Expanding on the logic riddle:

[CrusaderFrank]

Now answer using a 52-card deck.

1. 25/51
2. ? 25/51 or 100%
3. 25/51

 

[Toddsterpatriot]

Incorrect. It's 1/4. Four equally likely permutations exist, as you do not yet know at least one card is red:

BB
RB
BR
RR

As soon as you say one is red, you're down to three permutations.

 

[Toddsterpatriot]

Incorrect. Still 1/3. All you know at that point is that at least one card is red. So three, equally likely permutations remain:

BR
RB
RR

1/3, when you know at least one is red.
When you flip a red on the left, which ones are possible?
When you flip a red on the right, which ones are possible?

 

[Fort Fun Indiana]

As soon as you say one is red, you're down to three permutations.

Yes, correct!

You are getting so close.

When I said, "at least one card is red", for scenarios 3 and 4, I mean this is all the information you have, when it comes time to make your bet. You obtain this information when I flip a card, and it is red.

In scenarios 1 and 2, you are also given this information, when a red card is flipped. However, in these scenarios, you have MORE information: You know whether you are looking at the first card drawn from the deck or the second card drawn from the deck.

Permutations are ordered. In the following 4 permutations:

BB
BR
RB
RR

The first element is the first card drawn from the deck. The second element is the second card drawn from the deck.

So, in scenarios 3 and 4, if a flip a card, and it is red, you can only eliminate 1 permutation. Three remain:

BR
RB
RR

 

[Toddsterpatriot]

Yes, correct!

You are getting so close.

When I said, "at least one card is red", for scenarios 3 and 4, I mean this is all the information you have, when it comes time to make your bet. You obtain this information when I flip a card, and it is red.

In scenarios 1 and 2, you are also given this information, when a red card is flipped. However, in these scenarios, you have MORE information: You know whether you are looking at the first card drawn from the deck or the second card drawn from the deck.

Permutations are ordered. In the following 4 permutations:

BB
BR
RB
RR

The first element is the first card drawn from the deck. The second element is the second card drawn from the deck.

So, in scenarios 3 and 4, if a flip a card, and it is red, you can only eliminate 1 permutation. Three remain:

BR
RB
RR

When I said, "at least one card is red", for scenarios 3 and 4, I mean this is all the information you have, when it comes time to make your bet. You obtain this information when I flip a card, and it is red.

It's not all the information I have.

3) I pull one card, then another. Before I lay the two cards down face-down in front of you, I shuffle them behind my back. I lay them face-down 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?
.................

4) I pull one card, then another. Before I lay the two cards down face-down in front of you, I shuffle them behind my back. I lay them face-down in front of you, and I turn over the card to your right. It is red.

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

I'm not making my bet before the flip.

 

[Toddsterpatriot]

Yes, correct!

You are getting so close.

When I said, "at least one card is red", for scenarios 3 and 4, I mean this is all the information you have, when it comes time to make your bet. You obtain this information when I flip a card, and it is red.

In scenarios 1 and 2, you are also given this information, when a red card is flipped. However, in these scenarios, you have MORE information: You know whether you are looking at the first card drawn from the deck or the second card drawn from the deck.

Permutations are ordered. In the following 4 permutations:

BB
BR
RB
RR

The first element is the first card drawn from the deck. The second element is the second card drawn from the deck.

So, in scenarios 3 and 4, if a flip a card, and it is red, you can only eliminate 1 permutation. Three remain:

BR
RB
RR

So, in scenarios 3 and 4, if a flip a card, and it is red, you can only eliminate 1 permutation. Three remain:

BR
RB
RR

You eliminated permutation BB at the start. (At least one is red)
When you flip left red, you eliminate BR. Two left.
When you flip right red, you eliminate RB. Two left.

You're making the same error here that you did in the sibling riddle.

 

[Fort Fun Indiana]

When I said, "at least one card is red", for scenarios 3 and 4, I mean this is all the information you have, when it comes time to make your bet. You obtain this information when I flip a card, and it is red.

It's not all the information I have.

3) I pull one card, then another. Before I lay the two cards down face-down in front of you, I shuffle them behind my back. I lay them face-down 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?
.................

4) I pull one card, then another. Before I lay the two cards down face-down in front of you, I shuffle them behind my back. I lay them face-down in front of you, and I turn over the card to your right. It is red.

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

I'm not making my bet before the flip.

Correct, you are making your bet after the flip.

In scenarios 3 and 4, all the information you have is that at least one card is red, when you make your bet after the card flip.

In scenarios 1 and 2, the information you have is that the first card drawn from the deck is red or the second card drawn from the deck is red, when you make your bet after the card flip. Notice that both of the boldfaced statements also include the information that at least one card is red. You have more information in scenarios 1 and 2, when it is time to make your bet. This extra information raises your expectation.

 

[Fort Fun Indiana]

1. 25/51
2. ? 25/51 or 100%
3. 25/51

Read through the thread for the correct solutions.

 

[Orangecat]

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?

50/50, because the one card showing red would not effect the odds on an infinite deck.

 

[Fort Fun Indiana]

50/50, because the one card showing red would not effect the odds on an infinite deck.

Scenarios 1 and 2: 50% is correct

Scenario 3: 50% is incorrect.

But kudos to you for immediately understanding what the point and use of "infinite deck" was, in the OP.

 

[Fort Fun Indiana]

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

I sure do.

 

[Orangecat]

Scenarios 1 and 2: 50% is correct

Scenario 3: 50% is incorrect.

But kudos to you for immediately understanding what the point and use of "infinite deck" was, in the OP.

What is your logic behind scenario #3 not being 50/50? Drawn from an infinite deck of equal colors, every card drawn would have a 50% chance of being one color or the other, regardless of the scenario in which they were drawn.

 

[Fort Fun Indiana]

What is your logic behind scenario #3 not being 50/50?

Read through the thread a bit. I have posted it about 10 times, at this point. If you don't see it or get it, let me know.

 

[Orangecat]

Read through the thread a bit. I have posted it about 10 times, at this point. If you don't see it or get it, let me know.

I haven't read the entire thread. Gotta go hang some shelves for the girl. I'll catch up later, but, for now, I stand by my contention that every card drawn would have a 50/50 chance of being black or red, regardless of mixing them up behind your back. Of course, the odds of drawing 2 or 3 reds in a row may be different, but each individual card drawn would have the same 50/50 chance.

 

[Fort Fun Indiana]

I haven't read the entire thread. Gotta go hang some shelves for the girl. I'll catch up later, but, for now, I stand by my contention that every card drawn would have a 50/50 chance of being black or red, regardless of mixing them up behind your back. Of course, the odds of drawing 2 or 3 reds in a row may be different, but each individual card drawn would have the same 50/50 chance.

Quick rundown (scenario 3):

I have chosen two cards from the deck. 4 equally likely permutations of these two cards are possible:

{BB, BR, RB, RR}

The only information revealed to you in scenario 3 (when a card is flipped, and it is red) is that at least one of the cards is red.

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

{BR, RB, RR}

What is now the probability that both cards are red?

1/3

 

[Toddsterpatriot]

Correct, you are making your bet after the flip.

In scenarios 3 and 4, all the information you have is that at least one card is red, when you make your bet after the card flip.

In scenarios 1 and 2, the information you have is that the first card drawn from the deck is red or the second card drawn from the deck is red, when you make your bet after the card flip. Notice that both of the boldfaced statements also include the information that at least one card is red. You have more information in scenarios 1 and 2, when it is time to make your bet. This extra information raises your expectation.

In scenarios 3 and 4, all the information you have is that at least one card is red, when you make your bet after the card flip.

Wrong. You have the info that at least one is red AND you have the info that first flipped is red.

Draw order doesn't give you any useful info.

 

[Orangecat]

I have chosen two cards from the deck. 4 equally likely permutations of these two cards are possible:
{BB, BR, RB, RR}

The only information revealed to you in scenario 3 (when a card is flipped, and it is red) is that at least one of the cards is red.

In choosing whether the other 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

BR and RB are the same thing if you are talking sets of two cards.
Each card you pick has a 50/50 chance, regardless of the previous card chosen. Only because of the infinite deck, though.

 

[Fort Fun Indiana]

You have the info that at least one is red AND you have the info that first flipped is red.

In scenarios 3 and 4, those are equivalent.

Maybe this will help you get over the hump.

 

[Fort Fun Indiana]

BR and RB are the same thing.

No.

They are the same combination.

They are distinct permutations.

 

[Toddsterpatriot]

In scenarios 3 and 4, those are equivalent.

Maybe this will help you get over the hump.

Equivalent, both 50% chances.