It was two years ago today...

[jbjb]

I'm not going to drop my plans for a last second bet. Be patient, it'll happen.

And no, I'm not backing out!

[coach belly]

I'm not going to drop my plans for a last second bet.

You have plans that would keep you from this for the next 10 days?

[a2a3dseddie]

Coach you will lose this bet with jbjb, and this bet that jbjb proposed is NOT the same as the question with two dice and one showing a 2.

Let me go over jbjb's bet again for you.

Your chance of seeing 2-2 on any roll is 1/36.
Even when the dice are rolled and one 2 shows up, there is a 1/11 chance of another 2 showing up.

The difference with the two-dice question is that we know in advance that at least one die shows a two, and with that the odds of both dice showing two are reduced to 1/6.

But in jbjb's bet you don't know that at least one 2 will be rolled.

If jbjb would let you roll one die and if it comes up a 2 for a win, and if he's willing to pay you 7 to 6 for a 1 out of 6 possibility, take it.

Alan, you posted the original parameters with the Wizard that follow the original wording of the question exactly.

Gone are the "old rules". This is a new bet and quite frankly I like it.

Let me answer your questions specifically and I am sure this will be the understanding:

1. Two dice are in a cup or other device, shaken and in the cup placed on the table.
2. A witness will peek. If a 2 is shown the bet is on.
3. If a 2 is not shown, there is no betting.
4. In both cases, the cup will be removed and the dice can be viewed. This will prevent the original dice from showing 2-2 and the witness lying.
5. With one deuce the bet is on (#2) and if there is not a second bet the "player" will lose their bet.
6. With one deuce the bet is on (#2) and if there is a second deuce the "bank" will pay either 9-to-1 or 9-for-1 (the Wiz doesn't care.)

My own personal thought: I can't imagine why the Wizard agreed to this? There is a 1/6 chance that when one die shows a 2 that the other will also be a 2 yet he is willing to pay 9-for-1 or 9-to-1.

Is it possible he misread this bet the same way I suspect he misread the original question? Somebody wake me up from this (bad) dream.

This will be twice now that you have warned coach belly against taking the bet, and twice now that you are trying to introduce something different than what the original question asks.

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 does it say "roll one die"?
Where does it identify which die is the 2?
Where does it say set one die aside?

Both dice are rolled together in the cup. One of them, and we aren't told which one, is a 2.

The question has always asked for the probability of a specific outcome of 2 dice. 2 2.

Take it from a friend:

simply telling you that one of the dice shows a 2 does not remove it from its numerical relationship with the die with the unknown number showing. Whereas rolling 2-2 is a 36-1 possibility, knowing half the outcome, ie. one of the dice is showing a 2, reduces the odds to 11-1 since it is no longer a single die event. You now have to consider all combinations of the two dice where either one of them shows a 2.

[a2a3dseddie]

Alan, when 1 die is a 2, of course the odds of the other die being a 2 is 1/6. There are only 6 sides to a die. That is not what we disagree on. We are disagreeing on the question initially asked which is :

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

Your chance of seeing 2-2 on any roll is 1/36.
Even when the dice are rolled and one 2 shows up, there is a 1/11 chance of another 2 showing up.

Here it seems like you finally get it, but then...

The difference with the two-dice question is that we know in advance that at least one die shows a two, and with that the odds of both dice showing two are reduced to 1/6.

[a2a3dseddie]

The reason why your initial bet proposed 2 years ago with the Wizard is a loser for you, is because you realise there are 10 losses for every win.

How can this be if the bet is only on when there is at least one 2? If there is at least one 2 the odds are 1/6!

I don't know if it's been explained like this before, but I'm giving it a shot.

redietz mentioned something at the beginning of this thread about the question using tenses and sequential vs simultaneous events.

I'm going to throw in dependent probability.

Using red and green dice what is the probability you'll start with the red dice (or green) as the 2? Remember, we are NOT factoring in any of the non-2 containing dice combinations.

1. There are 11 combinations, and 6 of those 11 could start out with the red die as the 2. This is the first probability you have to overcome.

2. Once the red die is a 2, of course there is a 1/6 chance the green die could be a 2.

3. These 2 probabilities are multiplied together to get 6/11 X 1/6 = 6/66 or 1/11.


You seem to be stuck on step 2, completely disregarding step 1.

Put another way, let's say you are the peeker and I am the one betting.

We shake 2 dice in a cup.

Alan: At least one of the dice is a 2. What is the probability that both dice are showing a 2?

Me: Okay, there is a 1/6 chance the other dice is a 2. I bet $100. Come on... $800 payoff!

You show me red 2 green 1. I lose.

Me: Let's play again.

We shake 2 dice in a cup.

Alan: At least one of the dice is a 2. What is the probability that both dice are showing a 2?

Me: Okay, there is a 1/6 chance the other dice is a 2. I bet $100. Come on... $800 payoff!

You show me red 1 green 2. I lose.

See where I'm going with this? You could show me 10 distinct losing combinations all in line with "At least one of the dice is a two"

Of course, the odds "the other dice" is a 2 is 1/6.

5 losses and 1 win.

But that's for each of the dice and you have 2 dice.

That's why there are 10 losses for every win.

[Rob.Singer]

Rob, I believe arci and others have done independent calculations of your martingale-type strategy and the "probability of being successful" is indeed 80-85% FOR ANY GIVEN SESSION. That's the way it's supposed to work. No surprises there.

Well hot damn. 80% win rate is good enough for me.

You've got it mixed up, which you wouldn't do if you were interested in understanding the strategy and how/why it wins consistently.

The 85% is the % of sessions that have been and can be expected to win going forward. What's the chance that I'll win any given session? 100% of course....or else I'd never have played the strategy. It does lose, infrequently so, so it isn't perfect. But it's better than anything else out there.

[Rob.Singer]

I'm not going to drop my plans for a last second bet.

You have plans that would keep you from this for the next 10 days?

I'm going to be at the Silverton one night only--tomorrow (Tues.) night. If you guys want to work your bet there which I can witness, I'm available anytime between midnight and 6am.

[RS__]

If jbjb would let you roll one die and if it comes up a 2 for a win, and if he's willing to pay you 7 to 6 for a 1 out of 6 possibility, take it.

You've posted some really stupid shit, but this takes the cake.

[mickeycrimm]

Alan, since you'll be filming the contest between coach and jbjb, coach slams the dice cup on the table, jbjb peeks under the under the cup, if one of the dice is a two jbjb announces "at least one of the dice is a 2" then sits the cup back down. Coach then lifts the cup up to see what the other die is showing. This meets the requirements of the original question.

I really don't think you want to see this contest go down. After coach gets beat perhaps you can blame it on a reading comprehension problem.

We will be looking forward to seeing the contest on film.

[Alan Mendelson]

I see that we are back at square one. So to summarize square one:

The odds of rolling 2-2 are 1/36.
When you roll two dice there are eleven combinations that will show at least one 2, and one of those 11 combinations will show 2-2. That's 1/11.
When you roll two dice and one die lands on a 2, the odds that both dice are 2-2 is 1/6.

Square one: it's in the wording and how you visualize the step or condition of the question.

[quahaug]

When you roll two dice and one die lands on a 2, the odds that both dice are 2-2 is 1/6.
No it isn't.

[regnis]

I have a legitimate question. The guy that peeked under the cup and said one of them is a 2. Would you say that for him the odds that the other one is a 2 would be 1 of 6. And if so, then are you telling me that his odds are different than the other guy's odds?
The guy that peeked under the cup knows which one is a 2. He may also know what the other die is which gets us back to my example. But assuming he only saw the one die that showed a 2, what are the odds for him that the other is a 2?

Sorry, the peeker cannot ALWAYS answer the question "at least one die is a 2" unless he sees both die. Hence, once he sees both die then he knows the exact answer already. The odds aren't 1:6 either.

How right you are, if the peeker sees two twos The odds are 100% there is another two. If he sees any other number the odds are 0% there is a another two, but that has nothing to do with the original question Regnis' picture notwithstandig. And I don't think I want him representing me against the IRS.

I am hurt Qua!!!!

[quahaug]

Sorry, the peeker cannot ALWAYS answer the question "at least one die is a 2" unless he sees both die. Hence, once he sees both die then he knows the exact answer already. The odds aren't 1:6 either.

How right you are, if the peeker sees two twos The odds are 100% there is another two. If he sees any other number the odds are 0% there is a another two, but that has nothing to do with the original question Regnis' picture notwithstandig. And I don't think I want him representing me against the IRS.

I am hurt Qua!!!!

Well maybe you could help me out in my next criminal trial.

[regnis]

I firmly believe that my advice to criminals cannot fail. DON'T GET CAUGHT!!!

[Alan Mendelson]

When you roll two dice and one die lands on a 2, the odds that both dice are 2-2 is 1/6.
No it isn't.

Well... I have a problem understanding how you think it's 1/11. So do me a favor. Take two physical dice and roll them. After at least one of the dice shows a 2, explain to me how you get your 1/11 answer. Now, keep in mind I know that once at least one of the dice lands showing a two that die cannot be changed.

If you don't have a video camera on your phone please explain the process as best you can so I understand how you count eleven possible faces to be considered.

I just can't get past six faces on one die.

[quahaug]

When you roll two dice and one die lands on a 2, the odds that both dice are 2-2 is 1/6.
No it isn't.

Well... I have a problem understanding how you think it's 1/11. So do me a favor. Take two physical dice and roll them. After at least one of the dice shows a 2, explain to me how you get your 1/11 answer. Now, keep in mind I know that once at least one of the dice lands showing a two that die cannot be changed.

If you don't have a video camera on your phone please explain the process as best you can so I understand how you count eleven possible faces to be considered.

I just can't get past six faces on one die.

You'll just have to accept it as the fact that it is. Maybe after all is said and done you just need to take it on faith. My personal belief is you're just trolling/flaming the thread. It's hard to believe someone could be as stubborn as you make yourself out to be. It's hilarious to me though so cognitive dissonance yourself away..

[quahaug]

I firmly believe that my advice to criminals cannot fail. DON'T GET CAUGHT!!!

That's the problem. I'm to honest to be a criminal, I'm really bad at it. They cops ask me if I did it and I say yes officer that was me. I need to change jobs.

[a2a3dseddie]

When you roll two dice and one die lands on a 2, the odds that both dice are 2-2 is 1/6.
No it isn't.

Well... I have a problem understanding how you think it's 1/11. So do me a favor. Take two physical dice and roll them. After at least one of the dice shows a 2, explain to me how you get your 1/11 answer. Now, keep in mind I know that once at least one of the dice lands showing a two that die cannot be changed.

If you don't have a video camera on your phone please explain the process as best you can so I understand how you count eleven possible faces to be considered.

I just can't get past six faces on one die.

This is how.

Using red and green dice what is the probability you'll start with the red dice (or green) as the 2? Remember, we are NOT factoring in any of the non-2 containing dice combinations.

1. There are 11 combinations, and 6 of those 11 could start out with the red die as the 2. This is the first probability you have to overcome.

2. Once the red die is a 2, of course there is a 1/6 chance the green die could be a 2.

3. These 2 probabilities are multiplied together to get 6/11 X 1/6 = 6/66 or 1/11.

You seem to be stuck on step 2, completely disregarding step 1.

[Alan Mendelson]

You'll just have to accept it as the fact that it is. Maybe after all is said and done you just need to take it on faith. My personal belief is you're just trolling/flaming the thread. It's hard to believe someone could be as stubborn as you make yourself out to be. It's hilarious to me though so cognitive dissonance yourself away..

Using red and green dice what is the probability you'll start with the red dice (or green) as the 2? Remember, we are NOT factoring in any of the non-2 containing dice combinations.

1. There are 11 combinations, and 6 of those 11 could start out with the red die as the 2. This is the first probability you have to overcome.

2. Once the red die is a 2, of course there is a 1/6 chance the green die could be a 2.

3. These 2 probabilities are multiplied together to get 6/11 X 1/6 = 6/66 or 1/11.

You seem to be stuck on step 2, completely disregarding step 1.

Now let's talk about two red dice that are shaken in a cup and slammed down on a table. One person peeks and sees that at least one of the two red dice is showing a 2.

Both of you are unable to show me 11 possible options. You say they are there and you ask me to trust that they are there but you can't demonstrate that they are there.

You can't demonstrate the 11 possible options because 11 possible options in the real world do not exist -- not with two dice with at least one of those two dice already showing a 2.

In order to show 11 options you have to rotate the dice showing a 2 to show other faces. Or you take Arc's out of the world approach and you claim that there is no proof that any die shows a 2 even though the original problem tells us that at least one die shows a 2.

Never has anyone from the 1/11 camp shown that there are 11 faces or combinations to consider. Never. Never.

And frankly you can't because that's the way it is in the real, physical world when you have two real physical dice.

You can show me all the graphs and charts you want with all of the combinations of two dice, and you can point to all of those combinations but when it comes to using two real dice, with at least one die showing a 2 all of your theory does not apply. Because the theory does not apply AFTER the roll of the two dice and AFTER one of the two dice is known to be a 2.

Your theory only looks at possible combinations and it looks at the possible combinations using all of the faces of two dice. Remove one die because the face is known on that die, and your theory is no longer applicable to the question.

Once again, I invite you to take two physical dice, roll them, and when at least one die is showing a 2, prove to me there are 11 faces to be considered. Do it. Do it. Do it.

You can't. You just can't.

[a2a3dseddie]

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?

Remove one die because the face is known on that die.

Where does it say to do that?

let you roll one die and if it comes up a 2 for a win.

Where does it say to do that?

Alan, you shoot craps. You know there are more combinations for a 7 than any other number. If I say 2 dice are rolled and one of them landed on 2 do you think there is only 1 way to make a 7?

Die 1=2
Die 2=5
?

What about

Die 1=5
Die 2 =2
?

The original question doesn't specify which die is the 2. Only that at least one of them is. Why can't you see that EITHER die could be the 2 the peeker sees? What in the original question is preventing you from seeing that?