Expanding on the logic riddle:

[Fort Fun Indiana]

Right. The first card was red.

What color is the second card?

Either red or black. 50/50

In scenarios 3 and 4? No.

 

[Toddsterpatriot]

In scenarios 3 and 4? No.

In scenarios 3 and 4, the first card wasn't red?

 

[CrusaderFrank]

Read through the thread for the correct solutions.

Yeah, thank you

Good thread

 

[Fort Fun Indiana]

In scenarios 3 and 4, the first card wasn't red?

Time wasting troll questions will be ignored.

 

[Toddsterpatriot]

Time wasting troll questions will be ignored.

You should run away.....some more.

 

[Fort Fun Indiana]

You should run away.....some more.

I understand that you are frustrated.

It wouldn't be a riddle, if it didn't take some work.

 

[Toddsterpatriot]

In scenarios 3 and 4, the first card wasn't red?

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?

Darn, the first card turned over was red.

 

[Toddsterpatriot]

I understand that you are frustrated.

It wouldn't be a riddle, if it didn't take some work.

You keep working.
Get back to me when you get it right.

 

[Fort Fun Indiana]

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?

Darn, the first card turned over was red.

Yep, in all 4 scenarios, the first car flipped over is red.

An arbitrary choice by me. I could have used black for the riddle. The math remains the same.

Notice that you know which card was drawn from the deck first, in scenarios 1 and 2.

 

[Toddsterpatriot]

Yep, in all 4 scenarios, the first car flipped is red.

Notice that you know which card was drawn from the deck first, in scenarios 1 and 2.

Yes, I noticed it doesn't matter which card is first drawn from your infinite deck.

 

[Fort Fun Indiana]

Yes, I noticed it doesn't matter which card is first drawn from your infinite deck.

Yes, that is your fundamental ideological error.

Escaping this error and discovering your mathematical error is the riddle.

 

[Fort Fun Indiana]

Yes, I noticed it doesn't matter which card is first drawn from your infinite deck.

When you run out of steam or get the solution, I have a similar riddle for everyone. .

 

[Toddsterpatriot]

Yes, that is your fundamental ideological error.

Escaping this error and discovering your mathematical error is the riddle.

First card flipped is red.
Why does it matter if the first flipped card is first drawn or second drawn?
Or if you don't know if it was first or second drawn?

 

[Toddsterpatriot]

When you run out of steam or get the solution, I have a similar riddle for everyone. .

Why don't you post your proof for this one before you post a new one?

 

[Fort Fun Indiana]

Why does it matter if the first flipped card is first drawn or second drawn?

Now you are homing in on the solution.

Because, when you do not know whether you are looking at the first or second card drawn from the deck, you can only eliminate one of the 4 equally likely permutations from the possibilities. What remains is:

BR
RB
RR

3 equally likely permutations.

If you know you are looking at the first card drawn from the deck, you can eliminate 2 of the 4 equally likely permutations from the possibilities.
What remains is:

RB
RR

2 equally likely permutations.

 

[Fort Fun Indiana]

And, for the record, this is not a "test of intelligence". Nearly everyone fails this riddle the first time, unless they have some sort of deep background in discrete mathematics.

I like it because it demonstrates that the world does not always work intuitively, and that the mathematical arguments force you to accept something that seems absurd to you.

Imagine how Schröedinger felt, when first devising the mathematics of quantum mechanics. Quantum mechanical behavior STILL does not make sense to scientists. But the results are what they are.

 

[Toddsterpatriot]

BR
RB
RR

3 equally likely permutations.

How is this any different from scenario 3 and 4, where you also eliminated BB
and flipped one card?

 

[Fort Fun Indiana]

How is this any different from scenario 3 and 4, where you also eliminated BB
and flipped one card?

That is scenarios three and four.

 

[Toddsterpatriot]

That is scenarios three and four.

So how is that any different than scenario 1 and 2, where you also eliminated BB
and flipped one card?

 

[Fort Fun Indiana]

So how is that any different than scenario 1 and 2, where you also eliminated BB
and flipped one card?

Because you know whether the flipped card was the first or second card drawn from the deck.

Remember, your "sample space" here is the four possible permutations of two cards drawn from the deck. The first element is the first card drawn, and the second element is the second card drawn:

BB
BR
RB
RR

If you know the first card drawn from the deck is red, this leaves only 2, equally likely possible permutations for the two cards:

RB
RR

If you only know that "at least one card is red", this leaves 3, equally likely permutations for the two cards:

BR
RB
RR