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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.
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.However, I believe the crux of the problem is as follows:If you and I are playing a face to face card game and we need to change the deck, we gather the cards together and shuffle the deck. Probability of an ace 1st card is 4/52,In online poker at change of deck the player apparently selects, at random, from a group of previously shuffled decks. This appears to be where Box derives the infinite number of decks.However, his concern is with the unselected decks. Let's say you select decks 2, 4 and 7 for your next 3 games. I would say that for each deck the probability of an ace 1st card is 4/52, but box does not. His view is that decks 1, 3, 5 and 6 might have given better hands, hence the probability 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.This is more confused by some of the other scenarios he gives, and his poor understanding of maths.See also his lack of understanding in this reply to you
....we do not select the decks, when our table finishes a hand, we get a random new deck from the already shuffled decks. So does every other table.
and yes ''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 5deck one deck 2 deck 3 deck 4 deck 5........................deck 6......................table 3 finishes their hand first so get the next deck.
should not take account of this 'infinite vertical y axis' and so the probability for a particular deck is not a complex combination of infinity, time and the x, y axes, but the probability of an ace 1st card is 4/52 (ie odds in favour 1:12) for any deck however selected.
So if we have a choice of either deck, our first odds would be, 8/104, and then once we have made that choice, the odds return to 4/52.This is to show you that I understand.
added to clarify, imagine looking at 2 decks of card, there is 8 aces in total in the two decks, and a total of 104 cards, there is an 8/104 chance that one of , or both of the decks, has an ace as the top card.
Quote from: Thebox on 29/06/2015 18:18:46So if we have a choice of either deck, our first odds would be, 8/104, and then once we have made that choice, the odds return to 4/52.This is to show you that I understand. Unfortunately you don't understand as much as you think you do.This is conditional probability. You don't play with 2 decks, so you choose one and the probability for that game is 4/52
of cause you can not see the values in reality, player 1 or player 2 has the better chance of picking an (n)?
Quote from: Thebox on 30/06/2015 16:42:17of cause you can not see the values in reality, player 1 or player 2 has the better chance of picking an (n)?Yes, the way you have 'rigged' it, player 2 has a better chance.But you have given y 6 times what would occur from a random distribution and as you play more and more games the sequence tends towards a random distribution.Also, the way the decks are shuffled you will tend to get 4/52 in the y direction as well as x, so in the long run ( which is what probability is all about) you will get the same result.PS, I think you know, but I forgot to say that probability is often shown as % ie 1=100%=certainty etcHave fun with your new theory