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if you recorded the results of the coin toss and kept the results hidden, then asked someone to pick a result, the chance is not 1/2 any more. the chance is ? out of ?oooooooHow many heads or tails are in the above number of throws? You cant answer it can you?

You are beginning to see the light. Please keep your eyes open. Nobody will accept a bet after the race is over. But I can tell you that if the coin is fair and you made a very large number of throws, about half of them would be heads, and the more throws you made, the closer you would get to 0.5. So my best guess at the number of heads in a finite number n throws would be n/2, and the probability of guessing the next throw would be 0.5 too, because the trials are fair and independent. The essence of Bayesian statistics is that if all trials are independent, the particular trial you are considering is not special. It actually goes further than that:You have captured one of the enemy's guns. It bears the serial number 1846. How many of these guns does the enemy have?

... what is the world trying to cover up?...........why is everyone lying to me?

You are not considering before and after and recorded results, before a coin toss the result is 1/2 , if you recorded the results of the coin toss and kept the results hidden, then asked someone to pick a result, the chance is not 1/2 any more. the chance is ? out of ?oooooooHow many heads or tails are in the above number of throws? You cant answer it can you?

What probability tells you is the likelyhood that when shuffled the first card might be ace of clubs

In a normal distribution we would receive 12121121121, in a random leap distribution we could receive 1111111111.

If the trials are independent, the odds do not change, by definition. QuoteIn a normal distribution we would receive 12121121121, in a random leap distribution we could receive 1111111111.There being 11 members in the first sequence, the probability of generating it is 1 in 2^11. The second sequence contains 10 members so the probability is 1 in 2^10, so you are twice as likely to receive the second sequence as the first.

I think you have misunderstood the point Alan was making.

You say that by random leaping you are twice as likely to receive the second sequence, So if I have 100 decks of cards and we choose a deck, and we will get the top card, are you saying we are twice as likely to receive the same value?

Nothing to do with random leaping. You are twice as likely to get the second set because it contained one less binary element!If you had said 1212112112, which contains 10 elements, it would have had exactly the same probability as 1111111111, which also contains 10 elements, i.e. 1 in 2^10. That said, even if you made 1024 trials of 10 coin tosses, there is no certainty that either sequence would turn up once or once only, you'd just be a bit surprised if it didn't. And you have no prior knowledge of where in that 1024 trials it would turn up, so looking at random or sequentially would have exactly the same probability of finding it. Here's a real-world example of the "significant number delusion". Many years ago I rebuilt a laboratory rig that originally included an analog voltmeter. For about 10 years, people had recorded its readings to 3 figures, e.g. 38.6V. I replaced it with a digital voltmeter and printer that displayed 5 figures, up to 99.999 volts. My boss complained that the rig wasn't working properly because it often reported "repeated digits" like 16.885, 17.793, 22.541 and so forth. It was working of course perfectly: if you take 5 random digits, the probability that the second matches the first is obviously 1/10, and that the third matches the second, 1/10.... i.e. there is a 40% probability that any 5 digit random number (such as a digital voltmeter reading) will include a repeated digit.There is a similar problem with "significant sequences" and "significant cards". It is noticeable that throughout these discussions you have concentrated on the probability of finding aces whereas form the point iof view of the cards themselves, and indeed of the dealer (who can't see the faces) an ace is no more significant than a 6. So whilst 123 or AAA may mean something to you, it means nothing to the shuffled pack. Furthermore, you are not signifcant to the dealer who is also supplying cards to Alf, Bill and Charlie

What I think you are trying to say by "P(a)/X(A)" is that the probability of a named card N being first in the first deal is 1/52.It is clear that the probability of N being the first card in the second deal is also 1/52.So the probability of N being the first card in both deals is (1/52)^{2}But if you don't name the card in advance, the first card in the first deal defines a value n, and the probability of n being the first card in the second deal is 1/52. yesThus if I want the ace of spades to be the first card, the probability of getting AS in any one deal is 1/52 and the probability of getting it in two consecutive deals is (1/52)^{2}. However suppose I get 10C onhe first deal, and manage to win the hand with it. 10C has now become a "special" card and the chance of getting "my special card" on the next deal is 1/52 yesNow I see your delusion. You think that by choosing "nonconsecutive" deals you can increase the probability because 1/52 + 1/52 = 2/52. There's the error. The probability of getting one AS in two deals is indeed 2/52 because the desired result is "a OR b". But the probability of getting two AS in two deals is 1/52^{2} because the desired result is "a AND b". George Boole is credited with laying the mathematical foundations for machine logic, which starts with the realisation that "+" describes "or" relations and "x" describes "and" relations.no

I'm sorry you disagree with Boolean algebra. It's the basis for all computer hardware, and seems to work for everyone else. Beyond that, I have no idea what you are talking about.

Maybe you will understand this, at any specific point in time whilst playing live texas holdem poker there is only ever 1 of 52 variants aligned to your seat. On the internet playing internet texas holdem poker, there is 1,000,000 unknown variants aligned to your seat at any specific time.

OK, I understand where 52 variants comes from in live game = 52 different card faces.I don't understand where 1,000,000 comes from in Internet game. If you only play with one deck at a time in any one game then it's the same as live. For it to be different you would need to play with a single deck of 1,000,000 cards each with a different face value. So no I don't understand.

Quote from: Colin2B on 05/09/2015 09:35:53OK, I understand where 52 variants comes from in live game = 52 different card faces.I don't understand where 1,000,000 comes from in Internet game. If you only play with one deck at a time in any one game then it's the same as live. For it to be different you would need to play with a single deck of 1,000,000 cards each with a different face value. So no I don't understand.Yes you do understand, because in bold on the internet that is exactly what you do by the alignment of seat to card order,

Quote from: Thebox on 05/09/2015 10:59:31Quote from: Colin2B on 05/09/2015 09:35:53OK, I understand where 52 variants comes from in live game = 52 different card faces.I don't understand where 1,000,000 comes from in Internet game. If you only play with one deck at a time in any one game then it's the same as live. For it to be different you would need to play with a single deck of 1,000,000 cards each with a different face value. So no I don't understand.Yes you do understand, because in bold on the internet that is exactly what you do by the alignment of seat to card order,No I don't understand.What do you mean by "the alignment of seat to card order"How does this differ from live game.You are surely not suggesting that you play with a deck of 1,000,000 cards?And please, describe this in words don't try to use maths.

Well actually if there is 1,000,000 top cards there is 52,000,000 cards in total, If I asked you to pick any card from any position of any deck, I have asked you to choose a card of 52,000,000 cards.

If I tell you to choose a card from the top cards, you are choosing from 1,000,000 cards effectively playing a 1,000,000 card dec

But there are only 52 face values, so the probability of getting any given card is 1/52

I will put it this way, you are dealt a single hand using a single deck, or you pick a hand from 1,000,000 already made hands.Do you think this sounds like the same game?

Quote from: Thebox on 06/09/2015 08:16:36I will put it this way, you are dealt a single hand using a single deck, or you pick a hand from 1,000,000 already made hands.Do you think this sounds like the same game?Let me ask some questions so I can understand.1) In a live game the dealer uses a single deck which he shuffles after each hand.2) If instead he preshuffles 100 decks which he then lays out on the table and asks the first player to choose a deck at random. Is this the same probability as 1)? no3) if the dealer does as 1), but instead of dealing off the top of the deck he deals from the bottom has that affected the game?yesAgain words only please, no maths.

Ok, I now understand what you are saying.From a probability point of view all the games are equal. I would view the 100 decks as following probability rules when they were dealt and equal to the individual shuffled deck. Over a large number of games they are equal. Probability deals with what is most likely to happen under various circumstances and degree of knowledge.Your theory involves predetermination, which is not part of probability. You are dealing with individual events.If you want to develop a maths for your theory you have to stop using probability terms as it will only confuse people.I don't know how you can handle this with maths. Will have a think, but not hopeful.

ColinI think I am trying to describe this - SL(n, Z) → SL(n, Z/N·Z)...sorry, you cannot view external links. To see them, please REGISTER or LOGIN

86400/86400=152/86400=0.0006018518551/86400=0.0005902777750/86400=0.00057870371/86400=0.00001157407this seems connected somehow?added - 52*2/86400=0.001203703752*100/86400=0.0601851851852*1661.53846154/86400=1

Can you explain why you think that and how you link the variables to your problem?because of what this says,In mathematics, the special linear group of degree n over a field F is the set of n × n matrices with determinant 1, with the group operations of ordinary matrix multiplication and matrix inversion. This is the normal subgroup of the general linear group given by the kernel of the determinant I don't know why you are dividing 52, 51, 50 etc. we agreed that you are not looking at probability but at what is 'written' so you need actual values as what is written is no longer random but fixed. Also your problem has more to do with orientation of the decks that individual cards.I have to repeat however that for me what is written is irrelevant, it is how it was written that is important and that is what defines the distribution of the cards over a large number of games. Thus until we know the outcome, the rules of probability apply and P(X)=P(Y).And yes, in your view of the game P(X) cannot =P(Y) because the card values are a single, specific outcome.

presently looking at this....sorry, you cannot view external links. To see them, please REGISTER or LOGIN

because of what this says,In mathematics, the special linear group of degree n over a field F is the set of n × n matrices with determinant 1, with the group operations of ordinary matrix multiplication and matrix inversion. This is the normal subgroup of the general linear group given by the kernel of the determinant ...presently looking at this....sorry, you cannot view external links. To see them, please REGISTER or LOGIN

Anything of block randomness is predictable to a degree, I can accurately predict that if you continued to spin a roulette wheel and spin the ball, turn after turn for 24 hrs, I will predict that zero will always make an appearance within 24hrs.

Quote from: Thebox on 07/09/2015 19:17:56because of what this says,In mathematics, the special linear group of degree n over a field F is the set of n × n matrices with determinant 1, with the group operations of ordinary matrix multiplication and matrix inversion. This is the normal subgroup of the general linear group given by the kernel of the determinant ...presently looking at this....sorry, you cannot view external links. To see them, please REGISTER or LOGINThese are moving you further from understanding your problem.You need to concentrate on understanding why P(X)=P(Y). Then you will have your answer.Let us know when you have done that.

In algebra would this be correct A+B=AB?

This equation is satisfied in the case of A =2 and B = 2. Also A = 0 and B = 0. In General: for any value A, there is a value B = A/(A–1) which will satisfy your equation.

you asked if A+B=AB makes sense algebraically. I understand that statement to mean: "there are two numbers, A and B, for which A plus B equals A times B."This is a perfectly valid algebraic statement, and can be rearranged to show that every number A has exactly one value of B such that A plus B equals A times B. We can even solve for what B must be for any A, which was I meant to show as B = A/(A–1), which you should understand to mean: "B must equal the ratio of A and A minus one."

so if I put x+y=xythat would mean the same as a+b?

I have been playing around with something, and wish for an opinion, X^100=XY=10cm²XY^100=XYZ=10cm³E=XYZ-^XYZ=0cm³would this make any sense to you?

Quote from: Thebox on 08/09/2015 21:45:02so if I put x+y=xythat would mean the same as a+b?x+y=xy is the same as A+B=AB, which is the same as g+q=gq (unless you have defined the letters or added other relationships to them--for instance, if your equation is also accompanied with x+y=6, then it is assumed that the x in one equation is the same as the x in the other, and then numerical values can be found for both x and y)Quote from: Thebox on 08/09/2015 21:45:02I have been playing around with something, and wish for an opinion, X^100=XY=10cm²XY^100=XYZ=10cm³E=XYZ-^XYZ=0cm³would this make any sense to you?I don't think it makes much sense. I don't know how you're using the ^ symbol. It is also odd to put units (cm) on one side of an equation, but not the other...

^ to the power of and if you notice I was subtracted the power of contracting XYZ.Time begun XYZ-^XYZ=0

Quote from: Thebox on 09/09/2015 07:13:48^ to the power of and if you notice I was subtracted the power of contracting XYZ.Time begun XYZ-^XYZ=0You can't raise a - sign to a powerAlso, how can you say 'time begun' = 0cm^{3}When you write things like this it looks as though you are just pulling our legs (polite version) or trolling.If that is not the case you really do need to concentrate on learning basic maths, it would make your life, and ours, less frustrating.

I am not raising , I am tasking the power of away. I am contracting space, or expanding it, try it , it works for me.

You can correctly state it as XYZ^-XYZ or . So that if XYZ = 10 then XYZ^XYZ then equals 10000000000. This then is 1/10000000000

Is JefferyH's answer what you are trying to do?Also, you still havent explained how can you say 'time begun' = 0cm3

Because time is a measurement based on another measurement, i.e 24 hrs = 0.0288m/s