Two Dice Puzzle: Part Deux

ME, I am sorry, but you are doing what everyone else has done: you are changing the question. The question is specific. Sure, if you change the question you can get 1/10, 1/11 and even 1/36. But the original question is specific and the answer to that specific question is 1/6.

All of you who tell me that 1/6 is not the answer are changing the question. I'm sorry to say it, but you are.

Quote: AlanMendelson

...The question is what is the probability...

There are 36 ways to roll a pair of dice, right?
How many of those rolls happen to be Deuce-Trey?

How many of those rolls happen to be Deuce-Anything?

Quote: AlanMendelson

ME, I am sorry, but you are doing what everyone else has done: you are changing the question. The question is specific. Sure, if you change the question you can get 1/10, 1/11 and even 1/36. But the original question is specific and the answer to that specific question is 1/6.

All of you who tell me that 1/6 is not the answer are changing the question. I'm sorry to say it, but you are.

Then you're answering "no, my restatement is not the same as the question you're asking." Okay, what's the question you're asking?

Quote: AlanMendelson

ME, I am sorry, but you are doing what everyone else has done: you are changing the question. The question is specific. Sure, if you change the question you can get 1/10, 1/11 and even 1/36. But the original question is specific and the answer to that specific question is 1/6.

All of you who tell me that 1/6 is not the answer are changing the question. I'm sorry to say it, but you are.

Well.....if I roll two dice, only 36 possible outcomes can happen of which ONLY 11 have "AT LEAST" ONE 2 SHOWING, of which ONLY ONE OF THESE 11 CAN BE 2-2, so therefore the answer is 1/11!!

I asked this same question over on my Facebook page and I want to copy and paste the response of Cyndie Randall who works with statistics and understands tha the answer is 1/6. She agrees that the question asks us to consider one die and she wrote:

"Since he told us one die is 2, we need to rule out that die when calculating the odds of the other die. Each die has an independent number of variables. We create odds by counting variables. Knowing the 2 of one die means we don't calculate that as a variable any longer because it is a known."

Ibeatyouraces is again changing the question: it's not a matter of rolling two dice and having 11 combinations to consider. We know one die is a 2 (maybe both are 2s but it doesn't matter) and so we have to look at only one die.

Look at what Cyndie wrote.

I have to say it again: you missed the question.

ME, again here is the original question:

=========================
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?
==========================

It's not about rolling two dice. It's about two dice having been rolled and at least one is known to be a 2. The question pertains to the unknown die. The unknown die has how many sides?

Quote: AlanMendelson

Actually, I think I did. Doesn't "if one of these two dice happens to show a two then our challenge is on" mean the same thing as:

======================
"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?

Yes. However, in your proof there was never a 2-2 roll so we don't see how you would record that scenario.

Quote: AlanMendelson

I asked this same question over on my Facebook page and I want to copy and paste the response of Cyndie Randall who works with statistics and understands tha the answer is 1/6. She agrees that the question asks us to consider one die and she wrote:

"Since he told us one die is 2, we need to rule out that die when calculating the odds of the other die. Each die has an independent number of variables. We create odds by counting variables. Knowing the 2 of one die means we don't calculate that as a variable any longer because it is a known."

Ibeatyouraces is again changing the question: it's not a matter of rolling two dice and having 11 combinations to consider. We know one die is a 2 (maybe both are 2s but it doesn't matter) and so we have to look at only one die.

Look at what Cyndie wrote.


I have to say it again: you missed the question.

"That die" has no meaning. "Specific die" does.

Quote: Wizard

Yes. However, in your proof there was never a 2-2 roll so we don't see how you would record that scenario.

Really? Do you need me to do another video showing two dice under the cup with both showing a 2 and then saying "okay, let's remove the die on the left -- what are the chances the other die is a 2 (even though it already is a 2)? And now, let's remove the die on the right -- what are the chances the other die is a 2 (even though it already is a 2)? I don't think it's necessary to do that. But if you want me to, I will. It's a two-dice question and we are asked to consider the odds of ONE die.

Quote: Ibeatyouraces

"That die" has no meaning. "Specific die" does.

It's a two dice question. It doesn't matter which of two dice shows a 2. Because whenever at least one die shows a two, the question is about the other die. In a two dice question that means only one die remains.

Let me give it to you this way: 2-1 = 1

You start with two dice, and you are told at least one die is a 2, and you subtract the die with a 2 and that leaves one die to consider.

2-1 = 1

Quote: AlanMendelson

It's a two dice question. It doesn't matter which of two dice shows a 2. Because whenever at least one die shows a two, the question is about the other die. In a two dice question that means only one die remains.

Let me give it to you this way: 2-1 = 1

You start with two dice, and you are told at least one die is a 2, and you subtract the die with a 2 and that leaves one die to consider.

2-1 = 1

Again, there are 11 different ways "at least one will be a 2." And only one of them will be 2-2. How hard is that to see??

Quote: AlanMendelson

ME, again here is the original question:

=========================
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?
==========================

It's not about rolling two dice. It's about two dice having been rolled and at least one is known to be a 2. The question pertains to the unknown die. The unknown die has how many sides?

I see where we are. You're not actually convinced that rolling two dice and you personally looking at both of them is equivalent to rolling two dice and having someone else look at both of them. But the probabilities for fair dice rolls don't change regardless of who looks at them, or indeed whether anyone looks at them at all. Do you agree?

There are certain facts that are true for fair tosses of standard six-sided dice regardless of who throws them or observes them. One is that the probability of a pair of 2s in a fair throw of two standard six-sided dice is 1/36. Another is that the probability of at least one 2 in a fair throw of two standard six-sided dice is 11/36. Do you agree?

Quote: AlanMendelson

I asked this same question over on my Facebook page and I want to copy and paste the response of Cyndie Randall who works with statistics and understands tha the answer is 1/6. She agrees that the question asks us to consider one die and she wrote:

"Since he told us one die is 2, we need to rule out that die when calculating the odds of the other die. Each die has an independent number of variables. We create odds by counting variables. Knowing the 2 of one die means we don't calculate that as a variable any longer because it is a known."

Ibeatyouraces is again changing the question: it's not a matter of rolling two dice and having 11 combinations to consider. We know one die is a 2 (maybe both are 2s but it doesn't matter) and so we have to look at only one die.

Look at what Cyndie wrote.

I have to say it again: you missed the question.

ME, again here is the original question:

=========================
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?
==========================

It's not about rolling two dice. It's about two dice having been rolled and at least one is known to be a 2. The question pertains to the unknown die. The unknown die has how many sides?

It's all about 2 dice being rolled AT THE SAME TIME.

Does Cyndie calculate the CPI(Consume Price Inflation) Index for the government? If so that would explain alot. She and you are omitting half the data to answer the question.

Possible results of two dice roll rolled simultaneously containing a 2.
2+1, 2+3, 2+4, 2+5, 2+6, 2+2, 1+2, 3+2, 4+2, 5+2, 6+2

Take it for what it's worth. I'm just a dumb carpenter.

I see 11 possible combinations, only one of which is a pair of 2's.

Alan's interpretation of the original question is where he differs from us from what I can perceive. Not the actual probabilities.

Quote: Wizard

Long time no see, ME. Nice to have you back. I hope you're able to educate some of the 1/6 believers. I think your odds are about 1 in 11.

Quote: Face

I second that.

Quote: beachbumbabs

Welcome back, ME!

Thanks all. I was busy with a case but it's done and I have a bit more time now. I'll be in and out. Oh darn, now I want a hamburger.

Quote: AlanMendelson

There you go again... flipping coins.

Actually, I think my video really does portray the problem.

And I am certainly not going to be flipping coins to solve this.

But why don't you write a script (narration) for how you think a video demonstration of the problem would be? Write it here, and I will record your script using two dice.

Thanks.

What's wrong with flipping coins? It's the exact same thing as rolling dice...yet you fail to recognize that. If by some miracle you were able to recognize the two problems are identical, you would see how entirely wrong you are on this 1/6 answer.

Cyndie Randall is wrong -- it is not independent, but a dependent event. If neither dice is a 2, then nothing happens. Statitians are famous for doing statistics poorly, by trying to connect 2 unrelated things (autism + vaccines - just because both increase at the same time doesn't mean one causes the other)...or by forcing the incorrect answer with trollish logic (ie: this problem).

Quote: MathExtremist

I'll be in and out.

love the sound of that from U

Quote: MathExtremist

Oh darn, now I want a hamburger.

i neither darn nor knit
love burgers but do not really want them

Ah, thank you for that nice sex though
time for a smoke

Quote: AlanMendelson

Well, if you read my website you would know that I do not agree with Rob Singer's special play strategy. I play conventional video poker strategy as taught by Bob Dancer. In fact, I attended his lectures and used his software. I also use the book written by John Grochowski which is also conventional play. I presented Singer's articles and videos on my site because so many people have an interest in what he says and when his own website was taken down I offered my site as a place for him to show his stuff.

I also have a section on my site for the articles of Professor Nelson Rose that he provides to me.

I also have a section on my site for the Las Vegas Convention and Visitors bureau so they can promote their monthly events calendar.

And there is a section on my site for the Vegas Realtors group.

But getting back to Singer, I make it clear his stuff is controversial and I don't agree with it myself. You can read that here: http://alanbestbuys.com/id194.html in the bold, italic font.

I was being facetious Alan. This go around ad infinitum reminds me of Robs round and round and................

Quote: AlanMendelson

I asked this same question over on my Facebook page and I want to copy and paste the response of Cyndie Randall who works with statistics and understands tha the answer is 1/6. She agrees that the question asks us to consider one die and she wrote:

"Since he told us one die is 2, we need to rule out that die when calculating the odds of the other die. Each die has an independent number of variables. We create odds by counting variables. Knowing the 2 of one die means we don't calculate that as a variable any longer because it is a known."

Ibeatyouraces is again changing the question: it's not a matter of rolling two dice and having 11 combinations to consider. We know one die is a 2 (maybe both are 2s but it doesn't matter) and so we have to look at only one die.

This is incorrect. There is a decision made based on the two dice. Thus we have to take that into account. The preconditions to looking at the second dice DO change the odds of it showing a two.

The dice has 6 sides, but one of those sides is half as likely to be showing as the others BECAUSE we invoked a rule to only look at it in a certain case.

Quote: AlanMendelson

But that's not the original question either. It's not a question of how often the other die is a two, or how often it is not a two. The question is what is the probability. Thank you for recognizing it is a ONE DIE problem. And that one die has six sides.

Alan, you are the one who's emasculating the question.
You're chopping it all the way down to "Throw one Die."
Of course the answer to that is 1/6 for each possible pip-count.
But that is decidedly not the scenario under review.

The scenario is "Throw two Dice."
You've got nearly twice as many chances of achieving a Deuce.
Eleven chances, to be exact.

• 2-2, the golden result that's going to win you 9-to-1 on your wager
• 2-1, 2-3, 2-4, 2-5, 2-6, and all their evil twin mirror images...
• 1-2, 3-2, 4-2, 5-2, 6-2.

Quote: AlanMendelson

...Cyndie Randall...

Good. She can come to our casino, which has the insanely generous 9-to-1 payout on Pair-of-Deuces, but loses on Deuce-Anything Else. She can bankrupt us... or perhaps lose her a$$.

Hey looksee.... The much respected Cynthia just posted on Alan's facebook page.

Quote: Alan

Cyndie Randall who works with statistics and understands tha[sic] the answer ...

I wonder if she knows something we don't? Like, maybe she knows her own name isn't Cyndie?
Actually His Facebook page is worth finding. He's getting shredded by his friends.

Quote: Cynthia Randall

Oh boy. Here's where the humble/mature side of me kicks in... You're right Rob, 1:11 is it. After having thought back to Statistics 101 (okay, again by thinking of the game Yatzee), it dawned on me that yes, on a SINGLE roll, the odds are 1:11 because it doesn't matter if the judge told us the number or not. The odds of ANY two numbers (if you were going to guess them ahead of the roll) coming up are 1:11. The judge telling us a 2 is there has no bearing on the original odds. So, it appears that I was over-thinking it. I think the "trick" part of the question is by having that extra info in there that really isn't relevant (the judge). If the question were, "What of the odds of the 2nd die coming up 2 if rolled AGAIN, then the answer would be 1:6. Sorry Alan, but I have to agree with the nay-sayers on this one.

I'd still be interested in Alan's answer to this...

Alan could only faithfully depict the original question if :-
1. He rolled the dice a very very large number of times and actually measured the probability 'Of a pair of twos' by observation. Rough and ready couple of hundred times would soon show that it's closer to 1/11 than 1/6.
2. He rolled the dice a number of times which represented all possible and equally likely outcomes and then actually measured the probability 'Of a pair of twos'

I would be interested in what answers Alan would expect to come up with from each of those experiments.

Hah, I was just looking that up to. This is so sad.... https://www.facebook.com/alan.mendelson.1/posts/10203057456277672?comment_tracking=%7B%22tn%22%3A%22O%22%7D&pnref=story

Usually these types of things go like this:

*smart person asks confusing question*
*everyone says "oh it's X" *
*smart person says "nope it's Y, here's why/how" *
*everyone: "oooh you're right" *


But in this case....it's...well, the exact opposite.

I just spent three hours on a YouTube video explaining why the answer is 1/11. It included a a test of 200 resolved rolls, with the following results:

One two = 180/200 = 90%
Two twos = 20/200 = 10%

Alan's camp says the probability is 1/6 = 16.67%
The correct camp says the probability is 1/11 = 9.09%

The results of the experiment come a lot closer to 1/11 than 1/6.

The audio quality of the video was so bad that I decided to delete the whole thing rather than embarrass myself with the awful production value.

Quote: Wizard

I just spent three hours on a YouTube video explaining why the answer is 1/11. It included a a test of 200 resolved rolls, with the following results:

One two = 180/200 = 90%
Two twos = 20/200 = 10%

Alan's camp says the probability is 1/6 = 16.67%
The correct camps says the probability is 1/11 = 9.09%

The results of the experiment come a lot closer to 1/11 than 1/6.

The audio quality of the video was so bad that I decided to delete the whole thing rather than embarrass myself with the awful production value.

Why can't other peeps do that? Oh. we did!

Maybe you weren't answering the original question which was about one dice because 2-1=1.... Erm hang on, I swear i read somewhere

Quote: Alan

It's a two dice question.

around the same time he turned it into a one die question.

Somebody, anybody, with Facebook.
Invite that Cynthia character to our Casino.
Explain our Cup-and-Dice game, with payouts, to her.
See what she decides, with the prospect of real money on the line.

Quote: indignant99

Somebody, anybody, with Facebook.
Invite that Cynthia character to our Casino.
Explain our Cup-and-Dice game, with payouts, to her.
See what she decides, with the prospect of real money on the line.

No Need. Cynthia has already seen the light and joined the 1/11 camp, as have all his current FB commenters.

There are a few who side with him on his own forum, but those few are openly hostile to WOV and it's members* and I'm damned if I'm giving him free content on his site, just so I can get name-called by what I perceive to be ignorant, non-listening thugs.

Even the most hostile possible member of his Forum, Rob Singer has told Alan it's 1/11

Quote: OnceDear

...Cynthia...

Temporarily. She back-flipped again to 1/6.

Once Dear: Yes, it is a two dice question but after you are told what at least one of the two dice shows then it is only a one die question. Keep reading on Facebook. Cynthia and Rob Singer both now say the answer is 1/6 because of the way the question is worded. So do a lot of other people. In fact, only two people on Facebook say it's 1/11 and both of them are going thru the same convoluted math that the rest of you are using. If you simply isolated one die, as the question asks, you will also say it's 1/6. In fact, you concede the answer is 1/6 when you isolate one die but for some reason you won't accept that and you continue to use two-dice combinations and I don't understand why.

Wizard: I shot my video using a Samsung cell phone. It took less than three minutes to upload to YouTube. I know you say the answer is 1/11 but I want to see how you reached the answer of 1/11.

So far I have seen various spread sheets showing the various combinations of two dice that include a 2. And I am sorry, but that is not what the question is asking to see or show. So, Wizard, are you using the same 11 different combinations of dice? If you are, you are also not answering the question.

Quote: indignant99

Temporarily. She back-flipped again to 1/6.

Oh yeah. I read the timeline wrong. Nonetheless, she is as wrong as King Canute.
She's Alan's friend. I wonder if she has any money that she would be willing to throw into my wallet on the Wizard's bet.
Link to the bet

I'd be happy to stake my pension fund.

I really don't understand what question Alan thinks he is answering with "1 in 6".

Quote: thecesspit

I really don't understand what question Alan thinks he is answering with "1 in 6".

Is it 'what is the best case scenario probability that Alan could back out of this stupid argument with any dignity.'

Quote: AlanMendelson

Once Dear: Yes, it is a two dice question but after you are told what at least one of the two dice shows then it is only a one die question. Keep reading on Facebook. Cynthia and Rob Singer both now say the answer is 1/6 because of the way the question is worded. So do a lot of other people. In fact, only two people on Facebook say it's 1/11 and both of them are going thru the same convoluted math that the rest of you are using. If you simply isolated one die, as the question asks, you will also say it's 1/6. In fact, you concede the answer is 1/6 when you isolate one die but for some reason you won't accept that and you continue to use two-dice combinations and I don't understand why.

Wizard: I shot my video using a Samsung cell phone. It took less than three minutes to upload to YouTube. I know you say the answer is 1/11 but I want to see how you reached the answer of 1/11.

So far I have seen various spread sheets showing the various combinations of two dice that include a 2. And I am sorry, but that is not what the question is asking to see or show. So, Wizard, are you using the same 11 different combinations of dice? If you are, you are also not answering the question.

"at least one of the dice is a 2". So we look at all the combinations of the dice where "at least one of the dice is a 2". How is that not in line with the question?


Last I heard from Rob, he was saying the answer was 1/11 (on your forum).

Quote: AlanMendelson

But that's not the original question either. It's not a question of how often the other die is a two, or how often it is not a two. The question is what is the probability. Thank you for recognizing it is a ONE DIE problem. And that one die has six sides.

Uh-oh. Now your reading comprehension may be going south to join your math comprehension. In gaming events, when do these not all mean the same thing?

• "frequency"
• "how often"
• "likelihood"
• "probability"
• "odds"

Quote: OnceDear

Is it 'what is the best case scenario probability that Alan could back out of this stupid argument with any dignity.'

Quote:

"You have two 6-sided dice in a cup (standard dice). You shake the dice, and slam the cup down onto the table, hiding the result. An independent judge peeks under the cup, and announces truthfully, "At least one of the dice is a 2."

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

Here's the question:

If it's 1 in 6, that means that 1 in 6 times that a judge is doing the peeking, there's a pair of twos.

Great. Lets pretend that's true

But we all agree, there's at least one two eleven times out of 36.

Which therefore tells me, the chance of rolling a double two is 1 in 6 of 11 in 36:

11 out of 216. -> 1 chance in 19.63.

But there's two dice! The chance of rolling a double two is 1 in 36. We agreed that too.

So we have a inconsistency.

If we instead say the result is 1 in 11 of the other die being a two, and we know we see this situation 11 times in 36:

11 out of 396 -> 1 in 36 chance of double twos overall. Perfectly consistent.

Alan seems to take great store in 'what does it matter that we don't know which dice it is, there's only two dice'. Exactly. Failing to understand that is EXACTLY the core of the problem.


Lets try this another way:

Say I roll three dice. The judge looks at them, and says "there is at least 1 two". What are the odds there is at least a pair of twos?

Well, if we take Alan's process... there's two dice, and 11 in 36 chance of them rolling at least one two.

BUT it's not that. There's 216 options. 91 of them contain at least 1 two. 16 of those, are a pair or tripple of twos. Thus it's 16 in 91 of rolling a pair of twos or better in three dice, if we get reported there's at least one two.

What's the chance of their being a tripple of two's now we know there's at least one two? 1 in 91, not 1 in 36.

What's the chance, if we are told there's at least 2 twos of there being a tripple of twos? Not in 1 in 6.

There's only 16 out of 216 rolls that contain a pair of twos. Only one of those contain a tripple : 1 in 16.

Quote: RS

"at least one of the dice is a 2". So we look at all the combinations of the dice where "at least one of the dice is a 2". How is that not in line with the question?

That is exactly in line with the question. And when we are told at least one of the dice is a two we only have to be concerned with what the other die is showing. And that is 1/6. As I wrote above, even if both dice are showing a 2 the "other die" is 1/6 for showing a 2.

Quote: RS

Last I heard from Rob, he was saying the answer was 1/11 (on your forum).

Rob was bounced from my forum again for his obnoxious attacks on members of the WOV forum. If you read his comments on my Facebook page you will see where he now recognizes that the answer is 1/6 BECAUSE OF THE WAY THE QUESTION IS WORDED.

Again, 1/11 is a valid answer if you are looking for the various combinations of two dice with at least one die showing a 2. But the question is asking you to isolate the problem to one die and only one die.

Early on in the original thread I pointed out how the original poster probably worded his question to intentionally cause confusion. He did it with his original "set up" of the problem when he said "Along the vein of the Two Coin Puzzle,..." That statement I believe made the "1/11 camp" look at the question using what I will call the "spreadsheet answer" with the various two-dice combinations. But when you look at the wording of the actual question it clearly asks you to look at one die and only one die.

When I looked at the original poster's "spoiler" and saw the rambling text, I immediately suspected the original poster was trying to cause confusion. You know what they say in poker? You can play your cards or you can play the player. In this case, after seeing how the original poster phrased the question and what he wrote in his spoiler, I chose to "play the player."

I also find it very curious that the original poster has not responded to his own thread. Not even once. I said it before -- he was probably having a good laugh over the confusion he caused.

If you insist on applying the "spreadsheet answer" to the problem/question you will always come up with the 1/11 answer. I can only interpret the question to intend the chance of one die showing a 2 and that is 1/6. And as I said before, if you consider two dice rolled simulaneously you can also say the answer is 1/36 and if you want to eliminate the one combination of two dice as viewed by the friend or judge then indeed the answer si 1/10.

I think I've said it a hundred times: how you read the question will define your answer. And I read the question to define the result on one die.

I am really curious how the Wizard read the question to come up with his 1/11 answer? In fact, for those of you who believe the answer is 1/11 why isn't it 1/10 eliminating the combination that the judge/friend is viewing under the cup? And why isn't it 1/36 which is the chance of any hard-4 being rolled?

Quote: AlanMendelson

Cynthia and Rob Singer both now say the answer is 1/6 because of the way the question is worded. So do a lot of other people. In fact, only two people on Facebook say it's 1/11 and both of them are going thru the same convoluted math that the rest of you are using. If you simply isolated one die, as the question asks, you will also say it's 1/6.

The question literally asks "What is the probability that both dice are showing a 2?"

Maybe role playing will help. Suppose you're the shooter at a craps dice table. On your first throw, one of the dice bounces out of the table into the pit. The one that stays in the table is a 2 but you can't see the other one.

The dealer retrieves the dice and you're still shooting, but this time you throw *both* dice out of the table. You can't see either of them, but the pit boss (who has been reading this thread) smirks and says "at least one of these dice is a 2."

Do you believe these two scenarios are identical with respect to the question "what is the probability that both dice are showing a 2"?

Quote: AlanMendelson

That is exactly in line with the question. And when we are told at least one of the dice is a two we only have to be concerned with what the other die is showing.

Ok, so you agree, all combinations of the dice where at least one die is a 2 is consistent with the question?

1-2
2-2
3-2
4-2
5-2
6-2
2-1
2-3
2-4
2-5
2-6

I count a total of 11 combinations where at least one die is a two. I count one combination where both dice (or "the other die") is a 2. And 10 combinations where the other die is not a 2.

Do you agree there are 11 combinations where a 2 is present and 1 combination where both dice are a 2?

Remember, you agreed this is in line with the question.

Quote: AlanMendelson

As I wrote above, even if both dice are showing a 2 the "other die" is 1/6 for showing a 2.

This is the root of the issue -- you don't know about (or understand) conditional probability.

If both dice are showing a 2, the individual probability of one or the other die being 2 is not 1/6. It is 1.

The premise behind conditional probability is that additional relevant information can update the evaluation of a prior probability. For example, while the dice are in the air, the probability of a hard 4 is 1/36. After they land, one of them comes to a stop while the other is still spinning. The one that stopped is a 5. The probability that the dice are a hard 4 is now zero. Every craps player knows this intuitively; when the shooter is going for a point like 4 or 5 and one of the dice stops on a number like 5 or 6 that makes the point impossible, you hear groans at the table. More formally, the probability of a pair of dice showing 2-2, given that one of the dice shows a 5, is zero.

So what's the probability of a pair of dice showing 2-2 given that one of the dice shows a 2? That's the question you're trying to answer.

.......This is a public service alert.........

Quote: RS

...facebook...

There's just a tsunami of dense-in-the-head over there on Alan's Facebook page.
And all the females commenting, are wrong, wrong, wrong!

• Lauren Linett
• Cathy Barr
• Cynthia Randall

Quote: MathExtremist

The question literally asks "What is the probability that both dice are showing a 2?"

Maybe role playing will help. Suppose you're the shooter at a craps dice table. On your first throw, one of the dice bounces out of the table into the pit. The one that stays in the table is a 2 but you can't see the other one.

The dealer retrieves the dice and you're still shooting, but this time you throw *both* dice out of the table. You can't see either of them, but the pit boss (who has been reading this thread) smirks and says "at least one of these dice is a 2."

Do you believe these two scenarios are identical with respect to the question "what is the probability that both dice are showing a 2"?

I used the example of a shooter throwing two dice with one coming to rest immediately showing a two. The second die is a spinner and spins like a top. While spinning you say to yourself "there is a 1/6 chance of the spinner landing on a 5 for a 7-out," or "there is a 1/6 chance of the spinner landing on a 1 for a 3" or "there is a 1/6 chance of the spinner landing on a 2 for a hard 4."

Quote: RS

Ok, so you agree, all combinations of the dice where at least one die is a 2 is consistent with the question?

1-2
2-2
3-2
4-2
5-2
6-2
2-1
2-3
2-4
2-5
2-6

I count a total of 11 combinations where at least one die is a two. I count one combination where both dice (or "the other die") is a 2. And 10 combinations where the other die is not a 2.

Do you agree there are 11 combinations where a 2 is present and 1 combination where both dice are a 2?

Remember, you agreed this is in line with the question.

I said that the 1/11 answer would be correct if the question were asked differently. See my "spinner" response to ME above which is what the question is asking.

Quote: AlanMendelson

I used the example of a shooter throwing two dice with one coming to rest immediately showing a two. The second die is a spinner and spins like a top. While spinning you say to yourself "there is a 1/6 chance of the spinner landing on a 5 for a 7-out," or "there is a 1/6 chance of the spinner landing on a 1 for a 3" or "there is a 1/6 chance of the spinner landing on a 2 for a hard 4."



I said that the 1/11 answer would be correct if the question were asked differently. See my "spinner" response to ME above which is what the question is asking.

The question is not asking what are the chances of an unknown dice landing on a 2. We have rolled the dice and gained some information via the judge. You just refuse to acknowledge that. The question absolutely is NOT asking what you claim it is. There is a process involved and you've rejected it.

Quote: AlanMendelson

I used the example of a shooter throwing two dice with one coming to rest immediately showing a two. The second die is a spinner and spins like a top. While spinning you say to yourself "there is a 1/6 chance of the spinner landing on a 5 for a 7-out," or "there is a 1/6 chance of the spinner landing on a 1 for a 3" or "there is a 1/6 chance of the spinner landing on a 2 for a hard 4."



I said that the 1/11 answer would be correct if the question were asked differently. See my "spinner" response to ME above which is what the question is asking.

Political spin is a calling

Quote: AlanMendelson

... throwing two dice with one coming to rest immediately showing a two. The second die is a spinner and spins like a top... "there is a 1/6 chance of the spinner landing on a 2...

And 5 times out of 6, you get no pair.

Okay, what about the times that the resting die doesn't show a deuce {1,3,4,5,6} but the spinner does land as a deuce? Do those rolls qualify for the announcement "a deuce showed"? And here are 5 more times you get no pair.

Quote: thecesspit

The question is not asking what are the chances of an unknown dice landing on a 2. We have rolled the dice and gained some information via the judge. You just refuse to acknowledge that. The question absolutely is NOT asking what you claim it is. There is a process involved and you've rejected it.

Excuse me, but where did you get the idea that I am not acknowledging the information that the partner/judge said at least one of the dice is a 2? I've been saying it all along -- that we know one die is a two. And it is that information which makes it clear that the question now revolves on the other die -- which has six sides.

I have been arguing this information all along -- that once the judge/partner truthfully says at least one die is a two the problem shifts to the remaining die (there are only two) and that is the 1/6 answer.

On the other hand, the 1/11 "faction" claims we need to know which of the dice is showing a two. How that can matter when we start with only two dice is what I can't understand. Perhaps the Wizard would like to explain? I asked him for his video to explain the process of coming up with the 1/11 answer, just as I posted a video explaining the 1/6 answer.

Quote: indignant99

And 5 times out of 6, you get no pair.

Okay, what about the times that the resting die doesn't show a deuce {1,3,4,5,6} but the spinner does land as a deuce? Do those rolls qualify for the announcement "a deuce showed"? And here are 5 more times you get no pair.

Really? If the judge/partner sees one die as a deuce -- and it doesn't matter which die -- the decision still rests with the other die which is 1/6.

Can't you see that I was using the "spinner" as an example of how this is just a question relating to how one die lands? I guess not. So I will reverse it for you:

A shooter throws two dice. One lands behind a stack of chips and you can't see the result. One die is a spinner and eventually lands on a 2. What are the chances that the die behind the stack of chips is also a 2? It's 1/6.

Happy now indignant?

Geesh.

And the game goes on.........

Quote: AlanMendelson

I used the example of a shooter throwing two dice with one coming to rest immediately showing a two. The second die is a spinner and spins like a top. While spinning you say to yourself "there is a 1/6 chance of the spinner landing on a 5 for a 7-out," or "there is a 1/6 chance of the spinner landing on a 1 for a 3" or "there is a 1/6 chance of the spinner landing on a 2 for a hard 4."

Your spinner example is equivalent to not having thrown the 2nd die yet. You could start with a 2 and then hold the other die in your hand and ask "what are the chances of a 2 with the second die when I throw it?" The answer is obviously 1/6.

But that's not equivalent to the original question. In your question, one die is decided and known to be 2 while the other die is undecided (and is therefore unknown). In the actual question, both dice are decided and one is known to be 2. The answer to "what is the likelihood the other die is a 2" is not the same between those two questions. Do you understand this?

To see why they are not equivalent, examine the probability of a first 2 in each scenario. In your example, what is the probability on any roll that you see a 2 with the die that has come to rest? And in the original question, what is the probability on any roll that the dealer peeks and says "at least one die is a 2?"

There's an old children's riddle that goes like this:
"I have two coins that total thirty cents. One of them is not a quarter. What are the two coins?"

Quote: MathExtremist

Your spinner example is equivalent to not having thrown the 2nd die yet. You could start with a 2 and then hold the other die in your hand and ask "what are the chances of a 2 with the second die when I throw it?" The answer is obviously 1/6.

But that's not equivalent to the original question. In your question, one die is decided and known to be 2 while the other die is undecided (and is therefore unknown). In the actual question, both dice are decided and one is known to be 2. The answer to "what is the likelihood the other die is a 2" is not the same between those two questions. Do you understand this?

To see why they are not equivalent, examine the probability of a first 2 in each scenario. In your example, what is the probability on any roll that you see a 2 with the die that has come to rest? And in the original question, what is the probability on any roll that the dealer peeks and says "at least one die is a 2?"

There's an old children's riddle that goes like this:
"I have two coins that total thirty cents. One of them is not a quarter. What are the two coins?"

Show Spoiler

The one that isn't a quarter is a nickel. The other one is a quarter.

There are two ways to get from point A to point B. The shortest route is a direct line. And then there is the "1/11 route."

I give up.

But I am still interested in finding out how the Wizard decided the answer is 1/11 ??

Quote: AlanMendelson

...Happy now indignant? ...

Not at all. You're separating the dice into two separate, distinct observations.
IT DON'T WORK THAT WAY! You get to view both simultaneously, and either one - either GD one - can obligate the announcement of "deuce."