It was two years ago today...

[a2a3dseddie]

it was two years ago today...

You are correct, Alan, this is a trick question. I want to explain why it's a trick question. It's something I think everyone is missing the point on. The question asks, "If you roll the dice, and at least one is a 2, what are the odds that BOTH dice rolled a 2?".

That's the wording of the original question, and it's worded pretty different than normal speech. Normally, you would ask the question, "If you roll the dice and at least one of the the die is a 2, what are the odds the OTHER die also rolled a 2?" You wouldn't ask about both dice, after explaining that one is already a 2, you would just ask about the OTHER die.

This is why it's a trick question. In the original question, the answer is 1/11, the odds of BOTH dice rolling a 2 is 1 in 11. And, that's what the question asks. It's not asking the odds of the other die rolling a 2, it's only asking the odds for BOTH dice rolling 2's.

In the second question, the one you would ask in normal speech, the chances of the OTHER die also rolling a 2 is 1/6. The odds of 1 particular die rolling a 2 is 1 in 6. And that's what the 2nd question asks. It doesn't ask the odds for BOTH dice rolling 2's, only the "other" die.

This is why the question is a trick question. It asks the question in a different way than you would normally speak, and by doing this it gives a different answer than would normally be given.

I think a lot of these math guys don't understand why it's a trick question, most would probably, incorrectly, give the same 1/11 answer to both wordings of the question. It just so happens, in this case they are correct, the question asks the odds of BOTH dice rolling a 2, that's 1/11. If the question was asked about the "other" die, you would be right with 1/6, and I'm guessing most of the math guys would give the wrong answer of 1/11. But, this is why it's a trick question.

If you do the experiment itself, you'll see that BOTH dice will roll 2's 1 in 11 times. And, the "other" die will have a 1 in 6 chance of rolling a 2, just like in your video. So, depending on whether you say BOTH dice, or if you say the OTHER die, you'll get 2 different answers.

The original question asks for BOTH dice, it does that on purpose, to get the 1/11 answer, because if it asked about the "other" die, it would get a totally different answer of 1/6 (and this would trick a lot of people who think it's 1/11). Hope this helps explain why this is such a tricky question for you.

Alan, here again is the problem:

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?


Where do you see "What are the odds the other dice is a 2?"

[jbjb]

If at least one is, and the "other" is too, isn't that exactly the same as "both"?

[a2a3dseddie]

Look at this photo posted by Alan:

https://vegascasinotalk.com/forum/sh...vs-1-11/page20

If we agree that ANY of those 11 possible results "where at least one of the dice is a two" could be under the cup isn't it obvious to see that only 1 time in 11 both dice are showing a 2?

[Alan Mendelson]

Alan, here again is the problem:

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?


Where do you see "What are the odds the other dice is a 2?"

Look back at what you wrote.

When you roll two dice you have 11 possible outcomes containing at least one 2. You have one possible outcome of 2-2.

But in our question, the roll has been completed, and at least one die has been identified as a 2. Therefore there is only one other die to consider, isn't there? Or are you going to continue to insist that you must consider both dice which is exactly what you are doing?

[Alan Mendelson]

Look at this photo posted by Alan:

https://vegascasinotalk.com/forum/sh...vs-1-11/page20

If we agree that ANY of those 11 possible results "where at least one of the dice is a two" could be under the cup isn't it obvious to see that only 1 time in 11 both dice are showing a 2?

Except that's not the question. We know one die is already showing a 2. We need to consider the second die in this two-dice problem.

How many times must this be said?

[Alan Mendelson]

If at least one is, and the "other" is too, isn't that exactly the same as "both"?

Please... don't go here.

You still haven't clarified your own question about the stack of spades and the stack of hearts.

[a2a3dseddie]

Alan, here again is the problem:

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?


Where do you see "What are the odds the other dice is a 2?"

Look back at what you wrote.

When you roll two dice you have 11 possible outcomes containing at least one 2. You have one possible outcome of 2-2.

But in our question, the roll has been completed, and at least one die has been identified as a 2. Therefore there is only one other die to consider, isn't there? Or are you going to continue to insist that you must consider both dice which is exactly what you are doing?

Alan, do you agree that any of the 11 possible outcomes could be under the cup?

[Biloxi Bill]

Christ, what an abortion of a thread

[Alan Mendelson]

Alan, do you agree that any of the 11 possible outcomes could be under the cup?

Any of 36 combinations can be under the cup until someone looks.

In this case, someone looked and found at least one of the two dice is showing a 2.

That leaves 6 possible faces on the second die.

How many times do we have to go over this?

[mickeycrimm]

If you set one die on the table showing a 2 then roll the other die then yes the answer is 1 in 6. But when you roll both the dice at the same time the answer is 1 in 11.

Alan, I wrote this is the 2nd post I made in this thread. And and now you think someone here is going to give you 8 to 1 on setting one die on a 2 then rolling the other die? Wise up. 4 to 1 is what you will get. So bring it on.

[mickeycrimm]

Coach, you would go broke with this. When rolling two dice simultaneously, 2-2 shows up 1/36 times.
There are 11 combinations showing at least one 2, and only one combination that shows 2-2. So you are facing 10 losing decisions for each winning decision.[/QUOTE]

Isn't this evidence enough that Alan doesn't even believe his own bullshit. He's advising coach that it's a sucker bet. LOL! The only thing funnier is coach actually thinks we can win. Of course, coach was chomping at the bit to get at that hundred play a couple of months ago. Coach, you haven't posted your results on that. How'd you come out?

[mickeycrimm]

Alan, I've got another dice game for you. It's played with ONE DIE and a cup.

Player A is betting he can roll a 6 within 4 rolls of the die.

Player B is betting against Player A rolling a 6 within 4 rolls.

The chance of rolling a 6 is 1 in 6. That's 5 to 1 against.

Which side of the bet would you take, Alan?

Everyone please give Alan a chance to respond to this before you put up any information.

[Alan Mendelson]

All I can say is thank God you have stopped pushing your 1/11 answer. Bless you.

[mickeycrimm]

When you roll two dice you have 11 possible outcomes containing at least one 2. You have one possible outcome of 2-2.But in our question, the roll has been completed, and at least one die has been identified as a 2. Therefore there is only one other die to consider, isn't there? Or are you going to continue to insist that you must consider both dice which is exactly what you are doing?

You've admitted that there are 11 combinations that include "at least a 2." When you roll the dice out on the table and see that "at least one die is a 2, at that point you have the information that at least one die is a 2. But does that change anything? NO! No matter what the other die is you are seeing just 1 of 11 possible combinations that include at least one 2. You are going to see either 2-1, 2-2, 2-3, 2-4, 2-5, 2-6, 1-2, 3-2, 4-2, 5-2, 6-2.

The answer is and will always be 1 in 11.

[mickeycrimm]

Alan, I've got another dice game for you. It's played with ONE DIE and a cup.

Player A is betting he can roll a 6 within 4 rolls of the die.

Player B is betting against Player A rolling a 6 within 4 rolls.

The chance of rolling a 6 is 1 in 6. That's 5 to 1 against.

Which side of the bet would you take, Alan?

Everyone please give Alan a chance to respond to this before you put up any information.

So which side of the bet would you take, Alan?

[Alan Mendelson]

When you roll two dice you have 11 possible outcomes containing at least one 2. You have one possible outcome of 2-2.But in our question, the roll has been completed, and at least one die has been identified as a 2. Therefore there is only one other die to consider, isn't there? Or are you going to continue to insist that you must consider both dice which is exactly what you are doing?

You've admitted that there are 11 combinations that include "at least a 2." When you roll the dice out on the table and see that "at least one die is a 2, at that point you have the information that at least one die is a 2. But does that change anything? NO! No matter what the other die is you are seeing just 1 of 11 possible combinations that include at least one 2. You are going to see either 2-1, 2-2, 2-3, 2-4, 2-5, 2-6, 1-2, 3-2, 4-2, 5-2, 6-2.

The answer is and will always be 1 in 11.

In order for the answer to be 1/11, the die that originally shows a 2 would have to rotate to show other positions. Perhaps when you're drunk the dice you see rotate. Those of us who are sober know that when a die settles on 2 it doesn't rotate... unless it's in California during a major earthquake.

Go have another drink.

[mickeycrimm]

When you roll two dice you have 11 possible outcomes containing at least one 2. You have one possible outcome of 2-2.But in our question, the roll has been completed, and at least one die has been identified as a 2. Therefore there is only one other die to consider, isn't there? Or are you going to continue to insist that you must consider both dice which is exactly what you are doing?

You've admitted that there are 11 combinations that include "at least a 2." When you roll the dice out on the table and see that "at least one die is a 2, at that point you have the information that at least one die is a 2. But does that change anything? NO! No matter what the other die is you are seeing just 1 of 11 possible combinations that include at least one 2. You are going to see either 2-1, 2-2, 2-3, 2-4, 2-5, 2-6, 1-2, 3-2, 4-2, 5-2, 6-2.

The answer is and will always be 1 in 11.

In order for the answer to be 1/11, the die that originally shows a 2 would have to rotate to show other positions. Perhaps when you're drunk the dice you see rotate. Those of us who are sober know that when a die settles on 2 it doesn't rotate... unless it's in California during a major earthquake.

Go have another drink.

No. It's just something you fundamentally don't understand. It doesn't make you a bad person though. Casting asperions on me about my drinking could be construed as mean spirited though. Ordinarily, I would tell you to kiss off if you don't like my drinking. But the fact is I'm in my 81st day of being clean and sober. I feel good. Really good.

[Rob.Singer]

mickey, it really has nothing to do with any type of fundamental misunderstandings by either side. Those who cannot move away from 1 in 11 only see the problem as a study in theory, while the 1 in 6 crowd look at it in terms of reality.

For instance: when the peeker truthfully proclaims that at least one of the dice shows a 2, you're taking his word for it and choose to proceed with the problem based on a theoretical assumption and trust. OTOH, some of us approach the issue as seeing the die showing the 2, because why should the peeker be the only one able to see it? Nowhere in the language of the question/problem does it absolutely say what the peeker sees cannot be seen by anybody else.

Thus the multiple ways to conclude. I also know Alan doesn't need to think so deeply about this when he knows the answer is 1 in 6, because in his mind he's simply deducing that the whole world is picturing the die showing the 2 is now out-of-play.

Going back to what redietz said about the WoV genius who presented this problem, the wording was incomplete, misleading, or both. It happens, with people who base their lives on theoreticals, and who know how the realists among them will react.

Good job staying off the sauce. There are a lot of very good articles online dealing with the potential devastating effects binge drinking has on your liver, and whether or not/how they can be reversed. And if you kick the smoking habit then I'd say you're making progress.

[a2a3dseddie]

When you roll two dice you have 11 possible outcomes containing at least one 2. You have one possible outcome of 2-2.

But in our question, the roll has been completed, and at least one die has been identified as a 2. Therefore there is only one other die to consider, isn't there? Or are you going to continue to insist that you must consider both dice which is exactly what you are doing?

I am because that is what the original question asks.

What is the probability that both dice are showing a 2?

[a2a3dseddie]

Alan, do you agree that any of the 11 possible outcomes could be under the cup?

Any of 36 combinations can be under the cup until someone looks.

In this case, someone looked and found at least one of the two dice is showing a 2.

That leaves 6 possible faces on the second die.

How many times do we have to go over this?

All that tells us is that the peeker has determined 1 of the 11 possible combinations containing at least one 2 is under the cup. That's all he tells us.

Of those 11 possible combinations that could be under the cup, only one combination is 2 2.

Ten combinations are not.

The question asks for the probability that both dice are showing a 2.