Expanding on the logic riddle:

[Fort Fun Indiana]

I have an infinite deck of cards. Half are red, half are black.

1) I pull one card, then another, laying them face down in front of you. The first card I chose is on your left.

I turn over the first card. It is red. What is the probability the other card is also red?
.........

2) I turn over the second card, first. It is red. What is the probability the other card is red?

...........

3) Before I lay the two cards down, I shuffle them behind my back. I lay them in front of you and I turn over the card to your left. It is red.

What is the probability the card on the right is also red?

 

[Fort Fun Indiana]

Now answer using a 52-card deck.

 

[Baron Von Murderpaws]

Why is it I care?

 

[Toddsterpatriot]

I have an infinite deck of cards. Half are red, half are black.

1) I pull one card, then another, laying them face down in front of you. The first card I chose is on your left.

I turn over the first card. It is red. What is the probability the other card is also red?
.........

2) I turn over the second card, first. It is red. What is the probability the other card is red?

...........

3) Before I lay the two cards down, I shuffle them behind my back. I lay them in front of you and I turn over the card to your left. It is red.

What is the probability the card on the right is also red?

1) 50%
2) 100%
3) 100%

 

[Hang on Sloopy]

I have an infinite deck of cards. Half are red, half are black.

1) I pull one card, then another, laying them face down in front of you. The first card I chose is on your left.

I turn over the first card. It is red. What is the probability the other card is also red?
.........

2) I turn over the second card, first. It is red. What is the probability the other card is red?

...........

3) Before I lay the two cards down, I shuffle them behind my back. I lay them in front of you and I turn over the card to your left. It is red.

What is the probability the card on the right is also red?

wanna see the most interesting card trick a 2nd grade teacher taught me?

 

[Dogmaphobe]

I have an infinite deck of cards. Half are red, half are black.

1) I pull one card, then another, laying them face down in front of you. The first card I chose is on your left.

I turn over the first card. It is red. What is the probability the other card is also red?
.........

2) I turn over the second card, first. It is red. What is the probability the other card is red?

...........

3) Before I lay the two cards down, I shuffle them behind my back. I lay them in front of you and I turn over the card to your left. It is red.

What is the probability the card on the right is also red?

How can you possibly shuffle cards if you have an infinite number?

 

[Winco]

If it is 'infinite' then half doesn't exist.

If you insist it does exist, then the probability is 1/2.
Because there is a one to one correspondence between the remaining cards.

Now answer using a 52-card deck.

If it's 52 cards, and you pull a red w/out replacement, then the probability of the 2nd card also being red is 25/51.

 

[Fort Fun Indiana]

How can you possibly shuffle cards if you have an infinite number?

I only shuffle the two cards I chose from the infinite deck.

 

[Fort Fun Indiana]

If it is 'infinite' then half doesn't exist.

Of course it does. For example, the set of positive integers. Half are even and half are odd.

Believe that fact or don't, you still understand the "infinite deck" assumption:

You have a 50% chance of dawing a red card every time you draw a card from the deck, no matter how many cards you draw.

"With replacement", so to speak.

 

[Dogmaphobe]

I only shuffle the two cards I chose from the infinite deck.

Why would you shuffle two cards both known to be red?

You must have a very short memory.

 

[Fort Fun Indiana]

1) 50%
2) 100%
3) 100%

No. But I get what you did there. Same mistake Dogmaphobe is making.

It's not a trick question. It's three different scenarios. They do not happen in sequence.

 

[Fort Fun Indiana]

Why would you shuffle two cards both known to be red?

I'm not.

I drew two cards from the deck. Before placing them face-down in front of you, I shuffled them.

Neither of us knows the color of either card, at this point.

 

[Fort Fun Indiana]

If it's 52 cards, and you pull a red w/out replacement, then the probability of the 2nd card also being red is 25/51.

In all 3 cases I mentioned? Just one?

 

[Winco]

Of course it does. For example, the set of positive integers. Half are even and half are odd.

Believe that fact or don't, you still understand the "infinite deck" assumption:

You have a 50% chance of dawing a red card every time you draw a card from the deck, no matter how many cards you draw.

"With replacement", so to speak.

So......

The set of positive integers that are multiple of 20 versus the set of positive integers that are multiples of 2.

If you had to pick a positive integer, from the set of positive integers that are multiples of 2, what would be the ODDS of:

1). Picking a multiple of 2 versus a multiple of 20?
2). Picking a multiple of 20 versus a multiple of 2?

 

[Winco]

In all 3 cases I mentioned? Just one?

1/2 in ALL 3 cases.

 

[Fort Fun Indiana]

So......

The set of positive integers that are multiple of 20 versus the set of positive integers that are multiples of 2.

If you had to pick a positive integer, from the set of positive integers that are multiples of 2, what would be the ODDS of:

1). Picking a multiple of 2 versus a multiple of 20?
2). Picking a multiple of 20 versus a multiple of 2?

10 to 1.

Easily proven, too, by unions of sets. In shorthand:

For any consecutive 100-member subset of the positive integers, the probability of randomly selecting an even number is 50% and the probability of selecting a multiple of 20 is 5%.

The set of all positive integers is the union of all such sets.

Therefore, these probabilities also apply to the set of all positive integers.

I.E., your probability of randomly selecting an even number from the set of positive integers is 50%.

 

[Fort Fun Indiana]

1/2 in ALL 3 cases.

No.

 

[Winco]

Easily proven, too, by unions of sets. In shorthand:

Each set is infinite.
There is a one to one correspondence between every number in each set.

There are NOT 10 times more multiples of 2 than than are multiples of 20.
They are both infinite sets.
For any multiple of 2, a corresponding multiple of 20 can be found.

It is a one to one correspondence.


If you think it's 10 to 1, then you are suggesting that there are 10 times more multiples of 2 than there are multiples of 20.

This would be true if we were talking about a selected interval, but we aren't.
We are talking about infinite sets.

You are incorrect.

 

[Fort Fun Indiana]

Each set is infinite.

No, each set contains 100 consecutive, positive integers.

E.G., {1, 2, ......,100}, or {2, 3, ....., 101}

The union of all such sets is the set of positive integers.

The probability you will randomly select an even number from the set of positive integers (the union of all such sets) is 50%. The probability you will randomly select a multiple of 20 from the set of positive integers is 5%.

It is essentially analogous to "with replacement", in that "infinite deck" yields the same result. If it will get us past this trifle, I will use that phrase when posting to you (for now). Deal?

 

[Winco]

No, each set contains 100 consecutive, positive integers.

E.G., {1, 2, ......,100}, or {2, 3, ....., 101}

The union of all such sets is the set of positive integers.

Now you are changing the problem to try to prove you are correct.
You can't ask a question about an infinite set, then use a finite set of 100 to prove your point.

It doesn't work that way.