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4/52 = 1/13If P = probability of something happening in one trial, the probability of it happening n times in succession in independent trials = P^n.So probability of the first card being an ace in two (and only two) independent trials = (1/13)^2 = 1/169 A posteriori analysis of a small number of trials does not give you the probability of an event. Choosing or constructing two or three possible sequences tells you nothing about the likely outcome of any actual trial, let alone the probability of a given outcome in sequential trials. For the very last time: if the shuffles are independent (a) the probability of an outcome is exactly the same in each shuffle, by definition of independent, and (b) the statistics of actual outcomes only tend towards the calculated probability for a very large number of trials, by definition of probability.Belief in anything else is the first step on the road to gambling bankruptcy.
I do not why you think I am agreeing with the maths
You have got 100 decks of shuffled cards in front of you, and you are asked to draw the top card from one of the decks, what is the chance it is an ace?i bet you say 4/52 which is incorrect.
..........column 1 to the small blind contains 7.69% worth of aces. compared to 4/52
I forgot science likes simple
nnnnnnnnnwe know left to right of each row is still 1/3, but we do not know the columns values.
Thank you Colin, interesting that probabilities may not be the answer I am looking for,
Quote from: alancalverd on 09/07/2015 00:04:51Quote from: Thebox on 08/07/2015 17:31:13, I would bet you could not throw another heads, Gotcha! You are ignoring the evidence in favour of your preconception. The perfect mug.You have not got me, I do not ignore solid evidence, I know the coin still has 1/2 chance.
Quote from: Thebox on 08/07/2015 17:31:13, I would bet you could not throw another heads, Gotcha! You are ignoring the evidence in favour of your preconception. The perfect mug.
, I would bet you could not throw another heads,
Thank you Alan and Colin for your persistence. Your maths is accurate and true if y1,y2 and y3 were all coming to the same table.
Quote from: Thebox on 13/07/2015 14:52:42Thank you Alan and Colin for your persistence. Your maths is accurate and true if y1,y2 and y3 were all coming to the same table.No, that's incorrect. As I explained in my post, deck skipping (spacing, interception or timing as you call it) does not affect the outcome.If you have an infinite number of decks and you wish to choose 3 for each table it does not matter which order you distribute them in or by what timing. You could take the 5th, the 10th and the 81st for one table; the 93rd, 4th and 100006th for the 2nd table etc. All are as equal as if you had taken them in order 1 2 3 4 5 6 etc.If you had understood my post you would realise that what determines the randomness is not the distribution in the y direction, but the random shuffle of the decks. This makes all decks equal from a probability point of view.Any other belief is not probability. And as I have said before you cannot divide a probability by time.So let's leave it that I believe in probability and maths, and you believe in something else.
This is not true,they are not equal. y1y2y1y2y1y2That is a completely different distribution pattern if you deck skip . lets change the order, y1y1y1y2y2y2lets change the angle for you y1y1y1y2y2y2t...................tcan you not see the difference to a unified distribution and skipping time distribution?
You receive the first deck out of a stack of 1,000,000 decks, you receive an ace, the next turn in this example 100,000 of the decks have an ace as the second card, what is the chance you will receive a deck that the ace is the second card?
You receive the first deck out of a stack of 1,000,000 decks, you receive an ace, the next turn in this example 100,000 of the decks have an ace as the second card,
Quote from: Thebox on 14/07/2015 16:54:23You receive the first deck out of a stack of 1,000,000 decks, you receive an ace, the next turn in this example 100,000 of the decks have an ace as the second card, If you know this, the decks have probably not been shuffled fairly. Out of 1,000,000 decks the expected number with an ace as second card is 76,923, not 100,000. A 24% discrepancy over a million trials is good evidence of an irregularity.
It is also interesting to note that if TB understood probability he would also understand that even assuming we know the higher incidence of aces, deck skipping has no effect on the probability of receiving an ace.
It doesn't matter where you sit or when the cards are dealt. The probability of receiving an ace in any chair at any nominated point in any deal is 1/13.
None of the numbers in the set you call y is 4. So what? If you write down any finite set of integers, you can always find an integer that is not a member of that set. And you can also find a number that is. So what?
You can't "play y". In poker you play one hand at a time. Obviously x is not equal to y because you are comparing apples and chickens.There is no certainty in a fair shuffle, but because y is infinite and x is finite, the actual distribution in y is more likely to be close to the calculated probability than the actual distribution in any one x. This kind of second-order statistics comes under the heading of confidence limits:the greater the number of trials, the greater the confidence we can have that the actual and expected distributions will converge.However if all the trials are independent random shuffles, the expected distributions must be identical.
the chance of receiving a deck that the second card is an ace using 1,000,000 decks, is obviously (76,923/1,000,000)/t= 0.076923/t and then once the choice is made of deck (4/52)/t=0.07692307692.This is all good as long as no other tables are involved.(0.07692307692/t)(/n)/t where n is table.added - sorry that does not work, P(x)=0.07692307692∩n..................t
What do you think? You could offer genuine scientific proof of your theory.