The Wizard will bank this bet: 1/6 vs 1/11

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Originally Posted by Alan Mendelson

kewl you just don't understand? This is not about a bag with 36 balls or any combination of 11 dice with a 2.

This is about TWO DICE and we know there is a 2 on at least one of them.

And the question is what would it take for us to have 2-2 showing on both dice? Really -- in plain, simple English that's the question.

And if we have 2 on one die, the other die in the problem has a 1/6 chance of also being a two.

That's all this question asked. You guys got all wrapped up in your math and statistics and probabilities and forgot to look at the simple question that has a simple answer.

And as proof I asked the WOV forum to show videos of your thought process with two dice. And there it was for everyone to see how you compounded the question and came up with the wrong answer.

Yes, tell the Wizard he's wrong too.

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I encourage you to go ahead and bet real money on your believe that it is 1/6.

Roll the dice and count the results. It might come as a shock to you, but if you do that you will actually find out empirically that the odds are indeed 1/11.
Then post here your thoughts.

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Originally Posted by Alan Mendelson

No I don't understand.

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Yes, I know.

Go ahead and roll the dice. It will give you your proof eventually.

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Originally Posted by Alan Mendelson

This is about TWO DICE and we know there is a 2 on at least one of them.

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Correct. And there are 11 unknown faces that could be up, only one of which is a 2!

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Originally Posted by OneHitWonder

Why do you insist on heavy-handed responses w/o any attempt at correction; let alone clarification of your own stance if any???

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You made a bald, unsupported claim (regarding horizontal exclusive-or vertical). Well, not only is it unsupported - it's wrong. And I made that bald counter-claim.

Here are all the ways you could have "at least one two" under the cup:

http://www.imageurlhost.com/images/k...vfowiygkmn.jpg
You are told by the peeker:

• "At least one of the dice is a 2." (verbatim)
• Not "the green die is a 2."
• Not "the purple die is a 2."
• Not "both dice are 2's."
• Not "one of the dice isn't a 2."
• Not "neither die is a 2."
• Not "one die is a 2, but the other one isn't."

Explain how you can eliminate any of those combos, from potentially being under the cup. Explain how you can restrict the possibilities to "ones with a green deuce." Explain how you can restrict the possibilities to "ones with a purple deuce." Remember, good buddy, you can't see them.

Look you guys are nuts. Are you trying to sell us snake-oil? Because that's what it sounds like when you come up with hocus pocus about there being 11 possible faces when we know that at least one die is showing a 2.

I am going to tell you this point blank. When either die is showing a 2 you can't count the other five faces on that die. End of discussion.

We deal with reality on this forum. Now go back to the WOV and bad mouth me there.

Alan (or anyone whom believes the answer to the question/bet as proposed in the first post is 1/6) could you please explain what is wrong with OnceDear's Excel spreadsheet?

I know there are a number of simulations out there…but I feel most should understand Excel.

I’m aware that Alan wrote…

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OnceDear's spread sheet and all those other spreadsheets looked at the combinations of 2 dice that contained at least 1 two, and from that they said the answer is 1/11. They did not respond to the question.

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But I’m not seeing how it’s not responding to the question. The spreadsheet picks two numbers from 1 to 6 (aka roll two dice). Then, it shows that when at least one of them is a 2 that both of them will be a 2 about 1/11 of the time. This IS the question.

So please, explain how his spreadsheet does not represent the question/bet.

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Originally Posted by Zedd

So please, explain how his spreadsheet does not represent the question/bet.

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Because Alan says so, plain and simple. The simulation is wrong and the question you wov-ers are asking is wrong and all is wrong. The only right thing here is Alan. And the answer is 1/6. Point blank.
There you go.

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Originally Posted by Zedd

Alan (or anyone whom believes the answer to the question/bet as proposed in the first post is 1/6) could you please explain what is wrong with OnceDear's Excel spreadsheet?

I know there are a number of simulations out there…but I feel most should understand Excel.

I’m aware that Alan wrote…

But I’m not seeing how it’s not responding to the question. The spreadsheet picks two numbers from 1 to 6 (aka roll two dice). Then, it shows that when at least one of them is a 2 that both of them will be a 2 about 1/11 of the time. This IS the question.

So please, explain how his spreadsheet does not represent the question/bet.

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The spreadsheet does not represent the question. The spreadsheet applies to any two dice and shows the various combinations of any two dice.

In the question we have isolated that at least one die is a 2. So -- picture that die frozen as a 2. It does not change.

Given that at least one die is a 2 there can only be six faces on the second die used to answer the question. That's the 1/6 answer.

We've heard various excuses and reasons from the 1/11 "camp" about how we can't freeze one die as a 2. Well it's going to be treated as 2 because it's a question involving two dice -- and at least one them is a 2. Deal with it.

Zedd, welcome to the forum, and thanks for posting.

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Originally Posted by Alan Mendelson

The spreadsheet does not represent the question. The spreadsheet applies to any two dice and shows the various combinations of any two dice.

In the question we have isolated that at least one die is a 2. So -- picture that die frozen as a 2. It does not change.

Given that at least one die is a 2 there can only be six faces on the second die used to answer the question. That's the 1/6 answer.

We've heard various excuses and reasons from the 1/11 "camp" about how we can't freeze one die as a 2. Well it's going to be treated as 2 because it's a question involving two dice -- and at least one them is a 2. Deal with it.

Zedd, welcome to the forum, and thanks for posting.

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But,but,but the spreadsheet does isolate one die is a 2?
See the column where it says "Count of "At Least One Deuce" Showing" ? That's isolating one die as a two right there?

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Originally Posted by redietz

OneHit, you've definitely grown on me. I think pretty much anything I had to say on this is in better hands. Thanks, and good luck.

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You're welcome, but thank you. Many of these sorts of "paradoxes" I have already thought about a lot before now in one form or another.

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Originally Posted by redietz

Math has its own language, but language has its own math, in a sense. Language is not the first language of many of these folks.

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Sometimes, we have "it" in the (sequential) language of our thoughts, and must struggle to put "it" into math; sometimes, the other way around. Sometimes, "it" isn't about language at all; but our (random-access or simultaneous) spatial senses instead.

I leave you with an anecdote. One of the curious things, discovered upon examination after his death, about Einstein's brain was that the two lobes - lingual left and spatial right - were closer together than normal. This could have facilitated his ability to perform accelerated calculations. Princeton had hired him just for that purpose later on in his life. It was noted as well that he had the brain of a much younger man without the usual plaque of someone of 75.

The only spreadsheet I want to see shows one die as a two with the following options for the second die in this two dice problem: 1, 2, 3, 4, 5, 6.

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Originally Posted by Alan Mendelson

The spreadsheet does not represent the question. The spreadsheet applies to any two dice and shows the various combinations of any two dice.

In the question we have isolated that at least one die is a 2. So -- picture that die frozen as a 2. It does not change.

Given that at least one die is a 2 there can only be six faces on the second die used to answer the question. That's the 1/6 answer.

We've heard various excuses and reasons from the 1/11 "camp" about how we can't freeze one die as a 2. Well it's going to be treated as 2 because it's a question involving two dice -- and at least one them is a 2. Deal with it.

Zedd, welcome to the forum, and thanks for posting.

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The spreadsheet doesn’t necessary show all the various combinations of two dice though. All it does is simulate the rolling of two dice. If at least one of them is a two then the wager is on. He even paid 8 units for deuce-deuce and lose 1 unit for when it’s not deuce-deuce.

The spreadsheet simulates exactly the bet you described in post one.

This is not a rolling of two dice problem. It is separate from the bet with the Wizard. Frankly, I concede the bet because the way the Wizard and I agreed to our own bet was a limited number of rolls.

So the bet is not an issue. Don't make it an issue.

The issue is the original question.

The reality is the two sides will never come to an agreement. We who say the answer is 1/6 read the question one way, and you who say the answer is 1/11 do some sort of picking 36 balls out of a bag, or looking at charts of 36 dice combinations, or looking at a graph showing 11 combinations with a 2 and come up with your answer.

Go in peace. And thanks for posting.

Don't you guys know that Alan is just trolling? Let it go...

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Originally Posted by Alan Mendelson

This is not a rolling of two dice problem. It is separate from the bet with the Wizard. Frankly, I concede the bet because the way the Wizard and I agreed to our own bet was a limited number of rolls.

So the bet is not an issue. Don't make it an issue.

The issue is the original question.

The reality is the two sides will never come to an agreement. We who say the answer is 1/6 read the question one way, and you who say the answer is 1/11 do some sort of picking 36 balls out of a bag, or looking at charts of 36 dice combinations, or looking at a graph showing 11 combinations with a 2 and come up with your answer.

Go in peace. And thanks for posting.

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So then are you acknowledging that the Wizard bet has a probability of 1/11 to pay out? And it’s a different question that we are answering?

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Originally Posted by Zedd

So then are you acknowledging that the Wizard bet has a probability of 1/11 to pay out? And it’s a different question that we are answering?

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No. I am not saying it has a probability of 1/11. But I am saying he will probably win.

Are you aware of the specifics of the bet that I have with the Wizard? I tried to meet up with him in Vegas a couple of weeks ago when I was in town to meet with a client but I could not make it in time.

First of all none of you in the 1/11 camp has yet to read the "original problem" in the same way that those of us who say it's 1/6 read the problem.

You guys can't even tell me which die is showing the 2. Those of us who say it's 1/6 say it doesn't matter which die is the 2.

So, when you guys tell me it doesn't matter which die is the 2, OR, if you can tell me which die it is, then we can discuss a different bet.

Zedd you came to this forum late in the discussion. Early in the discussion I said the original problem resembled this event at a craps table:

Two dice are thrown and one die immediately comes to rest showing a 2. The second die becomes a spinner and spins like a top. What are the chances that the "spinner" will also land on a 2?

If you don't view this as the same as the original question, we are going in different directions on the same road.

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Originally Posted by Alan Mendelson

Early in the discussion I said the original problem resembled this event at a craps table:

Two dice are thrown and one die immediately comes to rest showing a 2. The second die becomes a spinner and spins like a top. What are the chances that the "spinner" will also land on a 2?

If you don't view this as the same as the original question, we are going in different directions on the same road.

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This is clearly not the same as the original question.

The answer to your question is obviously 1/6, no doubt about that and everyone here will agree.

However, the question remains, how is it possible that you (and a few others obviously) can read the original question, then transform it in their mind to become your question, and even more so, boldly claiming it is the rest of us who don't understand the original question?

So kewl what do you think is the meaning of the original question? Do you think there are 36 dice in a bag, or eleven dice combinations showing a 2 and you have to pick from those 11 combinations?

Sorry. I don't buy it.

Alan, if it isn't obvious by now I'll give a helping hand: these guys all had no choice but to step in the same trap the wiz did when he jumped to a theoretical, more complicated and weakly supportable conclusion, as he rushed to be the first mensa to throw up an answer. All this group is doing is saving face for their mentor.

The question of course has a 1in6 answer when taken verbatim along with a picture in one's mind. But the intent of the question was to see who'd be the first to over-think it so they could incorporate as much theory as possible, thereby impressing the many members over there who only read math books and do not actually gamble. Revenge of the nerds, so to speak. But they never calculated on their slide rules that they'd ever run into folks here who are " Too smart for words, Too hot for nerds, Too tough for turds, & Too high for birds".

If the meaning of the question isn't clear to a fair number of the reading audience, and the question is intended for that reading audience, then the fault lies with the author of the question, not the reading audience. The responsibility to be clear lies with the author.

What about this is so difficult to grasp?

If the question can be misinterpreted, there is a problem with the question as written. Period. It doesn't take an IQ of 140 to understand this. It may take an obsession with one's IQ of 140 to ignore it.