Question for Math/Gambling/Craps Experts

[jbjb]

You don't even need a cup and/or to peek to see the results. Just roll two dice. If two 2's show up, you win. If 2-1, 2-3, 2-4, 2-5, or 2-6 show up, you lose. True odds are 10:1.

The wording of the question is the whole trick to the puzzle, agreed. This is exactly how and why casinos do these things. To create confusion.

[OnceDear]

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.

I didn't say whether it makes a difference or not. I asked if you agree that all of those outcomes from the landing of two dice is equally possible. You either agree that they were equally possible or you don't

[OnceDear]

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.

Red, Would he stop looking if the first die was not a two? Is there just the teensy weensiest possibility that he carries on looking and says to himself,
'blow me, there wasn't a two on that first die but by crikey "at least one of the dice is still a darned two" I never expected that.'

[Rob.Singer]

These guys are doing what they do on WoV and what one-way-blinded math people do all the time: divert & deflect in order to defend their silly theories. I'm still waiting for one of them to describe the parameters of the bet several of their brave anonymous members posted.

[Rob.Singer]

Red, Would he stop looking if the first die was not a two? Is there just the teensy weensiest possibility that he carries on looking and says to himself,
'blow me, there wasn't a two on that first die but by crikey "at least one of the dice is still a darned two" I never expected that.'

That's stupid. Who says he was looking for a deuce? Again, you are unable to support any of your theories.

[OnceDear]

That's stupid. Who says he was looking for a deuce? Again, you are unable to support any of your theories.

Try to keep up Rob, we are talking about a scenario from a few posts back, where the peeker would report if 'At least one dice is a two' if that was at all possible. To do that he would look at one die, maybe the one nearest to him. If that is a two (deuce) he would report 'At least one dice is a two' and his job would be complete.... But if, by some weird chance, that first die was not a two, he would keep looking, he would see that there is a second die there. Maybe, just maybe, that second die was a two and he gets another chance to say 'At least one of the dice is a two'. If that second die was not a two, he could pop the cup down, say nothing, and slam the dice down again.
What is slightly at issue is how many possible different dice outcome could result in him reporting 'At least one of the dice is a two'

[OnceDear]

Hey Rob,

We could use a new perspective on this. Would you help me out with some maths please. Seriously?
Nothing complex, just a bit of counting.

[OnceDear]

Question for Rob Singer, if he would be so kind as to help me out....

Rob, If we throw a pair of dice in every one of the 36 possible ways and we can see both dice land, how many times would be able to truthfully say. "Hey! At least one of those dice is showing as a deuce"?

[Rob.Singer]

You didn't get it...again. Any obfuscation having to do with two dice, spread sheets, and proposed bets that disappear in the wind after one of the dice has been eliminated from the problem, is testiness beginning to set in.

[OnceDear]

You didn't get it...again. Any obfuscation having to do with two dice, spread sheets, and proposed bets that disappear in the wind after one of the dice has been eliminated from the problem, is testiness beginning to set in.

So Rob cannot or will not help me with that bit of counting. It seems only junior members here really care to think for themselves. Ho Hum, it's your forum, your rules.

[synergistic]

Synergistic your application of Excel was just wrong.

Would you mind terribly looking through my steps and telling me which column or columns do not accurately represent your initial question?

[OnceDear]

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

DO IT. DO IT. DO IT
. If RS's demonstration will show the same thing, then sit back and smugly watch it do it. Or don't you have the bottle?

[OnceDear]

Would you mind terribly looking through my steps and telling me which column or columns do not accurately represent your initial question?

Sadly, Alan is not the kind of business man to be able or willing to do that. To quote him from earlier :-"Sorry, I am not using a regular computer and I am unable to see the spreadsheets. "

I seem to recall that one member here with admin rights had the courage to suggest that 1/11 is correct. Maybe that admin has a copy of Excel that he would care to share with Alan. Or maybe that admin would not do so for whatever reasons. I cannot see Alan reaching out to borrow the facility to use Excel.

[1in11]

That's stupid. Who says he was looking for a deuce? Again, you are unable to support any of your theories.

Who says he was going to call out the value of the first die that he sees? Again, you've added information to the question that doesn't exist.

If the person calling the dice decides that he's going to look at the first die, and then claim that "At least one of the dice is showing [whatever that first die is]", and you have that information, then you are absolutely correct that the probability of 2-2 is 1/6.

But, we don't know that this has happened here. All we know is that the caller has said that "at least one of the dice is showing a 2." If we know his strategy for calling the dice, then we'll better know the probability of 2-2.

For example, if we know that the caller is going to call out the smallest die, then the probability is 1/9. If we know that the caller is going to call out the largest die, then the probability is 1/3. If we know that the caller is going to call out a random die, then the probability is 1/6.

[1in11]

These guys are doing what they do on WoV and what one-way-blinded math people do all the time: divert & deflect in order to defend their silly theories. I'm still waiting for one of them to describe the parameters of the bet several of their brave anonymous members posted.

I feel like the bet has been clearly described a number of times, but I'll describe it again for you.

For each trial, we roll 2 dice.
If neither of the dice show a 2, then there is no action.
If at least one of the dice show a 2, then the bet has action: The bet will win and pay odds if the dice are showing 2-2. The bet will lose 1 unit if the dice are showing a 2 and any other number.

The only thing left is to determine what the proper odds for the payout should be. You believe the answer is 1/6, and so from your perspective, the fair odds would be 5 to 1. I believe the answer is 1/11, and so from my perspective, the fair odds would be 10 to 1. Let's set the payout odds exactly between those two, at 7.5 to 1 (or 15 to 2, if you prefer).

[OnceDear]

For example, if we know that the caller is going to call out the smallest die, then the probability is 1/9. If we know that the caller is going to call out the largest die, then the probability is 1/3. If we know that the caller is going to call out a random die, then the probability is 1/6.

And if we know that the caller is going to call out "At Least one of the dice is a two" whenever it is truthfully possible to say that "At least one of the dice is a two"????
For the record, Alan has already agreed with me that the partner looks at both dice every time.


Can anyone work out for me how many possible unique and equally likely ways that that can happen?

Rob S doesn't seem to want to. Alan doesn't seem to have the capability of doing so. Maybe they don't think it is relevant. Maybe they think this perfectly reasonable question is "Wordsmithing". Maybe they don't know how to.

Does anyone who hasn't already worked out and shouted out the answer care to have a try at working that out?

Work out if I'm "Wordsmithing" or indeed work out if the 'partner' in the original post is calling out by the procedure as I describe it?

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? was the question, if I recall.

I'm simply asking at this point, how many equally likely ways the partner can call this out out of what I believe to be the 36 different ways that the dice might have landed. I'm not even asking about probabilty stuff. Just a simple count up.
Like Alan, you could place a pair of dice in front of you and set them up in the 36 possible outcomes sequentially. No need for Excel. No need for a cup. No need for a partner. Hell you don't even need a pencil.

[RS__]

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?

My demonstration will not show the odds are 1/6, but that the odds are 1/11.

Seriously, just try it. [Or if you want to keep arguing the 1/6 point, I suppose trying this experiment will be detrimental to your argument.] I tried it earlier today (well, I observed a craps game). I counted 26 rolls that were 2-X and 3 rolls that were 2-2.

[OnceDear]

Alan,
We were doing so well. We had SO MANY areas of agreement. I even agreed that if you set aside one die as a deuce that the other die had a 1/6 chance of being a deuce.

But you seem to have stopped answering my questions when I asked, in what seemed a perfectly reasonable way...

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?

[OnceDear]

Question for Rob Singer, if he would be so kind as to help me out....

Rob, If we throw a pair of dice in every one of the 36 possible ways and we can see both dice land, how many times would be able to truthfully say. "Hey! At least one of those dice is showing as a deuce"?

I was being sincere. I'd made the assumption that you would agree that when working out probabilities where two dice are thrown, that there are 36 possible different and equally likely outcones. I'd hoped that you would then count up for me how many times we would be able to truthfully say. "Hey! At least one of those dice is showing as a deuce"?
I am not interested in any question or puzzle that sets aside one dice while throwing the other. I am only interested in the original question, where there were undoubtedly two dice WHICH WE cannot see, and which our partner has given us some limited info about.

[Alan Mendelson]

Try to keep up Rob, we are talking about a scenario from a few posts back, where the peeker would report if 'At least one dice is a two' if that was at all possible. To do that he would look at one die, maybe the one nearest to him. If that is a two (deuce) he would report 'At least one dice is a two' and his job would be complete.... But if, by some weird chance, that first die was not a two, he would keep looking, he would see tjhat there is a second die there. Maybe, just maybe, that second die was a two and he gets another chance to say 'At least one of the dice is a two'. If that second die was not a two, he could pop the cup down, say nothing, and slam the dice down again.
What is slightly at issue is how many possible different dice outcome could result in him reporting 'At least one of the dice is a two'

Unfortunately your theory doesn't effect the answer to the question. It's still 1/6. Your arguments bore me with their insignificance.