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

[OnceDear]

Okay, then I'm gone. May I suggest asking the math guys... When at least one 2 is rolled, what are the odds the "other" die also rolled a 2?

I bet they give your argument in reverse. I bet they'll say that the odds the "other" die rolls a 2 is 1 in 11, because if the "other" die rolled a 2, that would be a 2-2, which had a 1 in 11 chance of happening. So, they'd say, the "other" die had a 1 in 11 chance of being 2. They would be wrong, and you would be right saying 1 in 6.

I'm one of said Maths guys. I'm an English graduate, in England. English is my first language. I'm familiar with some American English.
I'd go with 1/6 to that latter question.
Thanks.
Thanks for trying to persuade Alan, but give it a rest now, as I have money at stake )

[RS__]

Alan, I'm being serious. Try to imagine that scenario I presented above (where I roll a bunch of dice [60] and remove some of the 2's).

Are you being stubborn or ignorant? You seem very uncooperative in all of this. Why? Try to think outside the box a bit, instead of disallowing any other scenario.

[OnceDear]

...Try to think outside the box a bit...

That's just the kind of maths guy ' overthinking' that led to all this trouble. He won't fall into that trap. He's not stupid. Ah! This forum doesn't support strikethrough font.

[kewl]

One more time:


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?


What do the 1/6-ers say?

[kewl]

When at least one 2 is rolled, what are the odds the "other" die also rolled a 2?

Its still 1/11. If you ask the question before you throw the dice and not after seeing which die has a two. And disregard all non 2 throws.

[kewl]

I should probably be making fun of your reading comprehension. My take on this entire waste of space has been as crystal clear from the start as your trick question. Some people (arci, Rob) got it. Some people didn't.

Gee, I must have divided the reading audience into those who got it and dullards. Mercy me, what a rush (yes, Sheldon, that was sarcasm). What a clever boy I am.

Sorry but for a guy with avid writing skills you really said nothing of substance here. What a waste of space.

[kewl]

A quote from ME from the WoV forum:

"What is the probability of a fair throw of two standard dice showing 2-2 given that at least one of the dice shows 2?"

No humans,intentions or any unintended ambiguity. Though IMHO the original question is written very well and not tricky or poor writing, or anything. The 1/6-ers just can't grasp the concept. Instead they're accusing all the rest in trickery,overthinking and what not. Its not like that at all. Its a simple probability question from our very real and physical world, a little counter intuitive but that's all.

Your answer people?

[Alan Mendelson]

If you want me to say "I'm wrong" you are going to have to demonstrate the 1/11 answer using two dice, with one die showing a 2. That is what I asked in the video. Neither miplet or the Wizard showed the 1/11 with one die showing a 2. Both showed a 2 and then proceeded to alter the faces on the die that showed a 2 in order to reach their 1/11 result.

I am not questioning the 1/11 result when you are asking how many combinations of two dice include at least one 2, and of those how many would result in 2-2. That would be 1/11.

But given that at least one of the dice has settled on 2 -- SHOW ME -- using two dice that the chance of 2-2 is 1/11 and not 1/6.

Please, you have to use two physical dice.

[kewl]

If you want me to say "I'm wrong" you are going to have to demonstrate the 1/11 answer using two dice, with one die showing a 2. That is what I asked in the video. Neither miplet or the Wizard showed the 1/11 with one die showing a 2. Both showed a 2 and then proceeded to alter the faces on the die that showed a 2 in order to reach their 1/11 result.

I am not questioning the 1/11 result when you are asking how many combinations of two dice include at least one 2, and of those how many would result in 2-2. That would be 1/11.

But given that at least one of the dice has settled on 2 -- SHOW ME -- using two dice that the chance of 2-2 is 1/11 and not 1/6.

Please, you have to use two physical dice.

I don't WANT you to do anything. It'd be cool though if you and the rest of 1/6 supporters could see the question and the real probabilities for what they are.

My demonstration is as follows:


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?


Or, you could roll a pair of dice a statistically significant amount of times,disregard all the non 2 rolls and count all the 2x outcomes in relation to all 2-2 outcomes.
What that does is - demonstrate and prove that whenever you have at least one 2 showing the odds for a pair of 2's is 1/11.

[Alan Mendelson]

I'm sorry but I am going to tell you what the problem is: this is not a question of probabilities the way you have been presenting it. This is more a problem of arithmetic. It's really very simple -- almost mechanical. But you have turned it into a question of probabilities.

If it were a question of probabilities the answer would clearly be 1/11. But we have more information that cannot be overlooked and cannot be dismissed:

We are told that there are only two dice, and at least one of those two dice shows a 2. And that takes this problem out of the realm of probabilities and brings it down to earth as a question of simple arithmetic. One die is a 2 -- and how many other faces on the other die would create 2-2?

As I have said so many times, you math guys have over-thought a simple problem and by overthinking you came up with an impossible answer. Really -- 1/11 is impossible when you know what one of the two dice is and when you clearly understand the question.

So if you can possibly take two dice and show me the answer is 1/11 without the flip flopping of faces which is what the Wizard and miplet did, then I will say it's not a simple matter of arithmetic. But right now that's all it is. And I am very tired of you and me and the others going over the same arguments so many times.

Show me a video.

[regnis]

In general, wall's aren't real bright.

try a semi-gloss

[kewl]

I'm sorry but I am going to tell you what the problem is: this is not a question of probabilities the way you have been presenting it. This is more a problem of arithmetic. It's really very simple -- almost mechanical. But you have turned it into a question of probabilities.

If it were a question of probabilities the answer would clearly be 1/11. But we have more information that cannot be overlooked and cannot be dismissed:

We are told that there are only two dice, and at least one of those two dice shows a 2. And that takes this problem out of the realm of probabilities and brings it down to earth as a question of simple arithmetic. One die is a 2 -- and how many other faces on the other die would create 2-2?

As I have said so many times, you math guys have over-thought a simple problem and by overthinking you came up with an impossible answer. Really -- 1/11 is impossible when you know what one of the two dice is and when you clearly understand the question.

So if you can possibly take two dice and show me the answer is 1/11 without the flip flopping of faces which is what the Wizard and miplet did, then I will say it's not a simple matter of arithmetic. But right now that's all it is. And I am very tired of you and me and the others going over the same arguments so many times.

Show me a video.

Ok, a video(a virtual one, like in my mind) -

I hold a pair of dice in my hand and I start to think:

What are the odds I'll roll 2-2? 1/36? Correct.

Now, what are the odds I'll roll at least one 2? 11/36? Correct.

Now, I'm not interested in all other rolls. I only care about the "at least one 2" and a"2-2".

So, (keep thinking and holding the dice in my hand) when I'm rolling at least one deuce, 1 out of those 11 times I'll have rolled a 2-2.

How come,then, when I actually throw the dice, the 1 in 11 chance of rolling a 2-2 when at least one 2 is rolled, somehow collapses to a 1 in 6 chance?

Can you explain that for me?

In the OQ I'm told at least one of the dice is a 2. That is exactly the same as what I am thinking before the actual throw - I have one out of eleven chance to roll 2-2 given that I roll at least one 2.

So,how does the probability collapse to 1/6?

Edit: Sorry,where it says "collapse" it should read "expand" or "rise". I'm not a math guy.

[redietz]

Sorry but for a guy with avid writing skills you really said nothing of substance here. What a waste of space.

You use a poorly written trick question to have intellectual orgasms over other people's failings to grasp your cleverness. It's pathetic, really. I'm hoping none of you folks are in teaching professions.

Glad to hear you think I have avid writing skills. I've never made such a claim, and never will, but it's good someone thinks I'm competent, at least.

[OnceDear]

Ok, a video...
So,how does the probability collapse to 1/6?

Edit: Sorry,where it says "collapse" it should read "expand" or "rise". I'm not a math guy.

Kewl,
Don't you get it? As Alan has pointed out :-

I'm sorry but I am going to tell you what the problem is: this is not a question of probabilities

Alan forgot to capitalise the T' after the colon.
I'm Just saying, because these things are important to we English.

[redietz]

One of the issues I'm having with the attitude of the WoV posters is that they're attempting to denigrate a public person, namely Alan, while (A) using language they wouldn't use face to face with any one of us and (B) maintaining anonymity. Academic journals don't publish papers from the anonymous. Step out, sign your real names, and stop being a high school math clique.

God knows, arci and I have gone head to head with Alan tooth and nail over many math issues, and I have been sarcastic and direct on many occasions. But you guys go way, way beyond the bounds of semi-formal civility and set records for being snide. So instead of these weirdly cliquish narcissistic responses, why not just sign your names and pretend there's an actual person facing you?

[Rob.Singer]

One of the issues I'm having with the attitude of the WoV posters is that they attempting to denigrate a public person, namely Alan, while (A) using language they wouldn't use face to face with anyone of us and (B) maintaining anonymity. Academic journals don't publish papers from the anonymous. Step out, sign your real names, and stop being a high school math clique.

God knows, arci and I have gone head to head with Alan tooth and nail over many math issues, and I have been sarcastic and direct on many occasions. But you guys go way, way beyond the bounds of semi-formal civility and set records for being snide. So instead of these weirdly cliquish narcissistic responses, why not just sign your names and pretend there's an actual person facing you.

That'll never happen red. These guys are making so many personal attacks, first because they're not used to anyone contradicting them or telling them why they're wrong (Heaven FORBID!!), and secondly because the admins on their forum are such big wussys that anyone and everyone will get banned for hurting their feelings in the slightest. Face is a penniless non-gambler, Mission saves up all year to make that big trip to LV to play quarters and lose, beachbum wishes he were gay, and Shack gets torn between using that stupid "nuclear option" & going into a dark corner to cry if someone gently insults his social or gambling beliefs. So they come over here to let off steam, but with wizard's blessing. And you have to remember, most of the "geniuses" over there gamble very little or not at all because they much prefer the safety of theory over reality.

If you want to know their real names you'll have a long wait indeed. I do know one of these guys since I've trained him. But there's so much sensitivity with the group overall that all you'll get is more insults the more you ask.

[arcimede$]

Look math guys -- there are two ways to attack this problem: using rational, common sense, or creating the blue print for a Rube Goldberg machine.

And stop with your "test" of 1/11. Why don't you try MY TEST: Set one die as a 2 and then roll the other die and see if a 2 shows up 1/6 times?

Because that is not the problem that was asked. Why do you continue to deny the English language?

[jbjb]

Great one! And that rant above coming from the insult king! But agreed, there shouldn't be any.

I've been kicked out of more casinos than you all have ever been in, COMBINED! Last thing I'd do is reveal my identity anywhere!

One last thing. You COULDN'T PAY ME to be "trained" by Singer!! That's asinine!!

[kewl]

You use a poorly written trick question to have intellectual orgasms over other people's failings to grasp your cleverness. It's pathetic, really. I'm hoping none of you folks are in teaching professions.

Glad to hear you think I have avid writing skills. I've never made such a claim, and never will, but it's good someone thinks I'm competent, at least.

Huh? I use nothing to have nothing dude. I just love the "oh it's that then, wow that was more satisfying then I thought,I really get it now,plus it will help me make much better decisions from now on" moments in my own experience and is good to see more of these around.

I know this thread is long and one can easily be confused, but you either mistake me for someone else or are just indeed a very strange dude. Nowhere did I try to act as if I'm(or we, or whatever) superior or stg. Pointing errors doesn't necessarily come with intellectual orgasms, get over yourself.
As for your writing skills I should have made it clear that it is you and Alan and others who point to them so I referred to them like in quotes and in the end you didn't say anything after all.

This is not your usual trick puzzle, when you know why the answer is 1/11 you really get stg from it. Stg substantial.

[kewl]

Kewl,
Don't you get it? As Alan has pointed out :-

Alan forgot to capitalise the T' after the colon.
I'm Just saying, because these things are important to we English.

I know, I'm trying to demonstrate it is a question of probabilities cause I thought I can come up with one or two better examples and mind experiments then what were given so far, but alas , they don't work either so far...