Question for Math/Gambling/Craps Experts

[regnis]

Rob--you just don't see it. If the first die is a 2, you can have 11 combos:

2-1
2-2
2-3
2-4
2-5
2-6
2 over easy
2 much bbq
2's company
2 can live as cheaply as 1
2 wrongs don't make a right

[1in11]

The opening statement does not clearly eliminate one of the dice. It only gives you information about both dice together.

[regnis]

Absurd--it says one is a 2. That only leaves one other die.

[Rob.Singer]

Rob--you just don't see it. If the first die is a 2, you can have 11 combos:

2-1
2-2
2-3
2-4
2-5
2-6
2 over easy
2 much bbq
2's company
2 can live as cheaply as 1
2 wrongs don't make a right

If you posted that on the so-sensitive wizard's forum that's 3 parts D-I-V-E-R-S-I-T-Y and 2 parts T-O-L-E-R-A-N-C-E, you'd discover just how hypocritical they are over there. You'd be immediately suspended by some loser called Face, for insults unbecoming for a genius.

[regnis]

If you posted that on the so-sensitive wizard's forum that's 3 parts D-I-V-E-R-S-I-T-Y and 2 parts T-O-L-E-R-A-N-C-E, you'd discover just how hypocritical they are over there. You'd be immediately suspended by some loser called Face, for insults unbecoming for a genius.

That's why I only post here with the sensitive, open minded, liberal Alan

[1in11]

Absurd--it says one is a 2. That only leaves one other die.

But it doesn't say which one is a 2. Only that one (or both) is a 2.

[synergistic]

Are people deliberately screwing with other people here? I'm so confused...

[redietz]

You know, none of this has any real business on a Las Vegas/Gambling forum. It's somebody writing unclear, torturous verbiage so as to create an apparent logical paradox where none really exists. The people presenting us with this fascinating conundrum (yes, that's sarcasm, Sheldon) may be giving us a little glimpse into their realm. It's such a lively, fascinating place.

[Rob.Singer]

But it doesn't say which one is a 2. Only that one (or both) is a 2.

Which one doesn't matter. What you espouse is the type of wordsmithing that runs rampant on the wizard's site, and why so many of those so-called self-annointed geniuses create obnoxiously lengthy threads over the simplest of questions and comments. Most of them have no REAL ability to gamble, so they live out their theoretical fantasies by corrupting thread after thread with spin and one-upmanship. Here, you'll run into the stone cold truth, like it or not.

[Alan Mendelson]

Forget the cup. One is a 2 so just set that die aside. Now let's try to determine the odds that the second die is a 2.

Give me 11-1 all day!!!!!

Let's give them a chance to make a profit. How about a measly 9 to 1?

[Alan Mendelson]

That's why I only post here with the sensitive, open minded, liberal Alan

I am not a liberal. I am a Rockefeller Republican. You can Google it. :-)

[1in11]

Which one doesn't matter. What you espouse is the type of wordsmithing that runs rampant on the wizard's site, and why so many of those so-called self-annointed geniuses create obnoxiously lengthy threads over the simplest of questions and comments. Most of them have no REAL ability to gamble, so they live out their theoretical fantasies by corrupting thread after thread with spin and one-upmanship. Here, you'll run into the stone cold truth, like it or not.

What I'm espousing here are the probabilities of a pair of dice, and that the probability of rolling an ace-deuce is twice that of rolling a hard 4 (deuce-deuce)

[Alan Mendelson]

What I'm espousing here are the probabilities of a pair of dice, and that the probability of rolling an ace-deuce is twice that of rolling a hard 4 (deuce-deuce)

OK. So what? Each time you have a 1 showing you have a 1/6 of getting a 2 on the other die. Each time the 2 shows it's 1/6 for the 1.

Be honest: have you ever played craps?

And please tell the Wizard if he really thinks it's 1/11 he has to rewrite the craps section on Wizard of Odds. His odds are all wrong.

[fly2rei]

Well that's post number eleven for 1in11, so I hope he's done
Posting on this subject.

[OnceDear]

nswer. If they would accept that ONE DIE is already showing a two, then the obvious answer is that the remaining die has SIX faces ONE of which is a two.

Alan, another agree/disagree question.
Did you yourself say on that other forum
"This is the problem. The ORIGINAL QUESTION deals with TWO DICE under a cup with at least one of them showing a 2. YOU MUST address the problem using a pair of dice with at least ONE DIE showing a 2. If you don't do this you are altering the question."?
Agree or disagree?

And does the original question say "The die on the left is a deuce"

I say No. Do you agree

And does the original question say "The die on the right is a deuce"

I say No. Do you agree

And does the original question say "At least one of the dice" without any clue as to which?

I say YES. Do you agree

So, From the info that we know, there is as much probability that it is the dice on the left as there is that it is the Die on the Right?

Agree or disagree?
So there are at LEAST two qualifying ways that 'One of the dice is a deuce': Left=Deuce or Right=Deuce?

We cannot eliminate either of those two qualifying ways from the information available?

Agree or disagree?

[OnceDear]

Let's give them a chance to make a profit. How about a measly 9 to 1?

If Gambling law is an issue. I'll take up the challenge for something other than money. I'll wager kudos points if you will agree to the actual wager in all other respects.

[OnceDear]

I'm out watching the rangers right now, so unfortunately I don't have my computer with me to try to solve this via brute force at the moment.

What I will do when I get home, is load up excel and make a formula that randomly chooses a digit between one and six (inclusive, of course!) - anyone disagree?

Here's one I prepared earlier.
http://oncedear.com/AlansFolly.xlsx
It faithfully reproduces the question where one of the dice is a deuce AND WHERE WE DO NOT KNOW WHICH ONE. Please only ONLY answer the original question as in the first post of this thread. The question on Bob's page here is different and others have also created different models of the question

[OnceDear]

Rob--you just don't see it. If the first die is a 2, you can have 11 combos:

There is no first die, there is no second die. There are two rested Dice and one of them, we don't know which of them, is a Deuce.
So, we started with 36 possible ways that these dice could have landed.
With the information that we have, we can eliminate 1,1 1,3 1,4 1,5 1,6 3,1 3,3 3,4 3,5 3,6 4,1 4,3 4,4 4,5 4,6 5,1 5,3 5,4 5,5 5,6 6,1 6,3 6,4 6,5 and 6,6
That's twentyfive possible landings that we have eliminated leaving any of the following possible ways that 'At least one of the dice was a 2
2,1 2,2 2,3 2,4 2,5 2,6 1,2 3,2 4,2,5,2 6,2

There are 11 possible ways that those pair of dice could have landed, Those possibilities are real and equal and cannot be eliminated with the question that set up the problem. THIS IS FROM THE ORIGINAL QUESTION as per message one in this thread.

[OnceDear]

The original question said somebody is peeking at one of two dice under a cup, and it's a two. Okay, just do the real world wager. Roll two dice in a cup and have somebody peek. When one of them is a two, then make the wager. We need an honest peeker -- I volunteer.

Bring lots of cash. Rob usually has lots of cash.

Halleluia.... What original question are you looking at Red? If you are looking at the original question on Bob Singers comments page it does say that one dice is looked at. In the original question as agreed explicitely with Alan, BOTH the dice are peeked at. That gives 5 more possible ways that we could have ended up with the assertion that 'At least one of the dice is a two.

[Alan Mendelson]

Once again my responses are in bold type.

Alan, another agree/disagree question.
Did you yourself say on that other forum
"This is the problem. The ORIGINAL QUESTION deals with TWO DICE under a cup with at least one of them showing a 2. YOU MUST address the problem using a pair of dice with at least ONE DIE showing a 2. If you don't do this you are altering the question."?
Agree or disagree? I AGREE AND I STAND BY THAT. IN ORDER TO ANSWER THE QUESTION OF 1/6 (OR IN YOUR CASE 1/11) YOU MUST CONSIDER BOTH DICE.

And does the original question say "The die on the left is a deuce"

I say No. Do you agree AGREE. THE ORIGINAL QUESTION DID NOT GIVE ANY POSITION FOR THE DICE UNDER THE CUP. AND I DID SAY IT DOESN'T MATTER.

And does the original question say "The die on the right is a deuce"

I say No. Do you agree YES I AGREE, THE ORIGINAL QUESTION DID NOT GIVE A POSITION AND AGAIN, IT DOESN'T MATTER.

And does the original question say "At least one of the dice" without any clue as to which?

I say YES. Do you agree YES I AGREE, THE ORIGINAL QUESTION DOES NOT INDICATE WHICH OF THE TWO DICE, AND AGAIN I SAY IT DOESN'T MATTER.

So, From the info that we know, there is as much probability that it is the dice on the left as there is that it is the Die on the Right?

Agree or disagree? YES, I AGREE.
So there are at LEAST two qualifying ways that 'One of the dice is a deuce': Left=Deuce or Right=Deuce?

We cannot eliminate either of those two qualifying ways from the information available?

Agree or disagree? IF YOU WOULD LIKE TO REFER TO THE DICE UNDER THE CUP IN THE ORIGINAL QUESTION I WILL ACCEPT THAT AS A WAY TO FURTHER THE DISCUSSION... NOT THAT IT MATTERS IN THE END, BUT GO AHEAD.