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I got into a rather heated exchange over on the Wizard's forum on this 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?
My answer was 1/6 or one out of 6. The Wizard had a different answer and so did just about everyone else.
What's your answer?
Alan: When I first looked at your question I also instinctively thought, "1-in-6 chance".
After reviewing the thread you mentioned, however, I'm convinced it's really a 1/11 chance.
They are right in that it all boils down to whether a specific die is selected or not.
I won't argue the point with you, myself, as it appears Wizard's forum has already done that sort of work.
Well, I think the answer of 1/11 is not correct because the question was not worded properly.
Let me offer this and then tell me your answer:
I throw two dice on a craps table. One die immediately comes to rest showing a 2, but the second die hits the wall and starts spinning like a top. While spinning you ask yourself, what are the chances it will settle showing another two?
Is your answer still 1/11 or is it 1/6 ??
Now, going back to the question on the Wizard's forum: we were also told that one of the two dice shows a 2. There are only two dice involved in this "puzzle." Do you still think the answer is 1/11 or do you also think it's 1/6 ? Does it matter which of the two dice already shows a 2 when there are only two dice to begin with?
In the case of the craps table, the odds become 1/6 for you because you can see exactly which one of the dice has landed with a "2" with definite finality. When you cannot see either of the dice the odds become 1/11. It's exactly why your original problem with both dice hidden was presented that way. Actually seeing the dice and/or not seeing them makes all the difference.Quote:
Originally Posted by Alan Mendelson
Analogy: If I am playing heads-up hold'em with pocket KK, my odds of winning the pot are much greater not seeing my opponent's hand than when I actually see his pocket AA.
EDIT: I have a bad feeling I'm going to enter an argumentative quicksand with Alan if I continue, LOL.
I do totally appreciate the deceptive nature of this problem, though!
You're not reading the question properly. You are told that at least one of the dice is a 2. That's the same thing as seeing the dice rolled on the table. It is not a lie that at least one of the dice is a two according to the original question.
Are you reading the question? Do you see where it says Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."
Because the "partner" truthfully says at least one of the dice is a 2, it is the same thing as rolling two dice on a craps table with one die coming to rest as a 2 and the second die spinning.
By the way, on the Wizard's forum there are now two others who understand that we are now talking about only one die -- the unseen die.
Alan: I respectfully submit to you that, no, the partner telling you that one of the dice is a "2" (even with ironclad truthfulness & integrity), is not the same thing as rolling two dice on a craps table with one die coming to rest on a 2 as the second one still spins....Quote:
Originally Posted by Alan Mendelson
You have more information available to you at that moment on the craps table (because you can see which of the two dice is definitely a "2") than when both dice are hidden from you. For all practical purposes, only one die is "hidden" from you at the craps table (the one still spinning).
I assure you I have read and understand everything about these two different settings you described. Again, we will have to respectfully agree to disagree.
...And, again, I fully appreciate how mind-blowingly deceptive this problem can be!
There's not much more I can say. I hope Dan Druff can stop by soon and provide his input on the matter.
EDIT: If it would help matters any, think of what that one poster said on Wizard's site: Picture the two dice having different colors (such as blue and yellow). You know whether it was the blue die or the yellow die having a "2" at the craps table, yet in the original scenario with your truthful partner you still don't know which color die has the "2".
There is new information that I just discovered and it appears that for the last several years that the information existed on the Wizard's site and no one else saw it. This info makes me wonder if the whole thing was a joke and the forum members fell for it.
Here is the link to the thread:
http://wizardofvegas.com/forum/quest...o-dice-puzzle/
Click on the "spoiler tag" in the original post and read what is hidden in the "spoiler." It makes me wonder if he wrote the question as he did to poke fun at the whole forum.
In the meantime, is it really more complicated than saying this:
One die is a 2, what are the chances that another die is also a 2? Does it really matter if there are only two dice but you don't know which of the two dice is the one with the 2?
Read the original post, the way the question is actually worded, and then click on the "spoiler". It might give you a different idea.
I am amazed that this debate continues. As Alan continually points out, one die was already identified as having landed on 2. Therefore, there is only one remaining die in question which has a 1 in 6 chance of also being a 2.
The question was not "what are the odds of rolling two 2's". The question was "....one die is a 2, what are the odds of the second die also being a 2...". Obviously--1 of 6.
What if his friend had peeked and said both dies are on 2. What would you say the odds are that one of them is a 2? I hope you don't tell me 1 of 11.
Thank you Regnis. It also amazes me that over on the Wizard's forum there is such a lack of reading comprehension. Clearly the question makes it known that at least one die has a 2. It says that in the question, yet they ignore it. Even the Wizard ignored it. What is their problem?
However there is a glimmer of hope. Several others now have started to say the same thing: one die is known. The odds pertain to the second die. And since there are only two dice in the problem, it doesn't matter which is the second die. And believe it or not, those who believe in the 1/11 answer maintain that both dice must still be considered.
It's crazy. There must be a language problem.
I'd suggest putting an end to this with an actual field trial. Could someone grab a friend or partner and have him/her quietly roll dice in the background until a "2" shows up? Just simply repeat the exercise over many trials and tabulate actual results to make sure of this?Quote:
Originally Posted by Alan Mendelson
EDIT: Alan, take those guys' wagers on Wizard's forum. You're being offered 7:1 odds, apparently. If you are correct, it's a huge overlay for you.
I've asked them to take two dice. Set one showing a 2 and then using the other die to count the faces and figure it's 1/6 but that has been deemed wrong because it's a two dice problem not a one die problem. I'm stunned.
That's not what needs to be done at all!Quote:
Originally Posted by Alan Mendelson
Have them do exactly as the original problem says: Have a partner roll both dice in secret. When at least one "2" shows up, s/he truthfully tells you so. Then both dice are revealed. Checkmark "YES" when both are 2's, "NO" when there is only one 2.
Tabulate a bunch of sample trials and see what happens..
Count Room that is NOT what was asked in the original post. And this "test" that you suggest doesn't answer whether or not you just need to look at one dice and if the answer is 1/6. I am sorry, but what you are suggesting is more "static."
Look, everyone who has ever played craps and sees one die comes to rest on a 6 knows that the second die must avoid a 7 so there won't be a 7-out. What every craps player has gone through is a simple application of what answers the question on the forum.
Frankly, I think the math guys on the Wizard's forum just never played craps.
By the way, in Rob Singer's new article he mentions it: http://alanbestbuys.com/id362.html
To answer your suggestion, specifically, Count Room, why don't you do this:
Take two dice. Set one as a 2. (Because we are told in the original post on the Wizard's site that at least one die is a 2.) Now pick up the second die and look at it. As you look at the second die ask yourself: what are the chances that when the second die is viewed it is also showing a 2? As you look at that second die count how many faces it has. Does it have 6 or does it have Eleven?
(I am basically reposting how this thread began for placeholder purposes.)
Alan, your original post in this thread said:
"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?"
You, Rob Singer, and regnis all say 1/6 (when done EXACTLY how it is described above in bold lettering).
I say 1/11 (when done EXACTLY how it is described above in bold lettering).
I do think it's time to have an exact simulation of what is written above so this matter can be laid to rest.
We'll just be blowing hot air and never getting anywhere otherwise...forever in pointless disagreement when it can truly be resolved once and for all.
Also, Frank Scoblete has been notably absent from this discussion...
Count Room and others, I just posted this on the Wizard's forum:
Do any of you play craps?
When the dice are thrown on the table and the first one comes to rest on a 4 don't you say to yourself as you watch the other die "no 3, no 3"?
Of course, you will say that is not relevant to the original post. But let's once again examine the question in the original post:
Quote: Dween
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, you tell me why looking at just ONE DIE would not be the right way to answer THIS QUESTION ??
Let me give you a hint:
THE QUESTION: Dice in a cup are shaken and come to rest under a cup on the table.
REAL CRAPS: Dice are thrown on a table and come to rest.
THE QUESTION: The friend looks and says that at least one of the dice is a 2.
REAL CRAPS: One of the dice is visible and shows a 2. But the second die is behind a stack of player's chips. The stickman calls out "well one of them is a two, Rob (a base dealer) you call it!"
THE QUESTION: What is the probability that both dice are showing a 2?
REAL CRAPS: The die behind the stack of chips has only six sides. The probability is that 1/6 is another 2, and 1/6 is a 5 for a 7-out.
How do you get 1/11 ???
Maybe it would be more obvious with two coins. Same process only this time your partner tells you if there is one head. Does that mean there a 1 in 2 chance that the other coin is a head? No. Look at the 4 possibilities .... HH, HT, TH and TT. Your partner will find one head in the first three cases. However, as you cam clearly see there's only a 1 in 3 chance that both coins are heads. You can lay out all the possibilities with the dice as well and see the same basic logic.
This has to be my final post for a while on this matter. I have other things I need to do!Quote:
Originally Posted by Alan Mendelson
You have a GREEN die and a BLUE die at the craps table.
You shoot the dice, the GREEN die stops at a "2" while the BLUE die keeps spinning.
I agree from this standpoint your odds are 1/6 for the BLUE die to land on "2".
`````````````````````````````````````````````````` ```````
Now the craps dealer puts a screen in front of both the GREEN and BLUE dice.
Another player rolls the dice behind the screen, and the stickman truthfully says, "At least one of these dice is a 2."
The possibilities here with both dice unseen?
GREEN BLUE
2 1 <---MISS
2 2 <---(((((((HIT)))))))))
2 3 <---MISS
2 4 <---MISS
2 5 <---MISS
2 6 <---MISS
1 2 <---MISS
3 2 <---MISS
4 2 <---MISS
5 2 <---MISS
6 2 <---MISS
10 Misses, 1 Hit......
1/11
Does this help at all? I gotta run for now..
No Arc, it's not about coins which is why there is such a bunch of craziness over on the Wizard's forum. They are using the "coin problem" to address the dice question. That is a big mistake.
Count Room you are also changing the original question when you talk about two dice unseen. In the original question the dice have been seen and at least one of them is a 2. Therefore it is the other die that answers the question.
Look what was just posted on the Wizard's forum in response to my question about two dice thrown on a craps table (posted above):
Quote: Dalex64
On the craps table, if you throw two dice, one of them lands on a two where you can see it, and one of them lands out of sight - the odds of the one out of sight being a two is 1/11
And my response:
Are you serious?
Hey buddy, there are only six sides on a die, and the six sides are numbered: 1, 2, 3, 4, 5, 6. That die that's out of sight can be ONE of those SIX numbers.
Get real.
They have really lost it.
With the bad rolls and even worse results lately in craps, I think they have been using 11 sided dice. But if not, the odds of any number on 1 die is 1 of 6.
If you rephrase the question to what are the odds of rolling 2-2, then the answer is not 1 of 6. But if 1 die is already a 2, there is a 1 in 6 chance the other will be a 2.
Plain and simple--not that hard.
Regnis it is so funny how they tried to deny that 1/6 is the proper answer. The most prominent was "but you don't know which dice is showing a 2." Like that is supposed to matter?
I really think the whole thing stems from:
a. they misunderstood the question
b. they got mixed up with the coin problem and used the methodology for solving the coin problem to solve the dice problem
But the Wizard also got it wrong. How? Did he also misread the question?
By the way, it's not over. There are still those claiming that the answer is 1/11 for the craps table example with one die hidden.
And to all the craps players out there who are not members of the forum: thank you for your emails of support. Please join the forum.
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.
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.Quote:
Originally Posted by RS__
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.
Thanks Alan. Do you agree to let me do this my way?Quote:
Originally Posted by Alan Mendelson
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.
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:Quote:
Originally Posted by 1in11
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.
Of course I want you to proceed. Thank you.Quote:
Originally Posted by OnceDear
Again my responses are in bold type.
Quote:
Originally Posted by OnceDear
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.
This.Quote:
Originally Posted by OnceDear
No matter what kind of proof we show Alan, we'll get the same response, something like, "One die is a 2. The other is unknown. The chance of rolling a 2 on that die is 1/6. Simple. Easy. That's it. Nothing else."
I think deep down inside Alan knows the truth to be 1/11, but he's dug such a deep hole for himself, he can't admit 1/11 is the correct answer. I don't know, but I suspect he came to this / his forum because he thought the Rob Singers and others alike would undoubtedly agree with 1/6. Furthermore, I think Alan knows 1/11 to be correct, which is why he's not willing to wager $$$. But, anyone (who gambles) truly believes in 1/6, would have to have a very good reason to deny such a lucrative bet (33% edge on resolved bets, 5.55% edge on a per-roll basis). And I'm not talking about betting peanuts, either (unless you'd like to bet using peanuts).
The exercise with the single die was exactly that. an exercise. We are working step by step to an exercise where YOU will prove that the answer to the original question is 1/11.
You do trust me don't you?
The exercises will get harder.
And before we progress. I will say, for the record, that the 1/11 answer seems totally absurd, ridiculous, stupid.
But the word is 'Seems' We have to really bite the bullet and grind step by step to a proof.
And again, if you trust me, REALLY trust me, I will tell you right now that I KNOW CATEGORICALLY that neither you nor I is misinterpreting the question. That's a boggler, but really.... we are not.
I'm going to disagree with you slightly. I suspect that Alan does realise that it's odd to find so many others cannot see what is so blindingly obvious to him. He might have thought that by posting on his own forum that his friendly locals would all show him how he was right.Quote:
Originally Posted by RS__
One day Alan MIGHT address the FACT that the original question has been badly mis-quoted on the Bob Singer comment page. But that's for later.
The bigger mystery to me is what relevance this has to a forum discussing Las Vegas and gambling.
The whole problem here hinges on one piece of information: which die shows a 2.
If I tell you that at least one die is a 2, can you tell me with absolute certainty which die is showing a 2?
Alan, The Q and A are drawing to a close and would proceed onwards to some simple exercises to build trust in me and to establish all the facts needed for you to prove to yourself that 1/11 is correct. I'm going to refuse to 'state my case' unless you proceed with me. I'm not inclined to persuade your forum readers. I'm inclined to help you to prove it to yourself. We proceed my way or not at all. It's no skin of my nose.Quote:
Originally Posted by Alan Mendelson
Quote:
Do You DARE debate this here on your territory on my terms?
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.
Simple answer. Alan's forum, Alan's choice.Quote:
Originally Posted by redietz
There are many people prepared to offer Alan a wager which he believes give him an astonishing potential profit based on his belief that an event has a 1/6 probability where we know it to have a 1/11 probability. He's being offered odds of between 7:1 and 8:1
He's also being offered PROOF that he would be foolish to take those odds and PROOF that his calculation of the odds is deeply flawed with PROOF as to how and why he is wrong.
So far he has not shown much inclination to listen and even less inclination to take up a fantastic wagering opportunity. Oh hum. He proceeds to let me help him or he doesn't. If anyone else believes in Alan's logical prowess, there are some, and probably many, who are prepared to wager with you on similar terms.
Why is it important which die in particular shows a 2?Quote:
Originally Posted by 1in11
I am not interested in any side wagers and I said that on the Wizard's forum as well. I am interested in having you present your position and please do so.Quote:
Originally Posted by OnceDear
I am interested in your reasoning so I will respond to questions 12 and 13 and again my responses are in bold type:
Quote:
Originally Posted by OnceDear
ok. Thanks Alan, I will continue tomorrow, as time permits.
14. Can you see that Probability of Deuce result from a single die, could be expressed as "We expect that if we throw the single die a lot of times, it will land as a deuce approximately 1 in 6 times"
Do you agree with and fully understand that?
15. Now we need to explore this a tiny bit and get back to the concept of 'A Qualified Event'
E.g. 'If the roll of a single die is an even number, then what is the probability of it being a two?
To calculate that, we would use . . .
Probability = P(DeuceWhenEven) = (count of the number of qualifying ways that it might land as a deuce and be an even number)/(count of all the qualifying ways it might land as an even number)
We could work that out correctly as the 'probability of it being a deuce ( When we know it's even)' = Probability = 1/3
Are we still fully in agreement?
We would NOT work it out and come up with P(DeuceWhenEven) = 1/6, because to do so would be to ignore the qualifier in the question.
Do you agree with that?
Do you agree so far? We covered some ground there.
Once again my responses are in bold type.
EDITED TO ADD: I just had a chance to review questions 14 and 15 and Yes is my answer to those. But I really need you to start putting this into some sort of context so I know where you are headed?Quote:
Originally Posted by OnceDear
I'm out watching the rangers right now, so unfortunately I don't have my computer with me to try to solve this via brute force at the moment.
What I will do when I get home, is load up excel and make a formula that randomly chooses a digit between one and six (inclusive, of course!) I'll fill column a and b with that up to, I dunno, row 46656. I dint actually know if that's enough trials to approximate an answer, but given the huge difference between 1/6 and 1/11, I think it'd be obvious which camp is right. Anyway, columns A and B will represent almost fifty thousand rolls of two dice in a cup.
Column D (because I like spacing in my spreadsheet) will be a formula that returns a value of 1 if EITHER column A or B contains a two, and a value of 0 if NEITHER column contains a two - this represents my trustworthy friend telling me if there is a two under the cup.
Column F will be a formula that returns a 1 if BOTH column A and B contain a two.
Way down at the bottom, in row 46658, column D, I'll put a sum formula that adds up all the values in column D - that represents how many times, in 46656 trials, my friend told me there was a two under the cup.
In row 46658, column F, I'll put a sum formula that adds up all the values in column F - that will represent how many times both dice landed on two.
If I'm not drunk - I am, by the way, so PLEASE correct any logical fallacy I've fallen into - the sum at the bottom of column D should represent how many times my friend and I gave this a go. The sum at the bottom of column F should represent how many of those times BOTH dice ended up as twos. I THINK this is an accurate, simple way to simulate this - anyone disagree?
"Once dear, "RS", and any other wizard form members now here--the clear answer to the OP's scenario/question, is of course 1 in 6. All you math enthusiasts try to do in these type of "problems" is wordsmith your ways around the very simple question just so you can present more of a theoretical event than it really is. Which die is "peaked at" is irrelevant. The true intention of the question is to determine what the odds of the other die being a deuce is after knowing that one of them is a deuce. But being that it was purposely asked in a loose way--very similar to how the Monty Hall problem created such a firestorm--all the theory-heavy mensa aspirationists on wizard's forum, led by him, chose to lay this issue out into a more complicated calculation. Egos are very heavy over there. I saw that when all the geniuses ran away from my challenge a few years ago, then convinced the wizard to permanently ban me so I couldn't make them look foolish any longer.
Guys, think of it this way. If the problem was presented as "two six-sided dice were shaken & rolled, and when they were peaked at we were told that both dice showed a deuce" with the corresponding question being "what are the odds that both dice would show a deuce?"--Those who look at the question as if we're being asked about "right now" would obviously say 100%. That's because we all know that reality is far more important & meaningful than theory. But the ones who choose to overthink the issue and want to create conflict just so they can believe they're the smartest ones in the room, would say 1 in 36. Why? Because to that group they would rather talk about theory than reality, so they spin the words to fit their agenda.
And I understand that. Wizard's, of course, has the very foolish & misleading tag line on each of his posts about how making a "good bet" is more important than whether the bet is won or lost. Again, math people choose theory over reality because it's the safer choice. But it's not surprising seeing that self-proclaimed AP wizard, who gives the THEORETICAL perception that AP's are winners, had to instead beg for donations, then sell out to online casino foreigners. What will it take and when will you people learn that theory only yields phantom bucks? And you can't use those to buy groceries.
I saw some offers for a bet on the 1 in 6 vs. 1 in 11 issue and I'd like to get involved with this. But what exactly are the parameters, and who gets to interpret the OP's wording and /or intention?
Can I pose a different question: Can you tell me why the answer is NOT 1/6 ?? So, instead of trying to tell me WHY it is 1/11 can you tell me why the answer is NOT one out of six? Would that make this process easier?
Synergistic others have already posted answers using columns of possible dice combinations. They can be seen on this forum and on the Wizard's. So before you go to the trouble, please be sure you saw those. Thanks.
Rob,throw two dice. If one two and any other number shows you give me one dollar. If two twos show I give you 7 dollars. If no twos show its a push. Simple, no cup, no third person needed.
Alan, I believe you also need to see him or them explain exactly what their interpretation of the OP's question was as presented to the forum. You, I, regnis and maybe others see it as a very clear "what are the odds, now that we know one of the dice is a deuce, that the other die is also a deuce?"
What does this yield? And the problem only asks about the one die, not both, simply because we are given a value for the first die.Quote:
Originally Posted by quahaug
I'm sorry but I don't see how this answers the question.Quote:
Originally Posted by quahaug
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?
No fair, the whole argument revolves around whether uncertainty about which die comes up two has an effect on the odds - your version is nearly as much a straw man as the proposal to just set one die to two and only roll one die.Quote:
Originally Posted by quahaug
I'd rather make it one thousand rather than one dollar but I don't have a Rob Singer bankroll so I have to start small.
Yes and I said on this forum and on the Wizard's forum that they are not responding to the question, or, they are questioning the question, or, they are applying math to the question that should not be applied to this question.Quote:
Originally Posted by Rob.Singer
I think those of us who play craps deal with this type of question every time we play -- and I said this several times. We see one die come to rest on a number and then we say to ourselves "no ____" for whatever number would be a 7-out, or "give us a ____" for the number that would give us a winner. When we say those things we are well aware that it is a one in six chance.
But now we have folks who are asked a similar question but are answering not with a 1/6 answer but with something more complicated, and now I am asking them to explain why it's not 1/6 and why they say it's 1 out of eleven??
I'd be happy if they just prove to me it's not one out of six.
That's the bet. Take it or leave it hotshot. You got a lot of mouth lets see the cash big boy.Quote:
Originally Posted by Rob.Singer
Hold it. If anyone wants to make a side bet outside of a casino, do it thru a private message. There are still gambling laws and even an office football pool is illegal in California. So please, no gambling or betting here.
Can we stick to the subject? Again, I don't think this bet or any other resolves the question about the answer being 1/11 or 1/6.
I'll take a bet if it makes any sense. I can't even figure out what he's getting at :)
Rob. Have you been told today?
I've been reading more of the responses on the Wizard's forum and clearly this response from "ThatDonGuy" illustrates the problem and why they reject the 1/6 response. I reminded ThatDonGuy what the original question was and he wrote this:
Also, if you are serious about this statement:
"The ORIGINAL QUESTION deals with TWO DICE under a cup with at least one of them showing a 2. YOU MUST address the problem using a pair of dice with at least ONE DIE showing a 2."
(which, by the way, is what pretty much every one of us who have figured out that the answer is 1/11 has already done)
then show us the possible rolls of the two dice, each with equal probability, with at least one die showing a 2.
Aren't the possible rolls 1-2, 2-1, 2-2, 2-3, 2-4, 2-5, 2-6, 3-2, 4-2, 5-2, and 6-2?
Am I missing any?
Do you see what is happening here? They are ignoring that one die already shows a 2 and they are referring to the eleven different combinations of dice that show a two. This is why they insist on the 1/11 answer. If they would accept that ONE DIE is already showing a two, then the obvious answer is that the remaining die has SIX faces ONE of which is a two.
Exactly what I said about them wordsmithing the question to make it more complicated....more theoretical....a problem than it really is. If it were really about keeping both dice in play once the value of one of them was known to be a deuce, the question would have been absolutely clear to that point. But by leaving it loosely stated to these guys but clear to rational thinking people, the mensa-laced math theory door was opened up.
Rob, you should have just kept your mouth shut a little bit and lured them into a real-life wager with one die under each of two cups. Alan could have filmed the action, and we all could have thrown a party with the proceeds.
The wizard's bunch is indeed a gullible bunch. It doesn't take them much to be lured into making them whine to the admins. to get you suspended. So you're right.
Lets just put one die on 2, roll the other one, and take all the action they want, using their odds, on the second one being a 2.
wonder what they think the odds of neither being a 2 are, even though 1 already is a 2. They are calculating the odds of the second 2 as if there weren't a first 2.
They've already proposed such a bet.Quote:
Originally Posted by regnis
I accept that one of the dice is showing a two, but I do not know which die shows a two. This is the distinction here.Quote:
Originally Posted by Alan Mendelson
If I have the information of which die shows a two, and then you ask what is the chance that they both show a two, then the answer is clearly 1 in 6. But, I don't have that information. I only have the information that at least one of the dice shows a two.
That's fine 1in11, but it doesn't matter which die shows a deuce as long as we know one of them does. And once we know that info, that deuce-showing die is off the table. Please explain why you believe that's an important issue.
Knowing which die shows a 2 gives us 6 possibilities for the combination of the two dice: 2-1, 2-2, 2-3, 2-4, 2-5, 2-6.
Knowing only that at least 1 die shows a 2 gives us 11 possibilities for the combination of the two dice: 2-1, 2-2, 2-3, 2-4, 2-5, 2-6, 1-2, 3-2, 4-2, 5-2, 6-2.
This is why it is an important issue, as they are two fundamentally different conditions.
Rob, this is actually the problem with the guys on the Wizard's forum. Another just posted this:Quote:
Originally Posted by Rob.Singer
"Alan's group doesn't know the difference between a "certain" die being a 2 and a "random" die being a 2."
Somehow they can't understand that there are only two dice to start with, and it doesn't matter which of the two dice is showing the two. At this point I have to wonder why the Wizard himself hasn't commented -- even to say we don't know how to read and the 1/11 crowd does know how to read.
I am sorry but you need to look at two dice. When one die is a two -- there are only six possibilities on the other die -- a one, a two, a three, a four, a five or a six. What are you guys smoking?Quote:
Originally Posted by 1in11
These ain't Schrodinger's dice.
You are correct that there are only six possibilities for the other die. However, these don't happen with equal probability, given the constraint of the problem.
2-1, 2-3, 2-4, 2-5, 2-6 all happen with 2/11 probability, and 2-2 happens with 1/11 probability.