Question for Math/Gambling/Craps Experts

[RS__]

Ultimately Alan has to PROVE it to himself. At the moment he has the sticking point that he is answering his own question to which the correct answer is 1/6

This.

No matter what kind of proof we show Alan, we'll get the same response, something like, "One die is a 2. The other is unknown. The chance of rolling a 2 on that die is 1/6. Simple. Easy. That's it. Nothing else."


I think deep down inside Alan knows the truth to be 1/11, but he's dug such a deep hole for himself, he can't admit 1/11 is the correct answer. I don't know, but I suspect he came to this / his forum because he thought the Rob Singers and others alike would undoubtedly agree with 1/6. Furthermore, I think Alan knows 1/11 to be correct, which is why he's not willing to wager $$$. But, anyone (who gambles) truly believes in 1/6, would have to have a very good reason to deny such a lucrative bet (33% edge on resolved bets, 5.55% edge on a per-roll basis). And I'm not talking about betting peanuts, either (unless you'd like to bet using peanuts).

[OnceDear]

The exercise with the single die was exactly that. an exercise. We are working step by step to an exercise where YOU will prove that the answer to the original question is 1/11.

You do trust me don't you?

The exercises will get harder.

And before we progress. I will say, for the record, that the 1/11 answer seems totally absurd, ridiculous, stupid.

But the word is 'Seems' We have to really bite the bullet and grind step by step to a proof.

And again, if you trust me, REALLY trust me, I will tell you right now that I KNOW CATEGORICALLY that neither you nor I is misinterpreting the question. That's a boggler, but really.... we are not.

[OnceDear]

No matter what kind of proof we show Alan, we'll get the same response ...

I think deep down inside Alan knows the truth to be 1/11, but he's dug such a deep hole for himself, he can't admit 1/11 is the correct answer.

I'm going to disagree with you slightly. I suspect that Alan does realise that it's odd to find so many others cannot see what is so blindingly obvious to him. He might have thought that by posting on his own forum that his friendly locals would all show him how he was right.

One day Alan MIGHT address the FACT that the original question has been badly mis-quoted on the Bob Singer comment page. But that's for later.

[redietz]

The bigger mystery to me is what relevance this has to a forum discussing Las Vegas and gambling.

[1in11]

The whole problem here hinges on one piece of information: which die shows a 2.

If I tell you that at least one die is a 2, can you tell me with absolute certainty which die is showing a 2?

[OnceDear]

Go ahead and state your case. I don't see the need for any more Q&A about conditions. I am sure there are others here who would like to see your reasoning and please post it.

Alan, The Q and A are drawing to a close and would proceed onwards to some simple exercises to build trust in me and to establish all the facts needed for you to prove to yourself that 1/11 is correct. I'm going to refuse to 'state my case' unless you proceed with me. I'm not inclined to persuade your forum readers. I'm inclined to help you to prove it to yourself. We proceed my way or not at all. It's no skin of my nose.

Do You DARE debate this here on your territory on my terms?

Take up the challenge. Yes or no.
It's your forum, you could mess about with my posts, bar me or do whatever.
However, I will be happy to PROVE to you that the answer to the question as posed is 1/11. Will you let me lead you to the proof, inch by painful inch, word by painful word? Will you dare to do that without.... Ever mentioning Craps? Without ever mentioning Coins? Without ever mentioning spinning dice? Without setting aside one dice as a two and insisting it is the same question? Whether you believe that to be the same question or not, we will answer the question as posed and not your preferred equivalent.* I will walk you to the truth inch by inch on the condition that you do not sidetrack with any assertion that is outside of scope of the question as posed. I will also VERY HAPPILY supply you with a faithful simulation of the question as posed and PROVE with that that the answer is 1/11. I'll ask only that you challenge any assertion that I make with PROOF.

[OnceDear]

The bigger mystery to me is what relevance this has to a forum discussing Las Vegas and gambling.

Simple answer. Alan's forum, Alan's choice.
There are many people prepared to offer Alan a wager which he believes give him an astonishing potential profit based on his belief that an event has a 1/6 probability where we know it to have a 1/11 probability. He's being offered odds of between 7:1 and 8:1
He's also being offered PROOF that he would be foolish to take those odds and PROOF that his calculation of the odds is deeply flawed with PROOF as to how and why he is wrong.

So far he has not shown much inclination to listen and even less inclination to take up a fantastic wagering opportunity. Oh hum. He proceeds to let me help him or he doesn't. If anyone else believes in Alan's logical prowess, there are some, and probably many, who are prepared to wager with you on similar terms.

[Alan Mendelson]

The whole problem here hinges on one piece of information: which die shows a 2.

If I tell you that at least one die is a 2, can you tell me with absolute certainty which die is showing a 2?

Why is it important which die in particular shows a 2?

[Alan Mendelson]

Simple answer. Alan's forum, Alan's choice.
There are many people prepared to offer Alan a wager which he believes give him an astonishing potential profit based on his belief that an event has a 1/6 probability where we know it to have a 1/11 probability. He's being offered odds of between 7:1 and 8:1
He's also being offered PROOF that he would be foolish to take those odds and PROOF that his calculation of the odds is deeply flawed with PROOF as to how and why he is wrong.

So far he has not shown much inclination to listen and even less inclination to take up a fantastic wagering opportunity. Oh hum. He proceeds to let me help him or he doesn't. If anyone else believes in Alan's logical prowess, there are some, and probably many, who are prepared to wager with you on similar terms.

I am not interested in any side wagers and I said that on the Wizard's forum as well. I am interested in having you present your position and please do so.

[Alan Mendelson]

I am interested in your reasoning so I will respond to questions 12 and 13 and again my responses are in bold type:

Consider a single die in isolation
12. When we discuss the probability of A SINGLE DIE landing as a deuce, we work this out as
P(Deuce)=(count of the number of qualifying ways that it might land as a deuce)/(count of all the qualifying ways it might land)

Do you agree with and fully understand that? YES

13. Thinking about a single die : We can do the real world math for one die as above where . . .

(count of the number of qualifying ways that it might land as a deuce)=1
(count of all the qualifying ways it might land)=6

Therefore Probability of a deuce = 1/6

I use the term 'qualifying' to mean landing properly, in play and as prescribed by the rules of our 'experiment'. Probabilities are all about 'experiments' and 'events'


Do you agree with and fully understand that? YES

[OnceDear]

ok. Thanks Alan, I will continue tomorrow, as time permits.

14. Can you see that Probability of Deuce result from a single die, could be expressed as "We expect that if we throw the single die a lot of times, it will land as a deuce approximately 1 in 6 times"

Do you agree with and fully understand that?

15. Now we need to explore this a tiny bit and get back to the concept of 'A Qualified Event'

E.g. 'If the roll of a single die is an even number, then what is the probability of it being a two?

To calculate that, we would use . . .

Probability = P(DeuceWhenEven) = (count of the number of qualifying ways that it might land as a deuce and be an even number)/(count of all the qualifying ways it might land as an even number)

We could work that out correctly as the 'probability of it being a deuce ( When we know it's even)' = Probability = 1/3

Are we still fully in agreement?

We would NOT work it out and come up with P(DeuceWhenEven) = 1/6, because to do so would be to ignore the qualifier in the question.


Do you agree with that?




Do you agree so far? We covered some ground there.

[Alan Mendelson]

Once again my responses are in bold type.

ok. Thanks Alan, I will continue tomorrow, as time permits.

14. Can you see that Probability of Deuce result from a single die, could be expressed as "We expect that if we throw the single die a lot of times, it will land as a deuce approximately 1 in 6 times"

Do you agree with and fully understand that? YES

15. Now we need to explore this a tiny bit and get back to the concept of 'A Qualified Event'

E.g. 'If the roll of a single die is an even number, then what is the probability of it being a two?

To calculate that, we would use . . .

Probability = P(DeuceWhenEven) = (count of the number of qualifying ways that it might land as a deuce and be an even number)/(count of all the qualifying ways it might land as an even number)

We could work that out correctly as the 'probability of it being a deuce ( When we know it's even)' = Probability = 1/3

Are we still fully in agreement? HERE'S WHERE I HAVE TO SAY I CAN'T RESPOND. I WOULD HAVE TO LOOK AT THIS MORE CLOSELY SINCE IT'S NOT SOMETHING I AM FAMILIAR WITH, AND MY TV SHOW JUST COMPLETED AND I HAVE TO LOOK AT EMAILS AND WEB TRAFFIC DATA, BUT PLEASE PROCEED BECAUSE I AM FOLLOWING YOU.

We would NOT work it out and come up with P(DeuceWhenEven) = 1/6, because to do so would be to ignore the qualifier in the question.


Do you agree with that? I DON'T KNOW WHAT THE RELEVANCE OF THIS IS, BUT PLEASE PROCEED.

Do you agree so far? We covered some ground there.

EDITED TO ADD: I just had a chance to review questions 14 and 15 and Yes is my answer to those. But I really need you to start putting this into some sort of context so I know where you are headed?

[synergistic]

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!) I'll fill column a and b with that up to, I dunno, row 46656. I dint actually know if that's enough trials to approximate an answer, but given the huge difference between 1/6 and 1/11, I think it'd be obvious which camp is right. Anyway, columns A and B will represent almost fifty thousand rolls of two dice in a cup.

Column D (because I like spacing in my spreadsheet) will be a formula that returns a value of 1 if EITHER column A or B contains a two, and a value of 0 if NEITHER column contains a two - this represents my trustworthy friend telling me if there is a two under the cup.

Column F will be a formula that returns a 1 if BOTH column A and B contain a two.

Way down at the bottom, in row 46658, column D, I'll put a sum formula that adds up all the values in column D - that represents how many times, in 46656 trials, my friend told me there was a two under the cup.

In row 46658, column F, I'll put a sum formula that adds up all the values in column F - that will represent how many times both dice landed on two.

If I'm not drunk - I am, by the way, so PLEASE correct any logical fallacy I've fallen into - the sum at the bottom of column D should represent how many times my friend and I gave this a go. The sum at the bottom of column F should represent how many of those times BOTH dice ended up as twos. I THINK this is an accurate, simple way to simulate this - anyone disagree?

[Rob.Singer]

"Once dear, "RS", and any other wizard form members now here--the clear answer to the OP's scenario/question, is of course 1 in 6. All you math enthusiasts try to do in these type of "problems" is wordsmith your ways around the very simple question just so you can present more of a theoretical event than it really is. Which die is "peaked at" is irrelevant. The true intention of the question is to determine what the odds of the other die being a deuce is after knowing that one of them is a deuce. But being that it was purposely asked in a loose way--very similar to how the Monty Hall problem created such a firestorm--all the theory-heavy mensa aspirationists on wizard's forum, led by him, chose to lay this issue out into a more complicated calculation. Egos are very heavy over there. I saw that when all the geniuses ran away from my challenge a few years ago, then convinced the wizard to permanently ban me so I couldn't make them look foolish any longer.

Guys, think of it this way. If the problem was presented as "two six-sided dice were shaken & rolled, and when they were peaked at we were told that both dice showed a deuce" with the corresponding question being "what are the odds that both dice would show a deuce?"--Those who look at the question as if we're being asked about "right now" would obviously say 100%. That's because we all know that reality is far more important & meaningful than theory. But the ones who choose to overthink the issue and want to create conflict just so they can believe they're the smartest ones in the room, would say 1 in 36. Why? Because to that group they would rather talk about theory than reality, so they spin the words to fit their agenda.

And I understand that. Wizard's, of course, has the very foolish & misleading tag line on each of his posts about how making a "good bet" is more important than whether the bet is won or lost. Again, math people choose theory over reality because it's the safer choice. But it's not surprising seeing that self-proclaimed AP wizard, who gives the THEORETICAL perception that AP's are winners, had to instead beg for donations, then sell out to online casino foreigners. What will it take and when will you people learn that theory only yields phantom bucks? And you can't use those to buy groceries.

I saw some offers for a bet on the 1 in 6 vs. 1 in 11 issue and I'd like to get involved with this. But what exactly are the parameters, and who gets to interpret the OP's wording and /or intention?

[Alan Mendelson]

Can I pose a different question: Can you tell me why the answer is NOT 1/6 ?? So, instead of trying to tell me WHY it is 1/11 can you tell me why the answer is NOT one out of six? Would that make this process easier?

Synergistic others have already posted answers using columns of possible dice combinations. They can be seen on this forum and on the Wizard's. So before you go to the trouble, please be sure you saw those. Thanks.

[quahaug]

Rob,throw two dice. If one two and any other number shows you give me one dollar. If two twos show I give you 7 dollars. If no twos show its a push. Simple, no cup, no third person needed.

[Rob.Singer]

Alan, I believe you also need to see him or them explain exactly what their interpretation of the OP's question was as presented to the forum. You, I, regnis and maybe others see it as a very clear "what are the odds, now that we know one of the dice is a deuce, that the other die is also a deuce?"

[Rob.Singer]

Rob,throw two dice. If one two and any other number shows you give me one dollar. If two twos show I give you 7 dollars. If no twos show its a push. Simple, no cup, no third person needed.

What does this yield? And the problem only asks about the one die, not both, simply because we are given a value for the first die.

[Alan Mendelson]

Rob,throw two dice. If one two and any other number shows you give me one dollar. If two twos show I give you 7 dollars. If no twos show its a push. Simple, no cup, no third person needed.

I'm sorry but I don't see how this answers the question.

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?

[synergistic]

Throw two dice. If one two and any other number shows you give me one dollar. If two twos show I give you 7 dollars. If no twos show its a push. Simple, no cup, no third person needed.

No fair, the whole argument revolves around whether uncertainty about which die comes up two has an effect on the odds - your version is nearly as much a straw man as the proposal to just set one die to two and only roll one die.