Expanding on the logic riddle:

[Fort Fun Indiana]

42

Accepted. Yet wrong.

 

[Fort Fun Indiana]

You're half right.

No, those are the correct answers. And they are correct for the same reasons the correct answer to the sibling riddle was 1/3.

 

[Toddsterpatriot]

No, those are the correct answers. And they are correct for the same reasons the correct answer to the sibling riddle was 1/3.

Feel free to explain.

 

[Fort Fun Indiana]

Feel free to explain.

I will. Give it time.

 

[Dogmaphobe]

To answer the question as to a 52 card deck, if the first card picked were red, that would mean 25 of the remaining 51 cards were also red, thus indicating the odds of pulling another red card would be approximately 49.02%

 

[Fort Fun Indiana]

To answer the question as to a 52 card deck, if the first card picked were red, that would mean 25 of the remaining 51 cards were also red, thus indicating the odds of pulling another red card would be approximately 49.02%

In all 4 cases? No.

In cases 1 and 2? Yes.

 

[Dogmaphobe]

What is infinity-1/infinity and how would that be expressed as a decimal?

It is nonsensical because of the very concept of infinity.

 

[Fort Fun Indiana]

What is infinity-1/infinity and how would that be expressed as a decimal?

Not relevant to this problem..

But, (infinity - 1) = infinity, and (infinity - 1)/(infinity) = 1, if Cardinality is held constant in all cases.

 

[Fort Fun Indiana]

Using a 52 card deck, the frequencies of the permutations are effected.

So, considering pairs of individual draws from a 52-card deck (new deck for each pair of draws), the probability of each permutation is:

BB: (26/52) * (25/51) * 100% = 24.50980392%

BR: (26/52) * (26/51) * 100% = 25.49019608%

RB: (26/52) * (26/51) * 100% = 25.49019608%

RR: (26/52) * (25/51) * 100% = 24.50980392%

So, the answers for a 52-card deck:

1) 2 * 24.50980392% = 49.01960784%

2) Same, 49.01960784%

3) 24.50980392 / [(2 * 25.49019608) + 24.50980392] * 100% =32.4675324647799%

4) Same, 32.4675324647799%

So, not much difference between an infinite deck and a 52-card deck. The "House advantage" decreases, the smaller the deck. But it has only decreased slightly, when going from an infinite deck to a 52-card deck. A clue to gambling and to house advantage...

 

[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) You pull one card, place it on my left, flip it over, it's red.

You pull a second card, put it on my right. What is the probability this card is also red?

50%

2) You pull one card, place it on my right, flip it over, it's red.

You pull a second card, put it on my left. What is the probability this card is also red?

50%

3) Before you lay the two cards down, you shuffle them behind your back.
You lay them in front of me

Left card 50% chance it's red, 50% chance it's black.
Right card 50% chance it's red, 50% chance it's black.

You turn over the card to my left. It is red.

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

50%

 

[Fort Fun Indiana]

1) You pull one card, place it on my left, flip it over, it's red.

You pull a second card, put it on my right. What is the probability this card is also red?

50%

2) You pull one card, place it on my right, flip it over, it's red.

You pull a second card, put it on my left. What is the probability this card is also red?

50%

3) Before you lay the two cards down, you shuffle them behind your back.
You lay them in front of me

Left card 50% chance it's red, 50% chance it's black.
Right card 50% chance it's red, 50% chance it's black.

You turn over the card to my left. It is red.

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

50%

No. Partial credit given.

 

[Toddsterpatriot]

Using a 52 card deck, the frequencies of the permutations are effected.

So, considering pairs of individual draws from a 52-card deck (new deck for each pair of draws), the probability of each permutation is:

BB: (26/52) * (25/51) * 100% = 24.50980392%

BR: (26/52) * (26/51) * 100% = 25.49019608%

RB: (26/52) * (26/51) * 100% = 25.49019608%

RR: (26/52) * (25/51) * 100% = 24.50980392%

So, the answers for a 52-card deck:

1) 2 * 24.50980392% = 49.01960784%

2) Same, 49.01960784%

3) 24.50980392 / [(2 * 25.49019608) + 24.50980392] * 100% =32.4675324647799%

4) Same, 32.4675324647799%

So, not much difference between an infinite deck and a 52-card deck. The "House advantage" decreases, the smaller the deck. But it has only decreased slightly, to go from an infinite deck to a 52-card deck.

So, considering pairs of individual draws from a 52-card deck

You're not drawing a random pair.
You're drawing a red card and a random card.

 

[Fort Fun Indiana]

You're not drawing a random pair.
You're drawing a red card and a random card

No.

I am drawing one card at random, then another. Both might be black cards.

Until we know at least one is not a black card.

And then, the other card may or may not be a black card.

 

[Toddsterpatriot]

No.

I am drawing one card at random, then another. Both might be black cards.

Until we know at least one is not a black card.

Wrong.

Drawing 2 cards

1st card 50%-red / 50%-white
2nd card 50%-red / 50%-white

You flip the first card, it's red.
The odds the second card is red is still 50%

You flip the second card, it's red.
The odds the first card is red is still 50%

You shuffle the cards before you put them down.
You flip one, it's red. The odds the second is red, still 50%.

 

[Fort Fun Indiana]

You shuffle the cards before you put them down.
You flip one, it's red. The odds the second is red, still 50%.

No. It is 1/3. 33.333...%

 

[Toddsterpatriot]

No. It is 1/3. 33.333...%

Nope. Because you already said one is red.
The color of the first card doesn't change the odds of the second one being red.

 

[Fort Fun Indiana]

Toddsterpatriot and here is why:

I have chosen two cards. 4 equally likely permutations are possible:

{BB, BR, RB, RR}

The only.informatiom revealed to you is that at least one of the cards is red.

In choosing whether the second card is red, you are now forced to choose between three equally likely choices as one possible permutation has been removed from the set:

{BR, RB, RR}

What is now the probability that both cards are red?

1/3

 

[Fort Fun Indiana]

Nope. Because you already said one is red.
The color of the first card doesn't change the odds of the second one being red.

The reason your odds decrease is that you don't know which card was pulled.first.

 

[Toddsterpatriot]

Toddsterpatriot and here is why:

I have chosen two cards. 4 equally likely permutations are possible:

{BB, BR, RB, RR}

The only.informatiom revealed to you is that at least one of the cards is red.

In choosing whether the second card is red, you are now forced to choose between three equally likely choices as one possible permutation has been removed from the set:

{BR, RB, RR}

What is now the probability that both cards are red?

1/3

The only.informatiom revealed to you is that at least one of the cards is red.

The first card I chose is on your left. I turn over the first card. It is red.

^
Your first scenario, in the OP.

Here are the possible permutations.

{RR,RB}

I turn over the second card, first. It is red.

^
Your second scenario, in the OP.

Here are the possible permutations.

{RR,BR}

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.

^
Your third scenario, in the OP.

Here are the possible permutations.

{RR,RB}

50% chance the second card is red in all three scenarios.

 

[Toddsterpatriot]

The reason your odds decrease is that you don't know which card was pulled.first.

You told me you flipped a red one.

{BR, RB, RR}

If you said the left card was red, the first permutation you list here, in post 57, isn't possible.

If you said the right card was red, the second permutation you list here, in post 57, isn't possible.