I ask again ,

We now have 100 decks

numbered 1-100

We individually randomly shuffle all 100 deck's

I ask you to take the top card from each deck

What is the chance the top card of deck 1 is the exact same value as the top card of deck 2,3,4,5,.....100 ?

1/52 is the answer , you are adding them all together , I asked about each card. At this stage I have not removed the dependency of the cards independent deck.

i.e top card 88 is dependent to deck 88.

OK, I now see what you are asking at this stage. The top card on deck 1 has a 1 out of 52 chance of being the same as the top card on deck 2. Separately for deck 1 and deck 3 it is also 1 out of 52, and for deck 1 and deck 4 it is 1 out of 52. This continues for all decks.

Yes that is correct, before we move on and to the second part, to show the paradox, can you please confirm that so far we agree -


We will call this far, Event A
We have established of 100 independent decks of cards, that each independent top card after a random shuffle of each independent deck, has a 4/52 chance of being any ace, and a 1/52 chance to be any specific variant, a constant independent ratio, and also all/each of the top cards have a 1/52 chance of being the same card/cards.

Are we in agreement so far?

The way I read this I do not agree.

If you select two of the decks, the top card on one has 1/52 chance of being the same card as the other deck. This can be repeated for all combos of two decks.

d1/card 1 has a 1/52 chance of being the same card value as d2/card 1, d2/card 1 as a 1/52 chance of being the same as d3/card 1, d1/card 1 also has a 1/52 chance of being the same value as d3/card 1,


You have agreed with this , they are equally as likely to be the same value

''Quote:
Originally Posted by Didace View Post
OK, I now see what you are asking at this stage. The top card on deck 1 has a 1 out of 52 chance of being the same as the top card on deck 2. Separately for deck 1 and deck 3 it is also 1 out of 52, and for deck 1 and deck 4 it is 1 out of 52. This continues for all decks.''

We will call this far, Event A
We have established of 100 independent decks of cards, that each independent top card after a random shuffle of each independent deck, has a 4/52 chance of being any ace, and a 1/52 chance to be any specific variant, a constant independent ratio, and also each of the 100 independent top cards have a 1/52 chance of being the same value card as any other card of the 100 top cards.

Are we in agreement so far?

I would agree to this:

We have established of 100 independent decks of cards, that each independent top card after a random shuffle of each independent deck, has a 4/52 chance of being any ace, and a 1/52 chance to be any specific variant, a constant independent ratio, and also each of the 100 independent top cards have a 1/52 chance of being the same value card (suit and rank) as the top card on any one specific remaining deck.

Yes this says the same thing, so we are in agreement. Moving on -

Ok thank you, let's try it this way.

We now have 100 decks

numbered 1-100

We individually randomly shuffle all 100 deck's

I ask you to take the top card from each deck

I then ask you to place these cards in a box,
re-moving independent deck, dependency ,of each card,
and shake the box shuffling the 100 cards.

What is the chance the first card out is any ace?

What is the chance the first card out is any specific card?

I would expect each suit/rank to be in the box 100/52 or ~1.923 times. Since there are four suits of aces I would expect there to be 7.692 aces in the box. So -

The chance of any Ace would be 7.692 out of 100..
The chance of any specific card (suit/rank) would be 1.923 out of 100.

An interesting answer and not the answer I expected you to give. But an answer based on present probability. This answer would be wrong, the answer is 1 of 52. You have not took away the dependency.

Ok thank you, let's try it this way.

We now have 100 decks

numbered 1-100

We individually randomly shuffle all 100 deck's

I ask you to take the top card from each deck

I then asked you to place these cards in a box, removing independent deck, dependency ,of each card, and shake the box shuffling the 100 cards.


I draw a card and look to reveal the six of clubs, a 1/52 chance, then discard the card leaving 99 cards in the box.
I ask you to draw a card from the remaining cards in the box,
what is the chance that your card is the six of clubs?

Which answer are you saying is wrong?

''I would expect there to be 7.692 aces in the box.''

chance of 1 ace 4/52

chance of a second ace 4/52

chance of a third ace 4/52

chance of a forth ace 4/52


.............................
.................................
................................
so on......

chance of 100th ace 4/52



where do you get 7.692 aces from?

There could be no aces, there could be 50 aces, there could 100 aces,

all the cards have a 1/52 chance of being any card. all the cards have a 4/52 chance of being an ace

Each card has a 4/52 chance of being an ace.
There are 100 cards.
Total aces are 400/52.
400/52=~7.692.

Of course there are not partial cards in the box, but this is the average of what we can expect over a large number of trials.

400 /5200 =0.07692307692
4/52=0.07692307692
Shouldn't your sum be this?

No. You don't increase the denominator when adding.

Adding?

You put

There are 100 cards.
Total aces are 400/52.

100 decks = 400 aces

100 decks =5200 cards

Aren't we only dealing with the 100 cards in the box?

Ok I am going bed now , to get your answer I did it this way,

400 /5200 =0.07692307692*100=7.69230769231


But this is the wrong answer and I will explain why tomorrow.

This answer is still dependent to the 100 decks that have been removed from the equation once we place the 100 cards in the box.

Good god...

I eagerly await this new math!

Didace - your patience is incredible.

... and people say satire is dead.

Well! the point is there is no accurate way to work out how many aces are in the cards that are in the box. The values are unknown and impossible to know,

In a equivalent question, I post 100 Christmas cards into a box, how many of them are addressed to you?

I can see now that this is the place where we will get stuck for agreement.

In basic principle , we know that there could be no aces, 1 ace, 2, aces, 3 aces, 4aces..........................100 aces in the box.

do you agree with this?

P(A)/X=4/52

P(A)/Y=0_1/100


To save time how about I just agree with your maths and say on average there will be 7-8 aces among 100 random cards in the box?

Even this will not solve the problem. Even if you knew the values in the box, it would still not solve the problem.

Added - I do not want to try to bamboozle you with maths , but to express in simplicity is this -

/=in

\=of

∆=change

1/52∆1\52
1/52∆1\52
1/52∆1\52
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1/52∆1\52
1/52∆1\52
1/52∆1\52
1/52∆1\52

This is the entire process we discussed and have done by placing the 100 top cards in a box and shuffling them, By shuffling the top cards we un-align the card dependency to its own deck.

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