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

[Alan Mendelson]

Unfortunately, OnceDear I have to think about two dice in a two dice question. It should be very easy for you to show, using two dice, why the chance of 2-2 with at least one die showing a 2, is 1-11. After all, it only involves two dice.

The coin flip problem is simple to illustrate.
So is the boy-girl problem -- simple to illustrate.
Certainly illustrating a problem with two dice must also be simple.

I am waiting for you and the Wizard (and the Wizard says he has video capability) to take two dice and show why -- when at least one die is a 2 -- the chance of showing 2-2 is 1/11.

This is simple to do, right? You've been showing me enough graphs and layouts of bets. Show me how the dice work to show 1/11 chances. Do it already.

[Alan Mendelson]

OnceDear and RS___ in case you missed this on the WOV forum I explained MY PROBLEM in understanding how the answer is 1/11 when there are just two dice with at least one die showing a two. I responded to Math Extremists' post. His statement is in bold and mine follows.

MathExtremist

Please answer:
a) How many combinations of two dice have "at least one die showing a 2"?
b) How many combinations of two dice have "both dice showing a 2"?

First I want to say that I am sure you are a very nice and intelligent and worldly man and I appreciate your help in making me understand.

The problem I AM HAVING is that I can't consider 11 different combinations with at least one die showing a 2 in a two-dice problem. And let me explain why.

With each die, when a 2 is showing, there can no longer be 11 different combinations. Let me show you:

2....1
2....2
2....3
2....4
2....5
2....6

So, when you tell me that at least ONE DIE is showing a 2, I look at my combinations above and see that there are only 6 -- not 11.

If we look at the mirror image of the combinations above I still get only 6 combinations -- not 11.

1....2
2....2
3....2
4....2
5....2
6....2

Now, it is also possible that when we say "at least one die" it can also mean that BOTH dice are showing a 2. Well, when I look at two dice, both showing a 2, I treat either of the dice as being the "variable die." And by that I mean, this:

2....2 (but it could have been a 1)
2....2
2....2 (but it could have been a 3)
2....2 (but it could have been a 4)
2....2 (but it could have been a 5)
2....2 (but it could have been a 6)

And what this indicates to me, is that even if both dice were showing a 2, you still have only a 1/6 of getting that 2-2 combination.

So this is why I ask: using two dice, show me how the answer is 1/11 ??

[OnceDear]

OnceDear and RS___ in case you missed this on the WOV forum I explained MY PROBLEM in understanding how the answer is 1/11 when there are just two dice with at least one die showing a two. I responded to Math Extremists' post. His statement is in bold and mine follows.

MathExtremist

Please answer:
a) How many combinations of two dice have "at least one die showing a 2"?
b) How many combinations of two dice have "both dice showing a 2"?

First I want to say that I am sure you are a very nice and intelligent and worldly man and I appreciate your help in making me understand.

The problem I AM HAVING is that I can't consider 11 different combinations with at least one die showing a 2 in a two-dice problem. And let me explain why.

With each die, when a 2 is showing, there can no longer be 11 different combinations. Let me show you:

2....1
2....2
2....3
2....4
2....5
2....6

So, when you tell me that at least ONE DIE is showing a 2, I look at my combinations above and see that there are only 6 -- not 11.

If we look at the mirror image of the combinations above I still get only 6 combinations -- not 11.

1....2
2....2
3....2
4....2
5....2
6....2

Now, it is also possible that when we say "at least one die" it can also mean that BOTH dice are showing a 2. Well, when I look at two dice, both showing a 2, I treat either of the dice as being the "variable die." And by that I mean, this:

2....2 (but it could have been a 1)
2....2
2....2 (but it could have been a 3)
2....2 (but it could have been a 4)
2....2 (but it could have been a 5)
2....2 (but it could have been a 6)

And what this indicates to me, is that even if both dice were showing a 2, you still have only a 1/6 of getting that 2-2 combination.

So this is why I ask: using two dice, show me how the answer is 1/11 ??

YOU MUST treat either of the dice as being the "variable die." You don't victimise and nail one of the little 6u99ars down.
You simply can't nail one down.
It's like having two naughty children and a broken window. At least one of them broke it, but you cannot just take one of them and give him a slap.

2....2 (but it could have been a 2...1 AND it could have been a 1...2)
2....2
2....2 (but it could have been a 2...3 AND it could even have been a 3...2)
2....2 (but it could have been a 2...4 AND it could easily have been a 4...2)
2....2 (but it could have been a 2...5 AND it could quite possibly have been a 5...2)
2....2 (but it could have been a 2...6 AND it could unless someone had nailed one specific die to the table have been a 6...2)

And what this indicates to me, is that even if both dice were showing a 2, you still have only a 1/11 of getting that 2-2 combination.

[Alan Mendelson]

Below is a photo of a photo of 11 combinations of 2 dice with at least 1 die a 2. This is supposed to prove to me that the answer to the original question is 1/11 and my comments above about there being only 6 combinations are wrong. This was posted on the WOV forum.

Well, here's what I wrote on the WOV forum (in bold): Not quite. You're not using TWO DICE. This is a two dice problem.

Now, show me 11 combinations with at least 1 die showing a 2, with a total of only TWO DICE.

[Alan Mendelson]

OnceDear wrote above: YOU MUST treat either of the dice as being the "variable die." Unfortunately, I can't do that. If at least one die is showing a 2, I have to treat the die showing a 2 as the 2. And if both dice happen to be showing a 2, then I can pick either one and still get 1/6 as my answer.

Again, this goes back to what I said before: you have a two dice problem and you can't shift the value of the cube with a 2 to fit your 1/11 answer.

You guys are nucking futs.

[OnceDear]

And as I said on that forum. Now you are just talking stupid!!!
Goodbye Alan.
May you either learn to read, or die wondering if you might ever have figured this out, perhaps by listening, looking, or God forbid, thinking.

[redietz]

One would think a cadre of geniuses could make a simple video. C'mon, dudes -- pool your gambling bankrolls for 2015 and get a Super Eight.

[Alan Mendelson]

redietz they wouldn't dare attempt to make a video using ONLY two dice. They keep on using multiple dice to come up with 1/11. Even worse: when a die is settled showing a 2 they still use the other 5 faces on that same die to explain their answer is 1/11.

[blount2000]

Hi Alan,

I don’t want to muddy the waters or anything, but I wanted to take a moment to describe how my own past personal experiences made me believe the 1/11 figure to be the answer to the question.

Although I feel like I have eventually come to understand the math that has been provided explaining the 1/11, it was my real world experience that had the most initial influence on my opinion (versus the math).

I have enjoyed craps for a while, so I thought back on my own time at the tables over the years. I thought about the many different rolls of the dice I have seen and then applied two things to my recollections:


1. I assumed any roll of the dice on the craps table that included one (or more) 2’s would count as satisfying the initial statement of “two dice are thrown and at least one die is a 2”.

2. I asked myself: “For all of those times I saw any roll on the table that had a 2 in it, do I feel like I saw a pair of 2’s one-sixth of the time?”


I didn’t think about cups, peekers, multiple combinations, specific die, etc., etc. I just applied the two items above to my memories of the tables.

It’s not very scientific (or mathematical) and is obviously based upon my own personal “feelings” and memories of playing craps, but I just didn’t feel like I saw a pair of 2’s every sixth time (on average) when there was a 2 on the table.

If my assumption #1 is felt to be correct, I’d be curious to hear what other people’s feelings are when applying the two criteria to their own craps table experiences & memories.

Anyway, just thought I’d share my own thoughts on the question.

Thanks for providing the forum!

[Alan Mendelson]

blount2000 thanks for joining and for posting.

I'd like you to do me this favor: take two dice -- any two dice will do -- and put them on your table or desk with at least one of the two dice showing a 2. Knowing that one of the two dice is showing a 2 ask yourself: what are the odds that the other die can also show a 2? What is your answer?

Thanks.

[Alan Mendelson]

BULLETIN NEWS. A video has been shot. Please view and comment:

[blount2000]

Hey, I had a hard enough time getting my mind wrapped around the original question! I’m not mentally prepared at this point for a follow up!

Like I mentioned, I had to apply my own personal experiences at the craps tables to the situation in order to make it understandable to me.

There has already been much back & forth discussion on the wording, interpretation, intent, etc., of the original question. My own interpretation of the original question is reflected in how I described looking back on my past craps play.

I don’t want to “make it all about me”, but in the spirit of keeping it within my own experiences (and hence my understanding) I’d like to know if my assumption #1 written above does (or doesn’t?) satisfy the statement in the original question (i.e. two dice rolled, at least one is a 2).

If my assumption does satisfy the initial question, then I come back to my feeling that I’m just not seeing a pair of twos 1/6 of the time that any two is showing up on the table.

If my assumption doesn't satisfy it, well, then I must have interpreted the question wrong and that will be that. It won’t be the first (or last) time I was wrong!

[Alan Mendelson]

Your statements 1 and 2 are correct, blount2000. Whether or not you saw 2-2 when at least one of the two dice is a 2 one-sixth of the time doesn't matter. Frankly, when you roll two dice the chance of getting 2-2 is 1/36.

The original question and the question to be answered is simply this: if you have at least one 2 showing, what are the odds for the second die to be a 2 as well. And that is the same as rolling one die with a 2 showing 1/6 times.

If you are not seeing 2-2 with one 2 showing one-sixth of the time you're just like me. Like I said, that would be a 1/36 chance.

[Alan Mendelson]

And here's what I said about the video shot my miplet on the WOV forum: he ignored the original question. He simply rolled two dice until "at least" one 2 was showing and then proceeded to show us that there are 11 combinations of two dice with at least one 2. He didn't answer the question.

The original question and the question to be answered is simply this: if you have at least one 2 showing, what are the odds for the second die to be a 2 as well. And that is the same as rolling one die with a 2 showing 1/6 times.

[Alan Mendelson]

the Wizard himself has posted a video:

Again, he shows the question but instead of answering the question with the information provided (at least one die shows a 2) he proceeds to once again refer to the eleven dice combinations including a 2.

Mike threw in the possibility that 2-2 shows on the initial throw and yes, Mike, as I said before -- it doesn't matter. Yes, you take one of the 2s arbitrarily because it does not affect the question. The question simply asks for you to figure the chance that another die in a two dice problem matches the number shown on the other die in the two dice problem.

Anyone else care to comment? I also asked to be suspended from the Wizard's forum. No one there will actually address the actual question asked. Even Mike has sidestepped the question in his video response.

[synergistic]

Well, at least I finally think I understand what you're saying now - yes, the odds of rolling a pair is 1/6. I still think you're parsing the question unnaturally at best, and find it hard to believe you don't think our interpretation is at least equally valid.

[Alan Mendelson]

Well, at least I finally think I understand what you're saying now - yes, the odds of rolling a pair is 1/6. I still think you're parsing the question unnaturally at best, and find it hard to believe you don't think our interpretation is at least equally valid.

Yes, the odds of rolling a pair with two dice is 1/6. Thank you.

synergistic your interpretation does not include that one die has been identified. And that's the big sticking point. The 1/11 answer is the correct answer to the question about dice combinations. But it doesn't answer a specific question when one particular number has been identified.

Again: the original question asks you to solve the problem of a 2 being rolled and the chances of another 2.

The original question does not ask for anyone to discuss all of the combinations showing a 2 and then to say how many of those combinations would show 2-2.

I think it was regnis who said something early-on in the first discussion about taking the simplest, most direct route and answer. And when you know there is a 2 showing you should take the shortest, most direct route -- and not do the mathematical gymnastics.

"Miplet" did the gymnastics and then the "Wizard" did the gymnastics. The gymnastics is "valid" for showing a solution to this kind of problem, but it didn't answer this particular problem.

As I said over on the WOV forum it's amazing how many "non math" people using common sense answered 1/6. Does that make "common sense" stupid? Or does it mean that the math guys overthought the problem? They overthought it and missed it.

[Count Room]

synergistic your interpretation does not include that one die has been identified. And that's the big sticking point. The 1/11 answer is the correct answer to the question about dice combinations. But it doesn't answer a specific question when one particular number has been identified.

Alan: I respectfully submit to you that the die was not identified on purpose! The whole point of the question was a trick designed to lull the casual and unsuspecting into believing the answer is 1/6. It really is not 1/6, and the question was intentionally worded that way to stimulate conversation from those who initially believed it was 1/6.

Think of it this way: If the creator of that original question was clever enough to know that many people would be fooled into thinking it was 1/6, then maybe the question was not poorly worded at all? Rather, it was intentionally worded that way to provoke the exact reaction you are providing right here on this thread and on WOV?

Judging from your reaction and the length of this thread along with the one over on WOV, I would say the original question did a beautiful job of stimulating that conversation.

Whoever authored that question should be collecting some royalties by now based on the number of collective cerebral cortexes it managed to churn....

[synergistic]

No, my interpretation does NOT account for one die being identified - because the way I parse the sentence, you haven't been told anything about a specific die, you've been told something about the two dice as a unit. Your version is, to me, a less intuitive way to read the question, and I believe it yields a misleading answer.

[Alan Mendelson]

Count room if you read some of my posts in the early discussion, I said the question was worded to fool people into getting the 1/11 answer because it mentioned the two-coin problem. I wrote that again today on the WOV forum. I think that set up the math guys to answer the question using the wrong methodology.

Here's what I wrote on the WOV:

As I said early-on in the discussion, the question was written to make you consider the 11 combinations because of this phrase that preceded the original question:

"Along the vein of the Two Coin Puzzle,..." It was that phrase that made everyone make the question more complicated than it really is.

It was that phrase "Along the vein of the Two Coin Puzzle,..." that had you all thinking you had to solve the question as you would with the two coin puzzle and looking at all of the combinations of flipping coins. But again, this wasn't a coin flipping question. There are two dice and at least one of those two dice landed on a 2.


synergistic when you say the original question "does not account" for one die being identified I am completely lost. When someone looks at two dice and "truthfully" tells you "at least one" is a 2 that to me is all the identification I need to proceed with the question and coming up with the answer of 1/6. Are you suggesting that the statement "at least one" is a 2 was a lie?

What really frustrated me with the entire WOV bunch was when they started to attack the meaning of English sentences and phrases. They even attacked the meaning of "at least one" and more than one WOVer actually said that made it okay for the 2 to move from one die to another. And then there were those who said "contained a 2" or "showed a 2" on the two dice meant different things.