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

[kewl]

Are you asking me? Yes, there are two choices for each die. Much different than the question about being a 2 and what are the odds for another 2.

The main thing you need to understand in the original question is that the probability of throwing 2-x with a fair pair of dice is 11/36.
From there, you have this portion of all possible throws representing 100% of your region of interest.
Eleven unique outcomes is all you got.
Now, when I tell you "at least one 2 is showing" that will be 100% of the times true. And only one time out of eleven will you be right if you call "the other die" as a 2.

[Alan Mendelson]

Ya know kewl, we've been through all this ... a thousand times? It's just that when you look at this particular problem of two dice with at least one showing a 2, the conditions don't call for this exercise of looking at 36 combinations or 11 combinations.

You won't be convinced because you are sticking to a regimen that just doesn't apply to the conditions of this one problem.

[kewl]

You won't be convinced because you are sticking to a regimen that just doesn't apply to the conditions of this one problem.

It is not a regimen, its common sense.
The common sense calls to look at the odds to arrive to "at least one 2" and compare them to the odds to arrive to 2-2. But you are sticking to "I am told one die is a 2, therefore the odds only apply to the other die now." Okay, but why does the simulation prove that when you have at least one 2, you have 2-2 only 1/11 of the time?

[Alan Mendelson]

Okay, but why does the simulation prove that when you have at least one 2, you have 2-2 only 1/11 of the time?

Does it? Does the simulation actually apply to the condition when you know at least one die is a 2? This is what we've been arguing.

[RS__]

Does it? Does the simulation actually apply to the condition when you know at least one die is a 2? This is what we've been arguing.

Yes, it does.

[indignant99]

Does it? Does the simulation actually apply to the condition when you know at least one die is a 2? This is what we've been arguing.

YES, the simulation is well aware that these are the conditions where "at least one die is a 2."



Meanwhile, you deny this; and say you only get 6 horizontal. Exclusive-or 6 vertical.
Sorry, Alan. You're flat incandescent, neon-flashing, glowing, flaming wrong.

[Alan Mendelson]

Attention everyone from the WOV forum: If any of you ever again use any vulgarity, with or without letters missing, I will block you from posting the same way the moderators on the WOV forum would block you. It's true I am a supporter of free speech but this is a family-friendly website and I don't want to see any more of it. Final warning.

Thanks for posting indignant99 and welcome.

[kewl]

Does it? Does the simulation actually apply to the condition when you know at least one die is a 2? This is what we've been arguing.

Yes, it is precisely what the simulation applies to.

It simulates a large number of random rolls( 50 000) of a pair of dice. It counts the times when either or both dice is a 2 - that's our "you know at least one die is a 2".
It then counts the number of times when both dice are 2 - that's our "the other die is also a two".
And then it calculates the ratio between the two results - that's our "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".

I've initiated about 20 different simulations (1 000 000 trials) and the ratio ranges somewhere between 1/10.4 and 1/11.7.
Right about the closest thing to 1/11. And pretty damn far from 1/6.

[OnceDear]

OnceDear's spread sheet and all those other spreadsheets looked at the combinations of 2 dice that contained at least 1 two....

.

No it didn't. Have you even looked at it?

I won't accept your simulation unless it looks at one die. Guess why I said that!

Because you are an ignorant buffoon?

[RS__]

By the way, I've done my own simulation (in Excel). It's very easy and straightforward. Instead of using yes's it uses 1's. Instead of using no's it uses 0's. I actually made the spread sheet because I wasn't 100% sure the answer was 1/11 and wanted to confirm [or disprove] that answer.



What do you think about this scenario:

I have 10 sports-bet tickets in my hand, they are each $100 to win $100 ($100 to pay $200). I have a 50/50 chance of each individual winning (ie: #1 has 50% chance to win, and not related is #2 which also has a 50% chance to win, etc.). You can't see the outcomes on which tickets won or which lost. But I can see the outcome of each ticket. I tell you, "Hey, I'll give you a random ticket for $100, Alan." After all, it would be a break-even no-one-has-an-edge scenario (if I really did give you a random ticket for $100).

I look through all my tickets, and pocket a few of the winning tickets (let's say I pocket 3 winners). Of the 7 remaining tickets, is it still fair (no-edge either way) for you to buy one of my tickets for $100?


Let me make it 1000000x easier for you to answer this.

I have a $10k TITO, $5k TITO, $1k TITO, and a $400 TITO in my wallet. The total is $16,400. The average is $4,100. I tell you "I'll sell you a random ticket at a good rate, just $3,500! And it's valued at $4,100!" Any smart person would do this (IF it was fair). So naturally, you agree -- you give me $3,500 and I give you a random TITO. But before I pick a random TITO (of which I cannot see when I pick it out at random), I FIRST remove the $10,000 ticket from my wallet and put that in my other pocket.

Would you still think buying the ticket at $3,500 is a good deal? Of course not.



It's the same thing with the two dice. By "removing" one of the dice with a deuce on it, you are lowering the chances of the other die (or both dice) being a deuce.

[indignant99]

So now, counter-argue the content.

Quit side-stepping, matador-style, out of the way of the bull's horns.

You can't counter-argue, so you tiptoe to the side, and cry over a pre-censored adjective. Avoid direct confrontation. You're a regular Ray Comfort. Some weekend, go down to Huntington Beach, and watch that charlatan preach. You'll pick up some tips on deflecting, straw-manning, clamming-up, Gish-galloping, and outright lying.

[Alan Mendelson]

Simulate this for me if it's not what you already simulated: you have two dice. One lands on a 2. What is the probability that the other die is also a two?

[kewl]

Simulate this for me if it's not what you already simulated: you have two dice. One lands on a 2. What is the probability that the other die is also a two?

What is this? One has landed and the other one is mid air or what?

[RS__]

Do you mean a specific die landed on a 2 and we're looking at the other die? Or that at least one die landed on a 2?

[OnceDear]

Simulate this for me if it's not what you already simulated: you have two dice. One lands on a 2. What is the probability that the other die is also a two?

That's so easy that Alan could do it himself with my workbook.

Just replace ALL OF COLUMN A with the value 2. Not a lot of excel prowess required.

The proportion calculation would stop giving 11 as its answer and would start giving 6 instead. Different question: Different answer.

We always told you that we are interested in when 'AT LEAST one die is a deuce' Not when 'ONE KNOWN die is a deuce'

You even agreed what 'AT LEAST' means.

So, when you modify the workbook and it gives you your different answer of 6, you have to admit that different question yields different answer.

Then YOU have to define why the original workbook is wrong? In what way is it NOT dealing with 'AT LEAST' one die is a deuce.

It's not acceptable to change the workbook and point at the 6 (which answers a different question)

It is absolutely necessary for YOU to show in what way the original workbook is not representing the question.

[indignant99]

Simulate this for me if it's not what you already simulated: you have two dice. One lands on a 2. What is the probability that the other die is also a two?

Either one lands on a 2. Big hairy difference there, Bubba.

[RS__]

I think it's hilarious how Alan keeps saying the 'math/WoVers' are asking a different question than what the original question was about. Original question was 2 dice, at least one dice a 2, what's the chance both dice are a deuce? The question is NOT "one die is set to a deuce, what's the chance the other die, when rolled independently, is a deuce?" Alan is the one changing the question.

[indignant99]

Alan is the one changing the question.

No ...champagne... Sherlock!

[OnceDear]

I'm really grateful to Alan and friends for giving me the opportunity to monetize this situation.

I've been running a book on this with a wager called 'The Educate the Donkey Derby.'
It's a simple setup.

We have a Donkey and a Trainer, unlimited training resources and a pair of furry dice.
All the trainer has to do is to train the donkey to roll both the dice at the same time. All the donkey has to do is fail! He can sit there and snooze, he can bray loudly, or he can pretend to pay attention to the Powerpoint presentation and flowcharts.

My money is on Donkey and I'm coining it.

[Rob.Singer]

Another day, and more WoV frustration, insults, & vulgarity....and all because they simply cannot read simple English, or more to the point, that it took someone not from their special little clicque to expose the question for what it was.

I wonder....if I went over there and called them names (like homos or atheists) just because they aren't thinking the way I want them to, what do you think those sensitive admins would do?

Funny stuff.