It was two years ago today...

[a2a3dseddie]

Now let's talk about two red dice that are shaken in a cup and slammed down on a table. One person peeks and sees that at least one of the two red dice is showing a 2.

Both of you are unable to show me 11 possible options. You say they are there and you ask me to trust that they are there but you can't demonstrate that they are there.

I watched the Wizard's video, and I don't think I would have used the wording "double count" even though I know exactly what he means. At the end of the video, he has 36 sets of dice laid out with all 36 possible combinations 2 dice can be arranged.

Why didn't you accept that?

You know the 11 possible 2 dice combinations that contain at least one 2.

red die 1=2 red die 2=1
red die 1=2 red die 2=2
red die 1=2 red die 2=3
red die 1=2 red die 2=4
red die 1=2 red die 2=5
red die 1=2 red die 2=6

OR

red die 1=1 red die 2=2
red die 1=2 red die 2=2
red die 1=3 red die 2=2
red die 1=4 red die 2=2
red die 1=5 red die 2=2
red die 1=6 red die 2=2

How can you exclude any of these combinations the way the question is worded?

Once again, take it from your friend:

Rob I am 1000% honest here. I haven't got a clue as to what you're saying. So let me be point blank with some questions:

1. If wo dice are rolled at the same time and one die shows a 2 do you think the chance that the second die rolled at the same time is 1/6 for also showing needa 2 ?

2. If you said "yes" to question #1 why won't you accept the wager?

And leave out all of your criticism about foreigners, WOV, the Wizard, the WOV forum members, Obama, and anybody else. Answer the two questions.

No and I never have, not under that premise, which is the same way wizard interpreted the problem and not my interpretation of it. You need to understand that if one die is a 2, it's not a function of simply how many faces that 2nd die has. They work TOGETHER to form one of the possible 11 combinations remaining where a 2 is on at least one of the dice. Arci laid them out for you. I don't know how else better to show it to you.

[Alan Mendelson]

Show me two dice with at least one die showing a 2 and then explain to me how there are 11 possibilities?

You can't.

Give up.

It has nothing to do with laying out 36 dice combinations like the Wizard did in his video. It's mathematical double speak.

Deal with reality.

[Rob.Singer]

The whole problem, again, is that these 1 in 11 people refuse to acknowledge that there is a peeker, and that peeker has already seen one die showing a 2. This automatically eliminates one of the dice from the problem because it is a real occurrence. But the 1 in 11 theorists are not willing to count that specific die out--opting instead to pretend that the die showing the 2 remains, in theory but certainly not in reality, in-play. This incredible scenario is being used by eddie, qua, etc. because they WANT the answer to be 1 in 11, even though in the real world it's absolutely 1 in 6.

ONCE A PEEKER IDENTIFIES THAT ONE OF THE DICE IS A 2, THAT DIE IS OUT OF THE PROBLEM 100%, LEAVING ONLY ONE 6-sided die to determine probability from.

[Alan Mendelson]

The whole problem, again, is that these 1 in 11 people refuse to acknowledge that there is a peeker, and that peeker has already seen one die showing a 2. This automatically eliminates one of the dice from the problem because it is a real occurrence. But the 1 in 11 theorists are not willing to count that specific die out--opting instead to pretend that the die showing the 2 remains, in theory but certainly not in reality, in-play. This incredible scenario is being used by eddie, qua, etc. because they WANT the answer to be 1 in 11, even though in the real world it's absolutely 1 in 6.

ONCE A PEEKER IDENTIFIES THAT ONE OF THE DICE IS A 2, THAT DIE IS OUT OF THE PROBLEM 100%, LEAVING ONLY ONE 6-sided die to determine probability from.

Thank you.

[RS__]

The whole problem, again, is that these 1 in 11 people refuse to acknowledge that there is a peeker, and that peeker has already seen one die showing a 2. This automatically eliminates one of the dice from the problem because it is a real occurrence. But the 1 in 11 theorists are not willing to count that specific die out--opting instead to pretend that the die showing the 2 remains, in theory but certainly not in reality, in-play. This incredible scenario is being used by eddie, qua, etc. because they WANT the answer to be 1 in 11, even though in the real world it's absolutely 1 in 6.

ONCE A PEEKER IDENTIFIES THAT ONE OF THE DICE IS A 2, THAT DIE IS OUT OF THE PROBLEM 100%, LEAVING ONLY ONE 6-sided die to determine probability from.

Thank you.

Agreeing with Rob on something containing probability or math is basically admitting you're wrong.

[Alan Mendelson]

Agreeing with Rob on something containing probability or math is basically admitting you're wrong.

If this is the best you could come up with, then we know Rob is correct.

[RS__]

Agreeing with Rob on something containing probability or math is basically admitting you're wrong.

If this is the best you could come up with, then we know Rob is correct.

I'm not going to sit here and waste more of my time arguing in this thread. All of your posts are filled with contradictions, it's not even funny. I honestly don't think you're being serious anymore. Getting paid 7:6 on a 1/6 game is a bet you should take (approx. 55% disadvantage).....saying there are 11 combinations but it's a 1/6 chance (because you fail to even acknowledge conditional probability).....you keep saying something may be true in theory but not reality, yet you haven't shown us a situation where this is the case (PLEASE come up with ONE concrete example where something is true in theory but false in reality.....FYI: Rob's shit system isn't a concrete example).

Why you keep falling for Rob's shit is beyond me. He's admitted he just trolls online to get people's goat. You admit he's much different in real life than he is in person. Something here is clearly not adding up.

[Rob.Singer]

That's really funny! I'd LOVE to compare my two degrees to his .... Then for good measure, we could compare work history. I don't quite expect working in casinos requires using the same applied math as does calculating and determining the correct amount of amplified power coastal based airport transceivers need to get their VHF & UHF signals to aircraft at multiple heights and distances.

Oh wait....how do we split $75 in tips four ways?! ROTFLMFAO!!!

You idiots are really pathetic.

[Alan Mendelson]

the only conditional fact is that one die is identified as a 2. that changes the entire problem so there is one answer: 1/6

[mickeycrimm]

the only conditional fact is that one die is identified as a 2. that changes the entire problem so there is one answer: 1/6

Which die is showing the 2?

[a2a3dseddie]

Agreeing with Rob on something containing probability or math is basically admitting you're wrong.

In all fairness RS__, Rob was on the right track. His post here from 2 years ago shows that.

You need to understand that if one die is a 2, it's not a function of simply how many faces that 2nd die has. They work TOGETHER to form one of the possible 11 combinations remaining where a 2 is on at least one of the dice. Arci laid them out for you. I don't know how else better to show it to you.

Then Alan banned him. Then he came back to suddenly and mysteriously support Alan's 1/6 opinion after Alan lifted his ban. Maybe Alan promised him he would use his slot card the next time he plays?

[a2a3dseddie]

Coach you will lose this bet with jbjb, and this bet that jbjb proposed is NOT the same as the question with two dice and one showing a 2.

Let me go over jbjb's bet again for you.

Your chance of seeing 2-2 on any roll is 1/36.
Even when the dice are rolled and one 2 shows up, there is a 1/11 chance of another 2 showing up.

The difference with the two-dice question is that we know in advance that at least one die shows a two, and with that the odds of both dice showing two are reduced to 1/6.

But in jbjb's bet you don't know that at least one 2 will be rolled.

If jbjb would let you roll one die and if it comes up a 2 for a win, and if he's willing to pay you 7 to 6 for a 1 out of 6 possibility, take it.

Alan, you haven't taken the time to edit this earlier post. You do see your error don't you?

If your so confident, take the 8:1 bet for $100 per roll. Win $800 when 2-2 shows and lose $100 when 2-1, 2-3, 2-4, 2-5 or 2-6 shows.

For jbjbs' bet you would only lose if 2-1, 2-3, 2-4, 2-5 or 2-6 shows. Win on 2-2. Other combinations aren't counted. Where do you see:

But in jbjb's bet you don't know that at least one 2 will be rolled.


?

I think jbjbs' payoff of 8-1 is very generous compared to your 7-6 payoff.

[Rob.Singer]

Agreeing with Rob on something containing probability or math is basically admitting you're wrong.

In all fairness RS__, Rob was on the right track. His post here from 2 years ago shows that.

You need to understand that if one die is a 2, it's not a function of simply how many faces that 2nd die has. They work TOGETHER to form one of the possible 11 combinations remaining where a 2 is on at least one of the dice. Arci laid them out for you. I don't know how else better to show it to you.

Then Alan banned him. Then he came back to suddenly and mysteriously support Alan's 1/6 opinion after Alan lifted his ban. Maybe Alan promised him he would use his slot card the next time he plays?

eddie, there's nothing mysterious going on. if you've read my posts they nearly all address how there can be several answers to this problem, depending on how the reader perceives the question. You guys perceive it one way, Alan perceives it the other, and I am able to see it both ways--but I am now giving more weight to the 1 in 6 answer because I'm giving the peeker more importance. He sees the die with the 2, and it then becomes a 1 in 6 answer to him. And being that he's the most knowledgeable personality in the problem, a wise man would always latch onto such intelligence.

[slingshot]

The only thing that could be more interesting than this make believe stuff would be Rob's third book on how to win. I realize he gave all the basic topics in his very last VP Truth article, but geez it was interesting stuff to read. BTW, a guy setting next to my quarter machine was playing triple play-$1-15 credits- and hit $12 k royal holding Q-10 spades. Fun to talk to him. Said it had been forever since he had a royal.

[Rob.Singer]

Sling, I've toyed with that idea of coming out with that third book, and one of the best parts of doing it would be the continuation of the envy as well as the hatred and the lies that would radiate out of the "AP" community. I mean, can you just IMAGINE how they'd handle the fact that I won more money per year on avg. the first three years of my retirement from professional play, than I did prior....and all by playing my simple ARTT strategy to the letter?!

I guess the best part, however, would be how I'd get even more requests for training. That's what I did overnight at the Silverton--I trained a local "AP" who teaches 9th grade math in LV, and a few hours later someone who was on vacation from Michigan. They both won as they played 25c thru $2 ARTT and BOTH took advantage of hitting four face cards on $2 SDBP--and collecting $1200 W2G's--as a vehicle for their winning sessions. Even I won earlier in the evening while waiting, but I won less than either of them, which I enjoyed. It was a worthwhile trip and a good night all-around. If I had been the typical LV vp player throughout my playing life---ie, an overweight fatass, a smoker, and a big beer or booze drinker---there's no chance I'd be able to stay up over 30 hours straight while training two decent players along with driving 300 miles each way.

[a2a3dseddie]

Sling, I've toyed with that idea of coming out with that third book, and one of the best parts of doing it would be the continuation of the envy as well as the hatred and the lies that would radiate out of the "AP" community. I mean, can you just IMAGINE how they'd handle the fact that I won more money per year on avg. the first three years of my retirement from professional play, than I did prior....and all by playing my simple ARTT strategy to the letter?!

I guess the best part, however, would be how I'd get even more requests for training. That's what I did overnight at the Silverton--I trained a local "AP" who teaches 9th grade math in LV, and a few hours later someone who was on vacation from Michigan. They both won as they played 25c thru $2 ARTT and BOTH took advantage of hitting four face cards on $2 SDBP--and collecting $1200 W2G's--as a vehicle for their winning sessions. Even I won earlier in the evening while waiting, but I won less than either of them, which I enjoyed. It was a worthwhile trip and a good night all-around. If I had been the typical LV vp player throughout my playing life---ie, an overweight fatass, a smoker, and a big beer or booze drinker---there's no chance I'd be able to stay up over 30 hours straight while training two decent players along with driving 300 miles each way.

I called the executive offices at the Silverton, a place that won't allow me to play vp there any more because of being "too lucky". I asked if my situation was a result, in any way, of an OSN analysis. The person I talked to denied having knowledge of that company, and simply said he wasn't allowed to discuss my situation with me over the phone because he had no way of knowing if it was me or not. OK, so I set up a tentative visit with him in Oct. It's no big deal, but I'd like to know more.

The point is, Dancer's saying there's a fear that he would be tossed if he won, either by being lucky or knowledgeable. So what's wrong with walking away with their former money in his pockets? I certainly am not angry for taking probably less than $10k from the Silverton in 8 straight successful visits and not being able to take any more.

[arcimede$]

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?

This is why you look so foolish, Alan. Nothing in the question identifies the value of either die. Only someone who could write the absolute nonsense you just wrote above would think that it does.

Nothing identifies the value of either die, Arc? Are you insane or just senile?

Alan, assume one die is red and the other green. After the toss and verification that at least one die is a 2, tell us the value of the red die.

[a2a3dseddie]

Alan, assume one die is red and the other green. After the toss and verification that at least one die is a 2, tell us the value of the red die.

At one point, it looks like he understands:

In English, the statement "at least one of the dice is a 2" means that either Die A or Die B is a 2 or Both Die A and Die B are showing a 2.

But then he says something like this...

Saying that at least one die shows a 2 and setting aside one die as a two means the same thing in the language I speak.

That last statement is why he's stuck on 1/6. It's also why he keeps trying to introduce other wording or conditions into the original question to arrive at the 1/6 answer...

[Rob.Singer]

Sling, I've toyed with that idea of coming out with that third book, and one of the best parts of doing it would be the continuation of the envy as well as the hatred and the lies that would radiate out of the "AP" community. I mean, can you just IMAGINE how they'd handle the fact that I won more money per year on avg. the first three years of my retirement from professional play, than I did prior....and all by playing my simple ARTT strategy to the letter?!

I guess the best part, however, would be how I'd get even more requests for training. That's what I did overnight at the Silverton--I trained a local "AP" who teaches 9th grade math in LV, and a few hours later someone who was on vacation from Michigan. They both won as they played 25c thru $2 ARTT and BOTH took advantage of hitting four face cards on $2 SDBP--and collecting $1200 W2G's--as a vehicle for their winning sessions. Even I won earlier in the evening while waiting, but I won less than either of them, which I enjoyed. It was a worthwhile trip and a good night all-around. If I had been the typical LV vp player throughout my playing life---ie, an overweight fatass, a smoker, and a big beer or booze drinker---there's no chance I'd be able to stay up over 30 hours straight while training two decent players along with driving 300 miles each way.

I called the executive offices at the Silverton, a place that won't allow me to play vp there any more because of being "too lucky". I asked if my situation was a result, in any way, of an OSN analysis. The person I talked to denied having knowledge of that company, and simply said he wasn't allowed to discuss my situation with me over the phone because he had no way of knowing if it was me or not. OK, so I set up a tentative visit with him in Oct. It's no big deal, but I'd like to know more.

The point is, Dancer's saying there's a fear that he would be tossed if he won, either by being lucky or knowledgeable. So what's wrong with walking away with their former money in his pockets? I certainly am not angry for taking probably less than $10k from the Silverton in 8 straight successful visits and not being able to take any more.

eddie, you're just like the wounded Democrats are when they keep phishing for old info to try and alleviate their pain that never goes away.

Did you check the date of my post about Silverton? Have you also checked to see how many times their mgmt. has changed since? And did you remember my publication of my ban letter from Bellagio? Do you also remember (with sheer PAIN, I am sure) that my $25 royal was hit there two years after?

You seem to put large efforts into spinning different details about me. You might want to check your capabilities so your credibility doesn't keep taking nosedives My guess is that you're flailing because you're frustrated over not being able to talk Alan into agreeing with your silly, minority-solution position on the dice problem.

[a2a3dseddie]

Rob I am 1000% honest here. I haven't got a clue as to what you're saying. So let me be point blank with some questions:

1. If wo dice are rolled at the same time and one die shows a 2 do you think the chance that the second die rolled at the same time is 1/6 for also showing needa 2 ?

2. If you said "yes" to question #1 why won't you accept the wager?

And leave out all of your criticism about foreigners, WOV, the Wizard, the WOV forum members, Obama, and anybody else. Answer the two questions.

No and I never have, not under that premise, which is the same way wizard interpreted the problem and not my interpretation of it. You need to understand that if one die is a 2, it's not a function of simply how many faces that 2nd die has. They work TOGETHER to form one of the possible 11 combinations remaining where a 2 is on at least one of the dice. Arci laid them out for you. I don't know how else better to show it to you.

So... 1/11?

Or is it...

Day-in/day-out the prevailing answer of 1 in 6 will be correct.

1/6?

silly, minority-solution position on the dice problem.

What is YOUR interpretation of this, exactly:

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?

Obviously, the answer is 1in36. You see, WoVers thrive on spoofing and arguing ad nauseum. And your egos are thru the roof. Well, now you have to realize that while all you geniuses were busy solving mathematical problems and theorizing about this and that, the rest of the world was busy experiencing reality. So the answer to your question is I took the question in its purest of forms, unaffected by any of the aforementioned assumptions simply because none of those assumptions were mentioned in the question.

1/36? WINNER! WINNER!