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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.
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?
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 C5=J five times out of five is very small (1/55 = 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,