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

[OnceDear]

Alan I swear you're a f****** r*****....

Are you taking bets on that RS. Can you give me odds?

[arcimede$]

Alan I swear you're a f------ retard....

What I think is happening is Alan doesn't truly read any of the comments. Even the ones he responds to. He just skims them at best. Hence, he doesn't have a clue what you said. Since he's already made up his mind I think he just assumes it is a waste of his time. The reason I said he was lying last night was kind of a slap in the face. I wanted to get his attention. It did for a brief moment and you could see from his response he hadn't read my previous comment at all ... and then he went right back into ignore mode.

So, I can see where you get that impression. You really are talking to a wall. In general, wall's aren't real bright.

[OnceDear]

So, I can see where you get that impression. You really are talking to a wall. In general, wall's aren't real bright.

When watching someone trying to train a donkey to count, who's the least intelligent? Donkey? Teacher? Cheering onlooker?.... or the bookmaker taking bets on the Donkey

[Alan Mendelson]

There's never a change after the dice are rolled. The result is the result. Not sure where you are going with this.



No one has suggested changing dice values on any given throw. The only changes that occur are the result of more throws. Every throw is a unique and independent event just like playing one hand of VP.

Thanks Arc I just wanted to be sure I understood what you said. Walls do listen.

[Alan Mendelson]

I'm trying my best, but hey who could I blame for my losses?
So, you agree on my proposed - zero variance - bet then, eh?

No. And how many reasons do you need?

[Alan Mendelson]

Alan I swear you're a fucking retard....

Can you say this on the WOV forum? You can't say it here either. Don't do it again.

[kewl]

No. And how many reasons do you need?

I don't know, you didn't give any. What are your reasons?

[OnceDear]

I don't know, you didn't give any. What are your reasons?

I'd venture that would be far too many dice. A different question.
When will you get back to the original question as asked? Alan's trying his best to get idiots like me to understand. If I can see the light, there is hope for you too.

[Alan Mendelson]

I don't know, you didn't give any. What are your reasons?

Here's one: when a 2 appears on a die it can't be changed to another number. If that were possible I would never seven-out.

Yes, on charts and graphs the answer is 1/11. With two real physical dice that are not rotated for convenience the answer is not.

[RS__]

When watching someone trying to train a donkey to count, who's the least intelligent? Donkey? Teacher? Cheering onlooker?.... or the bookmaker taking bets on the Donkey

Touche. I'm thinking donkey is the least intelligent.

Can you say this on the WOV forum? You can't say it here either. Don't do it again.

My apologies. I will try to not do it again. It's just that sometimes when you post, it makes me say things I normally would not say [see one of my previous posts].

[kewl]

I'd venture that would be far too many dice. A different question.
When will you get back to the original question as asked? Alan's trying his best to get idiots like me to understand. If I can see the light, there is hope for you too.

But its just one pair of dice. We set them one combo at a time as if we are throwing a pair of perfectly-distributed-dice and examine if there is a two showing. And then the bets are on.
I'm trying to see the light as well and if Alan can really state his point clearly maybe I'll convert too, who knows..."cough" doubtful "cough"

[arcimede$]

Walls do listen.

Only with simple straight forward comments. If the comment requires you to think you just dismiss it. So, no, the Alan Wall does not listen.

[kewl]

Here's one: when a 2 appears on a die it can't be changed to another number. If that were possible I would never seven-out.

Yes, on charts and graphs the answer is 1/11. With two real physical dice that are not rotated for convenience the answer is not.

But I'm not claiming the 2 can be changed to another number.
Please read this and comment on the questions raised if you care:

But if you follow the original question word by word you'll get the inevitable 1 in 11 ratio, so how can it be?

Let's make it even more childish.

Imagine a pair of dice which ,for the purpose of better visual explanation, always rolls a perfect distribution of all 36 combinations on every 36 rolls.
So that, when you roll them 36 times you get all 36 combos one after the other, no repeats. I'm guessing we can all agree that is not how it is in reality, but it is a simple representation of what the distribution is expected to be in infinity since its derived from the pure fixed probability for each and every die.

So you roll those dice 36 times and you get 25 "no deuce"'s and 11 "at least one deuce"'s. With exactly one "deuce-deuce" among those 11 "at least one deuce"'s, of course.

And again, new 36 rolls and the result is again - 11 "at least one deuce"'s and exactly one 2-2.

So, if you are to bet $1 on those 11 "at least one deuce"'s ,which the peeker meticulously announces, at odds of 8 for 1 that a pair of two's will be under the cup , you'd have lost $2 total.

So how can that be?
How can it be that a 1 in 6 probability of "the other die" being a 2 actually yields an 1 in 11 actual results?

Doesn't this tell us that the dice themselves actually agree with the 1/11 bunch?
Doesn't this mean that 1 in 6 is actually the answer to a different question?

[PharoahsWin]

You are correct, Alan, this is a trick question. I want to explain why it's a trick question. It's something I think everyone is missing the point on. The question asks, "If you roll the dice, and at least one is a 2, what are the odds that BOTH dice rolled a 2?".

That's the wording of the original question, and it's worded pretty different than normal speech. Normally, you would ask the question, "If you roll the dice and at least one of the the die is a 2, what are the odds the OTHER die also rolled a 2?" You wouldn't ask about both dice, after explaining that one is already a 2, you would just ask about the OTHER die.

This is why it's a trick question. In the original question, the answer is 1/11, the odds of BOTH dice rolling a 2 is 1 in 11. And, that's what the question asks. It's not asking the odds of the other die rolling a 2, it's only asking the odds for BOTH dice rolling 2's.

In the second question, the one you would ask in normal speech, the chances of the OTHER die also rolling a 2 is 1/6. The odds of 1 particular die rolling a 2 is 1 in 6. And that's what the 2nd question asks. It doesn't ask the odds for BOTH dice rolling 2's, only the "other" die.

This is why the question is a trick question. It asks the question in a different way than you would normally speak, and by doing this it gives a different answer than would normally be given.

I think a lot of these math guys don't understand why it's a trick question, most would probably, incorrectly, give the same 1/11 answer to both wordings of the question. It just so happens, in this case they are correct, the question asks the odds of BOTH dice rolling a 2, that's 1/11. If the question was asked about the "other" die, you would be right with 1/6, and I'm guessing most of the math guys would give the wrong answer of 1/11. But, this is why it's a trick question.

If you do the experiment itself, you'll see that BOTH dice will roll 2's 1 in 11 times. And, the "other" die will have a 1 in 6 chance of rolling a 2, just like in your video. So, depending on whether you say BOTH dice, or if you say the OTHER die, you'll get 2 different answers.

The original question asks for BOTH dice, it does that on purpose, to get the 1/11 answer, because if it asked about the "other" die, it would get a totally different answer of 1/6 (and this would trick a lot of people who think it's 1/11). Hope this helps explain why this is such a tricky question for you.

[RS__]

But its just one pair of dice. We set them one combo at a time as if we are throwing a pair of perfectly-distributed-dice and examine if there is a two showing. And then the bets are on.
I'm trying to see the light as well and if Alan can really state his point clearly maybe I'll convert too, who knows..."cough" doubtful "cough"

I don't think Alan quite understands how dice even work. Really.

There was a discussion before on the WOV forum about how if a dice controller could change the house edge. Alan's opinion was that no -- you could not change the edge, because every number has a 1/36 and 2/36 chance of being rolled, simply because each die has 6 faces x 6 faces.

Even after I gave an example, like, what if you could throw a 7 1/10 times and a 6 1/4.5 times and an 8 1/5 times....would the edge change on the 6 and 8? Alan said no, it does not, because there are 5/36 ways to roll a 6, 5/36 ways to roll an 8, and 1/6 [6/36] ways to roll a 7.


The one game Alan thought he was a master of (craps)....he really knowing nothing about.

[OnceDear]

The original question asks for BOTH dice,

Exactly and you know one of them is a two. So we are not interested in both dice anymore, just 'the other one'
I see where Alan's coming from. It's the same question. It has to have the same answer.

[RS__]

Wait wouldn't it be 1/36 for both dice to be a deuce? :claps_hands:

[OnceDear]

Wait wouldn't it be 1/36 for both dice to be a deuce? :claps_hands:

It's ALWAYS 1/36.

[RS__]

It's ALWAYS 1/36.

But the chance of rolling a 2 is 1/6.

(1/6) * (1/36) = 1/216


So the answer is really 1/216. Any1 interested in a side wager?

[OnceDear]

But the chance of rolling a 2 is 1/6.

I think you are wrong.
Take two dice.
Look at those dice.
Each die has 6 faces
That's 12 faces all together
So that's twelve different ways that they can land. Count them.
1 of those ways is Deuce Deuce.
So you can throw Deuce Deuce ON AVERAGE 1/12 of the time.
1/36 is just a trick played by the casino to maximise its edge.