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

[Alan Mendelson]

Indignant, as long as you tell me the answer is 1/6 I say "thank you." But I still can't figure out your graphic.

[indignant99]

Indignant, as long as you tell me the answer is 1/6 I say "thank you." But I still can't figure out your graphic.

I can't make the graphic clearer, without showing duplicate pictures of those same (one red and one blue) dice. I wouldn't like to do that. You wouldn't, either.

[Alan Mendelson]

Thanks indignant, and if I ever do a TV show about gambling, I am going to hire you to do the graphics!!

[indignant99]

Thanks indignant, and if I ever do a TV show about gambling, I am going to hire you to do the graphics!!

I actually chuckled audibly. It struck me silly. Because the idea of me doing "professional graphics" is hilarious. I'm one of those people - hordes of people, actually - who describes himself as "the world's worst artist."

[Alan Mendelson]

Most TV and media graphics today are pre-produced and assembled using computer keyboards. Rarely is there any "original artwork" used.

[indignant99]

Most TV and media graphics today are pre-produced and assembled using computer keyboards. Rarely is there any "original artwork" used.

Well, I used a keyboard, mouse, and software to draw those graphics. (But they were not pre-produced.) I admit that I had to conceptually envision the graphic to draw. Creativity, maybe, there. But I will alter / edit the graphic, as I proceed, as I see it develop, to improve or hone the image.

[RS__]

Alan -- many pages ago on the original thread on WOV [I think it was there, not sure], you agreed to the "2-dice bet" and claimed the way the bet worked out would be identical to the question.

It was something like: 2 dice are rolled. If neither of them show a deuce, nothing happens, and re-roll. If at least one of the dice showed a deuce, then "the bet was on" -- and if both dice were a 2 then someone would pay you 9:1 (or 8:1 or w/e) in either peanuts or points (and points were to be used to whoever would buy the other a lunch). If the bet was on and both dice were not a 2, then you would pay the person 1 unit (peanut, point, whatever).

YOU agreed this bet/scenario would be identical to the original problem/question.

The simulations Wizard, I, and others have created all used this same scenario [the scenario you agreed to] -- and they've all reached 1/11. Hell, you even claimed this bet (against the Wizard or me or whoever'd bank it) was a GOOD BET. You're not just making yourself look like a fool; you're misleading others into making a poor bet.

[arcimede$]

To me the original question was no different than throwing two dice down a craps table and one coming to rest on a 2 while the other die is a spinner.

Yes, that is the wrong interpretation as been explained to you many times. You ignore simple logic and revert back to the basis of your belief .... "because" .....

The logic is trivial and so easy to understand that almost all 5th graders would see it in a minute. Your denial is emotional, not logical.

[OneHitWonder]

The logic is trivial and so easy to understand that almost all 5th graders would see it in a minute. Your denial is emotional, not logical.

Where is the "inclusive or" in the question as written?

The 'or' may be assumed inclusive, but that isn't necessarily always possible. There will be a 2 in one row or the other (of the chart) but not in both on a specific roll; or there will be two 2's but not both (one 2 and two 2's).

Still waiting for anyone to answer this question.

[redietz]

I have an old friend, a Shelley scholar with an advanced philosophy degree, who waxed eloquent on "inclusive or." I usually just nodded my head.

Anyway, OneHit makes a, maybe the, key point.

[Alan Mendelson]

Alan -- many pages ago on the original thread on WOV [I think it was there, not sure], you agreed to the "2-dice bet" and claimed the way the bet worked out would be identical to the question.

It was something like: 2 dice are rolled. If neither of them show a deuce, nothing happens, and re-roll. If at least one of the dice showed a deuce, then "the bet was on" -- and if both dice were a 2 then someone would pay you 9:1 (or 8:1 or w/e) in either peanuts or points (and points were to be used to whoever would buy the other a lunch). If the bet was on and both dice were not a 2, then you would pay the person 1 unit (peanut, point, whatever).

YOU agreed this bet/scenario would be identical to the original problem/question.

The simulations Wizard, I, and others have created all used this same scenario [the scenario you agreed to] -- and they've all reached 1/11. Hell, you even claimed this bet (against the Wizard or me or whoever'd bank it) was a GOOD BET. You're not just making yourself look like a fool; you're misleading others into making a poor bet.

Why don't you find the exact quote you are referring to because you will see that I did not agree to the bet but said the wording reflected the problem. My bet with the Wizard involved counting points and the loser buying lunch. No one accepted the actual bet and I don't blame them since in a short period of time anything can happen with two dice.

Regarding my lunch bet with the Wizard I'd love to do it.

[Alan Mendelson]

Yes, that is the wrong interpretation as been explained to you many times. You ignore simple logic and revert back to the basis of your belief .... "because" .....

The logic is trivial and so easy to understand that almost all 5th graders would see it in a minute. Your denial is emotional, not logical.

You say it's the wrong interpretation because you don't want it to be. In simple English it is the correct interpretation.

Two dice have been rolled and one is a 2. The chance of the other die being a 2 is 1/6 just as if that second die were a spinner on a craps table.

You guys cannot present the 1/11 answer with two real physical dice unless you combine results for two dice throws. You can't do it with one dice throw.

That's where your 1/11 answer does not apply to the original question the same way it doesn't apply to a spinning die on a craps table.

You were caught. Admit it. Or keep whining.

[OneHitWonder]

Anyway, OneHit makes a, maybe the, key point.

The Wizard's case is ours averaged out over many rolls. Ours is the seminal case from which "you have to start somewhere". If you put our two possible 5 to 1 against cases for a 2-2 roll (once we have been or are told that at least one die is a 2) together on average - necessarily by more rolls - then you end up with the Wizard's one overall possible 10 to 1 against case for a 2-2 roll.

But ours still has probability nature in that the other die is a 2 one time in six (once we have been or are told that at least one die is a 2).

To re-iterate, it comes down to comparing or talking about the two separate dies taken in separate rolls; or, comparing or talking about the dice together as one form of roll taken together with, or in the context of, a very large number of rolls. Are we asked the question in specific, or in general? We are asked about a specific roll (once we have been or are told that one die is a 2). And, we are not asked about any explicitly general probability. The general question for the 1/11 answer ought to ask, "Of the rolls of one 2, and the rolls of two 2's, what percentage are two 2's.

"One 2 or two 2's" on a specific roll of two separate dies means that the only die looked at is a 2, so there is still the chance of another 2. We must rule out the times when the die looked at is not a 2, because "one 2 or two 2's" has failed. Unless, we concoct a method of "peeking" which allows for the only die looked at is a 2 to apply from both the left die and right die sides. I think that this is Alan's approach instinctively. Which reduces to the 1/11 answer - necessarily by more rolls. No way to average out specific rolls taken separately, ie, one-time events in isolation.

If we had six clear-film slides represent each die with the 2's blackened, then the observer or "peeker" might superimpose the slides from their respective dies. If the combination appears black when held up to the light, then that sort of "at least one 2" would figure in the Wizard's favor. As if no one has to physically peer at the dice at all before counting chance results; and it were only a mental exercise. However, going at the 1/6 chance from both sides by separate Alan's, so to speak, is omnipresent in the math of the 1/11 answer by definition no matter how you look at it. Ours is the seminal result, which at some point is combined to give rise to the Wizard's.

Stuff like this, combined with other similar findings, tells me that the mind can't exist as defined completely separate from the body.

[OneHitWonder]

You can't do it with one dice throw.

Agreed...........

[RS__]

Why don't you find the exact quote you are referring to because you will see that I did not agree to the bet but said the wording reflected the problem. My bet with the Wizard involved counting points and the loser buying lunch. No one accepted the actual bet and I don't blame them since in a short period of time anything can happen with two dice.

Regarding my lunch bet with the Wizard I'd love to do it.

Bolded: that's my point.

So if you agree the wording of the bet reflects the problem....and we've shown the bet would have wizard win 10/11 times and lose 1/11 times......yet you're stuck on 1/6....where is the discrepancy?

Or do you really believe the bet you've made with him will reflect 1/6 as the answer?

[OneHitWonder]

Or do you really believe the bet you've made with him will reflect 1/6 as the answer?

What's the difference between losing 1 unit ten times, winning 10 units one time; and [losing 1 unit five times, winning 5 units one time] X 2 ? When the latter sees both sides paid for a roll of 2-2.

[arcimede$]

You say it's the wrong interpretation because you don't want it to be. In simple English it is the correct interpretation.

Nope. "At least" has a specific meaning. You want to interpret it to mean one specific die. That is not what it means.

Two dice have been rolled and one is a 2. The chance of the other die being a 2 is 1/6 just as if that second die were a spinner on a craps table.

Your answering a different question. Your assuming one specific die is a 2, that is not what the question asks.

You guys cannot present the 1/11 answer with two real physical dice unless you combine results for two dice throws. You can't do it with one dice throw.

Odds are based on multiple dice throws. That is what odds refers to.

That's where your 1/11 answer does not apply to the original question the same way it doesn't apply to a spinning die on a craps table.

Nonsense. All you ever see on one dice throw is one result which doesn't give you either 1/6 or 1/11. The question is not related to a single throw. The question relates to the overall odds across multiple throws.

[Alan Mendelson]

Arc we went thru this before: "At least" means that when you have two dice, one of the dice can show a 2 or both of the dice can show 2. That's what "at least" means.

When you have two dice and at least one die is a 2, your choices for the second die (whichever die that is) is either another 2 or it's a 1, 3, 4, 5, 6.

In the original question we are NOT asked to consider multiple dice throws. We have a specific condition and all we are asked to do is find the odds for THAT INDIVIDUAL dice throw. And that is the error you and the others who say the answer is 1/11 made: you ignore the actual question being asked.

[RS__]

Arc we went thru this before: "At least" means that when you have two dice, one of the dice can show a 2 or both of the dice can show 2. That's what "at least" means.

When you have two dice and at least one die is a 2, your choices for the second die (whichever die that is) is either another 2 or it's a 1, 3, 4, 5, 6.

In the original question we are NOT asked to consider multiple dice throws. We have a specific condition and all we are asked to do is find the odds for THAT INDIVIDUAL dice throw. And that is the error you and the others who say the answer is 1/11 made: you ignore the actual question being asked.

And your nonsense continues...

[redietz]

How many in the viewing audience know the first definition of "probability?" (Note: My hand is up)

How many in the viewing audience know that the original question, as technically interpreted by dictionary definition, is not necessarily a math question at all? (Note: My hand is up)

How many in the viewing audience noted the lack of words like "odds" in the original question? (Note: My hand is up)

How many in the viewing audience realize that automatically interpreting a word by its second or third definition is inappropriate unless context rules out the first definition? (Note: My hand is up)