Rob. Stop. I said no mentioning others who do not post here in any attacks.
Rob. Stop. I said no mentioning others who do not post here in any attacks.
For the record I just posted this (again) on the WOV forum and no one has commented. They have criticized both me and Rob Singer but there has not been a response to this question. Perhaps OnceDear would like to remind them that this question was posted:
One more time: will anyone bank this bet? This is my third time asking. If you will bank it, state your limits and time and place. I think I have some customers for you.
Quote: rawtuff
A person puts two dices in a cup, shake and slam the cup on the table.
A second person peeks under the cup and truthfully announces if there is at least one deuce showing or not (he will later lift the cup and show to all he wasn't lying).
If "at least one deuce" is announced, the 1/6-ers will wager 1 unit on the premise that there are two deuces under the cup( a one in six probability according to them). If they're right, they get 9 units back. If they're wrong, they lose the 1 unit wagered.
Repeat.
FFS Bob, keep up. If the die with the two is not identified or eliminated, there is no action. No bets would be accepted at any odds.Originally Posted by Rob.Singer
So in other words, you're running away from this with crocodile tears. Next we'll hear how all you're frazzled nerves over this gave you projectile vomiting.Originally Posted by OnceDear
Rob: the Wizard has agreed to the bet that will monetize the belief that the second die is a 1/6 chance of being a 2. See the other thread.
As you may have seen, I put a video on YouTube about the question:
The Wizard has seen my video and I asked him if my video correctly shows the QUESTION being asked. He responded and I responded to the Wizard. Here it is from the WOV site:
Quote: Wizard
No, you didn't accurately depict the question. AT the 0:09 point you said, "If one of these two dice happens to show a two then our challenge is on." You never address what happens if BOTH dice are a two.
Actually, I think I did. Doesn't "if one of these two dice happens to show a two then our challenge is on" mean the same thing as:
======================
"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?
======================
You also raise the question "you never address what happens if BOTH dice are a two." Well, it doesn't matter. If both dice are a 2 then it is also a true statement that "at least one of the dice is a 2." And even if both dice show a 2, then in a 2-dice problem the answer still involves only 1 die.
And let me jump ahead: it doesn't matter which die shows a 2 or if both dice show a 2 in a two dice problem.
If die A shows the 2 then it's a question of what shows on the 6 sides of die B. If die B shows the 2 then it's a question of what shows on the six sides of die A.
And again, if both dice show a 2 then it's still a question of either die A or die B both of which have six sides.
And, to state the obvious: if the judge or partner peeked and didn't see "at least one of the dice is a 2" the "challenge" or "question" is no longer valid and the dice must be rolled again.
So, I think my video clearly illustrates the question, and explains the answer to the way the question was originally worded.
I want to add one more thing in case it somehow was missed: 1/11 is the correct answer to a different question. That question might be something like this: how many different combinations of two dice, with at least a 2 showing on at least one die, would show 2-2?" And here the answer would be 1/11 because there is only one 2-2 and 11 combinations of 2 dice showing at least one 2.
I've said it from the beginning: you have to interpret the language of the question. The language is specific.
Thanks.
For what it's worth, I talked to a journalist friend of mine who writes financial articles for a living - unfortunately not the English professor you asked about. He said that parsing the original question as worded yields an answer of 1/11. If you wanted to be tricksy and word the question so as to actually yield 1/6, you would want to change the tense: "What were the odds..."
What were the odds would imply a past occurrence. Rather, the odds are static---they never change. They are always 1 of 6 if the first one is a 2, past, present, or future. Alan is correct and very clear in both what he has written and his video presentation.
Agreed - even if you rewrote the question in past tense to trick someone, the odds of getting a pair of twos don't change. What changing the tense does is allow you to - technically correctly - ask the question Alan is trying to ask.Originally Posted by regnis
I wish I'd asked him to phrase it for me, I'm having a hard time getting it to not sound like an obviously trick question. The best I've come up with is, "What were the odds of the other die being a two?" You could argue that the answer to that is 1/6 - because before it was rolled, the odds of "the other die" being a two were indeed 1/6. Writing the whole thing in past tense allows you to parse the last sentence in a way that turns it into a one die trick question.
But the whole thing is PAST TENSE. The two dice have been rolled. One is identified as a two. In fact, both could be a two because we are told "at least one" is a two. Now the question clearly falls on the second die.
What is amazing is that even going over this, seemingly line by line trying to explain the exact way the question was phrased, the WOV "math experts" still do everything they can to twist the question so that their answer of 1/11 is the correct answer. They just won't accept the question as it is written.
But the whole thing is PAST TENSE. The two dice have been rolled. One is identified as a two. In fact, both could be a two because we are told "at least one" is a two. Now the question clearly falls on the second die.
What is amazing is that even going over this, seemingly line by line trying to explain the exact way the question was phrased, the WOV "math experts" still do everything they can to twist the question so that their answer of 1/11 is the correct answer. They just won't accept the question as it is written.
Alan, I'm curious as to whether you agree with this, but here goes. One thing I have learned after editing various papers is that "Writing Guys" usually acknowledge when they have made math errors. In fact, they usually have a "math guy" review their material so there is no confusion, and so there are no egregious errors. "Math guys," however, usually wouldn't even know a writing or communications error if it bit them in the ass, much less acknowledge it.
Basically, and it's a joke among editors, "writing guys" are aware of their math deficiencies, but "math guys" aren't aware of their writing deficiencies. "Writing guys" are writing for an audience. "Math guys" are usually writing for themselves. Therein lies the problem.
Last edited by redietz; 05-04-2015 at 10:30 PM.
There is no doubt in my mind that the "1/11 crowd" either can't read and understand the question, or they are ignoring the particulars and just coming up with their own mathematical way of answering any similar question. Many times I have written on the WOV site that I think the question is confusing and possibly was written to be confusing and to trick the "math guys." And not one of them has responded to that.
After putting up my video explaining the 1/6 answer, the folks on WOV have started to become more creative. The Wizard said he tried to put a video on YouTube but he said he didn't like the result. I still want to see it because I asked him to explain how he came up with 1/11.
Another WOV member put up this graphic to explain why the chance is 1/11. I took a photo of the computer screen because I was unable to copy the web page (computer problem). The WOV poster continues to argue there are 11 ways for a 2 to show up on two dice. I responded this way: Nice graphic. But if a 2 shows up on one die the chance of a 2 on the second die is still 1/6.
Possibilities/probabilities are the domain of the mind; whereas, actualities/values are physically determined.Originally Posted by Alan Mendelson
Read-look at one of the dies. If it shows a 2, then the other will show a 2 one-sixth of the time. But, if it does not show a 2, then there can not be a question as asked because there is no matter or question both dies - or, more accurately, also the other die - show a 2. The other die becomes irrelevant when the read-looked at die is not a 2.
Can not read-look at both dies simultaneously. If could, then there would not be any question at all. Read-looking to establish the "at least one die is a 2" condition forces the physical nature of the exercise.
In conclusion, the mind may be fooled; but not the physics. The answer in terms of the latter (when a question exits) is one-sixth of the time.
Exactly. OneHit, this is what I was referring to way back in this exercise when I said looking at the dice was sequential, not simultaneous. Human eyes do not take in the entirety of a scene all at once -- they take it in point-to-point, sequentially.
This was just a trickily written piece of garbage question, designed to separate a reading audience into various categories of readership. As I said earlier, it reminds me of the purposefully lousy writing on the logic part of the GRE before it was redone, and it reminds Alan of his "nonsense-language" questions on the law exams.
OneHitWonder thank you and welcome to the forum.
It's amazing that that on the WOV forum they continue to maintain it's a two-dice problem.
Yet when we simulate this experiment (on a computer or in real life), we CONSISTENTLY get rates that are closer to 1/11 than 1/6. What a coincidence!
RS___ I think it was you who asked on the WOV this:
Quote: RS
How do you only consider one die with 6 faces when you don't even know which die is the for-sure-deuce die? You have to consider both dice.
RS if you really believe that it makes a difference which of two dice shows a two, when there are only two dice in the problem, then there is nothing left for us to discuss. As I wrote many times before, if one die shows a two then the chance that a 2 shows up on the other die is 1/6. If you want a more specific answer: If die A shows a 2, the chance of die B showing a 2 is 1/6. And if die B shows a 2, the chance of die A showing a 2 is 1/6. And if both die A and die B are showing a 2, then when you look at either die A or die B the chance that either of them landed on 2 was 1/6.
Actually, the "experiment" is not the actual question that was posed. The experiment also includes a payoff of 9-for-1 each time a 2 appears on a "second die" after one die shows a 2. By the way, if you insist it matters which die the 2 appears on, how can you possibly conduct your experiment or have the bet? I think you guys will have to make up your mind about whether or not it matters if the 2 appears on a certain die.
Thank you, Alan, but I think that you should cut them a bit of slack. It might help prove out your own contention as far as also its confirmation.Originally Posted by Alan Mendelson
As far as I understand your apparently unique position: there is one die to show a two, either one; hence the other one will show a 2 one-sixth of the time.
If you follow my earlier argument - about there being no question or matter of any chance of both dies showing a 2 when the initially read-looked at die does not show a 2, ie, can't group the double-2 result with the set of possibilities which begin with a non-2 - then your contention becomes my argument taken over both dies simultaneously. However, we require a special type of read-looking at the two simultaneously.
If, and when "one side of the brain literally does not know what the other does". Then, there are two separate streams of consciousness for the exercise. One stream to watch the die on the left; and the other, the die on the right. In this scenario, game on for the left-brain when its die shows a 2; and, game on for the right-brain when its die shows a two. Note that now there are two times to announce game on instead of the one time by my argument.
The compelling reason for your contention is that with two games on, the Wizard's fair payout meshes with your own two separate but same payouts. For the specially modified left and right -brain win(s) on the double-2 result, the fair payout is 5 to 1 (for one-sixth win odds) on either, for a combined payout of 10 to 1 on both. Which, to confirm also your argument, is precisely the fair payout of the Wizard of 10 to 1 (for one-eleventh win odds). His argument re-worded contains your contention.
Your contention thus became equally a valid argument. Though this manner of argument supports the result as a draw. (Sometimes we just inexplicably perceive things in such an inverted manner that we turn out to be as right as the supposed experts.)
P.S. If you have trouble understanding any of this, I suggest that you post it up at the Wizard's for clarification and/or rebuttal.
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