Two Dice Puzzle: Part Deux
Quote: WizardI was hoping I'd get more comments on my video but here is the thrust of my "argument using only dice."
As you can see, these dice show the 36 possible ordered pairs of two dice.
Alan's prior response to this demonstration was:
Quote: AlanMendelson
Not quite. You're not using TWO DICE. This is a two dice problem.
Quote: AlanMendelsonHonest to God, BBB. Do you understand this is a question about TWO DICE with at least ONE DIE SHOWING A TWO and during this entire debate those who say the answer is 1/11 are using MORE THAN TWO DICE to prove their point?
In other words, his refutation of your image and argument would be that it depicts 72 dice, not just 2.
Quote: Ibeatyouraces...for the 6th, or is it the 11th :-), time...
Totally off-topic, but did you ever realize that 6th is pronounced the same as 1/6, 11th is pronounced the same as 1/11, 3rd is pronounced the same as 1/3, but 2nd isn't pronounced the same as 1/2?
"I'd like a second of a pound of cole slaw."
"Tampa is in half place in the AL East."
Quote: indignant99You're having a problem, son. The scenario does not have "the observed die." It has "an observed pair of dice." Never is the peeker restricted to observing only one die. He's obligated to see both, and if any deuce manifests, he's further obligated to say "at least one of the dice is a two."
The original problem is silent on the methodology. The peeker isn't "obligated" to behave as you specify. In fact, assuming the game only runs if the peeker checks both dice, and only runs the game if one of them is a two, is a glaring weakness in the 1/11 calculation. It is just as likely (and perhaps more so) that the peeker only checks one die, and announces it. In the absence of the actual rules, The single peek methodology is the most elegant fit for the events as described.
Quote: Ayecarumba
The original problem is silent on the methodology. The peeker isn't "obligated" to behave as you specify. In fact, assuming the game only runs if the peeker checks both dice, and only runs the game if one of them is a two, is a glaring weakness in the 1/11 calculation. It is just as likely (and perhaps more so) that the peeker only checks one die, and announces it. In the absence of the actual rules, The single peek methodology is the most elegant fit for the events as described.
If we assume that, then we do get 1/11. And I think we've already established what the chances are in other scenarios (always announces a die, only announces a deuce, sees only one, etc.).
If you're talking about our argument with Alan, he has already confirmed he agrees to "both dice are peeked, only announces a deuce".
Quote: MathExtremistIn other words, his refutation of your image and argument would be that it depicts 72 dice, not just 2.
Dang. I've been waiting many years for the day those 72 dice would come in handy.
Quote: AyecarumbaThe original problem is silent on the methodology. The peeker isn't "obligated" to behave as you specify. In fact, assuming the game only runs if the peeker checks both dice, and only runs the game if one of them is a two, is a glaring weakness in the 1/11 calculation. It is just as likely (and perhaps more so) that the peeker only checks one die, and announces it. In the absence of the actual rules, The single peek methodology is the most elegant fit for the events as described.
I agree. Except for the single peek theory. I stated way back in the original thread that in the absence of the 'calling rules' the answer could not be derived. That fact and the situation you suggest has been done to death too. ME did it very well.Different calling rules, different probabilities
For now, though, we are working on the premise of only playing when the announcement is about a two and where the thrower always declares a two when he is truthfully able to.
Quote: AyecarumbaThe original problem is silent on the methodology. The peeker isn't "obligated" to behave as you specify. In fact, assuming the game only runs if the peeker checks both dice, and only runs the game if one of them is a two, is a glaring weakness in the 1/11 calculation. It is just as likely (and perhaps more so) that the peeker only checks one die, and announces it. In the absence of the actual rules, The single peek methodology is the most elegant fit for the events as described.
I disagree. First, it doesn't matter whether the game runs only if one of them is a two because in the scenario that's a given. The real issue is whether the question as written was intended to be tricky or straightforward in its use of "at least one." The implication of that phrase in context is that both dice are examined, because it would be misleading to say "at least one" if you only examined one. The intent of the problem was to present a mathematical puzzle, not a linguistic one.
Also, it's pretty farfetched to suggest that someone looked under a dice cup with two dice underneath and only actually saw one of them.
Edit: to clarify, the specific problem statement I'm referring to is this one:
"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 argument may not apply to other rephrasings of the problem.
Quote: AyecarumbaQuote: indignant99You're having a problem, son. The scenario does not have "the observed die." It has "an observed pair of dice." Never is the peeker restricted to observing only one die. He's obligated to see both, and if any deuce manifests, he's further obligated to say "at least one of the dice is a two."
The original problem is silent on the methodology. The peeker isn't "obligated" to behave as you specify. In fact, assuming the game only runs if the peeker checks both dice, and only runs the game if one of them is a two, is a glaring weakness in the 1/11 calculation. It is just as likely (and perhaps more so) that the peeker only checks one die, and announces it. In the absence of the actual rules, The single peek methodology is the most elegant fit for the events as described.
The peeker is obligated to be truthful. However, not obligated to speak.
He can view ONE die, or TWO dice... Seeing a deuce, he can "clam up."
Not seeing a deuce, he can't lie and say he did.
WE ARE ONLY CONCERNED WITH ROLL-EVENTS WHEN A DEUCE IS ANNOUNCED (even though some deuces might NOT be announced).
What I'd say is "Terminate that dishonest peeker," and "Get an electronic cup, with optical reader & LCD display on the cup-bottom, to thwart dishonest utterances and dishonest silences."
Quote: indignant99Granted, that could be the peeker's policy for announcement.
What I'd say is "Terminate that dishonest peeker," and "Get an electronic cup, with optical reader & LCD display on the cup-bottom, to thwart dishonest utterances and dishonest silences."
That's why a better statement of the problem is "what is the probability of a fair throw of two standard dice showing 2-2 given that at least one of the dice shows 2?" No peekers, human intent, truthiness, etc. Just a math problem with certain things as given.
Quote: MathExtremistQuote: indignant99Granted, that could be the peeker's policy for announcement.
What I'd say is "Terminate that dishonest peeker," and "Get an electronic cup, with optical reader & LCD display on the cup-bottom, to thwart dishonest utterances and dishonest silences."
That's why a better statement of the problem is "what is the probability of a fair throw of two standard dice showing 2-2 given that at least one of the dice shows 2?" No peekers, human intent, truthiness, etc. Just a math problem with certain things as given.
1/6. Obvious. Stop altering the question :o)
Quote: pewAre you kidding? Your bud Singer. Ring a bell?
Somehow, this it's how I picture Rob Singer in his RV park that he supposedly lives at...
Quote: AlanMendelson
Alan, you criticize and rail against flip-flopping, rotating a die every-which-way (6 ways). Well, damn, isn't that exactly what YOU DID with one die in your video?
Now, I ask you, to justify why we CAN'T rotate BOTH dice!
(And you don't get to STEAL one die, like you did in your vid.)
Quote: AlanMendelson... I am not questioning the 1/11 result when you are asking how many combinations of two dice include at least one 2, and of those how many would result in 2-2. That would be 1/11.
But given that at least one of the dice has settled on 2 -- SHOW ME -- using two dice that the chance of 2-2 is 1/11 and not 1/6...
How are those two statements not exactly equivalent?
- combinations of two dice include at least one 2
- at least one of the dice has settled on 2
Quote: MathExtremist...the condition "one or more dice show 2" is met by both the cases "exactly one die shows 2" and "exactly two dice show 2."
After the fact.
The ONLY way that the condition "one or more dice show 2" can be met, as a question of probability beforehand, is when the ONLY die inspected is a 2. In which case, one of the dice is a 2; and, possibly another die is a 2. There is NO other way to meet this condition in advance. Then, it's 1/6 chance for another 2. At which point, there will be one 2, or two 2s... after the fact. Nor would I put the result as (the undetermined condition) "one or more dice show 2" by any stretch of the imagination.
Regardless the sort of "peeker" or who does the "peeking". But, a strange method of "peeking" may be construed as to allow for the answer of 1/11 chance.
Add on: I hope this isn't going to turn out as with Alan. Can neither the "internet experts" concede defeat?
Someone peeks under the cup. If either one of the dice is a two, or if both dice are a two, he announces that "at least one of the dice is a two"
How is that complicated, counter-intuitive, or not the most basic thing you would expect someone to do when looking to see if there is at least one two?
Quote: Dalex64I don't understand this strange method of peeking that keeps being brought up.
Someone peeks under the cup. If either one of the dice is a two, or if both dice is a two, he announces that "at least one of the dice is a two"
How is that complicated, counter-intuitive, or not the most basic thing you would expect someone to do when looking to see if there is at least one two?
Because they're not "math geniuses" as they would call some people over here.
Quote: IbeatyouracesBecause they're not "math geniuses" as they would call some people over here.
Lol.
Quote: KerkebetRegardless the sort of "peeker" or who does the "peeking". But, a strange method of "peeking" may be construed as to allow for the answer of 1/11 chance.
Ass-backwards. It requires a "strange method" of peeking/announcing to get 1/6:
- Only look at one die, not both, to make the announcement, or
- Announce every pair-of-deuces, but only half the appearances of deuce-other.
Quote: indignant99...Ass-backwards.
Please refrain; or, attack my argument on which my conclusions are based.
Regardless, it's not so AB to conclude that it doesn't take a genius to figure out that were there some good and easy money at blackjack, then the (much) much-more skilled and financed poker players would've gone that route a long time ago. Same for the professional street beggars, who earn hundreds of dollars a day, for the same amount of work as the winning poker players. But, the remaining few persons who cling to stories of hundreds of thousands of dollars a year by casual blackjack haven't figured out where where any/the real money is? Trapped.
The worst internet writers by far aren't the trolls, but the persons who make longstanding claims of "rocking your world" without any evidence, let alone proof. And, in my opinion, who go on with old and trivial math snippets at every opportunity for/while others refer to them as geniuses.
Sorry, "but thems the facts" of the gambling forums.
Quote: indignant99When an argument is ass-backwards, I will not refrain from calling it "ass-backwards."
Lol.
2 dice are rolled and placed under 2 different cups.
SCENARIO 1
The dealer peeks under ONLY the first cup and then announces at least 1 die is a Two.
The 1/6 answer.
SCENARIO 2
The dealer peeks under BOTH cups and then announces at least 1 die is a Two.
The 1/11 answer.
Surely it can be seen that these are different problems with different probabilities.
Quote: AceTwoAnother Simple approach to convince the Doubters.
2 dice are rolled and placed under 2 different cups.
SCENARIO 1
The dealer peeks under ONLY the first cup and then announces at least 1 die is a Two.
The 1/6 answer.
SCENARIO 2
The dealer peeks under BOTH cups and then announces at least 1 die is a Two.
The 1/11 answer.
Surely it can be seen that these are different problems with different probabilities.
Doesn't make any difference if at least one of the die is a two
Quote: IbeatyouracesI always thought it was "bass-ackwords."
Almost! It is, but your "o" should be an "a."
Quote: indignant99Almost! It is, but your "o" should be an "a."
Ahahaha, you're right!
No genius here! :-)
Quote: AlanMendelson$100 times 40 = $40,000
My casino encourages patrons who bet $1000 cheques, but think they're $100's, and are happy with payoffs appropriate for a $100 cheque.
Awwww. He fixed it. AFTER OnceDear highlighted it.
Come to think of it, I did just fire my accountant.
Quote: KerkebetAfter the fact.
Of course it's after the fact. Before the event, you don't have the fact yet. You just have the probabilities that certain facts will occur. That's the whole point of probabilities. After the fact, there are no probabilities. Just certainties - you either had two 2s or you didn't.
Quote:The ONLY way that the condition "one or more dice show 2" can be met, as a question of probability beforehand, is when the ONLY die inspected is a 2. In which case, one of the dice is a 2; and, possibly another die is a 2. There is NO other way to meet this condition in advance.
Of course that's not true. You can calculate the probability that "one or more dice show 2" prior to throwing them and without inspecting any dice. That probability is 11/36. Don't fall into the trap of thinking you need physical dice to solve this problem.
You can also calculate the probability that "exactly two dice show 2" which is more familiar to dice players; that's 1/36.
The ratio between those two probabilities is 1/11, which is what the problem is intending to ask.
Quote: AlanMendelson
Okay, Alan, now that you have plainly demonstrated that you can take the DEUCE, and super-glue it down as a deuce, and then take the notorious other die and rotate it 6 ways showing 6 different faces...
Show the equivalent but opposite scenario. Super-glue the notorious other die as a DEUCE, and rotate the one that AIN'T notorious, AIN'T glued, and likely AIN'T a deuce any more.
Quote: AlanMendelsonMiplet and others: if you froze the die showing a 2 with the 2 face showing, can you still justify your answer of 1/11?
Yes, I damn sure can. I can do it with a video - to debunk your garbage. How about you think for a little bit, and maybe figure out how?
Quote: AlanMendelson
Okay, here's a little exercise for you. The RED DIE is yours; the BLUE DIE is mine.
When your precious little RED DIE shows a deuce, it does not eliminate the other five faces from possibility.
It is allowed to show 1, 3, 4, 5, 6 or even 2, WHEN MY PRECIOUS LITTLE BLUE DIE shows a 2.
And vice versa. THAT'S "AND," not "OR."
Quote: indignant99Yes, I damn sure can. I can do it with a video - to debunk your garbage. How about you think for a little bit, and maybe figure out how?
Maybe you should take it little easier on the 1/6 supporters? You actually started (re-started?) this topic by bumping a 1.5 year old thread (the original two dice post). You boasted that the Wiz was wrong in the post (he answered 1/11 in the first thread).
https://wizardofvegas.com/forum/questions-and-answers/math/15508-two-dice-puzzle/#post448076
You then changed your mind a number of times and now are clearly in the 1/11 camp. Alan and others may disagree (or are confused). You have also disagreed with 1/11 (or have been confused)....
Quote: Keeneonenow are clearly in the 1/6 camp.
I am? You'd better go back and read the last 40 pages.
Quote: indignant99I am? You'd better go back and read the last 40 pages.
I have corrected my post. Sorry for the error.
Quote: KeeneoneI have corrected my post. Sorry for the error.
Apology unnecessary. This whole maelstrom w/ Alan (& his censored) illustrates why I eschewed being a teacher (though I had lots of low-key urgings to do so).
I guarantee I would have been firing a select few deserving pupils (and I don't mean merely awarding an "F"), whereas Alan publicly declared he would fire his "superiors" (and I don't mean "bosses").
Quote: MathExtremistOf course that's not true. You can calculate the probability that "one or more dice show 2" prior to throwing them and without inspecting any dice. That probability is 11/36. Don't fall into the trap of thinking you need physical dice to solve this problem.
You can also calculate the probability that "exactly two dice show 2" which is more familiar to dice players; that's 1/36.
The ratio between those two probabilities is 1/11, which is what the problem is intending to ask.
Okay, this discussion is slowing down. A good time to get serious.
The answer 1/11 chance of two 2's has to do with the roll-of-two-dice forms of {2_non-2 and non-2_2} and 2_2. Both forms are involved. Everything to do with and, and hence addition. Ie, how many distinctly possible rolls of the form {2_non-2 and non-2_2} compared with how many of the form 2_2? Ten to one, for 1/11 chance of two 2's. In general, the forms which contain "one 2 and up to the number of dice". The reason for the relevance of such charts. No or's are involved. Certainly nothing to do with "one 2 or more" on a specified roll.
Even were a specified roll of the form {2_non-2 and non-2_2} or the form 2_2, this would still not be the same as "one 2 or more" on the specified roll for whichever form in actuality. The form {2_non-2 and non-2_2} doesn't actually allow for "one 2 or more" on a specified roll any more than does the form 2_2. To allow for "one 2 or more" on a specified roll, these two forms must somehow remain possible at the same time as when one die shows a 2, and the other (unseen or unrolled) die has a 1/6 chance of showing a 2. On a specified roll of two dice, there are two separate dies to consider. In general, the conditions "one 2 or up to the number of dice". Certainly not two roll-of-two-dice forms.
Add on: Lord forgive me... especially if I made a typo or "brain fart" in the delivery of this reply.
Quote: indignant99Quote: AlanMendelsonMiplet and others: if you froze the die showing a 2 with the 2 face showing, can you still justify your answer of 1/11?
Quote: indignant99Yes, I damn sure can. I can do it with a video - to debunk your garbage. How about you think for a little bit, and maybe figure out how?
Quote: AlanMendelsonOops my bad. I should have said "roll the dice until you have at least one of the two dice showing a 2 and if both dice show a 2 that is okay to start your video."
But for those of you who want to save time for the purposes of shooting your video explaining your 1/11 answer, you can skip the rolling part and just "set" one die on a table showing a "2" and the other die on the table showing whatever face you'd like it to show. But remember -- in the real world that "2" cannot jump from one die to another...
You can bet your sweet *** my video will start with both dice showing DEUCE :
- The first half of the video will keep the right-hand side die (blue) frozen as a Frozen Blue Deuce. Then the left-hand side die (red), starting as a 2, will rotate to all the other red faces: 1, 3, 4, 5, 6.
-
The second half of the video will keep the left-hand side die (red) frozen as a Frozen Red Deuce. Then the right-hand side die (blue), starting as a 2, will rotate to all the other blue faces: 1, 3, 4, 5, 6.
- There will be no "Mexican Jumping Deuces."
Quote: IbeatyouracesI don't think Alan is reading this anymore.
(I think he is. He's making snide responses over on his own forum.)
But, true, he's not contributing any postings here, anymore.
Regardless, the postings here may be instructive for the audience, sans Alan.
I wish he was a proper salesman/marketer.
Think what a great advocate he could be for the bets on offer by the Wizard and by me.
He's not persuaded one solitary person to put money where his mouth is, on a wager that pays way over fair odds.
His web site is advertising and promoting 'Best Buys'. What could be a better 'buy' than someone offering $9 for only $6 with an unlimited stock for sale. He could build himself a huge reputation. What a publicity opportunity.
What could be a worse 'Best Buy' site than one that failed to trumpet such a bargain?
Nobody can. Not even himself.Quote: Dalex64I do not understand what you are trying to say.
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?Quote: AlanMendelson
- Which one of these slam-cup-results actually occurred?
- All of them would qualify for "At least one of the dice is a 2," right?
- Still wanna bet (at 9-to-1 pay) that it ended up Pair Of Deuces?
Quote: AlanMendelson...he will have to choose which set of results he wants to use when he rolls two dice: either the horizontal results or the vertical results from his chart...
Isn't the whole G.D. friggin' chart - all 36 outcomes - in play when rolling two dice?
Yes, the result itself lies on either the horizontal, or the vertical, or lo-and-behold the intersection. But none of the ELEVEN rolls are ruled out.
But before the cup is removed, and the dice revealed, is the result gonna be one of the horizontal ones? Or one of the vertical ones? Huh, genius?
(The answer is: Yes, it's gonna be either one of the horizontal, or vertical, ones.)
Before cup removal, but after "deuce" announcement, exactly which restricted zone - horizontal vs. vertical - have we been incarcerated in?
The peeker/announcer did not inform "it's vertical," nor "it's horizontal."
What he actually informed, was "it's in the criss-cross."
After the cup is removed to reveal the dice, we do not enter into some restricted fantasy zone (horizontal versus vertical). The outcome is done. It is what it is.
