Sorry typo

edit - The chance remains the same if there is NO new information known, except there is new information known.

{Y}=σ²{X} is known information like you have yourself just stated

And you are using the whole and not considering card 1 line ups to the small blind [Z] of each table.

Shuffle X setting a SB vector of card one, to player Z

shuffle several X creating a Y, SB vector to several outputs of Z

shuffle Y and distribute.

pkdk, was this you also: holdemace486?

I do not know that name sorry.

The maths for this is very simple,

φ{X}=1/52

≠

φ{Y}=0_52/52

∑{Y}≠∑{X}

I kind of think you do.

But if not, you might want to take a look at his posts. I think the "two of you" have quite similar ideas.

Please describe this new information as if I was a child.

A school bus travels a journey every day and stops ten times to pick up 5 boys and 5 girls, to take to school.

Every day is the same 5 boys and 5 girls , this is constant/continuous over time.

We will call this journey X which has a constant/continuous ratio of 5 boys and 5 girls . We will represent this constant with a golden constant ratio symbol φ

we can express

φ(Boys)=5

φ(Girls)=5


Ok so far?

Go on...

When arriving at the school, the bus driver ask's the children to leave the bus, the children leave the bus in no specific order.

We can determine that the first pupil off the bus has a 5/10 chance of being a boy or a girl.

We can express that the constant chance of a boy or a girl being the first pupil off the bus is

φ{first pupil off the bus boy or girl}=5/10


Ok so far , you can relate poker to this ?

Seems straight forward.

Ok, I will write further


The school is a big school and has many pupils, in total 100 buses are used by the school and each bus picks up the same amount of pupils and the same amount of boys and girls.

Each bus has a golden ration of φ=5/10 that also remains true for the first person off each bus when it arrives at the school.

I was referring to both burns and discards. And now I know you are either trolling or stupid. The thread is no longer interesting to me.

Continuing with bus ,

The school also has bus 101, this bus follows a different route to all the other buses, and travels route Y.

This bus is a special bus, like the other buses it picks up ten pupils, but this bus has no golden constant ratio other than 10.
This Bus picks up pupils that are late for school, so it is undetermined whether they will be boys or girls.

Ok so far?

You're losing me. How does bus 101 relate to the original question?

Bus 101 is a mixture of boys and girls that is unknown in comparative to
picking 52 random variants from 52 known sets of 52 known variants.

No. For it to be similar, the bus would need to pick up 100 passengers that were first off (randomly) from the 100 buses that had a known ratio. Your bus is picking up passengers from a population we know nothing about.

And people are stating the chance of an ace from 52 cards randomly chosen from 52 random decks, is 4/52, they are stating this when they know nothing about the 52 cards that are chosen.

Imagine you are the driver of the 101 bus, it is empty until you pick up your first late pupil, the pupil gets on wearing a cloak and face covered so you can't tell if It is a boy or a girl. Do this for ten pupils. How many of the ten pupils are boys or girls on your bus?

This is not true. We know that that each of the 52 cards came from a regulation deck. We have no idea what the population was for your cloaked pupils.

you have no idea of the population of 52 drawn cards from 52 decks,

You do not know how many of the 4/52 chances you picked landed.

f{X}{boys}=5

f{X}{girls}=5

f{Y}=σ²f{X}

You can check this maths it is correct.

You've made much more complicated than needed. That 1st card came from 1 ordinary deck. The other 51 cards of this constructed deck are completely superfluous to the question asked. It's all smoke that clouds peoples thinking. Even if the constructed deck had been shuffled (more smoke), the card selected would still be one card from one deck. Every card of the constructed deck shares this same property.

Now that we have cleared away the smoke and see that this problem is really effectively about picking just 1 card from just 1 ordinary deck, the answer is just 4/52 = 1/13.

It is only when we pick a 2nd card from the constructed deck that the story changes, but the OP did not ask about that

But we can determine the chances of every possible combination of drawn cards. And for each combination we can determine the odds of picking an ace.

You can check this math, it is correct.

1. What if we have 52 decks and we draw one card from each deck and we use these 52 cards to create a new deck; and then we draw 15 cards from the new deck and place them face down; and then we draw one card from these 15 cards? What are the chances the drawn card is an A?



2. What if we shuffled all 52 decks together to create one "megadeck" and then drew one card from the "megadeck". What are the chances the drawn card is an A?



3. What if we shuffled all 52 decks together to create one "megadeck" and put the "megadeck" on the right half of a table; and then drew 52 cards from the "megadeck" and put these 52 cards on the left half of the same table; and then drew one card from the 52 cards on the left half of the table.



4. What if we shuffled all 52 decks together to create one "megadeck" and put the megadeck on the right half of a table; and then drew 52 cards from the "megadeck" and put these 52 cards on the left half of the same table; and then drew one card from the 2652 cards on the right half of the table. What are the chances the drawn card is an A?



5. What if we shuffled all 52 decks together to create one "megadeck" and put the "megadeck" on the right half of a table; and then drew 52 cards from the "megadeck" and put these 52 cards on the left half of the same table; and then drew 20 cards from the 52 cards on the left half of the table and put them on the right half of a different table; and then drew 10 cards from the 20 cards on the right half of the new table and put them on the left half of the new table; and then drew one card from the 10 cards on the left half of the new table. What are the chances the drawn card is an A?



6. What if we shuffled all 52 decks together to create one "megadeck" and put the megadeck on the right half of a table; and then drew 52 cards from the "megadeck" and put them in a bag which we take on an airplane to Las Vegas; and when we get to Las Vegas we go up to a hotel room and take out the bag and pull out one card and put it face down on the table; and we say that if the card is an A, then Person X has to pay Person Y $10,000, but if it is not an A, then Person Y has to pay Person X $1,000; and you have to choose whether you are Person X or Person Y. Do you choose to be Person X or Person Y? Why? What are the chances the card is an A?



7. What if we took each card from 52 different decks and put each card on its own folding chair; and then let a cat frolic among the chairs; and the first time the cat jumps onto a chair we go over and look at the card on that chair. What are the chances the card is an A?



8. What if we took each card from 52 different decks and 338 different folding chairs and we put 8 cards from the 52 decks on each of the folding chairs; and then we let a cat frolic among the chairs; and the first time the cat jumps onto a chair we go over and draw one card from the 8 cards on that chair. What are the chances the drawn card is an A?



9. What if we take all 52 decks up in a hot air balloon and just as we get to 2,000 feet and keep a steady altitude we see a buddy of ours in another hot air balloon still rising and just exactly coming even with us at 2,000 feet; and as we go by we throw all 52 decks into his face; and as it happens, 52 cards land in his hot air balloon as the other 2652 cards are falling down to earth; and one of the cards that landed in his hot air balloon somehow slips through his collar and falls down his shirt; and he starts to pull the card out of his shirt. What are the chances the card is an A? If we pretend there is no wind resistance, how fast does our buddy have to pull the card out of his shirt and look at it and register the value to know what the card is before the 2652 cards that were falling to the earth hit the ground?



10. What if we take all 52 decks up in a hot air balloon and just as we get to 2,000 feet and keep a steady altitude we see a buddy of ours in another hot air balloon still rising and just exactly coming even with us at 2,000 feet and as we go by we throw all 52 decks into his face; and as it happens, 52 cards land in his hot air balloon as the other 2652 cards are falling down to earth; and one of the cards that landed in his hot air balloon somehow slips through his collar and falls down his shirt; and another of the 52 cards lands on the controls for the hot air balloon and somehow causes the balloon to stop rising; and our buddy reaches for the card on the controls to move it off the controls so the balloon will start rising again so he can get away from us. What are the chances the card is an A? If we pretend there is no wind resistance and that the hot air balloon will gain altitude at a steady rate of 200 feet per minute as soon as our buddy removes the card, how fast does our buddy have to pull the card off of the controls to get 100 feet above us before the 2652 cards that are falling to earth hit the ground?



11. What if we have 52 decks and we draw 1 card from each deck and we use these 52 cards to create a new deck and then we take the other 52 decks that each have 51 cards now and we mix all those decks together to create a "super deck"; and we draw one card from the "super deck". What are the chances the drawn card is an A?

1/13, but you need to be more careful. Your megadeck does not share all the same properties of the OP's constructed deck e.g. using the original constructed deck, if after selecting 51 cards and all are aces, your chances of the 52nd card being an Ace is still 1/13. With your megadeck after selecting 51 aces, the chances the 52nd card being an ace is (4*52 - 51)/(52² = 5.8% (approx).

I point this out since your post seems to imply equivalence when in reality, they only happen to share the same answer.

It is getting late now i will read the long post and other posts tomorrow when I am more focused again.

I can tell you I believe this maths is getting really close .



φ{X}=1:52

{X}²=φ{Y₁.....Y₅₂}=(<1.=1.>1):52

My argument is based on golden constant ratio.

I look forward to learning the answers to the questions in the 11 situations at the below link tomorrow.


http://forumserver.twoplustwo.com/sh...2&postcount=72