Expanding on the logic riddle:

[Fort Fun Indiana]

How is the info in those scenarios different?

In 1 and 2? You know the order the cards were drawn. This allows you to eliminate more of the possible permutations.

The information you have in each scenario is limited to:

1) The first card I drew is red

2) The second card I drew is red

3) At least one card is red

4) At least one card is red

Notice that you have more information, in scenarios 1 and 2.

 

[Toddsterpatriot]

In 1 and 2? You know the order the cards were drawn. This allows you to eliminate more of the possible permutations.

Knowing the order they were flipped allows me to eliminate.

If you only know at least one of my two cards is red, you only know I hold one of three, equally likely permutations:

BR
RB
RR

In post #22, you said for scenario 3:

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?

If you flipped the card to my left, and it's red, the top permutation, BR, is not possible.
Only two left, RB and RR. 50% chance the second card is red.


In post #22, you said for scenario 4:

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?

If you flipped the card to my right, and it's red, the middle permutation, RB, is not possible.
Only two left, BR and RR. 50% chance the second card is red.

I know you can get this.

 

[Meathead]

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

Yeah, behind his back too.

 

[Fort Fun Indiana]

Knowing the order they were flipped allows me to eliminate.

Yep!

 

[Fort Fun Indiana]

If you flipped the card to my right, and it's red, the middle permutation, RB, is not possible.

No. It is still possible (Scenarios 3 and 4). Because you don't know if you are looking at the first card drawn from the deck or the second card. So three, equally possible permutations still remain:

RB
BR
RR

 

[Fort Fun Indiana]

Yeah, behind his back too.

Because I only shuffled two cards. Pretty easy.

 

[Meathead]

Because I only shuffled two cards. Pretty easy.

Can two cards be shuffled?

 

[Fort Fun Indiana]

Can two cards be shuffled?

Yep! Point being, when they are laid down in front of you afterward, you don't know which one was drawn from the deck first.

 

[Toddsterpatriot]

No. It is still possible (Scenarios 3 and 4). Because you don't know if you are looking at the first card drawn from the deck or the second card. So three, equally possible permutations still remain:

RB
BR
RR

The order they're drawn from the deck or placed on the table doesn't change anything.

 

[Fort Fun Indiana]

The order they're drawn from the deck or placed on the table doesn't change anything.

False (regarding the "drawn from the deck" part). That's the crux of the riddle, in fact.

Your expectation is better in scenarios 1 and 2, because you have more information and can eliminate more of the possible permutations, once a card is revealed.

 

[Toddsterpatriot]

False (regarding the "drawn from the deck" part). That's the crux of the riddle, in fact.

Your expectation is better in scenarios 1 and 2, because you have more information and can eliminate more of the possible permutations, once a card is revealed.

Draw and place two random cards.
4 possible permutations.

BB
BR
RB
RR

You eliminate the top, BB, when you said at least one is Red.

3 possible permutations left

BR
RB
RR.

Scenario one, choose left, it's red.
You've eliminated the top, BR.

2 possible permutations left

RB
RR

50% chance second card is red.


Scenario two, choose right, it's red.
You've eliminated the middle, RB.

2 possible permutations left

BR
RR

50% chance second card is red.

Scenario three, shuffle then choose left, it's red.
You've eliminated the top, BR.

2 possible permutations left

RB
RR

50% chance second card is red.

Scenario four, shuffle then choose right, it's red.
You've eliminated the middle, RB.

2 possible permutations left

BR
RR

50% chance second card is red.

 

[Toddsterpatriot]

Your expectation is better in scenarios 1 and 2, because you have more information and can eliminate more of the possible permutations, once a card is revealed.

Info is identical. Order drawn is not relevant.

 

[Fort Fun Indiana]

Scenario one, choose left, it's red.
You've eliminated the top, BR

Yep! Only because you know the card on the left was the first card drawn from the deck.

Scenario three, shuffle then choose left, it's red.
You've eliminated the top, BR

False! The permutations denote the permutations drawn from the deck.

Permutation BR occurs when the first card you draw is black, and the second card you draw is red.

In scenarios 3 and 4, you don't know if you are looking at the second or the first card drawn. You only know at least one of the two is red.

 

[Fort Fun Indiana]

Info is identical. Order drawn is not relevant.

Wrong.

 

[Toddsterpatriot]

False! The permutations denote the permutations drawn from the deck.

The possible permutations are reduced the moment you flip a card.

 

[Fort Fun Indiana]

These problems of counting permutations can be exploited. Like, the old game show, "Let's Make A Deal".

In scenarios 3 and 4, this trick can be used to take someone's money. Their expectation of both cards being red is 33% However, they may THINK their expectation is 50%.

(Like Toddsterpatriot )

So, if he bets $100 every time, he thinks he is Making an even-odds bet (like betting red or black on roulette). However, if he starts with (x * $300), he will only have (x * $200) after (x * 3) bets.

If he starts with $300K and makes 3000 bets, he will leave with $200K.

House wins.

 

[Fort Fun Indiana]

The possible permutations are reduced the moment you flip a card.

Correct. In scenarios 1 and 2, 2 of the 4 possible permutations are eliminated.

In scenarios 3 and 4, only 1 of the 4 possible permutations is eliminated.

 

[Toddsterpatriot]

In scenarios 3 and 4, only 1 of the 4 possible permutations is eliminated.

3) At least one card is red

4) At least one card is red

At the start, 3 possible permutations.

BR
RB
RR

If you put 2 cards down and ask, what are the odds both are RED, it's 1/3.

As soon as you flip the first card and see it's RED, the odds are 1/2 the second is red.

 

[CrusaderFrank]

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 depends if I understand the question, but I'll go with 50%
3 50%

 

[Fort Fun Indiana]

If you put 2 cards down and ask, what are the odds both are RED, it's 1/3.

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 flip the first card and see it's RED, the odds are 1/2 the second is red.

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