Question for Math/Gambling/Craps Experts

[OnceDear]

Synergistic your application of Excel was just wrong.

Alan, Have you seen Syn's spreadsheet? Have you seen mine? It's freely available here http://oncedear.com/AlansFolly.xlsx Take a look at any cell formula. Take a look at all Cell formulae if you can be bothered to read all 50001 lines. Show me any single tiny error in any one cell.

Explain to me in the simplest terms possible why columns a and b do not replicate the posed scenario exactly.

[OnceDear]

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?

You did not say that you agree with that last assertion. Do you agree with that last assertion or not?

[OnceDear]

Synergistic instead of Excel take a pair of dice and figure it out the old fashioned way. Make one die a 2 and then do the math. Then change the dice putting the 2 on the other die and figuring the math again.

1in11 you do the same thing.

Alan, Which Die has Syn got to set as a two. He has surely got to account for the fact that he really doesnt know. He needs to set the pair of dice in all the ways that they might have landed 'With at least one two'.
That's ALL the possible ways.
shall we name them
Some of the possible ways
Left=2,Right=1
Left=2,Right=2
Left=2,Right=3
Left=2,Right=4
Left=2,Right=5
Left=2,Right=6

Do you agree so far?

[Alan Mendelson]

I am sorry but I think this was a bit awkwardly written and I really don't understand what you are saying or asking:

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

The way I read your question is this: from the original question that information is not available. But yes, if you want to say one die is on the left or one die is on the right showing a deuce that is okay with me to help forward the discussion.

Frankly, at this point, I am starting to become tired of your questions. It's time for you to give us some answers, or to be more exact, what you are driving at? Your questions remind me of two things: one, is the exam I took for entrance to law school which contained a lot of theory including the use of imaginary languages; two, is the set-up that some conmen gave their victims before selling them fraudulent investments. By the way, I helped put those conmen in prison.

[Alan Mendelson]

Sorry, I am not using a regular computer and I am unable to see the spreadsheets.

[OnceDear]

I am sorry but I think this was a bit awkwardly written and I really don't understand what you are saying or asking:

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

The way I read your question is this: from the original question that information is not available. But yes, if you want to say one die is on the left or one die is on the right showing a deuce that is okay with me to help forward the discussion.

Frankly, at this point, I am starting to become tired of your questions. It's time for you to give us some answers, or to be more exact, what you are driving at? Your questions remind me of two things: one, is the exam I took for entrance to law school which contained a lot of theory including the use of imaginary languages; two, is the set-up that some conmen gave their victims before selling them fraudulent investments. By the way, I helped put those conmen in prison.

OK. Sorry if you find it awkwardly written. I too am getting tired.
I'll rephrase.
Is it equally possible, in the absence of any other infoirmation except that given that "at least one of the dice is a two' that we have any one of these possible outcomes.


Right=2, Left-=1
Right=2, Left-=2
Right=2, Left-=3
Right=2, Left-=4
Right=2, Left-=5
Right=2, Left-=6

Right=1, Left-=2
Right=3, Left-=2
Right=4, Left-=2
Right=5, Left-=2
Right=6, Left-=2

Do you yet again agree that we do not know which dice is a deuce?

Do you agree that without knowing which dice is a two, that all of the above outcomes are equally possible?

[Alan Mendelson]

Alan, Which Die has Syn got to set as a two. He has surely got to account for the fact that he really doesnt know. He needs to set the pair of dice in all the ways that they might have landed 'With at least one two'.
That's ALL the possible ways.
shall we name them
Some of the possible ways
Left=2,Right=1
Left=2,Right=2
Left=2,Right=3
Left=2,Right=4
Left=2,Right=5
Left=2,Right=6

Do you agree so far?

Unfortunately it doesn't make a difference which of two dice has a 2 for the purposes of finding the odds for 2-2 or a hard four. If the 2 is on the left or if it's on the right, or if it's on the top die or the lower die, the odds are still one in six for the "other die."

I understand you are in England? Well, in American lingo you are barking up the wrong tree, or working with a dog that can't hunt, and need to go back to square one, or go back to the drawing board, because this ain't gonna fly, cause your reasoning ain't kosher.

As in the phrase that came out of the Hollywood movie industry: it's time to cut to the chase.

[RS__]

The necessary part of the original post on WOV:

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?

So, let's simulate this in real life the best we can.

Let's grab two dice, both equal/fair with 6 sides each.

Take a cup, put both dice in them, shake 'em up.

Slam the dice down, have a partner peek. If you'd like to do this on your own, I'm sure you can peek yourself if you'd like, as I don't think that changes the odds (maybe you think it does, but I do not).

If either of the two dice show a deuce (2), then record the result as either "2-2" or "2-X" [where X is a non-2]. "2-2" is in one column and "2-X" is in a second column.

Do this a bunch of times.

But what do you do when neither of the two dice is showing a 2? Well, nothing. The OP states nothing about a no-deuce roll. But it does state that two dice are rolled (or rather, shaken in a cup and slammed onto a table). And no, it does not state anything like, "Set one die to a deuce then roll/shake the other die."


DO NOT take one die and set it to a 2 then roll the other die. Where in the original question does it say anything like, "You have two dice, you set one die on a 2, and you roll [or shake or slam a cup] with the other die."



PS: I'm not 100% on this, but, I'm pretty sure friendly wagers are legal. The illegal type of gambling is where there is a house that takes a cut/rake and/or where you have no skin in the game (ie: You offer a poker game for people to play in, but you take in some % of the pot every hand for your own profit).

[Alan Mendelson]

RS what is the point of this exercise?

[RS__]

RS what is the point of this exercise?

It's the same thing presented in the original question. You roll two dice. You do not set one die to a deuce then roll the other die.

[Rob.Singer]

It's the same thing presented in the original question. You roll two dice. You do not set one die to a deuce then roll the other die.

Your entire premise is based on a need to keep both dice in play when clearly one of them, and it doesn't matter which one, has an identified value--and is therefore out of play--and the other die does not. I understand how the self-absorbed geniuses on WoV who rarely gamble because they can't, always require unnecessarily looking at these questions in their most complicated theoretical scenarios strictly for purposes of perception. But over here it's the real world with no egos beyond getting to the undeniable truth. And that's why no one's buying a need to keep both dice in play when it's irrelevant to do so.

All one needs to do is "peek" at one of the current threads over there, where yet another "mensa" is claiming that on Mars, 2+2 may not = 4. Nuff said.

[1in11]

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?

Yes, I have played craps. I have also dealt craps, supervised a craps pit, and helped to teach new craps dealers. I believe that it is fair to say that I understand the game.

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.

It is 1/11 given that at least one of the dice is a 2. The probability that at least one of the dice is a 2 is 11/36. There is no way to roll 2-2 without rolling at least one 2, so the conditional probability formula gives us:

Probability(rolling 2-2) = Probability(rolling 2-2, given that at least one of the dice is a 2)*Probability(rolling at least one 2)
Probability(rolling 2-2) = (1/11)*(11/36) = 1/36

[Alan Mendelson]

It's the same thing presented in the original question. You roll two dice. You do not set one die to a deuce then roll the other die.

Oh really? Well my demonstration shows that the odds are 6/1. And guess what? Your demonstration will show the same thing.

By the way we all know the hard 4 is 1/36 rolling two dice. That's 1/6 x 1/6. So where is that 1/11 come in?

[jbjb]

Oh really? Well my demonstration shows that the odds are 6/1. And guess what? Your demonstration will show the same thing.

By the way we all know the hard 4 is 1/36 rolling two dice. That's 1/6 x 1/6. So where is that 1/11 come in?

Corret Alan. Now eliminate all of the non-2 combos from the 36 and you'll be left with 11, hence the 1/11.

[Alan Mendelson]

Corret Alan. Now eliminate all of the non-2 combos from the 36 and you'll be left with 11, hence the 1/11.

Thanks for joining. But I just did a quick Google search for odds on hardways and they all say 1/6 x 1/6.

[jbjb]

Thanks for joining. But I just did a quick Google search for odds on hardways and they all say 1/6 x 1/6.

Certainly. But only if you consider ALL 36 combinations. When you eliminate some, the odds for others improves.

[jbjb]

Rather you or your members get suckered into a bad bet!

[Alan Mendelson]

Certainly. But only if you consider ALL 36 combinations. When you eliminate some, the odds for others improves.

Are you suggesting that using the 1/11 odds for figuring a hardways bet improves your odds on various bets in craps?

[1in11]

Are you suggesting that using the 1/11 odds for figuring a hardways bet improves your odds on various bets in craps?

Nobody is suggesting this, because we don't have a bet on the craps table that pays on a hard 4 and loses on any other combination of 2-X.

Let's look at the all-day hard 4 bet that we do have on the table: it wins on a hard 4, and loses on any other combination of 4, or on any 7.
What's the probability of a hard 4? 1/36
Probability of an easy 4? 2/36
Probability of a 7? 6/36
Now, given that the bet resolves (that the roll is either a 4 or a 7), what is the probability that I won? It's the probability of a win (1/36) divided by the probability of either a 4 or a 7 (1/36 + 2/36 + 6/36 = 9/36)
(1/36)/(9/36) = 1/9.

The probability of rolling a hard 4 is 1/36, and the probability of rolling a hard 4 given that the total is 4 or 7 is 1/9. These are not contradictory facts. And this is the same method used to answer the original question.

[redietz]

Of course there is a first die and a second die in the original description. The description says someone (a person) peeked under the cup and reported. A person would see the marking on one die before the second and process that information sequentially, not simultaneously.

Alan asked if the folks positing this question ever played craps. I'm more concerned with the weirdness of designing questions to demonstrate one's "intellectual superiority" over other human beings with badly written trivia. Here's my question, ironic as it is. Have you ever asked questions to demonstrate your "intellectual inferiority" in other forums? Or do you go around waving some mental dicks in the air and debating trivia as a means to pass the time?