Two dice puzzle -- Part Trois

Quote: billryan

Absolutely, although Mike has first dibs. If Mike passes, I'll gladly play the house.

Not interested in playing against Mike.

In-person exercise only...will you still offer?

Quote: TravisR

What if our opponent rolls the dice r---e----a----l s----l----o----w ...... and he announces whatever number the first die happens to settle down on..... which is as random as the initial problem guys choice

SO he anounces a 2 as the other die is still spinning thru the air...... what's the odds he rolled a double 2?
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1/6

Quote: TravisR

What if our opponent rolls the dice r---e----a----l s----l----o----w ...... and he announces whatever number the first die happens to settle down on..... which is as random as the initial problem guys choice

SO he anounces a 2 as the other die is still spinning thru the air...... what's the odds he rolled a double 2?
link to original post

With one die landed on a 2 and the other die still in the air it's 1/6.

And that is the same as the original question even though the "1/11 crowd" won't accept that it is.

Quote: unJon

Quote: TravisR

What if our opponent rolls the dice r---e----a----l s----l----o----w ...... and he announces whatever number the first die happens to settle down on..... which is as random as the initial problem guys choice

SO he anounces a 2 as the other die is still spinning thru the air...... what's the odds he rolled a double 2?
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1/6
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Unjon you surprise me.

What Travis just presented is the same as the original question.

Quote: AlanMendelson

Quote: unJon

Quote: TravisR

What if our opponent rolls the dice r---e----a----l s----l----o----w ...... and he announces whatever number the first die happens to settle down on..... which is as random as the initial problem guys choice

SO he anounces a 2 as the other die is still spinning thru the air...... what's the odds he rolled a double 2?
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1/6
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Unjon you surprise me.

What Travis just presented is the same as the original question.
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Disagree.

Quote: AlanMendelson

With one die landed on a 2 and the other die still in the air it's 1/6. And that is the same as the original question even though the "1/11 crowd" won't accept that it is.

No, that's not the same as the original question. Not at all.

ANY TIME you can make a distinction of the two dice, the answer is indeed 1/6.

If the RED die is a 2, the probability both dies are a 2 is 1/6.
If the LEFT die is a 2, the probability both dies are a 2 is 1/6.
If the die that has STOPPED SPINNING is a 2, the probability both DIES are a 2 is 1/6.

But if the original problem was presented this way: "After rolling the dice, he peeks under the cup and says, truthfully, "At least one of the dice is a 2."

...and thus the answer to THAT problem is 1/11. Because you don't know which die he is referring to. Because you don't know, there are 11 possible dice rolls that are still valid, out of the 36 possible ways to roll the dice. Only one of those is a 2, hence the correct answer is 1 out of 11.

Ed it is the same but you won't accept it.

The peeker saying at least one die is a 2 is the same as throwing two dice on a table and one landing on 2 while the other die spins

It's the same thing.

Quote: AlanMendelson

Ed it is the same but you won't accept it.

The peeker saying at least one die is a 2 is the same as throwing two dice on a table and one landing on 2 while the other die spins

It's the same thing.
link to original post

We can set up the Wizard wager that way.

Wiz rolls dice under cup. Every time he sees at least one die is a 2, he will say “at least one die is a 2.” On those occasions you wager $1 to win $8 that there’s a second 2 under the cup.

Quote: unJon

Quote: AlanMendelson

Ed it is the same but you won't accept it.

The peeker saying at least one die is a 2 is the same as throwing two dice on a table and one landing on 2 while the other die spins

It's the same thing.
link to original post

We can set up the Wizard wager that way.

Wiz rolls dice under cup. Every time he sees at least one die is a 2, he will say “at least one die is a 2.” On those occasions you wager $1 to win $8 that there’s a second 2 under the cup.
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No. But here's the bet I'll make.

You take a single die and show it to a fifth grader. Ask the fifth grader how many sides are on the die. Then ask the fifth grader how many of those faces is a 2.

I'll bet the 5th grader says there are 6 sides and one face is a 2.

Quote: AlanMendelson

Ed it is the same but you won't accept it.
The peeker saying at least one die is a 2 is the same as throwing two dice on a table and one landing on 2 while the other die spins
It's the same thing.

And that's why you're mistaken. It's not. Your mind just doesn't see it or understand.

When one of the two dice have stopped spinning, and you make a statement about that particular die, you have this graphic:
If both dice have stopped spinning, and you make a statement like, "At least one of the dice is a 2." You have this graphic:

He sees it Ed. He’s taking the p*ss. As soon as you phrase the same question as a wager he sees it.

Quote: coachbelly

Quote: billryan

Absolutely, although Mike has first dibs. If Mike passes, I'll gladly play the house.

Not interested in playing against Mike.

In-person exercise only...will you still offer?
link to original post

I'll be happy to host the game and have hundreds of casino die you can pick from. Let me know when you are in the neighborhood.

Quote: billryan

Let me know when you are in the neighborhood.

Forward the address or neighborhood. If still Queens, I can be there this week.

Quote: EdCollins

Quote: AlanMendelson

Ed it is the same but you won't accept it.
The peeker saying at least one die is a 2 is the same as throwing two dice on a table and one landing on 2 while the other die spins
It's the same thing.

And that's why you're mistaken. It's not. Your mind just doesn't see it or understand.

When one of the two dice have stopped spinning, and you make a statement about that particular die, you have this graphic:
If both dice have stopped spinning, and you make a statement like, "At least one of the dice is a 2." You have this graphic:
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What's the difference between:

"One die shows a two"

And

"At least one die shows a 2"

When there are only two dice??

Answer: you still have one of two dice showing a 2.

And if you have one die showing a 2 the chance that the other die in this two-dice question is a 2 is 1/6.

If you disagree then we disagree.

And this has nothing to do with the "challenge question" which is another problem. And that answer is 1/11.

Quote: Wizard

I will decline to start a part trois to that topic. I invite you to explain how the wording was bad as a post in part deux. There, I'll be happy to entertain your complaint on the wording.
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The wording in that post is perfect. It gives multiple examples of wording, some of which result in odds of 1/6 and some result in 1/11.

However, somewhere along the line, the wording got muddied to the point where it is unclear why the peeker is reporting about the quantity of 2's, and an assumption that too many WoV members are making about what the peeker is doing.

I.E. Are all rolls where there are no 2s ignored? If so, the odds of two 2s is 1/11.

However, if every roll is live, and the peeker is obligated to report that there is at least one die showing <something>, then the odds are 1/6.

I hate to say it, particularly since I've waffled on this topic myself, but Alan might be right. It's all about the wording.

here's how to explain the 11 "different rolls"

there's 11 different rolls but when the opponet says "2" we know there is only a 50% chance he called out a 2 if he rolled somoething like a 2-5

but there is 100% chance he calls out a 2 if he has double 2's

so it's like a weighted average when you use all 11 rolls with the double 2 rolls getting twice the weight because of the 100%/50% ratio

..... just trying to think of different ways to explain how rolling doubles is 1 out of 6.... but all it does it make it more complicated than it needs to be ,lol

It's like the problem with the census taker.... if the census taker asks a person who has 2 kids if one of them is a boy and the answer is yes, the odds the other one is a boy is only 33%


but if someone with two kids just blurts out for no appearant reason "I have two kids and I have at least one boy" , then the odds the other kid is a boy is 50%

yeah but the opponent knows which die he's refering to, lol.

Either way there may be 11 rolls but not all 11 rolls get "2" announced with the same frequency

for 10 of the rolls it's announced 50% of the time

for 1 of the rolls it's announced 100% of the time

This difference in frequency is important

Of course it's about the wording..... everyone on this thread seems to have good math skills..... most problems like this have more to do with the wording than the actual mathematics.... but the wording wasn't tricky.... it is what it is..... if somebody "muddled" the wording along the way they unwittingly created a more challenging brain teaser than the boring one that has the 1 in 11 answer.

I didn't see other versions of the problem as it was presented to me..... but depending on the wording the answer will vary

Quote: TravisR

yeah but the opponent knows which die he's refering to, lol.

Either way there may be 11 rolls but not all 11 rolls get "2" announced with the same frequency

for 10 of the rolls it's announced 50% of the time

for 1 of the rolls it's announced 100% of the time

This difference in frequency is important
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Why?

Quote: DJTeddyBear

I hate to say it, particularly since I've waffled on this topic myself, but Alan might be right. It's all about the wording.
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Thank you. Yes, it's all about the wording.

1/11 is the correct answer to the "challenge question."

Quote: Dieter

pi has been voted to be 3 and 22/7.
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I throw the challenge flag at that.

There is an urban legend that Alabama's legislature voted to redefine pi as 3, to be in agreement with the bible. However, it isn't true.

Source = Snopes.com

You might say another governmental body did make such a vote, to which I respectfully ask for evidence. Thank you.

Quote: Wizard

Quote: Dieter

pi has been voted to be 3 and 22/7.
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I throw the challenge flag at that.

There is an urban legend that Alabama's legislature voted to redefine pi as 3, to be in agreement with the bible. However, it isn't true.

Source = Snopes.com

You might say another governmental body did make such a vote, to which I respectfully ask for evidence. Thank you.
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Indiana Pi Bill, according to Wikipedia, which I know is not the best source.

Quote: Wizard

Quote: Dieter

pi has been voted to be 3 and 22/7.
link to original post

I throw the challenge flag at that.

There is an urban legend that Alabama's legislature voted to redefine pi as 3, to be in agreement with the bible. However, it isn't true.

Source = Snopes.com

You might say another governmental body did make such a vote, to which I respectfully ask for evidence. Thank you.
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Indeed, I had misremembered numerous anecdotes over the years.

Bill #246 of the 1897 Indiana General Assembly included a proposed method for squaring the circle.
Wiser heads seem to have prevailed.

https://journals.iupui.edu/index.php/ias/article/view/4753/4589


I am sorry for throwing a wrench in the works with bad memory.
I stand by the principle that popular vote cannot change how math works.

Quote: billryan

Absolutely, although Mike has first dibs. If Mike passes, I'll gladly play the house. I may no longer be bound by math, but I know a great opportunity when I see it.
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Thank you for offering me first dibs. It's a free market, so you two may have a challenge on your own. I'll just say that I'll offer 8 to 1 odds on two 2's to anybody, subject to mutual agreement on the bet amounts and payment terms. As far as I'm concerned CB, or anybody, may choose whom (or is it "who"?) to do business with.

Once I came to realize how flexible math can be, I realized everyone was wrong about this.
Two dice are thrown, and one is a six. What are the chances the other die is a six. Some of you are claiming 1/11, others are claiming 1/6. I say you are all overthinking it.
The other die is a six or it isn't. 50-50.

The key is making sure you understand that math is not static.
If I write 2+2= 5, you'd say I was wrong. Most would say you need to add a one to the equation to get it right 2+2+1=5. I choose to add a letter. 2+2A=5. If I claimed 50 + 311 = 672,567 people would mock me, but write 50A+ 331X=672,567 and people accept you as one who knows.
The New Math was right all along. No wonder people hated it.
Go ahead and cling to your guns and your old math.

As the prophet wrote:
Don't criticize what you can't understand
Your sons and your daughters
are beyond your command.
Your old math is rapidly changing

Here is a different way to illustrate the logic of the two dice problem.

I understand that AlanMendelson will not agree with anything that contradicts his own idea, so this is aimed at others in the forum. I am trying to innovate on how to explain the logic of the solution of the 2-Dice problem.

Given: Two dice have been rolled in a cup, Joe's assistant has been instructed to say "at least one of the dice is two" after looking at both dice in the cup if and only if that statement is indeed true. Joe's assistant does look at both dice and then he does indeed make that statement.

Now, we blindfold our contestant Joe. Joe's assistant reaches under the cup and without looking at either of the dice, grabs one of the dice, removes it from the cup and places it in Joe's outstretched hand. Joe makes a fist around the die in his hand, so it cannot be seen. His blindfold is then removed.

Now we ask: "Joe, what is the probability that the dice that is still in the cup is a 2?"

And Joe replies: "if the die in my hand is a 2, I know the other die has a 1 in 6 chance of being a two. And therefore the odds of both dice being 2s is 1/6. I have already posted that statement a zillion times during the past week."

And, the assembled Elder Council of WOV Math Geeks chime in and say : "We agree with every part of that statement."

"BUT" the Wizard asks "What if the die in your hand is not a 2? What, then is the chance that the other die in the cup is a 2?" Joe starts to answer, then stops and thinks carefully. He then responds "the chance of the other dice being a 2 is 100% -not 1/6 - because I have been told that at least one of the dice must be a 2 and so it must be the die in the cup. And, of course, I also conclude that there is ZERO probability that both dice are a 2, because in this scenario the die in my hand is not a 2 !!!!!"

The Elder Council of WOV Math Geeks gently point out "Remember that the die in your hand was randomly removed from the cup and you have no further information about it. You can't just assume that it is a 2 and that the other die has a 1/6 chance of being a 2, because you must also take into account that the die in your hand may NOT be a 2, and that therefore the other die (in the cup) has a 100% chance of being a 2."

"Because to assume that a die in your closed hand is a 2 is to assume knowledge that you were not given in the statement of the problem."

Joe says "But I want to start with the knowledge that one of the dice is a two and make this a one die problem, that is, it is a problem of whether the other six-faced die is a two."

And the Elder Council responds: 'But there is no way to actually start with a die that is a 2 without peeking at it to determine it is a 2. You must accept that any die you select has a 6/11 chance of being a 2 and a 5/11 chance of not being a 2. Therefore, the other die will, respectively have either a 1/6 chance of being a 2 or a 100% chance of being a 2."

There is a 6/11 chance that any die you pick will be a 2, and in that scenario there is a 1/6 chance that the other die will also be a 2.

Thus, 6/11 x 1/6 = 1/11.

(Note: edited the name of the imaginary participant.)

This thread reminds me of a Happy Days episode where Fonzie couldn't get the words "I was wrong" out of his mouth

Humanity beats conformity.

It's creative, but it's different situation than somebody rolling two dice, looking at them, announcing that the number "x" is showing on at least one of the dice and then asking what the probability that the other dice also shows "x".

I mean, the number "2" is a nice number and all but if you substitute in an "x" instead it makes all the confusion go away and you can then see that the odds of the other dice being "x" is 1 out of 6

Quote: gordonm888

Here is a different way to illustrate the logic of the two dice problem.

I understand that AlanMendelson will not agree with anything that contradicts his own idea, so this is aimed at others in the forum. I am trying to innovate on how to explain the logic of the solution of the 2-Dice problem.

Given: Two dice have been rolled in a cup, Joe's assistant has been instructed to say "at least one of the dice is two" after looking at both dice in the cup if and only if that statement is indeed true. Joe's assistant does look at both dice and then he does indeed make that statement.

Now, we blindfold our contestant Joe. Joe's assistant reaches under the cup and without looking at either of the dice, grabs one of the dice, removes it from the cup and places it in Joe's outstretched hand. Joe makes a fist around the die in his hand, so it cannot be seen. His blindfold is then removed.

Now we ask: "Joe, what is the probability that the dice that is still in the cup is a 2?"

And Joe replies: "if the die in my hand is a 2, I know the other die has a 1 in 6 chance of being a two. And therefore the odds of both dice being 2s is 1/6. I have already posted that statement a zillion times during the past week."

And, the assembled Elder Council of WOV Math Geeks chime in and say : "We agree with every part of that statement."

"BUT" the Wizard asks "What if the die in your hand is not a 2? What, then is the chance that the other die in the cup is a 2?" Joe starts to answer, then stops and thinks carefully. He then responds "the chance of the other dice being a 2 is 100% -not 1/6 - because I have been told that at least one of the dice must be a 2 and so it must be the die in the cup. And, of course, I also conclude that there is ZERO probability that both dice are a 2, because in this scenario the die in my hand is not a 2 !!!!!"

The Elder Council of WOV Math Geeks gently point out "Remember that the die in your hand was randomly removed from the cup and you have no further information about it. You can't just assume that it is a 2 and that the other die has a 1/6 chance of being a 2, because you must also take into account that the die in your hand may NOT be a 2, and that therefore the other die (in the cup) has a 100% chance of being a 2."

"Because to assume that a die in your closed hand is a 2 is to assume knowledge that you were not given in the statement of the problem."

Joe says "But I want to start with the knowledge that one of the dice is a two and make this a one die problem, that is, it is a problem of whether the other six-faced die is a two."

And the Elder Council responds: 'But there is no way to actually start with a die that is a 2 without peeking at it to determine it is a 2. You must accept that any die you select has a 6/11 chance of being a 2 and a 5/11 chance of not being a 2. Therefore, the other die will, respectively have either a 1/6 chance of being a 2 or a 100% chance of being a 2."

There is a 6/11 chance that any die you pick will be a 2, and in that scenario there is a 1/6 chance that the other die will also be a 2.

Thus, 6/11 x 1/6 = 1/11.

(Note: edited the name of the imaginary participant.)
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This is a Rube Goldberg answer that missed the initial question.

How about handing a single die to a 3rd grader (we don't need a fifth grader or a fourth grader) and ask them to count the sides on the die?

Length doesn't make an answer correct.

Quote: AlanMendelson

This is a Rube Goldberg answer that missed the initial question.

How about handing a single die to a 3rd grader (we don't need a fifth grader or a fourth grader) and ask them to count the sides on the die?

Length doesn't make an answer correct.
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My four year old can count sides on a single d6 and get to 11. (She is precocious and cute as a button. We're still working on counting.)

Unfortunately, counting only one dice isn't going to get to the right answer for the right reason.

There seems to be a gap between what is known and what is sought, bridged by an assumption.

Quote: Dieter

Quote: AlanMendelson

This is a Rube Goldberg answer that missed the initial question.

How about handing a single die to a 3rd grader (we don't need a fifth grader or a fourth grader) and ask them to count the sides on the die?

Length doesn't make an answer correct.
link to original post

My four year old can count sides on a single d6 and get to 11. (She is precocious and cute as a button. We're still working on counting.)

Unfortunately, counting only one dice isn't going to get to the right answer for the right reason.

There seems to be a gap between what is known and what is sought, bridged by an assumption.
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And what happens when one assumes?

You end up with making an ass out of you and me.

Quote: TravisR

It's creative, but it's different situation than somebody rolling two dice, looking at them, announcing that the number "x" is showing on at least one of the dice and then asking what the probability that the other dice also shows "x".

I mean, the number "2" is a nice number and all but if you substitute in an "x" instead it makes all the confusion go away and you can then see that the odds of the other dice being "x" is 1 out of 6
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Math is hard. Logic is harder.

Your top graphic assumes that the first die to stop spinning is always going to be the "2".... ... That's going to be problematic I think

Quote: TravisR

Your top graphic assumes that the first die to stop spinning is always going to be the "2".... ... That's going to be problematic I think
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What are you looking at?

I was trying to reply to a post that had two graphics of dice combinations .... and a discussion about spinning dice.... I'm still learning how to navigate the thread so my apologies for any confusion

There are no spinning dice. The dice were thrown and landed and the correct and honest answer is there is at least one of your particular number. You know one of two dice is a Five. You don't know which die is the five.

You are answering a different question.
Everyone agrees that if we know the left die is a Five, then the right die will be a Five one time in six.
Go back and examine the question. Once you understand it, you won't keep answering the wrong question.

Well, I answered the correct question correctly about 20 times already, so I was just looking around for other side-coversations that were going on

It helps to hit quote instead of reply.

Quote: unJon

It helps to hit quote instead of reply.
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ah.... thank you unJon !

Quote: TravisR

Well, I answered the correct question correctly about 20 times already, so I was just looking around for other side-coversations that were going on
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Okay. Some of the finest math people in the country are wrong, the math is wrong and you are right.
Fair enough. It couldn't be you keep answering a question no one is asking.

Posts about the challenge have been moved to a new thread: Two Dice Challenge. Please keep that thread limited to the challenge itself and not to the math of the question at hand.

Quote: AlanMendelson

Okay. I'll agree that "Chance of rolling a 2 on at least one die is 11/36
Chance of rolling a pair of 2's where at least one die is a 2 is 1/11"

And the reason I wont take part in the "bet" is that I'd lose $100 to win $80.

But if you rolled a single die.... ah!
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Sorry to backtrack on you.... I sorta read the end of the book first and now am looking at the past discussion.

My thoughts here are that what you say is correct....but what seems to be overlooked is that the statement

Chance of rolling a pair of 2's where at least one die is a 2 is 1/11"

may be true, but that it not applicable to the problem where the guy announces he "sees at least one 2" before you are asked what's the chance of double 2's under his cup.

And I know I'm preaching to the choir but what throws the wrench into the mix is that the guy may not call a 2 if the roll was a 2-5.... half the time he'd call a 5..... and you'd be asked if there was double 5's under the cup and the answer would be 1 in 6 and when he takes the cup away you are looking at a 5-2......

So in the 36 possible rolls, not every number gets called out 11 times..... each number would get called out 6 times...... whcih makes sense... 6x6=36.... and for each of the 6 times each number is called out.... one of those times there will be doubles of that number.

you can't call 11 of each number in 36 rolls.... that is 66 call-outs and you only have 36 announcement to make..... 1 per roll :) okay that was all

Travis it's 1/6.

You know it. I know it. There are others who know it but they don't want to be called... well never mind.

Quote: TravisR

Quote: AlanMendelson

Okay. I'll agree that "Chance of rolling a 2 on at least one die is 11/36
Chance of rolling a pair of 2's where at least one die is a 2 is 1/11"

And the reason I wont take part in the "bet" is that I'd lose $100 to win $80.

But if you rolled a single die.... ah!
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Sorry to backtrack on you.... I sorta read the end of the book first and now am looking at the past discussion.

My thoughts here are that what you say is correct....but what seems to be overlooked is that the statement

Chance of rolling a pair of 2's where at least one die is a 2 is 1/11"

may be true, but that it not applicable to the problem where the guy announces he "sees at least one 2" before you are asked what's the chance of double 2's under his cup.

And I know I'm preaching to the choir but what throws the wrench into the mix is that the guy may not call a 2 if the roll was a 2-5.... half the time he'd call a 5..... and you'd be asked if there was double 5's under the cup and the answer would be 1 in 6 and when he takes the cup away you are looking at a 5-2......

So in the 36 possible rolls, not every number gets called out 11 times..... each number would get called out 6 times...... whcih makes sense... 6x6=36.... and for each of the 6 times each number is called out.... one of those times there will be doubles of that number.

you can't call 11 of each number in 36 rolls.... that is 66 call-outs and you only have 36 announcement to make..... 1 per roll :) okay that was all
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Why would he call out a five? You don't seem to understand the circumstances.

Post removed/transferred - I think I was meant to post it in this thread instead.

Quote: AlanMendelson

Travis it's 1/6.

You know it. I know it. There are others who know it but they don't want to be called... well never mind.
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A little further back in the discussion I saw where the situation unfolded slightly differently.... It said You ask the guy who just finished shaking the dice if there is at least one 2 and he says "Yes"..... I think at that point the odds of it being double 2's is 1:11.... becuase you've taken away the versions where he rolls something like 2-5 and instead of having any chance to call out a 5.... all 11 rolls now come into possibility equally......

I saw you had origianlly said 1 in 6 for this set-up as well and I'm wondering if you still think 1 in6?

Quote: TravisR

Quote: AlanMendelson

Travis it's 1/6.

You know it. I know it. There are others who know it but they don't want to be called... well never mind.
link to original post

A little further back in the discussion I saw where the situation unfolded slightly differently.... It said You ask the guy who just finished shaking the dice if there is at least one 2 and he says "Yes"..... I think at that point the odds of it being double 2's is 1:11.... becuase you've taken away the versions where he rolls something like 2-5 and instead of having any chance to call out a 5.... all 11 rolls now come into possibility equally......

I saw you had origianlly said 1 in 6 for this set-up as well and I'm wondering if you still think 1 in6?
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Pick up a pair of dice. Any dice.

Set at least one of the dice to show a 2.

Ask yourself: what are the odds both of the dice are 2s?

What would you do?

What I would do is pick up one die and count the faces. I would see six. One of the six is a 2. I would say 1/6.

Now this is NOT the actual question in the original problem. But this is the simple interpretation of the original question.

The original question uses additional words and conditions to make it more complicated.

But no where does the original question ask you to consider that there are 36 combinations of two dice and of those 36 combinations there are 11 combinations showing at least one two, and of those 11 combinations one combination shows 2-2.

Now I'm going to ask you:

I'm looking at two dice. One of the two dice is a two. But maybe both dice are 2s. But let's just start with one die as a 2. What are the odds that I'm looking at 2-2?

What are you visualizing in your mind from that question?

Are you visualizing in your mind 36 combinations of dice? Or are you visualizing two dice, one of which is a 2?

And here is the exact question again:

"You have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."

What is the probability that both dice are showing a 2?"

Now I'll ask you what I asked before. Are you visualizing in your mind two dice or 36 combinations of two dice?

Final question. If by chance you are visualizing 36 combinations of two dice WHY are you visualizing 36 combinations? What words in the question asked you to visualize 36 combinations?

Because some math teacher said there are 11 combinations with at least one 2?

Or do you visualize two dice because the question tells you to visualize two dice?

It's time to remind the forum what the question at hand is. Here is the exact wording.

You have two dice in a shaker. You ask a friend to shake the dice and peek at the outcome, which he does. You ask him, "Is at least one of the dice a two?" He says, "Yes." What is the probability both dice are a two?

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