Question for your Christmas Dinner Guests

[LarryS]

if you agree that the probability of the OTHER die being a "2" is 1 in 6........If the other die has a "2" then it can be said both dice have a "2"

so if theother die does have a "2"///then Both dice has a "2"

[a2a3dseddie]

In English, the statement "at least one of the dice is a 2" means that either Die A or Die B is a 2 or Both Die A and Die B are showing a 2.

So Alan, which scenario does the peeker see when he says truthfully "At least one of the dice is a 2."?

arcimede$ asked you that several times back in May. You never answered.

I know... you're going to say it doesn't matter.

We all agree there are 11 possible 2 dice combinations where at least 1 dice is a 2.

How do you determine probability when events are dependent? They are multiplied.

In order to start with Die A as a 2, the odds are 6/11. Agreed?

Once that is established the odds Die B is also a 2 is 1/6. Agreed?

Multiplying, you get 6/66 or 1/11.

Alan, you are skipping 6/11 odds first and going straight to the 1/6 part.

[a2a3dseddie]

if you agree that the probability of the OTHER die being a "2" is 1 in 6........If the other die has a "2" then it can be said both dice have a "2"

so if theother die does have a "2"///then Both dice has a "2"

Which die is "the other die"? Couldn't either die start out as the 2?

[LarryS]

I am getting the feeling that this centers around the meaning of "both dice" and "the other die". If thats the case then its just the parsing of words,,,since if the "othjer die" has a 2.....then both dice have a 2.

And after seeing the face of one die having any number.....the chance of the other die having the same number is 1 in 6....and any bets would involve the 1/6 as a starting point....not 1/11, Wheter a die is stationary and out of view...or spinning in view....they both have unknown results on the upper face of the die. And thats the issue.....with an unknow result of the second die...what are the chances of it being a similar nimber to the one die that is showing.

so with one die visible.....what are the chances the other unseen die whether stationary or spiining marches the viewable die..... I say i in 6....and anyone willing to give me10-1..i will gladly take it..

let me ask this.....under what circumsatances in this scenario would an 10-1 bet be accetable based on the 11-1 odds. Maybe that would help me understand.
give me the scenario where 10-1 makes sense in a bet.

[a2a3dseddie]

Larry, look again at the "bet" Alan first liked then later declined. That matches the conditions of the original question.

If you find someone willing to give you 10 - 1 on the 6 - 1 scenario you are suggesting let me know too.

[Alan Mendelson]

Set one die on a 2 then roll the other die the answer is 1 in 6. Roll both die at the same time the answer is 1 in 11. If there is a spinner its the same as setting one die on a 2 then rolling the other die, 1 in 6.

Max the 1 in 6 contingent here is not a crowd. Just a couple of dimwits that live in an alternate universe.

You would think the statement above would permanently end the discussion since it's a perfect summary, but somehow I doubt it will.

If you roll both dice at the same time getting 2-2 is a 1/36 possibility. There is no 1/11.

1/11 is the answer to a totally different question and that question might be this: how many combinations of two dice contain at least one "2"? Answer 11.
Of those possibilities, how many contain 2-2? Answer 1/11.

I'm sorry to say this, but you can't read english. You might be good at math... but the verbal part of your SATs? LOL

[tableplay]

Set one die on a 2 then roll the other die the answer is 1 in 6. Roll both die at the same time the answer is 1 in 11. If there is a spinner its the same as setting one die on a 2 then rolling the other die, 1 in 6.

Max the 1 in 6 contingent here is not a crowd. Just a couple of dimwits that live in an alternate universe.

You would think the statement above would permanently end the discussion since it's a perfect summary, but somehow I doubt it will.

If you roll both dice at the same time getting 2-2 is a 1/36 possibility. There is no 1/11.

1/11 is the answer to a totally different question and that question might be this: how many combinations of two dice contain at least one "2"? Answer 11.
Of those possibilities, how many contain 2-2? Answer 1/11.

I'm sorry to say this, but you can't read english. You might be good at math... but the verbal part of your SATs? LOL

Alan, as Eddie already pointed out, you are ignoring the pathway that the die which is seen takes to get to two on its top face in the 1st place. The probability that the seen die lands on a two during the rolling process is not 100%.

So here is the crux of the matter (IMHO):
1) The probability that the unseen die has a two as its top face given that the seen die has a two on its top face is 1/6

2) The probability that the unseen die has a two as its top face is 1/11 (since one of the dies must first land on the proper number, two, which is not guaranteed,
in order for the second die to be observed matching it)

[Alan Mendelson]

Tableplay the peeker truthfully tells you at least one die is a two. Is that too difficult for you and the others to understand? That leaves the second die in this two dice problem. The second die has six sides, one of which is a two. Still follow?

Where do you get eleven possibilities?

You get 11 possibilities from a chart that shows the 36 combinations of two dice. You don't get 11 possible answers to this one die problem.

[Alan Mendelson]

if you agree that the probability of the OTHER die being a "2" is 1 in 6........If the other die has a "2" then it can be said both dice have a "2"

so if theother die does have a "2"///then Both dice has a "2"

Which die is "the other die"? Couldn't either die start out as the 2?

This is the most ridiculous post yet. If neither die is a two the peeker couldn't truthfully tell you at least one die is a 2. And yes both dice could be a two. It still doesn't matter in a two dice problem.

Someone take two dice and act out the problem and without changing the face of the die that is a two (or one of the twos if both show a two) COUNT ELEVEN POSSIBLE COMBINATIONS. YOU CAN'T.

[RS__]

Stop responding to Alan. He's doing the same exact shit Rob Stringer would do. It's been how many posts or even months (years?) on this subject? He will not concede.

He's a troll and/or completely fucking retarded. Don't give him the attention.

[tableplay]

Stop responding to Alan. He's doing the same exact shit Rob Stringer would do. It's been how many posts or even months (years?) on this subject? He will not concede.

He's a troll and/or completely fucking retarded. Don't give him the attention.

Done. Will do.

[Alan Mendelson]

It's about time you 1/11 people concede you had the wrong answer.

[a2a3dseddie]

if you agree that the probability of the OTHER die being a "2" is 1 in 6........If the other die has a "2" then it can be said both dice have a "2"

so if theother die does have a "2"///then Both dice has a "2"

Which die is "the other die"? Couldn't either die start out as the 2?

This is the most ridiculous post yet. If neither die is a two the peeker couldn't truthfully tell you at least one die is a 2. And yes both dice could be a two. It still doesn't matter in a two dice problem.

Someone take two dice and act out the problem and without changing the face of the die that is a two (or one of the twos if both show a two) COUNT ELEVEN POSSIBLE COMBINATIONS. YOU CAN'T.

No Alan. This is one of your most ridiculous posts yet. Considering only the 11 combinations where "at least 1 of the dice is a 2.", you agreed either dice could be a 2. Then you state right afterwards that you can't rotate the dice to show either die could start out as the 2.

[a2a3dseddie]

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 deuce 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.

So why didn't you take this bet Alan? Is it because you realised the answer is 1/11? Is it because you can't roll the dice separately? Set one die on 2 and reroll the other die? Set one die on 2 and spin the other one?

Not going to respond to this Alan?

[a2a3dseddie]

Stop responding to Alan. He's doing the same exact shit Rob Stringer would do. It's been how many posts or even months (years?) on this subject? He will not concede.

He's a troll and/or completely fucking retarded. Don't give him the attention.

I'm about at that point RS__.

We need to put together a poll. Why is Alan still arguing about this problem 3 years later?

1. He's trolling.
2. He's unsuccessfully trying to dupe others into thinking it incorrectly is 1/6 to keep him company.
3. He really doesn't understand why it's 1/11 and is truly looking for help.
4. He knows it's 1/11 and the arguing is just posturing on his part to try and save face.

[Alan Mendelson]

Which die is "the other die"? Couldn't either die start out as the 2?

This is the most ridiculous post yet. If neither die is a two the peeker couldn't truthfully tell you at least one die is a 2. And yes both dice could be a two. It still doesn't matter in a two dice problem.

Someone take two dice and act out the problem and without changing the face of the die that is a two (or one of the twos if both show a two) COUNT ELEVEN POSSIBLE COMBINATIONS. YOU CAN'T.

No Alan. This is one of your most ridiculous posts yet. Considering only the 11 combinations where "at least 1 of the dice is a 2.", you agreed either dice could be a 2. Then you state right afterwards that you can't rotate the dice to show either die could start out as the 2.

Yes either dice or both dice could be a two BUT THAT IS HOW THEY LANDED. THEY AREN'T ROTATED OR CHARGE ONCE THEY LAND. Changing or rotating a die is the only way to get your 1/11 answer. It doesn't happen in reality and it only happens when you look at a chart of 36 possible combinations.

You applied the wrong answer to this simple one die question. Sorry but I really can't get past that a die only has six sides.

[a2a3dseddie]

This is the most ridiculous post yet. If neither die is a two the peeker couldn't truthfully tell you at least one die is a 2. And yes both dice could be a two. It still doesn't matter in a two dice problem.

Someone take two dice and act out the problem and without changing the face of the die that is a two (or one of the twos if both show a two) COUNT ELEVEN POSSIBLE COMBINATIONS. YOU CAN'T.

No Alan. This is one of your most ridiculous posts yet. Considering only the 11 combinations where "at least 1 of the dice is a 2.", you agreed either dice could be a 2. Then you state right afterwards that you can't rotate the dice to show either die could start out as the 2.

Yes either dice or both dice could be a two BUT THAT IS HOW THEY LANDED. THEY AREN'T ROTATED OR CHARGE ONCE THEY LAND. Changing or rotating a die is the only way to get your 1/11 answer. It doesn't happen in reality and it only happens when you look at a chart of 36 possible combinations.

You applied the wrong answer to this simple one die question. Sorry but I really can't get past that a die only has six sides.

Oh... They landed now?

One isn't spinning?

Since you are ignoring the initial 6/11 might as well ignore the second 1/6 since they both landed.

If you insist on asking the peeker the question and are ignoring all the odds to get to the result what would his answer be? 100% or 0%?

[a2a3dseddie]

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 deuce 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.

So why didn't you take this bet Alan? Is it because you realised the answer is 1/11? Is it because you can't roll the dice separately? Set one die on 2 and reroll the other die? Set one die on 2 and spin the other one?

Not going to respond to this Alan?

Not going to respond to this Alan?

[Alan Mendelson]

No Alan. This is one of your most ridiculous posts yet. Considering only the 11 combinations where "at least 1 of the dice is a 2.", you agreed either dice could be a 2. Then you state right afterwards that you can't rotate the dice to show either die could start out as the 2.

Yes either dice or both dice could be a two BUT THAT IS HOW THEY LANDED. THEY AREN'T ROTATED OR CHARGE ONCE THEY LAND. Changing or rotating a die is the only way to get your 1/11 answer. It doesn't happen in reality and it only happens when you look at a chart of 36 possible combinations.

You applied the wrong answer to this simple one die question. Sorry but I really can't get past that a die only has six sides.

Oh... They landed now?

One isn't spinning?

Since you are ignoring the initial 6/11 might as well ignore the second 1/6 since they both landed.

If you insist on asking the peeker the question and are ignoring all the odds to get to the result what would his answer be? 100% or 0%?

You really amaze me. I told you before the spinning die was a way to explain that one die has been decided and one die is left to be decided.

And so it is with the dice under the cup with the peeker seeing one die.

When you have a spinner on the craps table the answer is 1/6 and it doesn't matter which of the two dice it is.

When the peeker sees one die the other die is 1/6 and it doesn't matter which die it is.

What's even more amazing is this belief that there are 11 possible answers for a die with six sides. You can only get 11 by changing the die with a two.

What a strange game that is. Patent it and sell it. Call it mystery craps. Or Wild Craps. You might have a winner.

[mickeycrimm]

Alan initially liked the bet but later, forgetting he had written about playing poker several times, begged off saying he didnt gamble with individuals. Shack talked him into playing for lunch. The challenge was very convenient for both parties as Shack lived in Vegas and Alan visited Vegas frequently. But Alan never showed. I'm sure Shack is still willing but I don't think Alan is. He's afraid of what the outcome will be.