The Wizard will bank this bet: 1/6 vs 1/11

[Alan Mendelson]

It seems that probability allows for using the other five faces on the die showing a 2 but chance might isolate the second die.

[OnceDear]

Let me ask the question differently: if instead of using the word probability and we used the phrase "what are the chances that the other die is also a 2" would the 1/11 answer still hold?

I'll answer, but please Alan, try to comprehend the answer FULLY.

We are NOT EVER asked what the chances or PROBABILITY is of the other die being a two. We ARE asked IN THE ABSENCE OF KNOWING WHICH DIE IS A TWO, what the probability is that we have a pair of twos. That is different. WAYYYYY different.

You need to get your head around this rather odd set of facts.

For each die, the odds of that die being a deuce would be 1/6. You can use the words odds, chance and probability without too much disagreement.

You, not unreasonably, assert that the odds of 'the other die being a two' is 1/6

Are you happy to agree with me on that? It's not really what's being argued. Really. it's not.

BUT. GREAT BIG BUT....
That is not the question.
We are NOT asked what the probability of one other die being a deuce. You adamantly believe that's what is asked. But the question is absolutely explicit. We are asked what the probability is of A PAIR of Deuces! YOU can NOT extract one die, the die that is a deuce, because only the peeker knows which die that is, and he is not telling.

Your assertion would only follow from the original question if we could identify which die is already a deuce.

We know only one thing and that 'AT LEAST one of the dice is a deuce'.

Until the cup is lifted, we cannot ELIMINATE any die. We cannot eliminate any face on any die. All we can eliminate is outcomes of dice where there is no deuce. All we can eliminate is...
1-1
1-3
1-4
1-5
1-6
3-1
3-3
3-4
3-5
3-6
4-1
4-3
4-4
4-5
4-6
5-1
5-3
5-4
5-5
5-6
6-1
6-3
6-4
6-5
6-6

UNTIL we lift the cup, none of the above outcomes could sit under the cup, but there are still those pesky 11 other ways that the dice under the cup COULD have come to rest:-
1-2
2-1
3-2
2-3
4-2
2-4
5-2
2-5
6-2
2-6
2-2

UNTIL the cup is lifted. That is the sum total of all that we know. Having a red die and a blue die, or a left die and a right die or a big die and a small die can all help us to visualise that.

If we have a big die and a little die, you absolutely cannot rule out which one is a deuce. The peeker could put it in his pocket, but until you see it, you are none the wiser.

Once you see the cup lifted, probabilities are no longer at issue: bets are off. But until then..... 1 of 11 outcomes is hiding under that cup and you and I don't know which it is. THAT is the very nature of 'chance' or 'probability' or 'odds'. the reason no-one shows you this in a video is because once the cup is lifted, we can only see one fixed outcome of both dice. You, yourself rotated the 'other' die in some effort to show the six ways it could have landed. But you would have needed to show how both dice could have landed and for that you would have to acknowledge that until the cup is lifted, you don't know which one to rotate. Your video would have had to show you rotating both dice.

By 'setting' one specific die as a deuce, you would be imposing a condition that is not imposed.

I don't know much about craps, but would the casino let you set a die and roll just the other? Well this is why, because it is breaking the rules and the payout table would be shot to hell.

Now, the more serious question. Why has no-one taken up wagers offered by Wizard, myself and others.

Alan seems to assert that the bet offered DOES model the question appropriately.
Alan does agree that our peeker saw both dice and would only declare deuces or no deuces. It's agreed that that was an original omission from the question, but this thread is very specific.

With agreement on the peeking and declaring rules, we have a pretty darned explicit question. It has only one correct answer, unlike the very original question in a distant memory far far away.

Why is not one person taking the offered wager. It's either going to favour the player or the offerer and after a few hundred rolls, variance would become insignificant. If you want more confidence, stake $5 per resolved bet over 100,000 bets. The money is there for the taking. Maybe some 'maths guy' could tell you what you could expect to win or lose and even within what reasonable margin of error.

You could even use my spreadsheet, It's free and it even tells you what I estimate that you would lose.

Even Rob Singer said

But, if as I suspect, we don't get to see which die has the original 2 until after the bet, then 9-1 is a punk bet because two 2's, after knowing half the outcome at that point, is an 11-1 event in this case.

[quahaug]

let me ask the question differently: If instead of using the word probability and we used the phrase "what are the chances that the other die is also a 2" would the 1/11 answer still hold?

yes!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!yes!

[OnceDear]

It seems that probability allows for using the other five faces on the die showing a 2 but chance might isolate the second die.

NOTHING isolates the second Die and NOTHING isolates the first die.
They are a PAIR of Dice and the one guy who can tell them apart, and who can tell you which one is a deuce is NOT telling. THAT is the nature of probability, betting on, or taking chances on what you don't know.

[arcimede$]

Let me ask the question differently: if instead of using the word probability and we used the phrase "what are the chances that the other die is also a 2" would the 1/11 answer still hold?

As long as either die can be the 2 then the answer is yes. The critical part of the question is whether you lock one die into being the 2. As in the case of the red and blue die ... If you ask what are the chances the blue die is a 2 when the red die is a 2 then the answer is 1/6.

[Alan Mendelson]

OnceDear wrote above: "We are asked what the probability is of A PAIR of Deuces!" That answer is 1/36.

Unfortunately I have to deal with real life and not some imaginary fantasy about probability. In real life if you tell me that at least one of two dice is a 2, in my mind I define one of two dice being a 2. Yes, it is possible that both dice are a 2, but one die is definitely a 2.

Also, my real life experience tells me that given two dice and with at least one die being a 2, that the chance for the other die in the problem can only be 1/6.

Now, if this were a textbook exercise and you set up the question as a textbook exercise explaining that there are 36 dice combinations using two dice and that there are 11 different combinations with at least one die showing, then I could go for your theoretical probability gymnastics.

But I deal with reality and physical dice which is why I created my video explanation that shows the answer is 1/6.
The Wizard also did a video explanation and in it broke all the rules of reality -- because he did not accept a die showing a 2 as a 2 and flipped it around to fit the 1/11 answer.

So here is how I would like to sum it up:

Those of you who say the answer is 1/11 get a gold star for answering a theoretical probability question that was not asked but had it been asked you would have the right answer.

For the rest of us who say the answer is 1/6 we also get a gold star for keeping our minds in this real, physical world and reasoning what is real and not imaginary.

[OnceDear]

In the real physical world where odds of 9 to 1 are available on unlimited wagers where probability is 1/6 there would be a stampede of takers.

Roll up. Roll up You can wear your gold star on your dunces cap if you stay away or if you show up.

[OneHitWonder]

So here is how I would like to sum it up:

Those of you who say the answer is 1/11 get a gold star for answering a theoretical probability question that was not asked but had it been asked you would have the right answer.

For the rest of us who say the answer is 1/6 we also get a gold star for keeping our minds in this real, physical world and reasoning what is real and not imaginary.

Not me.

I would rather know the single correct answer as summed up by the experts in this field. The one that at least doesn't involve the words nonsense and dunce cap.

No one ever got anywhere by giving up.

[Alan Mendelson]

I would rather know the single correct answer as summed up by the experts in this field.

Unfortunately I'm afraid that the experts in this field will fail to read English and will not correctly reason the question. All of the 1/11ers have failed to read and understand the question and have simply applied their "probability textbook answer" to the question when in reality they shouldn't.

I'm sorry but the "math guys" get a fail for understanding the actual question, but I am sure they learned their textbook probability lesson well because that's all they parrot.

[kewl]

A specific roll which is (an element) of a defined set of rolls is different from a roll which is a defined set of rolls. A major difference, but one which may seem an unobvious nuance to someone with virtually no algebraic experience.

The problem as written here doesn't address a specific roll which "is (an element) of" a set of rolls. It addresses the specific roll as, "At least one of the dice is a 2." It does not mention the words "is (an element) of" the "at least one of the dice is a 2" set of rolls of the 1/11 chance answer.

The roll in the problem by which "at least one of the dice is a 2" means a roll by which one or two of the dice is a 2. This definitely doesn't include a non-2 with a 2, or 2-2. Eg, an end roll of 6-2 isn't included in one or two of the dice is a 2, because 6-2 has no possibility of two of the dice is a 2. However, a specific roll of 2-X, where X has yet to be determined, has the possibility of one or two of the dice is a 2.



The set of rolls defined by "at least one 2" definitely does include {a non-2 with 2} or {2-2} but is different from a specific roll defined by "at least one 2". The latter doesn't include a non-2 with 2, or 2-2. It's the set of rolls of 2-X, where the X-die has yet to be looked at: {2-(1), 2-(2), 2-(3), 2-(4), 2-(5), and 2-(6)}. An end roll of this set is either 2-1, 2-2, 2-3, 2-4, 2-5, or 2-6.


You apply the nonsense word as "a crutch" to not make the effort to understand what I and others have written, and then to not write an intelligent reply. Even a nonsense idea deserves an intelligent reply, as I continue to reply to you.

In my opinion, this is, foremost, a math question. Not a gambling forum, or court of law, exercise. To this end, I will post it up at one of the math forums, along with a summary of my own arguments of course, when I have some more free time.

It doesn't surprise me that neither the Wizard nor Alan has chosen such a form of "binding arbitration". They aren't the math and formal logic experts. Aside from redietz, and perhaps Singer, I get the feeling that a lot of this went over everyone's head. Same at the Wizard's.

I'm using "nonsense" as a short and to the point way to express my believe you are barking the wrong tree here with your desire to dismantle a simple logical math question into philosophical argument about the meaning of a phrase, detail twisting and
challenging the very structure of math-semantic-logic up to the point where every effort
to make sense of anything becomes futile.
I can do that as well. You tell me something simple, such as "Good day." and I can start questioning, arguing and wooden philosophizing the "real" meaning of the phrase to the end of all days. But I choose not to, because to me, such exercise is an utter waste of time and demonstrates perfectly how wooden philosophizing on simple things can lead to total paralysis of the mind and breaks all possible ways of communication. That's all it accomplishes.

Why don't you bring the question to a math experts( or semantics experts or both) dedicated forum or blog and see how they interpret it and answer it? I don't think discussion as the one you seek belongs in a gambling forum.

I bet the math experts there will first clear one thing - the intention of the peeker, and then will give you the correct answer in identical manner as this answer:

"There are 36 combinations of two 6-sided dices, only 11 of them show at least a 2, from which one is 2-2. Hence the probability would be 1/11.

However, this would only hold if the player would say nothing when no 2 were spotted.

If (before peeking) your friend is determined to say "at least one of the dice is a X" where X is one of the numbers he may spots, the probability for a pair would be different.

If the friend sees any pair (1/6), probability of "X-X" where X is the named number, is one.
If the friend sees a non-pair (5/6), probability of "X-X" is zero whatever he choses for X.

Hence, like the Monty-Hall-Problem, if your friend must always say "At least one of the dice is a X" (while not lying), probability of a pair is 1/6."

[regnis]

I'm going to try to make this real easy AGAIN!!!

I roll one die. It is a 2.

What were the odds that it was going to be a 2----1 of 6.

What are the odds that it is a 2----100%.

It is all about understanding plain simple english.


This has been argued for months now and here it is in plain simple english. We (1-6ers)see the 2 and therefore only consider the odds on the remaining die. You (1-11ers)are viewing the probabilities of something that the plain english does not ask for.

[RS__]

I'm going to try to make this real easy AGAIN!!!

I roll one die. It is a 2.

What were the odds that it was going to be a 2----1 of 6.

What are the odds that it is a 2----100%.

It is all about understanding plain simple english.


This has been argued for months now and here it is in plain simple english. We (1-6ers)see the 2 and therefore only consider the odds on the remaining die. You (1-11ers)are viewing the probabilities of something that the plain english does not ask for.

Your post perfectly describes the 1-in-6ers' folly.


Let me try this:

I have 4 cards: Ace and king of clubs (black). Ace and king of hearts (red). I deal you one black card and then one red card. At least one of the cards is an ace. What is the probability/chance both cards are an ace?

1/2
1/3
1/4
1/8
1/16
Other


PS: I am using the same "at least one..." requirement as in the dice problem, ie: The peeker doesn't say something about every roll or every hand, but only when a deuce (or in this case an Ace) is present.

[kewl]

I'm going to try to make this real easy AGAIN!!!

I roll one die. It is a 2.

What were the odds that it was going to be a 2----1 of 6.

What are the odds that it is a 2----100%.

It is all about understanding plain simple english.


This has been argued for months now and here it is in plain simple english. We (1-6ers)see the 2 and therefore only consider the odds on the remaining die. You (1-11ers)are viewing the probabilities of something that the plain english does not ask for.

Try to rationalize on the example with the 36 cards. It demonstrates in a very simple terms why the odds of 2-2 are to be considered only in conjunction with what the odds of 2-x were prior to the throw.

We draw a random card and disregard all non 2 cards. In fact we are drawing one card out of eleven cards (that's the number of the cards that matter to us).
We know the outcome. Its 2-x. What are our odds it is 2-2?

[kewl]

Also, why are all 1/6-ers forgetting/not commenting (excluding Alan's ridiculous views) the fact that the bet from page 1, based word by word on the question is yielding 1/11?
Is it because they believe the question is supposed to be a one time event? What if so? This doesn't change the odds one bit.

[kewl]

In the real physical world where odds of 9 to 1 are available on unlimited wagers where probability is 1/6 there would be a stampede of takers.

Any thoughts on this?
What does the real, physical world say about the odds?
What does it say when we experiment one time across many experimenters?

How come all of them agree it is 1/11?
Where does the 1/6 come from???

Oh, yeah, it is because Alan says so. What about any proof? Oh, yeah a die has six faces... Nevermind he says that in a totally irrelevant context - when the throw is made and he comes in to pick the other die and "conclude the real odds". A total brain mess...

[OnceDear]

.... We (1-6ers)see the 2 and therefore only consider the odds on the remaining die.

No you don't. Only the peeker see's the dice. Only the peeker can eliminate one die, and that peeker could eliminate probability completely because he knows the full outcome.

Unless YOU can see through the cup, you cannot eliminate the red die, or blue die or left die or right die or big die or small die. All dice faces are in play and unknown to you.

First re-read the wager offered in post 1 of this thread !!!

Take on the damned wager. Big stakes or little stakes or roll the dice yourself. This is not about English. Just damned well do it. Try two dice of different size or colour. After you've rolled the dice a dozen times you will realise that you were abjectly wrong to eliminate one die. Unless you are the kind of idiot who would just roll one die. That's why I said re-read post 1 !!!

If you want to use identical dice the outcome will still tend towards 1/11 it will just be a tiny bit harder to visualise why.

[regnis]

I don't eliminate a red die, a blue die, a left die, or a right die. I eliminate the one with a 2 on it. That leaves one die for which the odds are 1 of 6.

I clearly explained the difference between a die and cards or marbles. Once one die is eliminated, you have eliminated all the other options on that die.

[regnis]

No you don't. Only the peeker see's the dice. Only the peeker can eliminate one die, and that peeker could eliminate probability completely because he knows the full outcome.

Unless YOU can see through the cup, you cannot eliminate the red die, or blue die or left die or right die or big die or small die. All dice faces are in play and unknown to you.

First re-read the wager offered in post 1 of this thread !!!

Take on the damned wager. Big stakes or little stakes or roll the dice yourself. This is not about English. Just damned well do it. Try two dice of different size or colour. After you've rolled the dice a dozen times you will realise that you were abjectly wrong to eliminate one die. Unless you are the kind of idiot who would just roll one die. That's why I said re-read post 1 !!!

If you want to use identical dice the outcome will still tend towards 1/11 it will just be a tiny bit harder to visualise why.

The peeker said one is a 2. The other one has 6 faces. Can we please drop this nonsense.

[regnis]

Try to rationalize on the example with the 36 cards. It demonstrates in a very simple terms why the odds of 2-2 are to be considered only in conjunction with what the odds of 2-x were prior to the throw.

We draw a random card and disregard all non 2 cards. In fact we are drawing one card out of eleven cards (that's the number of the cards that matter to us).
We know the outcome. Its 2-x. What are our odds it is 2-2?

Please address my example of one die.

Again--cards are different because removal of a card leaves the other options. Removal of a die removes the other options.

[Alan Mendelson]

Regnis they cannot drop their nonsense because they were not taught to think in real terms and only theoretical terms. Wizard proved this in his video when he rotated the die with a two. They cannot deal with the question in a real, physical world.