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

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Originally Posted by Alan Mendelson

No. I am not saying it has a probability of 1/11. But I am saying he will probably win.

Are you aware of the specifics of the bet that I have with the Wizard? I tried to meet up with him in Vegas a couple of weeks ago when I was in town to meet with a client but I could not make it in time.

First of all none of you in the 1/11 camp has yet to read the "original problem" in the same way that those of us who say it's 1/6 read the problem.

You guys can't even tell me which die is showing the 2. Those of us who say it's 1/6 say it doesn't matter which die is the 2.

So, when you guys tell me it doesn't matter which die is the 2, OR, if you can tell me which die it is, then we can discuss a different bet.

Zedd you came to this forum late in the discussion. Early in the discussion I said the original problem resembled this event at a craps table:

Two dice are thrown and one die immediately comes to rest showing a 2. The second die becomes a spinner and spins like a top. What are the chances that the "spinner" will also land on a 2?

If you don't view this as the same as the original question, we are going in different directions on the same road.

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I’m aware of the specifics of the bet…which, as far as I know, is basically what you had described in your original post. These are the same specifics that OnceDear used in his spreadsheet.

I’m also aware of the original question. And you’re right; it can be read in different ways. The original question is ambiguous because it’s unclear how the condition of ‘at least one deuce’ was satisfied. Therefore, we must make assumptions on what sample space to use in our calculations. With this particular problem—one assumption produces an answer of 1/6 while another assumption produces 1/11.

BUT…this ambiguity is eradicated in your original post. We now know that the condition is satisfied only when EITHER one of the two dice is a deuce. This is different from when a specific die is a two—which is a more difficult condition to meet. Therefore, we must look at all two-dice combinations with either dice being a deuce (there are 11 of them).

1/11 is the absolute, no doubt about it, correct answer to the question in your OP.

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Originally Posted by Alan Mendelson

So kewl what do you think is the meaning of the original question? Do you think there are 36 dice in a bag, or eleven dice combinations showing a 2 and you have to pick from those 11 combinations?

Sorry. I don't buy it.

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The meaning? Its a very straightforward question, there is no hidden meaning to it.

You roll pair of dice. Mind you, they both come to rest peacefully, no spinners, no nothing.

Given that at least one of them shows 2, what are the odds they are both 2.

What is your proof the odds are 1/6? Its the fact that a die (the other die lol) has 6 faces.
Now, go ahead and show this "proof" to any person you trust is good in probabilities.
No, scratch that , I don't trust your judgment.

Show this "proof" to any person which has a proven record in dealing on at least mid level with probabilities.

You don't have a proof (you can't have one, cause your conclusion is incorrect, very incorrect). You are only running your mouth here, honestly. You have no idea how do pair of dice work in reality.

It is not a matter of ambiguous wording or misunderstanding the question. No. As I earlier quoted a perfectly worded question you answered the same - 1/6.

The odds of a fair throw of two standard dice showing 2-2 given that at least one of the dice shows 2 are?

You do understand that the Rawtuff's bet was word by word based on the original question, right? What made you disregard it as a proof to the answer then?
You understand that the simulation with the OnceDear's spreadsheet does exactly like the bet would've and what the results will look like after 50 000 trials, right?
What made you say it is not applying to the question then?

Listen this is the last time I am going to say it:

IT DOESNT MATTER WHICH OF THE TWO DICE IS A TWO. If Die #1 is a 2, then Die #2 has a 1/6 chance of being a 2.
If Die #2 is a 2, then Die #1 has a 1/6 chance of being a 2.
If both Die #1 and Die #2 show a 2, then the chance that either Die #1 or Die #2 was a 2 was still 1/6.

1/11 does not answer the question.

The spread sheet does not matter.
The graph of 36 dice combinations does not matter.
The graph showing all 11 combinations of dice with at least one die showing a 2 does not matter.
The bet only tried to monetize the problem but it doesn't matter either because anything can happen when you throw two dice.

So, it's over. And it has been over since the day the Wizard showed that video with him rotating the die with the 2 to show the 1/11 answer.

Alan, it's been over for a long time. Your denial does not change reality. You can continue to deny the facts forever just like you always do. It doesn't change a thing. 1-11 is the right answer and will always be the right answer.

Alan! Eureka! You're Right!
You had the key. Ignore the impossible faces.

I now have undeniable evidence and proof that your evaluation - one out of six chances (1/6) - is the correct answer to the pair-of-deuces puzzle.

http://www.imageurlhost.com/images/d...b1j3qw7vvc.jpg

Dotted cubes (dice) are utterly impossible, when that cube shows a Deuce.
My picture shows that you have 2 winners out of 12 combinations.
So pay attention, folks.
2 out of 12 is exactly the same as 1/6. Alan was right; is right; always will be!

You did it again, Indignant. You used FOUR sets of dice instead of just TWO.

We caught you. And you're not going to get it by us.

It's still 1/6 no matter how many pretty pictures you post.

Give it a rest here and go back to WOV. Thanks for coming by.

What? I'm agreeing with you.

I show a Red Deuce, with all its six partners.
I show a Blue Deuce, with all its six partners.

I'm only showing exactly one Red Deuce (with five red impossible ghosts, not real).
I'm only showing exactly one Blue Deuce (with five blue impossible ghosts, not real.)

The extra cubes are for illustration purposes only. (What coulda been, but was not in reality).

Dotted cubes (impossible) cannot show up.
Solid cubes (possible) are only to show what could have happened. (The solid ones ain't real either.)

Did you just react to my name? Or did you focus and read the post?

I'm only showing ONE RED DEUCE, and the six-ways going along with it.
I'm only showing ONE BLUE DEUCE, and the six-ways going along with it.
Of course, as you say, it doesn't matter which die is showing the deuce.

The only real combinations, possible, are at the ones at the ENDS OF THE ARROWS.

I didn't show 4 sets of dice.
I show 24 total cubes. If anything, that's TWELVE pairs of dice.
Can't you notice the difference between theoretical, versus actual?

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Originally Posted by Alan Mendelson

IT DOESNT MATTER WHICH OF THE TWO DICE IS A TWO. If Die #1 is a 2, then Die #2 has a 1/6 chance of being a 2.
If Die #2 is a 2, then Die #1 has a 1/6 chance of being a 2.

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And that's exactly right.

So, to visualize:

Die #1 will be a deuce 1/6 of the time. Die #2 will be a deuce 1/6 of the time also.

Now if we were only interested in scenarios where Die #1 is a deuce we would have Die #2 be also a two 1/6 of those scenarios, thus giving us a 1/6 chances we see 2-2 when, and only when Die #1 shows 2.

And if we were only interested in scenarios where Die #2 is a deuce we would have Die #1 come as deuce 1/6 of those scenarios, thus giving us a 1/6 chances we see 2-2 when, and only when Die #2 shows 2.

But we are not interested in those separate scenarios. It indeed does not matter which die is a 2.

So, we now have both dice equally likely to show a deuce (1/6 each) and we are interested in all of those scenarios where either of them is 2.

That makes twice as many chances we see a 2-x --> 1/3 of the time. Well almost.
We need to subtract the one of the cases where Die #1 is deuce and Die #2 is deuce, or where Die #2 is deuce and Die #1 is deuce because we've just double counted the same roll. Since each of these rolls has 1/36 chance to happen we subtract 1/36 from 1/3.
That gives us: 1/3-1/36=11/36.
In other words the chances we see 2-x are exactly 11/36 of the time.

But we didn't increase our chances to see exactly 2-2. They remain 1/36 for each throw.
The fact that it doesn't matter which die is a 2 gives as (almost) twice the odds on seeing 2-x, while changing nothing else.
So when we see 2-x that means that either Die #1 is a 2 or Die #2 is a 2 and we still have 1/36 chance of seeing exactly 2-2 (both dice come up 2).

Well, simplifying our results gives us 11/36 chances for 2-x versus 1/36 chance for 2-2, or one time out of eleven 2-x we will see 2-2.

The very fact that it doesn't matter which die is a 2 makes it 1/11. If we'd cared which die is 2 and were only interested in one particular die to show 2 first and only then examine the second die we would have 1/6 chance that the other die is also a 2.

Kewl when you roll two dice ONE TIME you'll have to choose ONLY ONE of the three scenarios. You can't have two and you can't have three.

We have all our chances decided before we threw the dice. Not after the throw. Remember this.

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Originally Posted by kewl

We have all our chances decided before we threw the dice. Not after the throw. Remember this.

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What are you saying?

Two dice have been thrown. You know at least one die is a two. The second die in the problem has a 1/6 chance of also being a 2. That's it. Finished.

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Originally Posted by Alan Mendelson

What are you saying?

Two dice have been thrown. You know at least one die is a two. The second die in the problem has a 1/6 chance of also being a 2. That's it. Finished.

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Let me give you an analogy.

How do you know that upon throwing a single die the chances it will come up as 2 (or whatever) is 1/6?
You've derived it from your knowledge in probability that when there is one outcome you are interested in and six possible outcomes total, your chance to get a particular outcome is 1 divided by 6.

You don't throw the die one time and conclude from the outcome anything - its impossible.

Therefore, you have all the information about your chances based on how many outcomes total are there and how many outcomes (combos) you are interested in.

Two dice have been thrown. You know at least one die is a two.
Same thing.
You have such and such ways to arrive to at least one two and such and such ways to arrive to 2-2.

You don't come midstream and conclude since the outcome is 2-x you can base your odds now on one die.

Both the bet from post#1 and the simulations are based verbatim on the original question. Yet you refuse to acknowledge that.

Kewl let me ask you a question: are black helicopters circling your house?

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Originally Posted by Alan Mendelson

Kewl let me ask you a question: are black helicopters circling your house?

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No, not ones I can see at least. Why, you seeing any?

I was thinking - this post here:

http://vegascasinotalk.com/forum/showth...5054#post25054

does indeed have some good points. Well put :), admittedly some vulgarity was used, but nevertheless it really reflects what you are doing here (and over there in that thread).

Kewl just consider who wrote it.

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Originally Posted by Alan Mendelson

Kewl just consider who wrote it.

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That would be MickeyCrimm, the world-class drunk who was banned from also the Wizard's -reluctantly I might add - because there seemed no subject or manner of phrasing thereof, or ridiculous claims, off limits to him. It was becoming so obviously pathetic that the Wizard didn't want the intellectually challanged over there spotting it and mistaking the Wizard's mentality for the same. Why would the Wizard be afraid of such? I don't get it, lol.

Anyway, Kewl, you had better spend at least one a day a week in the casinos making that $100,000 or $200,000 free and clear a year. Or the Wizard might have to ban you as well.

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Originally Posted by OneHitWonder

That would be MickeyCrimm, the world-class drunk who was banned from also the Wizard's -reluctantly I might add - because there seemed no subject or manner of phrasing thereof, or ridiculous claims, off limits to him. It was becoming so obviously pathetic that the Wizard didn't want the intellectually challanged over there spotting it and mistaking the Wizard's mentality for the same. Why would the Wizard be afraid of such? I don't get it, lol.

Anyway, Kewl, you had better spend at least one a day a week in the casinos making that $100,000 or $200,000 free and clear a year. Or the Wizard might have to ban you as well.

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Lol, dude, you come of as a totally inadequate poster to me, what the hell are you trying to say?

Are you by any chance assuming I am "KewlJ", the member on WoV? Cause I'm not lol.

I'm really having a hard time truing to decipher what are you saying/your point actually.
Most of the time it sounds like you actually farted instead of say something meaningful, no offence lol.

Alan's very good at being a troll.