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

[Alan Mendelson]

They're still talking about me over on the WOV forum and I am really surprised at what they're saying.

First from miplet who did one of the videos:

"I think Alan's missing the fact that once you roll two dice at the same time until there is at least one 2, there are only 11 possible outcomes. Just one of those outcomes is a pair of 2s. If he does understand this, I don't know how he is getting 1 in 6."

I'm sorry miplet but once you do that roll with two dice and you know "there is at least one 2" there are only 6 possible outcomes -- which are the six faces on the other die.

And from the Wizard:

"I'm sure deep down Alan knows he is wrong. Otherwise, he would be betting thousands of dollars as opposed to lunch money. However, I truly believe he doesn't understand the flaw in HIS argument. I tried to explain it in my video. He is simply can't get past a die having six sides."

That's right, I can't get past a die having six sides.

And it gets funnier. From Ibeatyouraces:

"12 faces on the dice. One is known, but since we can't know exactly which one, you CANNOT discount the other 11.

I'm not a "math guy" and I can clearly see this."

No, I can't discount all of the other 11. But I can eliminate from consideration the five other faces that are on the die with the 2.

[Rob.Singer]

Alan, they're still talking about and making fun of you on WoV because they indeed have recognized you exposed a flaw in both the original question wording and how they all quickly & very unwisely followed the wizard down the tunnel of mis-interpretation with their blinders firmly on. They now have no choice but to dig in. And the fact that RS keeps showing up here on orders from Shack just to repeat the same verse, is pure gold, if you know what I mean.

[redietz]

I think the original question refers to one throw and one peek. It's part of the verbal illusion.

You know, Shackleford making that "Alan can't get past a die having six sides" comment is pretty funny. Maybe it's deserving of multiple threads and 300 responses and loads of debate. Was Shackleford being unbelievably stupid or was he making a cogent comment about Alan's psychology? That line could be taken either way.

Perhaps we "writing guys" should repeat ad nauseam how intellectually superior we are to anyone who could stupidly write something that could be interpreted, at face value and in isolation, as really, really dumb.

Naaawww, we don't have the time or inclination.

[Alan Mendelson]

The show continues. Now the participants on the WOV are arguing about the rules and the conditions in the original question:

"Ayecarumba" says: What I am proposing is that the peeker would call out whatever was on the observed die.

"Indignant" says: You'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."

Let me remind everyone on the WOV what the original problem said:

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?

I don't think the peeker was obligated to say anything but the truth. And when he said "at least one of the dice is a 2" it could have been that he saw one 2 or a pair of deuces.

[RS__]

I said when you see a 2 playing craps, the chance of seeing 2/2 is a 1/36 event.

THIS is gold.

[arcimede$]

One die is a 2. For both to be a 2, the other has to be a 2. What are the odds of that----1 of 6. The one that was a 2 can't suddenly become a 1,3,4,5, or 6 to create more combinations. I once got beaten pretty badly for having dice like that!!!!

Once a 2, always a 2. Arci's dice are chameleons.

Yes, my dice are chameleons. They can be either red or blue to help folks understand the problem. On any given throw the red die can be 2 or the blue die be a 2 (or both). If you ask what are the odds of the blue die being a 2 when the red die is a 2 the odds are 1-6. Or, if you ask what are the odds of the red die being a 2 when the blue die is a 2 the odds are 1-6. But, those are NOT the question. The question allows EITHER die to be a 2 which means all those situations could occur.

[Alan Mendelson]

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?

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.

[OnceDear]

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?

What does that even mean, Alan?
On any roll of the dice, any slamming of the cup, there will be no change of anything. The outcome would be the outcome. The probability of that outcome, until it's revealed, of being a pair of deuces, does not change. When it's been revealed what the full outcome is, by lifting the cup, then the probability of the dice being a pair of deuces will either be 0% (if it's not a pair) or 100% (if it is a pair). But until the great reveal, the probability of the pair is 1/11.
Are you now at the Eureka moment when you realise that the original question has the one correct answer of 1/11?
If so, what persuaded you, and are you grateful?

[Alan Mendelson]

OnceDear we'll have to wait for Arc to respond. Thanks for dropping by. And I hope you collect on your bet with the Wizard. Send my regards.

[OnceDear]

And welcome back to the forums Rob. Unbanned how often now?
Gotsta say Alan is a generous host.

()

[Rob.Singer]

And welcome back to the forums Rob. Unbanned how often now?
Gotsta say Alan is a generous host.

()

That would be "forum" once dear. If you haven't heard or can't tell by how often my name appears on the forum now owned by online casino activists, I'm too feared over there to be "welcomed". BTW I saw where the collection of genius moderators banned mickeycrimm. You mean they finally figured out his stories of gambling-prowess were nothing more than what one would get when visiting Scotty of Scotty's Castle lore? A little slow for the self-proclaimed Walter O'briens of the world, wouldn't you say??

[OnceDear]

That would be "forum" once dear.

I stand corrected Rob. Oh, and it's Oncedear, all one word.
But I respectfully withdraw my welcome because of your hostile tone. 'Gotsta' was bad English too, do you wish to pull me up on that? Maybe I misused an apostrophe somewhere.

I know that there is history and hostility twixt you and that other forum. Not my history: Not my hostility.

I wouldn't say anything about Scotty or Walter, because those names are not familiar to me and I'm not inclined to google them. My opinion on the banning of MC is all by the by.

[kewl]

Wouldn't all 1/6-ers be persuaded by the fact that if they put money on a bet which follows the OQ exactly, i.e. roll dice,peek and announce if at least one deuce is showing and if yes,then bet on 8 for 1 odds for a pair of deuces, they will lose in long term?

Doesn't that mean that the real,physical dice concur that indeed the probability of a pair of deuces when at least one deuce is showing is 1/11 in the real,physical world?

Or do they still believe the bet will win them money?
Cause it was proved and demonstrated (and even the 1/6-ers know it) in multiple ways that the 11 combos of showing at least one deuce are "in play" before any particular throw and they WILL come out 11/36's of the time. And only 1 out of those 11/36's will there be a pair.

[OnceDear]

Wouldn't all 1/6-ers be persuaded by the fact that if they put money on a bet which follows the OQ exactly, i.e. roll dice,peek and announce if at least one deuce is showing and if yes,then bet on 8 for 1 odds for a pair of deuces, they will lose in long term?

What Kewl, you mean like when I offered to stake my entire and not unsubstantial pension fund? I got no takers for that. Wife is grateful.

[Alan Mendelson]

When I roll two dice I get 2-2 1/36 of the time -- not 1/11 and not 1/6. When I know one of two dice is a 2 the chance of another two on the other die is 1/6. What bet do you want?

[kewl]

When I roll two dice I get 2-2 1/36 of the time

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?

[redietz]

OnceDear,

Yeah, I'll bite. In my opinion, misusing an apostrophe is usually a more egregious intellectual error than getting the dice problem "wrong." What makes you think your dice problem trivia is somehow more important?

[regnis]

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.

[regnis]

And welcome back mr. Singer!!!!!!!

[jbjb]

The bet is as follows:

Roll 2-Anything, lose.
Roll 2-2, get paid 9:1.
Roll anything else, bet pushes.

Only a moron would take this bet!