How do you interpret the "dice problem"?

[1in11]

I don't know what language you "math guys" speak, but English is English.

True, English is English. Unfortunately, English is full of ambiguities. For example:

In English, "one or the other die is a 2" means that either Die A is a 2 or Die B is a 2. Both cannot show 2.

Your interpretation of this statement is different from OneHitWonder's interpretation. You understand this statement as using the exclusive-or, where OneHitWonder (and I) understand this statement as using the inclusive-or.

Now, where is that video with at least one die showing a 2 that proves the 1/11 answer when the die showing a 2 isn't changed?

I'm sorry, Alan, I can't in good faith make a video in which I arbitrarily ignore nearly half of the possible outcomes.

[Alan Mendelson]

Thanks for your response 1in11 and there is nothing left for me to discuss with you on this subject. When you find those magical dice that settle showing one number but allow you to utilize the other faces please let me know because I'd like to buy some.

[Alan Mendelson]

I made a mistake. I read Indignant's post on the WOV forum in which he called me a "liar" for the following post here:

I have to laugh... after quoting Ibeatyouraces here he removed this post from the WOV forum (his typo remains):

They keep thinking it's a once die problem. It's not! People, there are 12 TOTAL faces. Only one face gets eliminated. Not the whole Damn die! There are 11 faces left. The damn answer is 1 in 11. How GD hard is that to understand???

After seeing that, I saw that in fact, the original quote by Ibeatyouraces was not removed from the WOV, so that was my error. Yes, that same nonsensical comment is still there because the 1/11ers still want to believe that when a die settles on a 2 you can still use the other five faces on that same die to figure their equations.

As I told 1in11 that if those magical dice are ever available for sale I would like to buy them. And I would love for the casinos to use them at craps, too.

[Rob.Singer]

Sign me up for a couple dozen too. And they better come with instructions on how you can fit six die faces on a single face. But I'm sure the mensas can figure it out.

[regnis]

Hmmmmmm-can we invent a game where the player gets to change one die after the roll and still have a house edge???

[OneHitWonder]

Clearly it means there is at least one prize and there can be more than one prize. If there is only one prize then it can be both deceptive and false. As a matter of fact, "deceptive" statements are "false" statements

And certainly when such statements concern the impossible or unreasonable.

To claim that a lottery involves a prize or prizes - or that dice involve the number 2 - is quite different from setting out specific scenarios' ranges. Probabilities can be assigned to each of the "or more" parts, to calculate the EV's for such dice and lotteries. Now we're getting serious? AP fun.

de·cep·tive

adjective
giving an appearance or impression different from the true one; misleading.

https://www.google.com/search?q=dece...sm=93&ie=UTF-8

You could have been a fine lexicographer, Alan. I'm not surprised that they attack you at your strengths. The natures of the AP's after all: attack the casinos; and employ multiple hasty 5-cent simulations and 10-cent equations.

Besides all of this, one item remains from the start of the original dice problem thread which hasn't been addressed at all with respect to the original problem. A number isn't agreed upon in advance of the roll. I think it was Mango who called it "forcing the number" when all of the rolls are used or none is discarded. He concluded that if a number is forced for each roll, then the chance of two numbers the same on a roll is 1/6.

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?

http://wizardofvegas.com/forum/quest...o-dice-puzzle/

[Rob.Singer]

Alan can you please explain what he was attempting to say?

[Alan Mendelson]

Rob... I think he agreed with me. :-)

[OneHitWonder]

Alan can you please explain what he was attempting to say?

Rob, this problem has been mangled to death by both sides. The Wizard's, of which who ought to know better; and, Alan's, of which who have the right answer but haven't expressed it fully.

[OneHitWonder]

Rob... I think he agreed with me. :-)

Yes, I agree completely with your response, and some of your reasoning. I now also agree that this forum is a much better one in which to present ideas.

Besides all of this, one item remains from the start of the original dice problem thread which hasn't been addressed at all with respect to the original problem. A number isn't agreed upon in advance of the roll. I think it was Mango who called it "forcing the number" when all of the rolls are used or none is discarded. He concluded that if a number is forced for each roll, then the chance of two numbers the same on a roll is 1/6.

Alan's video shows him concerned with waiting for any 2 to occur in a roll. That is a different version of this problem in general than the one originally presented at the Wizard's in the link I posted above. Mango pointed out right away that if we take each roll and say "at least one die is an X, where X is one of the numbers 1 to 6" that the answer is 1/6. Mango is correct. Eg, if the roll comes 1-3, then saying that "at least one is a 3" means that the other die will be a 3 with a 1/6 chance. The Wizard gave his answer right after the question, and everyone forgot about Mango's answers for the two popular cases of the problem.

If you have to wait on a 2, usually after many rolls, then the correct answer would be 1/11. Unless you ask it as "at least one two", which would make it about the 1/6 answer in some way. It has to be "one or the other die is a 2", which means only that the number two is involved. (So that one 2, two 2's, one or more 2's, or another concoction of these, is not specified about the specific roll.)

[1in11]

Besides all of this, one item remains from the start of the original dice problem thread which hasn't been addressed at all with respect to the original problem. A number isn't agreed upon in advance of the roll.

This information is not on the original problem. We do not know the peeker/caller's "strategy." If we knew the "strategy" then we could analyze that.

For example, if the peeker picks a die at random and then says thay there is at least one of (whatever that die is showing) then then answer is 1/6. If the peeker picks the largest die and says that there is at least one of those, the answer is 1/3. If the peeker looks at the smallest die and calls that, the answer is 1/9.

[OneHitWonder]

This information is not on the original problem." If we knew the "strategy" then we could analyze that.

We know that the "peeker" didn't in advance say, "I am concerned only with rolls of some sort of 2, so keep rolling until we hit some sort of 2."

The in-advance condition allows for the chart to build up, at least in theory, even with the number 2 appearing on the first roll.

We do not know the peeker/caller's "strategy." If we knew the "strategy" then we could analyze that.

This problem has more wrinkles than a 100-year-old woman.

[regnis]

This information is not on the original problem. We do not know the peeker/caller's "strategy." If we knew the "strategy" then we could analyze that.

For example, if the peeker picks a die at random and then says thay there is at least one of (whatever that die is showing) then then answer is 1/6. If the peeker picks the largest die and says that there is at least one of those, the answer is 1/3. If the peeker looks at the smallest die and calls that, the answer is 1/9.

?????--can't be.

[Rob.Singer]

On vpFree now a different poster is believing some hi-rolling craps player at Bellagio named Alan, went on a hot streak at a reserved-for-one table, made all kinds of money, then lost it all in the same session.....is AM!

[Alan Mendelson]

On vpFree now a different poster is believing some hi-rolling craps player at Bellagio named Alan, went on a hot streak at a reserved-for-one table, made all kinds of money, then lost it all in the same session.....is AM!

The last time I played craps for any serious money at Bellagio was 9 or ten years ago -- when I was told to leave the table because I was having too good of a roll and my dice were not bouncing off the back wall far enough. And then a few months later a floorman at the Bellagio rudely threw my players card at my hand, hitting it, and strongly suggested I leave.

About three years ago I went to a $10 table at Bellagio and bought in for $40 just to see if some facial recognition software equipment would nail me. No one said a word. But it might have been because I bought in for only forty dollars.

[Rob.Singer]

Yes I remember you said that. I believe you didn't have a problem because of the same reason I didn't after being banned from playing vp--then several years later going in there and hitting a $25 royal: change of ownership. I was banned when Wynn owned it, then went in after it was sold.

[Alan Mendelson]

Yes I remember you said that. I believe you didn't have a problem because of the same reason I didn't after being banned from playing vp--then several years later going in there and hitting a $25 royal: change of ownership. I was banned when Wynn owned it, then went in after it was sold.

I don't follow, Rob? I had a real problem. There was a shouting match with the dealers and me, and then the dealers shouting at the floor person who said I did nothing wrong. And then when I returned to Bellagio a few months later it was obvious that there was a notation in their computer system and that's when I was pressured to leave.

Remember that at both MGM and NYNY they took the dice away from me, and NYNY told me to leave the craps pit.

[Rob.Singer]

No not that problem. The problem of whether you could go back in or not after they threw you out, without being recognized.

BTW, one of the six posters who regularly pontificates to impress on vpFree (Harry Porter, a nerd's nerd) just posted how wrong you are on the 2 dice problem, without identifying an iota of your interpretation.

[Harry Porter]

Just curious ... can we agree that the original question is equivalent to asking, "If you roll two dice, for those rolls on which at least one "2" appears, how frequently will both dice show a "2"?

If not, what's different?

[Alan Mendelson]

Harry Porter thanks for joining and for posting.

Yes that's what the question asks. But what's your answer?