just going to quit science now and not bother any more , It is like talking to a brick wall at times.

I know very well that ∞ represents infinite, ..

Anybody who knows what ∞ means knows better than to treat it like a number and thus make any attempt to try to divide one by it.

I have a Y axis that is infinite ...

and contains an infinite number of variant x's, where my x axis only contains 52 variants of x.

I am trying to associate maths, and represent 52 variants of x, and 52 variants stacked for infinite time.

The worst part about your attitude in this particular thread is that you asked us if the math was correct and when we explain that it isn't you claim that we're wrong.

shuffle a deck of cards, the odds of the top card being an ace is 4/52 every single shuffle.shuffle 100 decks of cards, the odds of any of the singular decks top card is 4/52This is the problem, lets say for example purposes, that in 100 decks of pre-shuffled cards , 15 of the top cards out of the 100 decks, was an ace.

I assume you know what it means without explanation.

X is any one of the 52 variants of x axis.

1: manifesting variety, deviation, or disagreement 2: varying usually slightly from the standard form

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.By random timing this could be intercepted of the distribution. Do you agree?

1/3, obviously. Since the rows are independent, it doesn't matter which you choose.

You only make one choice. None of the groups knows anything about any of the other groups.Consider a simpler example. I have tossed a coin 1,000,000 000 times. What are the odds that it will come down heads next time? Now I roll a die 1,000,000,000 times. What are the odds of getting a six on the next roll?

Why don't I get the 1st and 7th?

123231123132all x axis would be 1/3 where y axis you can clearly observe is different.

The sequence of the first ten throws arehhththttthThis 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?

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.

12345231455324112345In the above I have a 2/4 chance of receiving a 1 if my go is first of the distribution.

Your last shuffle and deal was a bad one. The probability of C_{5}=J five times out of five is very small (1/5^{5} = 0.00032) if the shuffles are truly random.

Can you confer my scenario is the correct logic and maths?........Would you agree that the player could intercept values by chance and that the maths is player (a) intercepts x,y over timeP(a)∩(x,y)=0-∞/tand x≠y/t

Quote from: Thebox on 25/06/2015 04:41:39Can you confer my scenario is the correct logic and maths?........Would you agree that the player could intercept values by chance and that the maths is player (a) intercepts x,y over timeP(a)∩(x,y)=0-∞/tand x≠y/tI don't know about the logic because you haven't fully described the scenario.The maths is wrong. You are saying the probability of player (a), what does that mean? If you have a player (a) then probability value is 1, if you don't have a player it is 0. So P(a) is confusing me!(x,y) what does that mean? You have cards in x,y. So it is really not valid to put these together with the probability of having a player!What does 0-∞ mean? It has been pointed out before that using ∞ in equations is not valid, to be honest the closest you could say for this is that it is -∞ , what does that mean? Then you divide -∞ by t. This doesn't make sense either, is -∞/60 seconds really any different from -∞/100seconds??Time is not an issue here. You can't divide a probability by time.It would be better if you took the trouble to learn maths rather than stringing random maths symbols together! In most other forums, you would be ridiculed for these equations.As far as the scenario goes, you will need to describe it in more detail. But let me make some suggestions.Your game obviously involves cards. Let's say, to follow one of your examples, that you have 5 packs of cards each shuffled.If each player takes 5 cards from separate packs, clearly they both have equal probability of any sequence of 5 cards. If they both move on to new packs for the next hand, or they return their cards to the packs and shuffle, then again on the next draw they both have equal probability. You don't need to assume an infinite number of packs.The only reason there might be a difference between the players is if one is expected to make a selection which has a different probability to the other.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.Unless you can clearly explain your scenario, without the false maths, it is impossible to know whether you are right. For example, this sequence is not clear:added -consider spacing and random time of deck distribution1.akj..time.34.akj...time.107.akjIt seems to have something to do with selection of decks?

What about if we used 1,000,000 decks, and the same scenario, how many hands out of 1,000,000 decks would contain a J in the 5th position?

akj akjjkaplay 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.

the first card, the odds of an ace are 4/52. 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 shuffled

X is 52Y is repeats etcX is not equal to yOne 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

Hi Colin you are getting colder and away from the thinking

Quote from: Thebox on 26/06/2015 18:10:08Hi 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.Let me propose a game. Alan has gone home, so just the 2 of us.We will each have a deck of cards in front of us and we will turn over the top card and highest wins (aces high), best out of three.I take a new pack and shuffle it, put it down in front of me.In front of you are 10 shuffled packs, and you choose one at random, place it in front of you.We play. I then return my card and shuffle my pack. You discard your pack and choose another from the remaining 9.We play, I shuffle, you pick from the remaining 8.We play, we have a winner.This is an equal, fair game. We both have equal chance of winning. If we played 100 games we would, most likely, both win around 50.Do you agree?I suspect not, because you will say what about the other 7 decks, they might have held better hands for you.That is irrelevant to probability.Probability only deals with likely outcomes over a large number of games.Probability theory says that all the decks were well shuffled, all had 52 cards and all had equal probability of a winning hand. Forget the other 7 decks, they are irrelevant.

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?

You have a 4/52 per every shuffle 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,

what is the odds that a pick a deck that as already been shuffled that as an ace aligned to my seat?AjkKjaKajAkjKajI 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 , ................................., x is short term and y is long term I do not have two lifetimes,

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. it would be would it not?a={x+y}={4/52}²~t

I will show you I understandif I rolled a number 1 with a dice, the next go my odds are 1/6^2 to roll another number 1.

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.

if I rolled a number 1 with a dice, the next go my odds are 1/6^2 to roll another number 1.I understand.

Quote from: Thebox on 27/06/2015 10:06:22if I rolled a number 1 with a dice, the next go my odds are 1/6^2 to roll another number 1.I understand.Apparently not. The odds of you rolling a 1 on the next throw, and indeed any throw, are 1/6, because the throws are independent.So here's a little puzzle for you. The odds of throwing six successive 1's in 6 throws are clearly (1/6)^{6} but what are the odds of throwing (a) at least and (b) exactly one 1 in 6 throws? I somehow think this is the problem you are trying to solve.

the wind took the note.gone with the wind.don't play online poker.

If the second road is infinite, anything and everything can happen as you travel it, but the probability of any one thing happening before you die is negligible.Online gambling is an industry, not a charity.

what are the odds of throwing (a) at least and (b) exactly one 1 in 6 throws?

I somehow think this is the problem you are trying to solve.

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.