Expanding on the logic riddle:

[Toddsterpatriot]

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

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

Yup. 50/50 that the second card drawn is red.

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

BR
RB
RR

Yup. Keep going.....you're getting close.

 

[Fort Fun Indiana]

Yup. 50/50 that the second card drawn is red.

Yes, in scenarios 1 and 2. You have that part down.

Of course, the riddle is to figure out why it is 1/3 for scenarios 3 and 4.

 

[Toddsterpatriot]

Yes, in scenarios 1 and 2. You have that part down.

Of course, the riddle is to figure out why it is 1/3 for scenarios 3 and 4.

Of course, the riddle is to figure out why it is 1/3 for scenarios 3 and 4.

Show me.

BR
RB
RR

Scenario 3, which of the above is impossible?

 

[Fort Fun Indiana]

Of course, the riddle is to figure out why it is 1/3 for scenarios 3 and 4.

Show me.

BR
RB
RR

Scenario 3, which of the above is impossible?

None of them are. All three remain as equally likely possibilities, after a card is flipped and revealed to be red.

 

[Toddsterpatriot]

None of them are. All three remain as equally likely possibilities, after a card is flipped and revealed to be red.

BR
RB
RR

You said the one to my left is red.
How can the top one (BR) still be possible?

 

[Fort Fun Indiana]

You said the one to my left is red.
How can the top one (BR) still be possible?

Because the permutations are ordered by the order drawn from the deck. Not by how they are laid in front of you. Before a card is flipped, the permutation has already been chosen, from the 4 possibilities. That choice has already been made.

The card flip simply reveals information about the permutation to you.

In scenario 1, the card flip reveals that the first card chosen from the deck (first element of the permutation) is red.
Leaving:

RB
RR

In scenario 3, the card flip reveals ONLY that at least one of the cards is red. Leaving:

BR
RB
RR

 

[Toddsterpatriot]

In scenario 3, the card flip reveals ONLY that at least one of the cards is red. Leaving:

BR
RB
RR

No. You told me first, that at least one is red. Then you flipped the left one in 3 or the right one in 4.

In #3, you've eliminated BR, because the left card is red.
Leaving the 2 possibilities,
RB left red, right black
RR both red

In #4, you've eliminated RB, because the right card is red.
Leaving the 2 possibilities,
BR right red, left black
RR both red

 

[Fort Fun Indiana]

You told me first, that at least one is red.

No. Toss that out right now. Sorry for the confusion. That's just a way of restating that "you flip a card, and it is red", in scenarios 3 and 4.

"At leat one card is red" is the only information revealed to you, if you flip a card, and it is red, in scenarios 3 and 4.

 

[Toddsterpatriot]

No. Toss that out right now. Sorry for the confusion. That's just a way of restating that "you flip a card, and it is red", in scenarios 3 and 4.

"At leat one card is red" is the only information revealed to you, if you flip a card, and it is red, in scenarios 3 and 4.

You told me, in post #22....

Before the flip

{B} both black {B}
{B} <black left red right> {R}
{R} <red left black right> {B}
{R} both red {R}

Four possibilities.

If the left is red, you eliminate one and two.
Leaving.....

{R} <red left black right> {B}
{R} <both red> {R}

 

[Fort Fun Indiana]

If the left is red, you eliminate one and two.

No.

In scenarios 3 and 4, it does not matter which card is on the left and which is on the right. That is a red herring.

 

[Toddsterpatriot]

No.

In scenarios 3 and 4, it does not matter which card is on the left and which is on the right. That is a red herring.

It was your post that mentioned left and right.

Expanding on the logic riddle:

Now you are changing the problem to try to prove you are correct. No, that's what I originally said. You must have missed it. I am not even talking about "the problem" in the OP. I am explaining why I used "infinite deck", instead of "with replacement". So, let's roll with, "With...

www.usmessageboard.com

 

[frigidweirdo]

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%

 

[Fort Fun Indiana]

It was your post that mentioned left and right.

Correct. Intentional and necessary red herring (in scenarios 3 and 4), so as not to give clues to the correct answer.

 

[Fort Fun Indiana]

50%

1 and 2: yes

3: no

 

[braalian]

Now answer using a 52-card deck.

I got in a heated argument with someone over a similar riddle involving tossing a coin. Say you flip a coin five times, and all five times it’s tails. He was convinced that meant the sixth time had a higher probability of being heads. No! It’s still 50/50!

Also, people won’t believe that if you play the lotto with the numbers: 1,2,3,4,5,6,7 it has just as much chance of winning as any other single combination of numbers.

it’s amazing how most people do not understand the basics of probability.

 

[frigidweirdo]

1 and 2: yes

3: no

Well, yes to all. You shuffle a pack of infinite cards. That means they're equal number of red and black cards. Even after you shuffled them.

 

[braalian]

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

Is the key word here “also”?

 

[braalian]

Is the key word here “also”?

Because, in and of itself, each individual card draw will always have a 50/50 chance, but the chance that *two* consecutive draws will be the same color lowers the probability. Am I right?

 

[Fort Fun Indiana]

You shuffle a pack of infinite cards.

No. I only shuffled the 2 cards I drew from the infinite deck. An infinite deck does not require shuffling.

 

[Fort Fun Indiana]

Is the key word here “also”?

Well, not really. You could say, "What is the probability both cards are red". Equivalent.