Try it. You can be peeker and tosser.Quote:
Originally Posted by regnis
Remember now. Rob Singer is a supporter of Alan's, But even he says this... And even he asserts that this bet is a 1/11 proposition.
Quote:
Originally Posted by Rob.Singer
Printable View
Try it. You can be peeker and tosser.Quote:
Originally Posted by regnis
Remember now. Rob Singer is a supporter of Alan's, But even he says this... And even he asserts that this bet is a 1/11 proposition.
Quote:
Originally Posted by Rob.Singer
I have three questions for both the 1/6 camp and the 1/11 camp. What percent of an audience of general readers do you think would arrive at 1/6 after reading the original question as written? Not after 50 posts and commentaries and coaching, but after reading the original question. What percent of a general audience do you think would arrive at 1/11 after reading the original question as written? Do you think that a jury of your peers in a US courtroom would vote that the 1/11ers have proved their case beyond a reasonable doubt?
This is not an equation. It is a piece of writing. This is all about the writing, not the reading.
I'm not the least bit bit afraid. Here's that word least again, lol.Quote:
Originally Posted by Alan Mendelson
Those guys are sharks when it comes to semantics. With them, it's not a matter of agreeing with each other, whichever camp, so much as with the correct answer. And, they go at it until it is unanimously resolved however long it takes.
BTW, the correct answer is almost invariably more complex and complicated (to the remainder of us) then the wrong answers.
Not putting percentages to it. The original post was on a maths subforum of a gambling website. Of the readers there, there was a mix of opinions. However after an absolute wealth of explanation, only about three members on wov still believe the correct answer, after clarifying the peeking rules, would still assert 1/6. I pity a nation where Alan could sit on a jury. Even amongst his friends here he is not listening at all to any proofs or arguments.Quote:
Originally Posted by redietz
I believe you are quite wrong here.Quote:
Originally Posted by Alan Mendelson
The real problem for you is you think the people able to perceive 1/11 are deluded theorists and not "grounded" like yourself.
What the situation actually is, is that you are unable to see further than your nose and the people you're accusing in being theorists are just the ones who have been there where you are now long time ago and see no real benefit to go down to your level and explain things like to a child, because they can see it all from a higher perspective than you and see exactly what your thought process is.
They just see the things clearly and can shift through the levels as they like if they want. You can't.
I'm no theorists nor mathematician by far, but was able to understand "the grounds" of the problem.
Its a shame people like you would fire someone just for being up the level you are.
I've read their arguments on the WOV forum over semantics and they get ridiculous making up their own definitions sometimes.
Had I been with the Wizard when he shot that video and he started to rotate that die with the two I would have stopped him right there.
Kewl in the real physical world you cannot rotate that die showing a 2. Deal with it.
You are not allowed to remove one die. You only know "at least one of the dice" is 2. That makes it possible for any die to be 2. If you go ahead and remove one die , you don't see the problem in full.Quote:
Originally Posted by regnis
The example with the cards is EXACTLY the same as the original question.
You draw a card and the peeker tells you it is 2-x. Period.
Put some percentages on it. Where the original post appeared is irrelevant. I could argue, after all, the other way -- that the fact it appeared on a math subforum means the readers there are lousy writers and don't know anything about effective clear writing.Quote:
Originally Posted by OnceDear
So let's put those alleged IQs to work. What do you think the percentages are for a general audience arriving at 1/6? What are the percentages for arriving at 1/11?
The key element to this question, partly because it's a piece of writing and not an equation, is how and where it guides an audience. So where do you think it guides what percentage of a general audience?
You don't have to. The only thing you need to comprehend is the experiment begins with the throw of pair of dice. All consecutive probabilities stem from there.Quote:
Originally Posted by Alan Mendelson
Not with setting one die and now throwing a second die. Where do you guys get this from?
Try to deliver your logic while holding pair of dice before the throw. Not holding one die. If the question was what are the odds for a single die it would say so. it would read: Look, you have one die and the other is not of interest.
Any die is a 2--but one of them is, and it can't change its spots. Sheesh---the mysterious chameleon dice again. Where can I get me a pair of those.
That's why I tried to make it easy for you and I only used 1 die. You refuse to address my one die example.
Surprisingly, in my opinion, the ones who think about this the least (for whichever reasons) or the most will come to a 1/6 chance answer.Quote:
Originally Posted by redietz
"At least one die is a 2" has more of a one-die and consecutive nature or flavor to it than does "one or the other die is a 2". If the first die of a specific roll under the requirement of "at least one die is a 2" is a 2, then the requirement has already been met. If it's not a 2, then ""one or the other die is a 2" would be the requirement required to justify looking at the second die.
As well, we could get into how restrictive the possible or's are. The unannounced, general case of no restrictions when the "peeker" just looks at the first die, or randomly chooses one of the dice to look at. The least restrictive case when the "peeker" looks at both, and announces the smaller or larger dice-number, and discloses his own variation. And, the most restrictive case when a dice-number is chosen in advance, and the "peeker" follows suit.
If you rush to judgment as the Wizard did, or have a broader but still simpler notion of probability questions as seen or heard of before, then you are more likely to arrive at the 1/11 chance answer.Quote:
Originally Posted by redietz
If you go online and look for other instances or form of this exercise, you'll see the same rush to judgement by the same sorts of persons. A lot of the math fellas forget the general case of 1/6 chance, and fail to specify their dice number from the outset with the 1/11 chance answer. I guess this is what Allen might be afraid of with the math guys; but I would point out the other things brought up here and there. Many of the internet experts don't take the time to exhaust every little thing brought up. Scientific specialists in the field who do the actual research and reporting of results in the journals do but tend to stay where they are.
With the 1/11-ers, it's that Mark Twain quote over and over again which I noted earlier. Makes the Wizard's forum so very boring after a while. Nobody is getting rich at blackjack; or having any fun bickering with other players and the casinos. Old news. Yet seemingly idiotic experts who sell useless gambling books and offer advice about online casinos try to dig up some other way, like Markov chains, to relate one random distribution to another, to exactly determine how one might go ahead by 300 units while flat betting. The rest is all canned stuff. Nobody with half a brain and a life is ever going to sit there at a blackjack table year after year to wait for any stretch of wins to amount to 300 units while flat betting. Secondly, no one is going to actually write down and reduce those specific Markov chain probabilities. The real experts care only that it can be done in the theory. The suckers off the streets want rationales they don't want to understand to convince themselves that they can keep losing ridiculously but still come out ahead.
Of course not. There are so many untraveled avenues even with the usual number of dead ends.Quote:
Originally Posted by redietz
In my opinion, court procedure and administration is crazier than even poker. Math is a game of checkers. So, where do the gambling forums fit in?
I have noticed that gamblers are concerned with distinctly but simplistically defined notions of now, and past/future. So too, their forums. Everything has to be a quick and easy fix or "road to riches". Insults are poorly implied and inferred, and topped off with the mentality of "nuking" or parading offenders. Here, a bit of sensationalism with Alan, who is a well-known celebrity in the area.
Gambling can be a boring yet ever more addicting grind of a life, both mentally and physically. To have to come somewhere anonymously and spend also the remainder of one's time then talking about it is a whole new level. So many things to talk about in a more productive manner in the forums, even about gambling, yet so few are barely broached. So many PhD studies, yet no such PhD's on the gambling forums.
This is the only forum I've come across where no one seems to be locked into the gambling thing. And, few seem "to have something to prove".
formal logicQuote:
Originally Posted by redietz
n
1. (Logic) Also called: symbolic logic the study of systems of deductive argument in which symbols are used to represent precisely defined categories of expressions. Compare philosophical logic
2. (Logic) a specific formal system that can be interpreted as representing a fragment of natural argument
http://www.thefreedictionary.com/formal+logic
Okay…Quote:
Originally Posted by regnis
Yes, 1/6 and 1/1 are the correct answers to those questions. Now try this one…Quote:
Originally Posted by regnis
I roll two dice until at least one of them is a 2.
What were the odds that it was going to be 2-2?
What are the odds that it is 2-2?
It is all about understanding plain simple conditional probability (and knowing English helps as well).
This is not the original question that started this whole mess. If you can't see this there is no use talking any further.
The original question referred to one roll where a 2 was showing on 1 die. It made no mention of numerous rolls being made until there was a 2.
to answer your first question: 1/36Quote:
Originally Posted by Zedd
to answer your second question: 1/6
Read the very first post on this thread. It was agreed that the dice are rolled until there were a 2.Quote:
Originally Posted by regnis
CorrectQuote:
Originally Posted by Alan Mendelson
WrongQuote:
Originally Posted by Alan Mendelson
Actually, it's not wrong. It's the same thing we've been arguing about. One die is known to be a two, and the question is what is showing on the six sides of the second die.
Indeed, it is the same thing we’ve been arguing about (unlike the questions that regnis proposed).Quote:
Originally Posted by Alan Mendelson
So regnis (and others)…can you answer it?
Quote:
I roll two dice until at least one of them is a 2.
What were the odds that it was going to be 2-2?
What are the odds that it is 2-2?
I am only interested at this point in Redietz' question about the odds after Arci looks down the waitress' dress (and I don't even know if that was part of this thread anymore). The rest of this discussion has been exhausted.
I did address it, I believe.Quote:
Originally Posted by regnis
You roll one die - the odds of it to stop on 2 is 1/6. When it stops at 2 it is a 2 for eternity.
Now, add another die. Same thing - the odds of it to stop on 2 is 1/6. When it stops at 2 it is a 2 for eternity.
Now, what is the probability (see, we have combined probability now) of pair of dice to come 2-x?
And what is the probability to come 2-2?
If you perceive this as if a single roll is made and it happens to be 2-x, then the answer would be 1/6. But that's not the case. The case is it has to be a 2-x.Quote:
Originally Posted by regnis
Because if the roll happens to be a 6-4 for example, there wouldn't be any question to be asked and no answer. This was cleared long ago for the purpose of the old discussion and here on the first page and in many,many answers on WoV and here as well.
Ok---my final post on this subject. I roll a 2. What are the odds it becomes a 6 so that we can satisfy the 1-11ers?
I set my die on 2 one thousand times and it hasn't changed even once. Is my sample size too small??
Why would you (and Alan) expect a die to change it's value? Don't get ridiculous now.Quote:
Originally Posted by regnis
Instead, you need to only comprehend that you are dealing with two dice and either one of them can become a two. That doesn't give the ability to a die to change it's spots, don't get silly. Instead it makes it twice as possible to get a 2 (and twice as possible to miss a 2 because its "on the other die" lol), than with one die only (the error you all make when concentrate on one die and forget that there were two dice to begin with).
Kewl there were two dice at the start of the problem. At least one of them is a 2. We can subtract one of two dice from consideration. The remaining die has six faces. Consider this:
2(6 faces) - 1(6 faces) = 1(6 faces) or 1/6
Why can we remove one die? Because we are told at least one die shows a 2. Does it matter which of two dice we remove? No because either die would not effect the answer or the question.
That is what you do when you fail to freeze at least one die as a 2.Quote:
Originally Posted by kewl
And you also can't rotate the non-2 in the real physical world. But if I remember correctly, this is what you did in your video and is your "proof".Quote:
Originally Posted by Alan Mendelson
Failure in comprehension. What we do is give you a reminder there are two dice to consider and not one. You can't come after the throw and eliminate anything. That's the outcome now. You don't mess with the outcome. You calculate probability. And it's done before the roll.Quote:
Originally Posted by Alan Mendelson
Yes I did. To show that it had six faces giving us 1/6 for the second die. The die showing a 2 remained a 2.Quote:
Originally Posted by RS__
Obviously you must alter reality. Thanks for admitting that's what you do.Quote:
Originally Posted by kewl
Quote:
Originally Posted by regnis
Aww…I had answered your questions after you cried about people refusing to do so. Yet, you're refusing to answer my questions.Quote:
Originally Posted by regnis
Huh? What?Quote:
Originally Posted by Alan Mendelson
Give 'em heck, Alan.Quote:
Originally Posted by Alan Mendelson
Kewl you wrote "it's done before the roll." That alters reality.
The original problem specifically states that two dice were rolled and at least one is showing a 2. Whatever you thought prior to that no longer matters. Now you have to deal with at least one die showing a two. The question now centers on the second die in the problem which may or may not be a two and the chances for that second die to be a 2 is 1/6.
By figuring probability before the roll alters the reality of the question. You are changing the facts and the conditions of the problem.
As I have said a hundred times, your 1/11 is "good math" but for a different question -- perhaps for a question that asks: "how many different combinations of two dice have at least one 2 and of those combinations how many would show 2-2?" That would be your 1/11 answer.
The only way to resolve this problem is to actually have Alan bet some of his money on the 1/6 odds.
That won't resolve anything because anything can happen when two dice are rolled. I know that and you know that. In fact, since the bet with the Wizard that we decided on has a very limited number of rolls -- and only a $50 lunch was at stake -- I've already conceded defeat.Quote:
Originally Posted by arcimede$
More importantly, can you show the proof of 1/11 using two dice, with the "2" being fixed?
There is no "fixed" 2 in the original question. Are you trying to change the question to match your answer?Quote:
Originally Posted by Alan Mendelson
It's very simple. We could do it over skype.
- I roll two dice in a cup and peek under and tell you whether at least one 2 is present.
- If so, the bet is on.
- When uncovered if 2-2 shows, I owe you $80.
- When uncovered if 2-x shows (where x is a non 2), you own me $10.
- We run the test for 100 bets. That is 16 cycles of your 1/6 odds. The result should be very close to statistical reality. If you want to do more rolls that is fine with me. the more the merrier.
- When all is done the loser writes a check to the winner.
Arc I have something that won't cost you any money or much time. Why don't you make a video using two dice -- with at least one die coming to rest as a 2 and without changing that die showing a 2 show me how your answer is 1/11?
You can do it yourself. Set up your camera over a table.Quote:
Originally Posted by Alan Mendelson
- Throw a pair of dice
- Every time at least one of them is a 2 put a mark on a piece of paper (that meets your first criteria ... "with at least one die coming to rest as a 2)"
- Every time both of them are a 2 put a mark in a another column. Never is any die changed after coming to rest.
Tally your columns after 100 throws when a 2 is present and show us the video..
Is your reading comprehension that bad Arc? It appears so. Go ahead shoot the video with one die fixed as a two and show me your proof that there are eleven other possible faces of which 2-2 appears 1/11 times. Go ahead and do it. Yeah, do it just like the Wizard did it.Quote:
Originally Posted by arcimede$
This thread has been brought to the attention of users at http://math.stackexchange.com/ That being said, I am a user there and have come to give an argument in favor of the "1-11'ers." I have multiple degrees in mathematics and my field is in combinatorics. This is precisely the type of question we would ask students in an undergraduate class involving basic probability.
The scenario (as I understand): You have two dice. You roll both dice and someone checks to see if there are any twos. If there are *no* twos, then the person verifying reveals the results of the dice and has the dice cast again.
Eventually the verifier will truthfully state "There is at least one two" at which point you ask the question of "What is the probability that *both* dice are two?"
Let us reimagine this scenario somewhat. To keep track of what is going on a bit more easily, let us imagine that the two dice are *different colors*, one red, and one green. Clearly, adding or removing color to the mix will not in any way change the probability of the situation.
As the dice are *fair*, every result is equally likely. We have the following results:
1-1. 1-2, 1-3, ...
2-1. 2-2, 2-3, ...
...
For a total of 36 equally probable scenarios. As per usual, our definition of probability in an equiprobable setting (such as this one) is Pr(A) = |A|/|S|, where A is the event in question, |A| represents the size of the event in question, and |S| represents the size of the sample space.
This question however is one about "conditional probability," written mathematically as Pr(A|B) "the probability of event A occuring given that we know that B occured" or more simply as "probability of A given B."
Let A represent the event that BOTH dice show a 2. Let B represent the event that "AT LEAST ONE" die shows a 2. Let us go one step further and also let C represent the event that the red die shows a 2, to explain where your answer comes from.
The question being asked is: "What is the probability that both dice show 2 given that at least one die (be it the red one *or* the green one) is confirmed to be a two", i.e. Pr(A|B)=?
The *definition* of conditional probability states that Pr(A|B) = Pr(A and B)/Pr(B). which, since we are in an equiprobable setting can be simplified further to be = (|A and B|/|S|)/(|B|/|S|) = |A and B|/|B|.
We now try to calculate |A and B| and also calculate |B|. How many outcomes have it such that "both dice are 2 AND at least one die shows a 2", the only outcome is 22, so |A and B|=1, and the probability Pr(A and B) = 1/36. Calculating the size of |B|, we look at all outcomes where there is at least one two. Referring to our chart above, we see that occurs when either the red die is 2 or when the green die is 2 (or both), for a total of 11 outcomes. (specifically 2-1. 2-2, 2-3, 2-4. 2-5, 2-6, 1-2. 3-2, 4-2, 5-2, 6-2). Remember: Each of these events is equally likely to be the one that has been tossed.
Thus, we calculate Pr(A|B) = |A and B|/|B| = 1/11.
----
Why are you convinced that the probability is instead 1/6? That is explainable as well. You seem to be confused as to the nature of how the dice are acting. You think "The *FIRST* die is a 2, the second die could either be a 2 or not", however how do you define the "first die." With colors on the dice, it makes it easier to see whats going on.
You seem to have mistaken the situation as: You have two dice. You roll both dice and someone checks to see if THE RED DIE IS A TWO. If the red die is not a two, then the person verifying reveals the results of the dice and has the dice cast again.
Eventually the verifier will truthfully state "The red die is a two" at which point you ask the question of "What is the probability that *both* dice are two?" I.e. Pr(A|C)=?
We calculate this the same way as before, noting that |A and C| = 1, as before, and |C| = 6. You get then that Pr(A|C) = Pr(A and C)/Pr(C) = (1/36)/(1/6) = 6/36 = 1/6.
Yes, of course the results of the dice themselves are independent, however an event which uses information about both dice simultaneously ("at least one die is a two") is no longer independent of the results of the dice.
Jihkro thanks for joining and for posting. Unfortunately everything you wrote above has been discussed before.
Our difference is over the wording of the question.
Those of us who say the answer is 1/6 believe that the question simulates the following:
Two dice are thrown on a craps table and one comes to rest on the 2 face. The second die spins like a top. What are the chances that the "spinner" will land on a 2?
We -- the 1/6ers -- see nothing to indicate that given two dice with one frozen to show a 2, could possibly result in 11 faces to be considered.
If you are clearly able to distinguish between the "spinner" and the not spinner, then yes the probability will be 1/6. However, how could you have "chosen" one to be frozen and one to be a "spinner". What if the "frozen" die is not a two and the "spinner" eventually rests on a two? Why would the verifier not have stated "there is at least one two" in that situation. It is as I said in my previous post, you have somehow arbitrarily changed the wording of the question to be "Given that the *red* die lands on a two, what is the probability that both are a two", or "Given that the leftmost die lands on a two, what is the probability that both are a two", or in your wording "Given that only one die is stationary at the time and is a two, what is the probability that after the second comes to rest that both are a two". In each of these alternate wordings, there is some clear distinction being made between the dice, and you are ignoring the possibility that the second die could be a two with the first one not, and still be considered as a scenario where the verifier truthfully states "at least one die is a two."
In the interpretation the 1/11'ers are using, which appears to be the original wording of the question, is that "both dice are tossed" (and both come to rest), and the verifier sees the results of both dice before stating "at least one die is a two."
The two are very different scenarios, and have different probabilities (as explained in my previous post).
In which case there is no bet.Quote:
Originally Posted by Jihkro
There's still nothing in the question about one die "fixed as a 2". All we know is at least one die is a 2. No more, no less. Why do you keep trying to change the question? Why don't you want to test the situation and see what happens? I provided you a simple way to test it. If you want to prove all of us 1/11's wrong, do what I proposed.Quote:
Originally Posted by Alan Mendelson
I have a feeling Alan now knows he is wrong and simply won't admit it.
For the millionth time, the answer to that question is indeed 1/6. No one is arguing against that. Jihkro even explained that in his post (by the way, thanks for jumping in Jihkro).Quote:
Originally Posted by Alan Mendelson
Alan (and other 1/6ers)…you are failing to realize that the ‘spinner’ question is different from the bet you had described on the very first post of this very thread.
Not true.Quote:
Originally Posted by Alan Mendelson
To quote the first post…
So if the frozen die is not a deuce and the ‘spinner’ eventually rests on a duece…that DOES meet the condition of “at least one deuce”. In which case, the bet WOULD be on.Quote:
Originally Posted by Alan Mendelson
Whoa Zedd you are mistaken.
If the die that came to a rest is not a 2 and there is still a spinner, the "peeker" would not see a 2.
You see -- this is all about reading comprehension and understanding the question.
The original question IS the spinner situation at a craps table. We know that at least one die is a 2 and the question is the 6 faces on the other die which either is spinning or is yet to be revealed. That's why it's 1/6. The question gives that information.
Ironic how Alan claims others have poor reading comprehension....yet he's discussing something about a die being frozen/set to a 2.
Because we are told there is a die showing a 2.Quote:
Originally Posted by RS__
Then the peeker will wait until the spinner has come to a rest. It was agreed that the peeker would be able to see both dice before making the “at least one deuce” announcement.Quote:
Originally Posted by Alan Mendelson
I'm at a loss as to physically how one die could be "spinning" in the scenario described. Shaken and Slammed. Do dice spin after being slammed? Are both dice at rest normally after a cup is slammed opening down? In the realm of reasonable physics, even if in the split second between slamming the cup and lifting the cup to peek, one (or both) is spinning, if one comes to a stop only milliseconds sooner than the other, could the verifier's mind recognize which came to rest first? How would he decide which was the "spinner" in that case?Quote:
Originally Posted by Alan Mendelson
The sentence "The one that came to rest first is a two" is the same as "The red die is a two" is the same as "The die furthest to the left is a two" is the same as "The die we rolled (versus the die we spun) is a two" is the same as "The die in vancouver (versus the die in Bangkok) is a two" is the same as...
All of these are different than the following equivalent list:
"There is at least one two (be it the red or the green)" is the same as "There is at least one two (be it the stationary or the spinner's eventual result)" is the same as "There is at least one two (with both dice stationary at the time of announcing the fact)" is the same as "I slammed dice on the table and looked and saw at least one two."
What happens if the spinner has stopped spinning after he has made his announcement but before lifting the cup and the spinner has come to rest on a two? He will have lifted the cup and the audience/betters see that there is a two and a something else. To the audience's eyes, they won't know that the spinner was the two, and they will then conclude that the person was lying.Quote:
Originally Posted by Alan Mendelson
You're missing the point. The spinner situation is an analogy for the question. It is not the actual question.Quote:
Originally Posted by Zedd
In fact, if this happened at an actual craps table the peeker would wait for the spinner to come to a rest. And then if the spinner came to a rest showing a 2 what then? Well, the question would then center on where the first die came to a rest -- and that would be 1/6, wouldn't it?
Of course it would be -- you said that yourself. When one die is known as a 2 the situation rests with the other die's six faces.
This is why those of us who say the answer is 1/6 just can't figure out how you guys continue to say the answer is 1/11 ??
Read above. The spinner is an analogy.Quote:
Originally Posted by Jihkro
Read above yourself. You chose to focus on a single part of my statement to which you can respond (rather poorly I might add), and have not responded to any of the rest of the questions or comments I made in my latest post. You continue to work in the situation that the dice are somehow distinguishable (either by color, by spinning status, by location, or by time at which they stop spinning) and declare that the peeker will choose to completely ignore the result of one die (even if it is a two) and only announce the result of the other die.
Again, "What happens if the spinner has stopped spinning after he has made his announcement but before lifting the cup and the spinner has come to rest on a two? He will have lifted the cup and the audience/betters see that there is a two and a something else. To the audience's eyes, they won't know that the spinner was the two, and they will then conclude that the person was lying." There *is* a 1/6 chance that if the peeker was cheating by ignoring the result of the die that stopped second (or equivalent wording) that he will be caught in the act. If I were officiating the proceedings, I would conclude that the peeker was either paid off by the 1/6'er to increase his/her odds by giving an unfair advantage via additional information, or that he was paid off by the 1/11'er in order to make it appear that way. In either case, I would render all previous and future bets cancelled and all money returned until such a time as we could find a fair judge.
"When one (specific) die is known as a two" then yes, the probability is 1/6.
"When *at least one die* is known as a two (nonspecific as to *which*)" then the probability is 1/11.
Toasting in epic bread and feeding the trolls. I don't really care if I'm feeding the trolls. I'm just curious how you refute those statements. Oh, right, by vague responses "its an analogy" and by not directly responding to any specific logical statements.
At the very least respond to the question "Do you understand that if both dice are tossed and both dice are viewed and if the total number of twos shown is either 1 or 2 (regardless of which die shows the two if only one) that the probability is 1/11?" completely disregarding the scenario described in the analogy as having a "spinner."
Whether or not you choose not to respond, I recommend you read http://en.wikipedia.org/wiki/Boy_or_Girl_paradox as this problem is exactly the same as that one. In the boy/girl paradox the full statement reads:
"Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls?" The answer to this question is 1/2
"Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?" The answer to this question is 1/3
The question you think is being asked is equivalent to the first, "the older child is a girl" is like saying "the stationary die is a two"
The question that is actually being asked is equivalent to the second, "at least one of (disregarding whether it is the older or younger or both) the children is a boy"
Included in the wiki article (here is the link again: http://en.wikipedia.org/wiki/Boy_or_Girl_paradox please read it) is a full explanation of both the ambiguity and the solution, as well as references to textbooks, research studies, and several more similar examples.
Jihkro I really don't care to read your stuff. We've heard it all before. Just show me a video with two dice: one showing a 2 and tell me how you arrived at 1/11 without touching the die showing a 2.
thanks for stopping by.
Here is the assisting graphic that you're needing.Quote:
Originally Posted by Alan Mendelson
OneHitWonder can provide you an excellent explanation.
http://www.imageurlhost.com/images/i...6sh2no9dvr.jpg
Alan, you seem dug in on the idea that if the probability is 1/11, then someone should be able to show eleven equally likely outcomes using a single die. And no one can... but this is a red herring. Once you're told that one of the dice is a 2, the six possible outcomes of a single die are no longer equally likely.
I can understand Alan's intuition and common sense here. After all, the information you're being given is just that there's at least one 2. That's the exact *same* information you'd have if the 2 were set aside in the first place and only one die were rolled under the cup. So how can the probabilities be different in the two situations?? But this (like the Monty Hall problem) is one of those situations where what plain common sense tells you is wrong, and you need to do some work to understand how.
Here are a sequence of simple statements that are all true, and where each one (to me) follows pretty directly from the previous one.
- You roll two dice in plain view. Getting an 11 is twice as likely as getting a 12.
- Your friend rolls two dice under a cup. It is twice as likely that there's an 11 under the cup as a 12.
- Your always-honest friend rolls two dice under a cup and peeks. You ask, "Did you get at least 11?" He says, "yes". Only 11 and 12 are now possible, and it's still twice as likely that there's an 11 under the cup as a 12.
- He rolls and peeks. You ask, "Is it true that one of the dice is a 6 and the other is at least a 5?" He says, "yes". Only 11 (5&6) and 12 (6&6) are now possible; 11 is still twice as likely as 12. (So the probability it's a 12 is 1/3.)
- Same as the previous story, except you then say, "Show me a 6." He moves the cup to reveal a 6. The remaining die must be either a 5 (if the roll was 11) or a 6 (if the roll was 12), but these aren't equally likely: 5 is twice as likely as 6, and the probability there's a 6 under the cup is 1/3.
- He rolls and peeks. You say, "If it's true that one of the dice is a 2, show me a 2." He moves the cup to reveal a 2. The remaining die can be 1-6, but these aren't equally likely: 1 and 3-6 are each twice as likely as 2, so the probability there's a 2 under the cup is 1/11.
If you accept the first statement and reject the last, then there's some transition in there that you don't accept. Which one, and why? To me, the fourth statement is okay, and the fifth is counterintuitive... but put side by side, they are clearly equivalent statements. That particular transition helps me get past the (erroneous) common-sense feeling that the *information* should be enough to determine the *probabilities*. Maybe you get tripped up somewhere else.
This is pretty good Alan.Quote:
Originally Posted by Alan Mendelson
The only change I would make is to substitute "how many different combinations of two dice involve the number 2 and of those..." for "how many different combinations of two dice have at least one 2 and of those...". Exactly one 2 is (an element) of at least one 2 - which is the same as one 2 or two 2's - but one 2 isn't (defined, or normally referred to, as) at least one 2. {One 2} isn't {one 2 or two 2's}. Those are two different sets.
Alan, you did a good thing by pointing out that this problem isn't written for the 1/11 chance answer in theory. The original problem supplies only a specific roll, or rolls of a specific nature; and nothing has been restricted about the number 2 beyond the "peeker's" specific observation.
To view also a specific roll theoretically - in the probability or "how often" sense (given that the combinations of two dice involve the number 2) - ask how often will the left die involve a 2. Half the time; the other half of the time the right die will involve a 2. Hence, half the time the other die will show a 2 given the number 2 somewhere. When it's the left die, the other die will show a 2 with 1/6 chance; and, when it's the right die, the other die will show a 2 with 1/6 chance. Now, we put this into a calculation as did the 1/11 chance answerers. There are two parts to this calculation:
(1/2 X 1/6) + (1/2 X 1/6) = 1/12 + 1/12 = 1/6.
This is the specific roll or rolls theoretical counterpart to the 1/11 chance answer. The way to perform the calculation if going by a specific roll or rolls in theory. All specific rolls considered, in theory, in terms of which side the roll or rolls involve the number 2.
This calculation may be done also within the 1/11 chance calculation:
(5/11 X 1/6) + (5/11 X 1/6) + {(1/22 X 1/6) + (1/22 X 1/6)} = 1/6.
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Note that the only way to arrive at the 1/6 chance answer in and of itself for a specific roll or rolls is to find a 2 on the only die looked at. This is the exact manner in which the original problem is written.
Again, exactly one 2 is (an element) of at least one 2 - which is the same as one 2 or two 2's - but one 2 isn't (defined, or normally referred to, as) at least one 2. {One 2} isn't {one 2 or two 2's}. Those are two different sets. The reason to require "one die or the other is a 2" - the same as involving the number 2 - instead of "at least one two". "One die or the other is a 2" would allow for the specific roll or rolls to read the same as the general roll or rolls, though waiting on a 2 would still be a requirement in advance. This is not the original problem as written.
One more thing... one might argue that you *do* have extra information in the case where a 2 is set aside ("frozen") from the start: namely, you know *which* die is a guaranteed 2. Perhaps that extra information changes the probability to 1/6. This is sort of true, and bears on the business with spinners and so on; but it's also vague and a little misleading, in that it reinforces the misconception that "same information = same probabilities". Let there be a red die and a green die, and tell your friend to show you a 2 if there is one (with no color bias), and let your friend reveal a red 2. You now have *exactly* the same definitive information ("the red die is a 2") as you would if the red die were set aside as a 2 and only the green die were rolled. And yet (since the addition of the colors is purely cosmetic), the probability that the green die is a 2, in this situation, remains 1/11.
I find this scenario, if anything, even more counterintuitive than the original one. The assumption of no color bias is key here. If your friend shows you the green whenever he can, the probability that the green die is a 2 given that he showed you the red die is *zero*. If he shows you the red die whenever he can, then the probability that the green die is a 2 given that he showed you the red die is 1/6. This extra complexity is swept under the rug... and the no-bias condition is enforced... when the dice are indistinguishable.
No, one of either the 11 or the 12 is no longer possible.Quote:
Originally Posted by onyx99
The "peeker" saw either the 11 or the 12. To be honest, if the "peeker" saw the 11, then he can not say or the 12. To be honest, if the "peeker" saw the 12, then he can not say or the 11.
He may say that what he saw is an element of the set defined by at least the 11. But, he may not honestly say that the specific roll or element of the set defined by at least the the 11 which he saw is that set. Can't roll or get a set of rolls on one roll. Unless the definition of the set is specially worded to work for both the general and the specific.
You will have to spend a few days thinking about this, and elaborating more carefully.Quote:
Originally Posted by onyx99
It's not fair to the rest of us who have already put in the time on this to have to straighten all of that out.
Thanks.
I don't mean to be rude, but is English your native language? The definition of the set is, in fact, worded to work for both the general and the specific. The question "did you get at least 11?" (assuming 'get' is understood to mean 'generate by rolling the dice', and 'at least 11' is understood to mean 'a configuration of the dice with at least 11 total pips on the top faces') is a straightforward yes-or-no question... no tricks. The set defined by 'at least 11', intersected with the set of possible rolls, is { 11, 12 }. And saying "I got at least 11" is the same as saying "I got an element of the set defined by 'at least 11'," with the advantage that it's how people actually talk.Quote:
Originally Posted by OneHitWonder
I believe you’re one of the first 1/6ers to attempt actual math. Too bad you’re wrong.Quote:
Originally Posted by OneHitWonder
If it is given that one or both of the dice shows a deuce, then 6/11 times the left die will be a deuce. Likewise, 6/11 times the right die will be a deuce.
Then (to kind of follow your approach), when the left die is a deuce the other die will not show a 2 with 5/6 chance; and when it is the right die showing a deuce, the other die will not show a 2 with 5/6 chance. Then the calculation for NOT getting a pair of deuces:
(6/11 * 5/6) + (6/11 * 5/6) = 10/11
Therefore, the probability to get a pair of deuces:
1 – (10/11) = 1/11
I certainly don't dispute that "at least 11" is { 11, 12 }.Quote:
Originally Posted by onyx99
Then, people talk deceptively, or in ignorance. Claiming "one or more prizes" in a lottery of a single known prize is wrong. One might attach EV's to the "or more" part, and discover a constant deficiency in winnings.Quote:
Originally Posted by onyx99
Not so fast.Quote:
Originally Posted by Zedd
Yes, so I must have meant when the "peeker" goes from the left and the right sides at the same time. 6/11 = 6/11, so half the time from the left, and half the time from the right.Quote:
Originally Posted by Zedd
Depends how you look at it. In any event, the 1/11 is is made up of twice the basic 1/6 answer.Quote:
Originally Posted by Zedd
Five-to-one money odds paid twice on a roll of 2-2 is the same as ten-to-one money odds paid once on a 2-2. And, ten losses of 1 unit is twice five losses of 1 unit on the other rolls.
Okay. This is about specific rolls, whether summed by probability.
The double 1/6 chance game of what to do when 2-2 has been rolled is like the 1/11 chance answer chart. The single game or roll can do also the double game of the 2-2's roll; but the double game can not do the singles' games (by coming apart while remaining the double game).
Specific rolls happen consecutively. You have to work with that. And, in practice, it's as meaningless to do row and column of chart together as it is to speak of all the rolls of one 2 and two 2's.
This doesn't take a few days to figure out. As many have pointed out, it's an elementary exercise in a first-year probability course to show that the answer is 1/11. It might take a few days to convince you, or Alan, that you're wrong... but then again, it might take forever. I'm not planning to spend forever. I wrote down what I think is a logical progression that, if followed, may help you find the spot where your intuition is leading you astray. But you have to be *trying* to find that spot. I can't understand it for you.Quote:
Originally Posted by OneHitWonder
That being said, if you're still convinced you've got the right end of this particular stick, all I can say is... wanna bet?
Please... all of you who say the answer is 1/11:
Show me your video with at least one die with the 2 face up and how you can count 11 possible faces for consideration. You can use your cell phone and post on YouTube. It is free to do.
Have two dice on a table and with either die showing a 2 (remember the original problem says at least one die is a 2) count for me the 11 faces that fit your 1/11 answer. Remember, in the real world when a die settles on a 2 you cannot change the value or face on that die.
Show me the 11 faces. Now go to work.
There aren't 11 different faces for the hidden die to be showing, of course. There are six. That tells you nothing AT ALL about the probability of each face unless you assume they're all equally likely. (Spoiler: they're not.) Here's the million-dollar question: WHY do you think you can assume that?Quote:
Originally Posted by Alan Mendelson
Exactly. And that's the answer to the original question. The original question is not asking for your "probability gymnastics." Sure the word is used -- but it's not used in a strict context.Quote:
Originally Posted by onyx99
I get it. Regnis gets it. Redeitz gets it. Singer gets it. A dozen of my close and intimate friends on Facebook get it. Only your "math guys" don't get it. You see the word probability and you lunge for your textbooks. You can't see the forest for the trees.
As I said early on over on the WOV forum you guys "overthought" the question. In reality it is a simple math question in which you look at one die and count the faces on one die. No graphs, charts, spreadsheets, advanced theory is needed.
And the only way to justify your graphs, charts, spreadsheets, probability math, and so forth is by altering the main condition of the problem which is that you have at least one die already showing a 2 and that die should remain showing a 2.
You guys dug the Panama Canal when all you needed was a drainage ditch.
How about a thirty-six sided die? Rolling one thirty-six sided die is the exact same probabilistically as rolling two six-sided dice. We can show you all eleven faces of the thirty-six sided die without needing to touch.
If the thirty-six sided die has faces labeled (1,1), (1,2), (1,3), ... (1,6),(2,1), (2,2), ..., (3,1),...(6,1),...(6,6), then the eleven faces are those with either a 2 in the first entry or a 2 in the second entry. The probability of this thirty-six sided die landing on a specific face (say for example 2,2) is exactly the same as the probability of two distinct dice (with a clearly defined order) (say for example one red die and one green die) with the first distinct die (say for example the red die) corresponding to the first number and the second die (the green) corresponding to the second number.
If you complain about the fact that the thirty-six sided die doesn't have faces in a more usual labeling, I can remind you that there are several dice used in several common games which use nontraditional face labels, such as backgammon doubling dice, poker dice, etc. If you absolutely insist that the dice faces must be 1,2,...,36 that is fine too. Let the number on the d36 be x. Then x can correspond to a red d6 value of floor((x-1)/6)+1, and a green d6 value of ((x-1)mod6) +1. I.e. exactly the labeling described earlier (1,1), (1,2), (1,3),...
Roll the d36. If it lands on any of the faces (1,2),(2,1), (2,2), (2,3),...(3,2),(4,2),...,(6,2), then the bet is on as "there is at least one two." As each and every one of the 11 faces was equally likely, being told that the d36 landed on one of those 11 faces, the probability that it is in fact the face labeled (2,2) is... you guessed what I'd say... 1/11.
In using a d36, there is absolutely no way to have one "stationary" and one "spinner."
How do you refute this argument? In what way is the throwing of a d36 as opposed to two d6's somehow different? (its not for the record)
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also... "and you lunge for your textbooks"... implying that we don't have such elementary math memorized already...
I set one die on 2 another 1000 times. Still no change.Quote:
Originally Posted by regnis
Let me know when my sample size is enough.
Wait--while typing this I knocked it over. Now it is a 5.
Now I get it.
Go Hawks!!!!!
Have you run the simple test yet? At most a half hour of your time and you can prove us all wrong. What are you waiting for?Quote:
Originally Posted by Alan Mendelson
Arc, I ran the test. I have one die showing a 2. I have another die and 1 out of six times when I roll it, it also shows a 2.
What I enjoy seeing is the math guys getting so frustrated over Alan's repeated very simple common sense request that this collection of geniuses put their swollen heads together with just a little of that AP/+EV cash they so often theorize about, and come up with that video where at least one of the two dies shows a 2. Real world reality guys....stop stalling. Even EvenBob could do it while he's sitting around picking his toes. So enuf with the game of dodge ball. SHOW us what you're made of and stop stop talking probabilities and expectations. It's time you all graduated from that world of make believe.
Ironic you say this.Quote:
Originally Posted by Robsinger
Alan, that is not what the question asks. There are only two conditions in questionQuote:
Originally Posted by Alan Mendelson
"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?"
So, once again. Throw the dice, mark one row when the first condition is met, mark a second row when the second condition is met. When you've run 100+ tests then divide the first row by the secon row and tell us what you get. Couldn't be simpler. The only possible reason you are avoiding the test is you already know you are wrong.
More ironic appears to be how you (and arci, etc. etc) keep dodging the request for a simple video by--this time by deflecting. Alan has been asked to perform a test, and we know that will result in a 1in11 conclusion. However, he's asked you ego-driven, self-proclaimed know-it-all's to produce a simple video that supports the much more mundane & comprehensive interpretation of the question long ago and many times, and you can't even figure out how to do that.Quote:
Originally Posted by RS__
Theory: don't order it from any menu, and don't try to deposit +EV phantom bucks into any bank.
Since the analytical proof and our "EV phantom bucks" arguments aren't working, here is 20 minutes of me rolling the dice moving chips into or out of the box with the payouts as described.
https://www.youtube.com/watch?v=uv01RPDYJy0
I started with 50 blue chips in the box and 50 red chips on the table. If exactly two deuces rolled, I moved eight chips out of the box into the pile. If exactly one deuce rolled, I moved one chip into the box from the pile. If no deuces rolled, I did not move any chips. I intended to and was perfectly willing to continue the video until either all red chips were put away (the player was 50 units "in the red") or until all blue chips were in the pile (the player was 50 units "in the green"). I.e. a 9-1 payout. My phone ran out of storage in the process, but in the effort of keeping this as a single take (to avoid argument that I only wanted to show video that supports my argument), I'm uploading anyways.
If you don't wish to watch the video and hear me talk, the recording stopped at the player being 27 units in the red with that payout rate. I went back and counted how many times double-deuce rolled and used that to figure out how many single-deuce rolled. In this video double-deuce rolled a total of five times compared to 69 single deuce rolls and an estimated 240 total rolls of the dice. Had the probability been 1/6 as claimed, this result is a whopping 2.2 standard deviations below the mean. It is highly unlikely that anyone shooting a video would be patient enough to wait for such a result. Considering the actual probability of 1/11, this result occurred only 0.68 standard deviations below the mean, well within the realm of likelihood.
I admit that this is *not* conclusive proof. Lucky streaks (and unlucky streaks) can and always will eventually occur. Mathematicians are well aware of this fact, which is why we develop analytical tools and methods to approach problems. The only reason I chose to do it is because what *is* conclusive proof has been outright ignored, so a physical demonstration might at least make you rethink it a little bit.