How do you interpret the "dice problem"?

[Rob.Singer]

The mundane meaning behind the original question, once it is identified that "at least one of the dice is a 2", does indeed ask what the odds are of the other die being a 2. Similarly, there is nothing in the question that suggests this is nothing more than a one-throw event. However, I would not fault anybody who wants to believe the 1in11 interpretation, simply because it can be understood in that manner. So why can't the 1in11ers accept that 1in6 is also just as viable--or even more so--an answer as well as an interpretation? Does the pride take THAT much of a hit?

[redietz]

Evidently.

[Zedd]

The mundane meaning behind the original question, once it is identified that "at least one of the dice is a 2", does indeed ask what the odds are of the other die being a 2. Similarly, there is nothing in the question that suggests this is nothing more than a one-throw event. However, I would not fault anybody who wants to believe the 1in11 interpretation, simply because it can be understood in that manner. So why can't the 1in11ers accept that 1in6 is also just as viable--or even more so--an answer as well as an interpretation? Does the pride take THAT much of a hit?

Again, the original question is open to an interpretation that produces an answer of 1/6.

The problem is…Alan (and others) believes 1/6 still holds on an interpretation that ONLY produces 1/11 as the answer.

[arcimede$]

The mundane meaning behind the original question, once it is identified that "at least one of the dice is a 2", does indeed ask what the odds are of the other die being a 2. Similarly, there is nothing in the question that suggests this is nothing more than a one-throw event.

The event may be one throw but to compute the probability REQUIRES looking into multiple events. That is, all possible throws of the dice where at least one 2 is rolled. There are 11 such events and only one of them provides the case where the other die is a 2.

The question itself is asking for the "probability". It is truly amazing how many people continue to ignore this simple fact.

[RS__]

Harry, you're not gonna convince these guys of anything. You can't teach 'em because they don't know how probability is figured out (ie: "only 1 throw not multiple") nor what conditional probability is (ie: Alan's "set one die to 2, toss the other").

It's like trying to teach a 4 year old multiplication. If he doesn't know how to add numbers, the multiplication will go right over his head.


My advice is never look back....but if you must, just read. It is kind of funny just reading it.

[OneHitWonder]

Nonsense. You are making an arbitrary (and incorrect) assumption.

Explain please. And, while you're at it, correct my mistake.

Eg, how was Mango wrong?


P.S. Ask specific questions to get specific answers. Don't rush to the nonsense word, and imagine your own thoughts for mine.

[OneHitWonder]

You can't teach 'em because they don't know how probability is figured out.

This isn't about "how probability is figured out". Clearly, you want to make it about this because that's the sort of cheap "hammer" you carry.

Anyone can look up and apply simple conditional probability formulas. If you want to be a real math star in the real world, then go and unify Bayesian probability theory with the Frequentist theory. Quantum decoherence would be a good place to start.

What's that about "can't teach 'em because they don't know how probability is figured out"? Don't make us laugh.

[OneHitWonder]

Nothing in the question asks anything about "the other die".

Good grief.

Both can't be 2 if also the other isn't a 2.

[OneHitWonder]

It's like trying to teach a 4 year old multiplication. If he doesn't know how to add numbers, the multiplication will go right over his head.

BTW, from which college/university did you graduate? Wizard of Vegas?

[OneHitWonder]

But, the original question is indeed incapable of solution without making assumptions on how the condition of 'at least one deuce' was met.

No assumptions. The "peeker" didn't say we must wait for a 2-roll. Necessary information in that case. In any event, must use 'one or the other die', as in 2-rolls or rolls involving the number 2 to allow for the general roll in the specific roll. This is, after all, about also the specific rolls.

However, after we had cleared up this ambiguity and established that the number IS agreed upon in advance--Alan still believes the answer is 1/6. This is simply wrong.

The number isn't agreed on in advance.

Even were it agreed on in advance, Alan is still right. One 2 isn't "one or more 2's" or "one 2, or two 2's". Set A isn't the union of set A and set B, unless set B is the empty set. But, "or more" isn't the empty set.

Take a bow, Alan.

[regnis]

Here is the original statement again:

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?


The question says you have 2 dice, shake, slam, and the peeker says one of them is a duece. It doesn't say you shake , slam, and peek a million times. It says shake, slam, peek---1 is a duece. What is the probability both (i.e. the other) is a duece.

Where do you see anything more than a one shot deal.

[regnis]

Arci--we are not looking for the probability of multiple events. At least I'm not. What are the odds the other is a 2 is all the question asks. If you read it in that context, it can only be 1-6.

[Zedd]

No assumptions. The "peeker" didn't say we must wait for a 2-roll. Necessary information in that case. In any event, must use 'one or the other die', as in 2-rolls or rolls involving the number 2 to allow for the general roll in the specific roll. This is, after all, about also the specific rolls.


The number isn't agreed on in advance.

Even were it agreed on in advance, Alan is still right. One 2 isn't "one or more 2's" or "one 2, or two 2's". Set A isn't the union of set A and set B, unless set B is the empty set. But, "or more" isn't the empty set.

Take a bow, Alan.

Yea...and the peeker didn’t say that we must not wait for a two. Nor did he say that he was randomly calling out dice X. Nor did he say that he had waited until a fixed die was a two. This is why the original question is ambiguous. In order to give an answer (be it 1/6 or 1/11) you must make an assumption.


The number was agreed on in advance with the Alan/Wizard Lunch Bet thing. The ambiguity was cleared up—only rolls with at least one deuce were looked at.

Correct, one deuce isn’t “one or more deuces” or “one deuce or two deuces”. But at least one deuce IS “one or more deuces” or “one deuce or two deuces”.

[arcimede$]

Good grief.

Both can't be 2 if also the other isn't a 2.

The question is ... "What is the probability that both dice are showing a 2?" .... do you see any mention of "the other die"? No? LOL.

[arcimede$]

Here is the original statement again:

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?


The question says you have 2 dice, shake, slam, and the peeker says one of them is a duece. It doesn't say you shake , slam, and peek a million times. It says shake, slam, peek---1 is a duece. What is the probability both (i.e. the other) is a duece.

Where do you see anything more than a one shot deal.

It is inherent when the question asks for a "probability" response. Words have meaning.

Arci--we are not looking for the probability of multiple events. At least I'm not. What are the odds the other is a 2 is all the question asks. If you read it in that context, it can only be 1-6.

No, the question asks for a probability. Do you really think you can change the question to your liking? LOL.

[Alan Mendelson]

Yes Arc it is about probability... considering one die, one time.

It is you and the other 1/11ers who are changing the question to your liking.

[redietz]

Arci winks at the cocktail waitress, slams down his drink, then peeks down her blouse. What is the probability of a sexfest?

I fail to see how this is necessarily about multiple events.

[regnis]

I can probably speak for Arci, myself, and most of us here. One such event is enough. The odds of a multiple event are increasing daily.

[Rob.Singer]

The event may be one throw but to compute the probability REQUIRES looking into multiple events. That is, all possible throws of the dice where at least one 2 is rolled. There are 11 such events and only one of them provides the case where the other die is a 2.

The question itself is asking for the "probability". It is truly amazing how many people continue to ignore this simple fact.

You must've missed it. The probability part of Alan's interpretation is 1in6.

[OneHitWonder]

Yea...and the peeker didn’t say that we must not wait for a two. Nor did he say that he was randomly calling out dice X.

Doesn't have to then, doing so would make no difference.

If this exercise is about a specific number, and the number 2, we must know in advance. We must be able to verify what's happening along the way. The "peeker" can't make it up as he/she goes along, let alone after each roll.

The Wizard isn't allowed even fair money odds on a 1/11 chance for a 1/6 chance outcome. Ie, he can't just claim any number from each roll was what he was waiting for.

Anyone see a pattern here in the 1/11 chance answer people? This is how the Wizard seems to define "a good bet". The reason that he always seems to take the worst of it from people who not only passed the Mensa test, but who earned near genius rankings by it.

If there is a number from the start, it's 1/6 chance a 2. And, it's less than 1/3 chance of some 2 on the first roll. Furthermore, a 1/6 chance that the "peeker" awaits that number - assuming that the "peeker" was even waiting on a specific number. At least a 100 to 1 against your 1/11 scenario based on the odds alone. However, we still require the announcement of the number for the 1/11 chance answer to be verified, ie, hold up under any sort of scrutiny.

What the heck sort of bet is Alan supposed to make, and under such constant hounding and duress? Certainly not at the Wizard's level of authority.

With virtually everyone playing "follow the leader" over there, I see the reason now that no one pointed any of this out over there from the start of the thread. Are you the Wizard? It's very hard to tell with all the silly evasiveness in specific, and plain silliness in general. For a second there, I thought I was back reading the Wizard's forum. Maybe you should "get the heck out of here" like the other Wizlings? No one's going to truck such nonsense and shiftiness here.

You must make an assumption.

Understand yet? Guessers are losers.

The number was agreed on in advance with the Alan/Wizard Lunch Bet thing. The ambiguity was cleared up—only rolls with at least one deuce were looked at.

You still don't get it. Any 1/11 chance answer is made up of the 1/6 chance answer. The latter is the basic and, or, and "at least", of this.

Correct, one deuce isn’t “one or more deuces” or “one deuce or two deuces”. But at least one deuce IS “one or more deuces” or “one deuce or two deuces”.

You still don't get this either. The condition must read "a roll which is an element of those in which at least one is a 2"; not "a roll which is at least one 2 (or more)".

Wake up!