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

[Rob.Singer]

Regnis go post the common sense of how a 2 is always a 2 on a die that no longer moves, on the WoV forum, then watch as the super duper math conspiracy theorists all attack you at the same time. We already know of their thin skin as shown by the gamblers wizard sent over here for his own CYA needs.

Oncedear, Scotty was our most famous made-up story teller of all time. Walter O'brien was a true genius. From here--not the UK.

[kewl]

This is like a practical joke on you Alan. All these people know it's 1 of 6 but they are playing a joke trying to get you to switch.

This is how they get convictions here in Chicago Alan. You know you didn't do it but they will keep telling you that you did it until you can't take it anymore and give in and confess.

Stay strong Alan!!!! On my dice the 2's don't disappear and reappear. Once you roll a 2, it's a 2. The other die only has 6 options--1 to 6. I've got the proof to exonerate you.

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?

[OnceDear]

I'd like to offer my sincerest thanks to Regnis and Rob.

[jbjb]

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?

Don't help them.. you'll cost us money!

[OnceDear]

Don't help them.. you'll cost us money!

You helped them too jbjb. Even I helped them. We live and learn.
Now with cold hard cash at stake, I'm grateful to anyone who helps demonstrate that Alan is correct and to anyone encouraging him to 'stick it to' those maths guys from WOV.
I'm on Alan's side now.

$;o)

[Alan Mendelson]

Yep.
And you get at least one deuce 11/36 of the time, yes?
And no deuce 25/36 of the time, yes?

So, since you'll be interested only in those 11/36 of the time when at least one deuce is under the cup, there will be only one 2-2 for those 11 times,yes?

Regarding the cup question: when one die shows a 2 the other five faces are not eligible. That leaves 1/6.

[OnceDear]

Regarding the cup question: when one die shows a 2 the other five faces are not eligible. That leaves 1/6.

Way to go Alan. Of course they're not. You stand your ground mate!

[kewl]

You helped them too jbjb. Even I helped them. We live and learn.
Now with cold hard cash at stake, I'm grateful to anyone who helps demonstrate that Alan is correct and to anyone encouraging him to 'stick it to' those maths guys from WOV.
I'm on Alan's side now.

$;o)

https://vine.co/v/b9IXzIqe9F2

[kewl]

Regarding the cup question: when one die shows a 2 the other five faces are not eligible. That leaves 1/6.

But the question was:

So, since you'll be interested only in those 11/36 of the time when at least one deuce is under the cup, there will be only one 2-2 for those 11 times,yes?

I don't see how you answering this question?
So, yes or no?

[kewl]

Also please consider my reply to regnis from above.
What are your thoughts about the questions raised?

[OnceDear]

https://vine.co/v/b9IXzIqe9F2

Nice. But seriously, we have to admit that Alan has been right all along..... If I'm to collect on my advantage play.

[Alan Mendelson]

But the question was:

So, since you'll be interested only in those 11/36 of the time when at least one deuce is under the cup, there will be only one 2-2 for those 11 times,yes?

I don't see how you answering this question?
So, yes or no?

I think the bet you are proposing and has been proposed before is a good bet. But it involves the variance of dice so I could lose.

Since you're so sure the answer is 1/11 why don't you offer me even better odds?

Better yet-- let's skip rolling two dice and just set one as "2" and roll the "other die"? But wait-- you folks have been saying you can't tell which one the "other die" is.

[kewl]

I'm proposing even "better" bet - zero variance.
We set a pair of dice in all 36 possible combos one at a time and then start all over. Every time there is at least one 2 you bet and if there is 2-2 you win 8 for1 (alright , I'm giving you 9 for 1, I feel generous). If not 2-2 - you lose your bet (its a flat betting though,no cheating).

[OnceDear]

I think the bet you are proposing and has been proposed before is a good bet. But it involves the variance of dice so I could lose.

Indeed, one session of this game would involve a lot of variance.
Whoever claims victory could hardly call it a proof. I'm already relying on that variance factor to pull me back from the brink....Of p1551ng myself laughing

[OnceDear]

I'm proposing even "better" bet - zero variance.
We set a pair of dice in all 36 possible combos one at a time and then start all over. Every time there is at least one 2 you bet and if there is 2-2 you win 8 for1 (alright , I'm giving you 9 for 1, I feel generous). If not 2-2 - you lose your bet (its a flat betting though,no cheating).

That's just silly Kewl. This is a two dice problem.

But there is a way that Alan could eliminate variance while still using fair and random throws of the dice. He could hedge his bets.

[Alan Mendelson]

Regarding your bet: if one die shows a 2, eliminating five faces, how can you believe 2-2 is still 1/11? I want you to think carefully before betting.

[arcimede$]

Arc I think the point is this: I understand what you are saying now. You are just using colors instead of saying, for example, the die on the left or the die on the right. You are not saying that the numerical value will change.

For example on roll #1 if you say the Red=6 and Blue=5 there will not be a change for roll #1.

Do I have it correct now?

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

The problem was that others have suggested changing dice "values" to justify their 1/11 claim, and they have used "colors" to illustrate what they are saying -- and I think that is what caused confusion. I know it caused confusion with me.

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.

[kewl]

I want you to think carefully before betting.

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?

[kewl]

Regarding your bet: if one die shows a 2, eliminating five faces, how can you believe 2-2 is still 1/11? I want you to think carefully before betting.

You didn't respond on my request to examine this:

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?

Why is this happening when the probability for "the other die" is 1/6?

[RS__]

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