Expanding on the logic riddle:

[Fort Fun Indiana]

This is your sample space of combinations: {BB, RB, RB, RR} (note the repeated combination "RB", to denote it being 50% likely)


There are only three combinations of two red or black cards possible.

Both black, both red or one of each.

RB, RB is not two separate combinations.

They are separate permutations.

But you are correct that the "turned the card to your left" messes with the answer and makes it 50%, in scenarios 3 and 4 . But, this is for a slightly different (yet also similar) reason than the answers to 1 and 2 being 50%.

I should have said, "I tell you at least one card is red". Then, the odds both are red are 1/3. This would have made scenarios 3 and 4 analogous to rhe sibling riddle, which was my original intent.

Here is why:

When you flip one of the two cards to reveal it is red (doesn't matter which,like you said), you eliminate all of BB combinations and HALF of the RB combinations. Thus you are left with two equally possible combinations.

Same space of remaining, possible combinations: {RB, BB}

Why?

Because, if the two cards are randomly shuffled, then half the RB combinations would end up with their black card on the side where you just flipped a red card.

So you get extra credit.

 

[Toddsterpatriot]

They are separate permutations.

But you are correct that the "turned the card to your left" messes with the answer and makes it 50%, in scenarios 3 and 4 . But, this is for a slightly different (yet also similar) reason than the answers to 1 and 2 being 50%.

I should have said, "I tell you at least one card is red". Then, the odds both are red are 1/3. This would have made scenarios 3 and 4 analogous to rhe sibling riddle, which was my original intent.

Here is why:

When you flip one of the two cards to reveal it is red (doesn't matter which,like you said), you eliminate all of BB combinations and HALF of the RB combinations. Thus you are left with two equally possible combinations.

Same space of remaining, possible combinations: {RB, BB}

Why?

Because, if the two cards are randomly shuffled, then half the RB combinations would end up with their black card on the side where you just flipped a red card.

So you get extra credit.

But you are correct that the "turned the card to your left" messes with the answer and makes it 50%

Obviously.

I should have said, "I tell you at least one card is red". Then, the odds both are red are 1/3. This would have made scenarios 3 and 4 analogous to rhe sibling riddle, which was my original intent.

Which I said in post #121

Expanding on the logic riddle:

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 Until you flip over a red one. Now it's 50%.

www.usmessageboard.com

When you flip one of the two cards to reveal it is red (doesn't matter which,like you said), you eliminate all of BB combinations and HALF of the RB combinations. Thus you are left with two equally possible combinations.

Welcome to the party pal.

So you get extra credit.

And you get 50% credit. 75% for admitting you were wrong.

Now go back and find your error in the sibling riddle.

 

[Fort Fun Indiana]

But you are correct that the "turned the card to your left" messes with the answer and makes it 50%

Obviously.

I should have said, "I tell you at least one card is red". Then, the odds both are red are 1/3. This would have made scenarios 3 and 4 analogous to rhe sibling riddle, which was my original intent.

Which I said in post #121

Expanding on the logic riddle:

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 Until you flip over a red one. Now it's 50%.

www.usmessageboard.com

When you flip one of the two cards to reveal it is red (doesn't matter which,like you said), you eliminate all of BB combinations and HALF of the RB combinations. Thus you are left with two equally possible combinations.

Welcome to the party pal.

So you get extra credit.

And you get 50% credit. 75% for admitting you were wrong.

Now go back and find your error in the sibling riddle.

Wait a minute. I just corrected scenario three and four to match the sibling riddle. I see you actually are not getting this...

Which I said in post #121

No you didn't. You only said it was 50%, without saying why. But your reasons for arriving at that answer were wrong.

Which is why you are still getting the sibling riddle wrong.

So, I have two cards. I tell you at least one is red. The odds both are red is 1/3.

Right?

 

[Toddsterpatriot]

Wait a minute. I just corrected scenario three and four to match the sibling riddle.

No you didn't. You only said it was 50%, without saying why. But your reasons for arriving at that answer were wrong.

Which is why you are still getting the sibling riddle wrong.

So, I have two cards. I tell you at least one is red. The odds both are red is 1/3.

Right?

No you didn't. You only said it was 50%, without saying why.

I said it was 50%, because the odds of the second, unflipped card being red was 50%.

Which is why you are still getting the sibling riddle wrong.

You were wrong in the sibling riddle as well.

So, I have two cards. I tell you at least one is red. The odds both are red is 1/3.

Right?

Post #121.

 

[Toddsterpatriot]

Wait a minute. I just corrected scenario three and four to match the sibling riddle. I see you actually are not getting this...

No you didn't. You only said it was 50%, without saying why. But your reasons for arriving at that answer were wrong.

Which is why you are still getting the sibling riddle wrong.

So, I have two cards. I tell you at least one is red. The odds both are red is 1/3.

Right?

Which is why you are still getting the sibling riddle wrong.

You meet a girl. She tells you she has one sibling. What are the odds her sibling is also a girl?

[sibling] [sibling]

[girl] [sibling]

The sibling can be a girl, the sibling can be a boy.

50% chance it's a girl.

Riddles

First one: You meet a girl. She tells you she has one sibling. What are the odds her sibling is also a girl?

www.usmessageboard.com

You've already flipped the card.

 

[Fort Fun Indiana]

I said it was 50%, because the odds of the second, unflipped card being red was 50%.

That's is the same statement, twice. In post 121, you made that statement once. Saying it twice in a row doesn't help you. You are not saying why.

you did describe why earlier, and you were wrong.

You were wrong in the sibling riddle as well.

Ooops, no.

Just what I thought: You really are not getting this at all.

You say "flipping the left card" changes the problem, in scenarios 3 and 4. I remove that item. Then you say the problem has not really changed, and the answer is the same. Which is wrong.

 

[Fort Fun Indiana]

The sibling can be a girl, the sibling can be a boy.

50% chance it's a girl.

Wrong. That choice was made at fertilization. Now you are choosing from a sample space of permutations of pairs of siblings.

 

[Fort Fun Indiana]

Expanding a bit further on the riddle:

If I lay the cards in front of you and flip one -- and it is red -- the odds the other card is red are 1/2.

If I hold the cards behind my back, look at them, and show you one is a red card -- the odds the other card is also red are 1/3.

 

[Toddsterpatriot]

That's is the same statement, twice. In post 121, you made that statement once. Saying it twice in a row doesn't help you. You are not saying why.

you did describe why earlier, and you were wrong.

Ooops, no.

Just what I thought: You really are not getting this at all.

You say "flipping the left card" changes the problem, in scenarios 3 and 4. I remove that item. Then you say the problem has not really changed, and the answer is the same. Which is wrong.

That's is the same statement, twice. In post 121, you made that statement once. Saying it twice in a row doesn't help you. You are not saying why.

I had the correct answer to your post #22, in post #27. Why do I need to explain further?

You say "flipping the left card" changes the problem, in scenarios 3 and 4

No, I was explaining why shuffling the cards behind your back doesn't make scenarios 3 and 4 any different than scenarios 1 and 2

 

[Fort Fun Indiana]

I had the correct answer to your post #22, in post #27. Why do I need to explain further?

You don't. You explained in a different post. And your explanation was wrong.

Which you just highlighted by your most recent incorrect answer, and your incorrect answer to the sibling riddle.

 

[Toddsterpatriot]

Wrong. That choice was made at fertilization. Now you are choosing from a sample space of permutations of pairs of siblings.

Wrong, for the same reason as your post #22.

First one:

You meet a girl. She tells you she has one sibling. What are the odds her sibling is also a girl?

You met a sibling (flipped a card), she's a girl (the card is red), what are the odds the other sibling (card) is a girl (red)?

50%.

If you had said, "You meet someone, they tell you they have one sibling. What are the odds they are both girls?", then you would have been right ........

 

[Toddsterpatriot]

You don't. You explained in a different post. And your explanation was wrong.

Which you just highlighted by your most recent incorrect answer, and your incorrect answer to the sibling riddle.

You explained in a different post. And your explanation was wrong.

My explanation in all of them is, you flipped one card, the odds for the other card are 50/50.

 

[Toddsterpatriot]

Expanding a bit further on the riddle:

If I lay the cards in front of you and flip one -- and it is red -- the odds the other card is red are 1/2.

If I hold the cards behind my back, look at them, and show you one is a red card -- the odds the other card is also red are 1/3.

If I hold the cards behind my back, look at them, and show you one is a red card -- the odds the other card is also red are 1/3.

OMG!!!!

You still don't get it.

 

[Fort Fun Indiana]

You explained in a different post. And your explanation was wrong.

My explanation in all of them is, you flipped one card, the odds for the other card are 50/50.

If I hold the cards behind my back, look at them, and show you one is a red card -- the odds the other card is also red are 1/3.

OMG!!!!

You still don't get it.

Clearly you do not get it. In fact, not once have you even come close to the type of reasoning that would help you understand this problem. Oh well, I did my best to explain it to you. Moving on...

 

[Toddsterpatriot]

Clearly you do not get it. In fact, not once have you even come close to the type of reasoning that would help you understand this problem. Oh well, I did my best to explain it to you. Moving on...

Tell me again how shuffling cards behind your back changes the odds.

Hilarious!!!

 

[Fort Fun Indiana]

Tell me again how shuffling cards behind your back changes the odds.

Hilarious!!!

Shuffling would only matter if I could "divine" to you a red card. Say, if I could tell you at least one card is red, then you have to bet both are red.

Or, using card flipping: I get to look at the cards, and I always flip one that is red.

(Throw the "card to the left" part out.)

I set the pair of cards in front of you (shuffled). If there are any red cards, I flip a red card. I might flip the card to your right, or I might flip the card to your left.

If you know the card on the left is the first card drawn (no shuffle), your odds of the other card being red after I flip either card are 1/2

If you do not know which card was drawn first, after I flip either card to show a red card, your odds of the other card being red are 1/3.

 

[Toddsterpatriot]

Shuffling would only matter if I could "divine" to you a red card. That is, if I get to look at the cards, and I flip the one that is red. That's the error I made, that, when corrected, makes 3 and 4 analogous to the sibling riddle.

(Throw the "card to the left" part out.)

I set the pair of cards in front of you (shuffled). If there are any red cards, I flip a red card. I might flip the card to your right, or I might flip the card to your left.

If you know the card on the left is the first card drawn (no shuffle), your odds of the other card being red after I flip either card are 1/2

If you do not know which card was drawn first, after I flip either card to show a red card, your odds of the other card being red are 1/3.

Shuffling would only matter if I could "divine" to you a red card.

Shuffle or don't. Flip left or right. Flip the first one drawn or the second.
None of that matters once you flip one of the cards.
After you do that, the odds are 50/50.

 

[Fort Fun Indiana]

Toddsterpatriot Try this:

I draw four pairs from the deck. Heck I could draw them simultaneously.

In a perfect world, I draw the following combinations:

1 BB
2 RB
1 RR

I toss out the BB pair (I only flip a red card). We have three pairs of cards remaining. We make the bet 3 times with the three remaining pairs of cards.

Round 1) I lay down one of the mixed pairs and flip the red card. You bet the other card is red. Oops, you lose. House wins.

Round 2) I lay down the other mixed pair and flip the red card. You bet the other card is red. Oops. House wins again.

Round 3) I lay down the RR pair and flip a red card. You bet the other card is red. Yay! You finally win the bet.

You won 1 out of 3 rounds.

Do you suppose this changes, if we just draw one pair at a time? Nope.

Do you suppose this changes, if we do it 100, 1000, or 10000000 times? Nope.

 

[Toddsterpatriot]

Toddsterpatriot Try this:

I draw four pairs from the deck. Heck I could draw them simultaneously.

In a perfect world, I draw the following combinations:

1 BB
2 RB
1 RR

I toss out the BB pair (I only flip a red card). We have three pairs of cards remaining. We make the bet 3 times with the three remaining pairs of cards.

Round 1) I lay down one of the mixed pairs and flip the red card. You bet the other card is red. Oops, you lose. House wins.

Round 2) I lay down the other mixed pair and flip the red card. You bet the other card is red. Oops. House wins again.

Round 3) I lay down the RR pair and flip a red card. You bet the other card is red. Yay! You finally win the bet.

You won 1 out of 3 rounds.

Do you suppose this changes, if we just draw one pair at a time? Nope.

Do you suppose this changes, if we do it 100, 1000, or 10000000 times? Nope.

In a perfect world, I draw the following combinations:

1 BB
2 RB
1 RR

You're not drawing 4 combinations at a time, you're drawing one combination.
You look at both cards. If they're both black, you throw them out. Draw another.
Which combination do you want to pick next?

RB or RR?

 

[Toddsterpatriot]

Toddsterpatriot Try this:

I draw four pairs from the deck. Heck I could draw them simultaneously.

In a perfect world, I draw the following combinations:

1 BB
2 RB
1 RR

I toss out the BB pair (I only flip a red card). We have three pairs of cards remaining. We make the bet 3 times with the three remaining pairs of cards.

Round 1) I lay down one of the mixed pairs and flip the red card. You bet the other card is red. Oops, you lose. House wins.

Round 2) I lay down the other mixed pair and flip the red card. You bet the other card is red. Oops. House wins again.

Round 3) I lay down the RR pair and flip a red card. You bet the other card is red. Yay! You finally win the bet.

You won 1 out of 3 rounds.

Do you suppose this changes, if we just draw one pair at a time? Nope.

Do you suppose this changes, if we do it 100, 1000, or 10000000 times? Nope.

Still trying to understand your error?

Try this.

You have 2 six-sided dice and two dice cups. Put one die in each.
Shake one cup and flip it onto a table, so the die is covered.
Do the same for the second cup.

When you lift up one cup, the die shows one pip on top.

What are the odds the second die also shows one pip on top?