Question for Math/Gambling/Craps Experts

The craps table example - where one die lands on 2 and the other die is hidden - is not equivalent to the question posed. Once you know for certainty the value of a specific die is 2, then the probability of 2-2 is 1 in 6.

The question posed is "at least one of the dice is a 2." For this question, we do not know if it is the first die that is a 2, or if it is the second die that is a 2 (for how ever you want to define "first die" and "second die"). The person who viewed the dice knows this information, but we do not. This is where the 1 in 11 answer comes from - there are 11 ways to roll at least one 2 (2-1, 2-2, 2-3, 2-4, 2-5, 2-6, 1-2, 3-2, 4-2, 5-2, 6-2), and each of these has the same probability. Of these 11 ways, only 1 way satisfies the condition that "both dice are showing a 2." Hence, 1 in 11.

Take up the challenge. Yes or no.
It's your forum, you could mess about with my posts, bar me or do whatever.
However, I will be happy to PROVE to you that the answer to the question as posed is 1/11. Will you let me lead you to the proof, inch by painful inch, word by painful word? Will you dare to do that without.... Ever mentioning Craps? Without ever mentioning Coins? Without ever mentioning spinning dice? Without setting aside one dice as a two and insisting it is the same question? Whether you believe that to be the same question or not, we will answer the question as posed and not your preferred equivalent.* I will walk you to the truth inch by inch on the condition that you do not sidetrack with any assertion that is outside of scope of the question as posed. I will also VERY HAPPILY supply you with a faithful simulation of the question as posed and PROVE with that that the answer is 1/11. I'll ask only that you challenge any assertion that I make with PROOF.

Finally, It's your forum. Select who you wish or all your fellow members to adjudicate. I WILL NOT be sidestepped into addressing any issues raised by other posters unless they directly and exactly reference the original question and my analysis of it.

Do you DARE? On your own forum? Yes or no.

Incidentally, There are scenarios where the probability cannot be known and certain assumptions that MUST be addressed.
Also, I'd be forever grateful if you could acknowledge in this thread, that the poster from WOV that you have quoted on the topic related to the hidden die being the stack DID LATER RETRACT the assertions which you have quoted here. Quoted erroneous argument :-
http://vegascasinotalk.com/forum/showth...ll=1#post27116
Later withdrawn by Dalex, so not fair to quote it here.

For the case of the spinning second die, where the first die has come to rest on a 2, the odds are indeed 1/6 that the spinning die will also land a two. I steadfastly refuse to argue that with you, because it something that (it pains me to say) I agree with you about.

C'mon Alan, let's make a wager.

-Roll two dice. Any non-deuce roll is a push (where neither die is a deuce).
-Any 2-2 I pay you 7:1.
-Any counting roll (one die a deuce, other not a deuce) you lose your wager to me.

Let's do this a bunch of times, over and over, so you can laugh your way to the bank with all my money.

Quote:

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Originally Posted by RS__

C'mon Alan, let's make a wager.

Let's do this a bunch of times, over and over, so you can laugh your way to the bank with all my money.

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Hi RS, Nice to see you over here. We are both junior members in Alan's territory. Please lets keep it polite, factual, and lacking in any controversy, opinion or hostility. I'm trying to help Alan, I really am.

Though, for the record, I'm also prepared to wager money on similar terms, though I very, very strongly suspect that obstacles would find their way into that possibility.

Welcome to my forum guys. State your case.

Quote:

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Originally Posted by Alan Mendelson

Welcome to my forum guys. State your case.

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Thanks Alan. Do you agree to let me do this my way?

Ok.... Lets see what we can agree on. This will be a slow road.

Please answer 'Agree' or disagree to each of these assertions Or say that you agree with all and save a few keystrokes.

1. Was the question as posed exactly as follows:-

"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?"

2. Do you agree that there is no mention of craps?

3. Can we agree that we will avoid mentioning craps.

4. Do you agree that there is no mention of coins?

5. Can we agree that we will avoid mentioning coins?


I hope you can agree to all of the above.

Quote:

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Originally Posted by 1in11

The craps table example - where one die lands on 2 and the other die is hidden - is not equivalent to the question posed. Once you know for certainty the value of a specific die is 2, then the probability of 2-2 is 1 in 6.

The question posed is "at least one of the dice is a 2." For this question, we do not know if it is the first die that is a 2, or if it is the second die that is a 2 (for how ever you want to define "first die" and "second die"). The person who viewed the dice knows this information, but we do not. This is where the 1 in 11 answer comes from - there are 11 ways to roll at least one 2 (2-1, 2-2, 2-3, 2-4, 2-5, 2-6, 1-2, 3-2, 4-2, 5-2, 6-2), and each of these has the same probability. Of these 11 ways, only 1 way satisfies the condition that "both dice are showing a 2." Hence, 1 in 11.

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No, I am going to stick with the craps table example and once again I am going to post the original question on the Wizard's Forum:

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 doesn't matter which of the dice is a 2 -- the answer is still 1/6.
Even if both dice showed a 2 -- the answer is still 1/6.

This exercise of yours (and everyone else including the Wizard) where you list the 11 ways to roll at least one 2 (2-1, 2-2, 2-3, 2-4, 2-5, 2-6, 1-2, 3-2, 4-2, 5-2, 6-2) is meaningless.

When you have a 2 showing on one die -- there is only a 1/6 chance of a 2 showing on the second die. And again, it doesn't matter which die the 2 is showing on. And it doesn't matter if a 2 is showing on both dice because the answer is still 1/6.

I am afraid that all of you "overthought" this question.

I am going to copy this and post it on the Wizard's forum also.

Thanks for joining my forum.

OnceDear I agree to all five. Please read my response to 1in11.

Alan,
I'm going to agree with you on one thing, and only on one thing. If you answer a question about one die coming to rest as a deuce while the other spins, then the probability of the spinning die coming to rest as a deuce is indeed 1 in 6. I absolutely concur with you on that. If you take one die and set it aside as a deuce, then the probability of the other die being thrown and turning out to be the pair creating deuce is also 1 in 6. No-one is disputing that.

Do you agree with that?

You very firmly believe that your own model of the question is equivalent. You say it is equivalent because it is obvious and because I'm overthinking the problem.

Do you agree with that?

If you believe it is so obviously equivalent or 'the same thing', then you will have absolutely no problems picking holes in my ongoing attempt to answer the ORIGINAL question, regardless of what you believe to be obvious.

Do you agree with THAT?

Do you agree that it is the original question to be answered and proved?

Now, do you want me to show you why the correct answer to the original question is 1 in 11 or not. Agree to let me proceed or disagree forever.

Thanks.

I think our case has been presented rather thoroughly on the WOV forum...repeatedly.


You've brought up the dice-spin situation a few times (one dice lands on 2, the other is spinning, what's the probably the spinning dice lands on a 2?). But, you've forgotten about the other possibility:

The first dice settles and is a 1 (or 3, 4, 5, or 6). What is the probability the spinning dice lands on a 2? Let's see.

Firstly, there's a 5/6 chance the first die is a non-deuce. There is a 1/6 chance the spinning die will land on 2 [deuce]. This will happen 5/36 times.
From there, the chance of a non-deuce (for the spinner) is 5/6. This gives us a 25/36 (ie: 25 ways to roll the dice where neither are a deuce).
The one you presented is there is a 1/6 chance the first die is a deuce. What is the probability the second (spinning) die is a deuce [1/6]. This gives us 1/36.
Also, there is a 5/6 chance the spinning die is not a deuce. This gives us 5/36.

#1 has landed, #2 is spinner. X means non-deuce, 2 means deuce.
#1 | #2 | Probability
X | 2 | 5/36
X | X | 25/36
2 | 2 | 1/36
2 | X | 5/36

SUM probability: (5+25+1+5) / 36 = 1.0

Wait a second, RS. Why are you raising the question that the first die did not settle on a 2? The question is at least one die is a 2 and what is the chance that the second die is going to also be a 2.

It seems to me that this has been the problem from the get-go over on the Wizard's Forum. You questioned the question. And that led to overthinking about all the different dice combinations that contain a 2. And this is how you came up with the 1/11 number. And this is wrong.

Clearly the question is this (and quoted directly from the Wizard's forum):

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 has been established in the question that at least one die is a 2. Perhaps both dice are showing a 2 -- but at least one die is showing a 2. When one die is showing a 2 there is a 1/6 chance that the second die is also showing a 2. That's it. Plain and simple. You have to use the information provided in the question and the information in the question is that at least one die shows a two. I am sorry but you and the Wizard and the others seemed to have missed that basic piece of information and you "over-thought" the question.

Oh RS..... Being right won't score any brownie points with Alan. We have to PROVE we are right in tiny undisputable increments. Ultimately Alan has to PROVE it to himself. At the moment he has the sticking point that he is answering his own question to which the correct answer is 1/6

I'm prepared to prove only the answer to the original question. Everything else is distraction. It's all wrapped in "It doesn't matter which of the dice is a 2"

If he lets me prove the answer to the original question, to his satisfaction, then it might be easier to have an epiphany and to see what he has missed. I propose to lead him to his own conclusive proof of the 1/11 answer.

But, I fear he may pass up on that opportunity for his own reasons.

Alan,
I'll ask again.
Alan,
I'm going to agree with you on one thing, and only on one thing. If you answer a question about one die coming to rest as a deuce while the other spins, then the probability of the spinning die coming to rest as a deuce is indeed 1 in 6. I absolutely concur with you on that. If you take one die and set it aside as a deuce, then the probability of the other die being thrown and turning out to be the pair creating deuce is also 1 in 6. No-one is disputing that.

Do you agree with that?

You very firmly believe that your own model of the question is equivalent. You say it is equivalent because it is obvious and because I'm overthinking the problem.

Do you agree with that?

If you believe it is so obviously equivalent or 'the same thing', then you will have absolutely no problems picking holes in my ongoing attempt to answer the ORIGINAL question, regardless of what you believe to be obvious.

Do you agree with THAT?

Do you agree that it is the original question to be answered and proved?

Now, do you want me to show you why the correct PROOF to the original question is 1 in 11 or not.

Agree to let me proceed or disagree forever.

OnceDear -- again this is an open forum. Fire away.

Please confirm that you have read and agree with all the above assertions in my latest post. It is absolutely the best way to proceed. I need you to be with me every painful extra thought of the way.

The ball is in your court Alan. 'Fire Away' is not quite what WE need. Please, I ask you politely, to humour me. Confirm that you agree or indeed disagree with the assertions in my post.

I'll be off to bed soon. UK time.

Here are a few more things to ponder. I WILL NOT PROCEED until you confirm that you have read all of my questions and responded with agreement or disagreement with all of them.

Now some more tricky ones.


6. Can we agree that these are just plain and simple ordinary 6 faced dice?

7. Can we agree to refer to 'Two' as being 'deuce' for the avoidance of confusion?

8. Can we agree that from the meaning of the question, that we can reasonably assume that the 'partner' saw both dice? It's implied, but central to the question.

9. Can we agree that as implied by the question, that both dice had come to rest before the partner peeked at the dice? No 'Spinning dice'.

10. Can we agree that if EITHER one dice or the other dice, OR both dice landed as a two, that the partner would ALWAYS say "At least one of the dice is a 2."? Again, it's implied but important.

11. Can we agree that purely for the interests of clarity that one of the dice falls slightly to the right and that one falls slightly to the left as viewed by the 'partner'?
It's not a big sticking point, but we might later choose to refer to the dice by name.


I really hope that we can agree to the above as they are all essential stepping stones. If you disagree with any of the above, please say so now.

My responses to your question are in bold type.

Quote:

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Originally Posted by OnceDear

Alan,
I'll ask again.
Alan,
I'm going to agree with you on one thing, and only on one thing. If you answer a question about one die coming to rest as a deuce while the other spins, then the probability of the spinning die coming to rest as a deuce is indeed 1 in 6. I absolutely concur with you on that. If you take one die and set it aside as a deuce, then the probability of the other die being thrown and turning out to be the pair creating deuce is also 1 in 6. No-one is disputing that.

Do you agree with that? YES, I AGREE

You very firmly believe that your own model of the question is equivalent. You say it is equivalent because it is obvious and because I'm overthinking the problem.

Do you agree with that? YES, I AGREE.

If you believe it is so obviously equivalent or 'the same thing', then you will have absolutely no problems picking holes in my ongoing attempt to answer the ORIGINAL question, regardless of what you believe to be obvious.

Do you agree with THAT? OBVIOUSLY IT APPEARS THAT WE HAVE A DIFFERENCE OF OPINION ABOUT WHAT THE ORIGINAL QUESTION ASKS AND I THINK THIS MAY BE THE REASON WHY YOU AND THE OTHERS SAY 1/11 WHILE I SAY IT'S 1/6.

Do you agree that it is the original question to be answered and proved? YES, I AGREE AND I THINK WHAT YOU ARE GOING TO TELL ME IS THAT I HAVE MISINTERPRETED THE ORIGINAL QUESTION OR THAT YOU HAVE INTERPRETED THE ORIGINAL QUESTION DIFFERENTLY. WELL, GO AHEAD, AND STATE YOUR CASE.

Now, do you want me to show you why the correct PROOF to the original question is 1 in 11 or not. PLEASE GO AHEAD.

Agree to let me proceed or disagree forever.

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Of course I want you to proceed. Thank you.

Again my responses are in bold type.

Quote:

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Originally Posted by OnceDear

Now some more tricky ones.


6. Can we agree that these are just plain and simple ordinary 6 faced dice? YES

7. Can we agree to refer to 'Two' as being 'deuce' for the avoidance of confusion? YES

8. Can we agree that from the meaning of the question, that we can reasonably assume that the 'partner' saw both dice? It's implied, but central to the question. YES, BUT IT DOESN'T MATTER FROM MY POINT OF VIEW. JUST SEEING ONE DIE IS ALL THE INFO I NEED.

9. Can we agree that as implied by the question, that both dice had come to rest before the partner peeked at the dice? No 'Spinning dice'. YES, BOTH DICE CAME TO REST UNDER THE CUP.

10. Can we agree that if EITHER one dice or the other dice, OR both dice landed as a two, that the partner would ALWAYS say "At least one of the dice is a 2."? Again, it's implied but important. ABSOLUTELY. I READ THE ORIGINAL QUESTION TO MEAN AT LEAST ONE DIE CAME TO REST AS A 2.

11. Can we agree that purely for the interests of clarity that one of the dice falls slightly to the right and that one falls slightly to the left as viewed by the 'partner'?
It's not a big sticking point, but we might later choose to refer to the dice by name. SURE, IF IT AS A MATTER OF CONVENIENCE YOU WANT TO REFER TO THE TWO DICE UNDER THE CUP AS LEFT DIE OR RIGHT DIE THAT IS OK WITH ME.


I really hope that we can agree to the above as they are all essential stepping stones. If you disagree with any of the above, please say so now.

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Great Alan.

Thanks for humouring me. Try not to assume what I'm going to say
Just please read, understand, and hopefully agree with each of my assert I will get into more complex logic and maths. I want you to see clear as day that I'm not trying to trip you up or fool you or misguide you.

I'm concentrating only on the question as asked. not my interpretation of it, not the wizards, not even yours.

OK. Now we have so much that we agree about, we need to define a few things. Forgive me, but I'll assume that probability and maths need some definition.


Again, Agree or Disagree as appropriate. I value your opinion. It's a bit of an exercise for me as much as for you.

Consider a single die in isolation
12. When we discuss the probability of A SINGLE DIE landing as a deuce, we work this out as
P(Deuce)=(count of the number of qualifying ways that it might land as a deuce)/(count of all the qualifying ways it might land)

Do you agree with and fully understand that?

13. Thinking about a single die : We can do the real world math for one die as above where . . .

(count of the number of qualifying ways that it might land as a deuce)=1
(count of all the qualifying ways it might land)=6

Therefore Probability of a deuce = 1/6

I use the term 'qualifying' to mean landing properly, in play and as prescribed by the rules of our 'experiment'. Probabilities are all about 'experiments' and 'events'


Do you agree with and fully understand that?

OnceDear regarding your latest questions I fully understand what you are talking about with a single die. But let's stick to the original question in the Wizard's Forum. It doesn't matter what one die shows individually. What matters is what one die shows in conjunction with a second die -- in other words a pair of dice. Go ahead and state your case. I don't see the need for any more Q&A about conditions. I am sure there are others here who would like to see your reasoning and please post it.

Thanks.

Edited to add: by the way, I have a TV show on in 2 hours (KCAL channel 9, 5:30pm) and I might not be able to get back to you till late tonight or tomorrow. Thanks.