Then accept the 8:1 bet. I'm 100% certain you'll go broke.
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Then accept the 8:1 bet. I'm 100% certain you'll go broke.
Hard to understand. First the WoV forum badmouths Frank for being a member of this forum, he's rudely mocked along with me for having enough money to be able to give away our books while AP wizard had to beg for money when his skills just weren't there, then Frank's criticized for not opining on the issue.Quote:
Originally Posted by jbjb
You guys need to get not only the original problem straight--you need to get your stories straight.
Alan, at this point I am trying to understand how you read the question. I understand fully well, that based on your understanding of the question, the answer is 1/6. Based on my understanding of the question, the answer is 1/11. Neither of these two facts are arguable.
You contend that the wording is unclear, and I contend that it is not. The wording as presented in the problem is exactly the wording that I would expect to see, and is the wording that I would use in a simple conditional probability exercise. If this wording is unclear, I would like to understand why so that I can avoid this issue.
I would like to understand how you would word the problem with the answer 1/11 in a clear and concise manner.
I'll accept that bet. And I'll bet you 10X what I win or lose that I won't go broke. And no....this time I will not pay for anyone's airfare to LV, hotel, or food.Quote:
Originally Posted by jbjb
I've already answered this: the 8:1 bet is a lousy bet for the player. There is no way in hell I would make that bet. The reality of the situation is that throwing 2-2 is a 1/36 event.Quote:
Originally Posted by jbjb
And once again: the damn bet has nothing to do with the original question/problem.
Wake up -- we're not stupid. And we might be almost as smart as you are.
Thank you. This is probably the first sensible post from those who are in the 1/11 "camp."Quote:
Originally Posted by 1in11
Honestly, I don't know how to phrase the question so that the appropriate answer would be 1/11, but I am sure there is a properly phrased question that would elicit that answer as the correct answer.
I can only say this:
When someone tells me that there are two dice, and at least one die shows a 2, then common sense and my knowledge of what is on a six sided die tells me that the chance of a 2 showing on the "other die" has to be 1/6. To put it another way (which I have said before) the information provided to me only allows me to answer 1/6. The answer is defined by the question.
That doesn't exclude 1/11 as being an answer -- it's just that it can't be the answer given the information that was provided in the stated original question.
Thank you again for asking this.
I take a bit of offense here, I think my asking you if you would tell me which of my steps don't match up your original question was plenty sensible. I'd still appreciate your doing so, actually.Quote:
Originally Posted by Alan Mendelson
Please don't be offended. But the original question made it clear that the only thing to be considered was the "other die" and with the other six-sided die it was a 1/6 answer.Quote:
Originally Posted by synergistic
Over on the Wizard's forum they are continuing their mantra about 1/11 with all of the combinations of two dice. So one of you guys who is here reading this please send them this message:
I understand that there are 11 combinations of two dice that include 1 die with a 2. But that wasn't the question that was asked and so, those 11 combinations of two dice do not fit what is needed to answer the question. The question doesn't ask for the number of dice combinations that contain a two. The question asks: "What is the probability that both dice are showing a 2?"
Now you can take the "long way" around and explain that there are 11 different combinations of dice that include a two, but I am answering the specific question: What is the probability that both dice are showing a 2? And the answer to that specific question is 1/6.
I am sorry that such a simple answer is all that is needed. Now go figure out when that big meteor is going to crash into earth.
edited to add:
Over on the other forum there are several who posted the various combinations of two dice that include a two. And then there is this fellow Ibeatyouraces who wrote:
I can't fathom how hard it it's to understand that you can roll 1-2, 2-1, 2-2, 2-3, 3-2, 2-4, 4-2, 2-5, 5-2, 2-6, and 6-2. That's ELEVEN possible combos, only which ONE CAN BE 2-2. Just pathetic!!
On each roll. AT LEAST ONE DIE IS A 2!!!
Clearly this is overthinking the problem. And by overthinking they make a mistake because it's not a question of any two dice combination -- it's what are the odds of a 2 showing on the other die.
Well. Alan, I never thought that I would hear you say that. Thank you.Quote:
Originally Posted by Alan Mendelson
Do you actually know what probability is and how it is worked out? You do know that probability is a numberr don't you, generally expressed as one number divided by another number?Quote:
Originally Posted by Alan Mendelson
Do you know where the top number comes from and the bottom number comes from?
I'll give you a clue.
The probability of an event=Count of the ways that event can happen/Count of all possible events that can happen
We agreed that there is one way that a 2-2 can happen so that makes the top bit, the numerator, a 1.
The count of all possible events that can happen.... well, you just admitted it. there are 11 possible ways that we could have ended up in this stupid mess where at least one of the dice is a two. We might reasonably conclude that probability in this instance, and not in any other instance, not with 5 dice or one dice or one spinning dice, but with two resting dice under a cup...
that probability = count of number of ways 2-2 can happen/ count of number of ways two dice can be at rest under a cup with at least one dice showing
= 1/11
Now.... explain it to me. what part of my logic is flawed? It should be possible without any bluster about knowing anything else.
OnceDear your logic is fine, but you are not answering the original question. The original question makes it clear that at least one of the two dice is a two. And once again (how many times, Lord, how many times?) given that information you only have to consider the six faces on the other die.
Tell your buddies over on the Wizard's forum who enjoy insulting me and the others here who also say it's 1/6 that we appreciate your logic but your logic does not answer the question and the conditions posed in the question.
And frankly, I wish the original poster would come out of the bushes, desert, closet, Siberia, or wherever else he might be to clarify why he asked the question that way? I don't know if it was in error, an accident, or intentional to mislead? But whatever he did and whatever his motive, he sure created a mess out of a simple exercise.
Alan, a WoV poster called dalex (who may also be posting under a different anonymous handle here) has put up a very good, very comprehensive post about how he gets to the 1 in 11 conclusion. I guess he did it with the assumption that people who know the answer to the original question to be 1 in 6, just can't figure out how the geniuses arrived at 1 in 11.
Well, just as we've been saying from moment one, we all have always known where 1 in 11 came from, except that it is not answering the OP's question. The way they interpret it is as if there always needs to be two dice in the formula even though the OP clearly eliminated one die from the problem. I can only summize that the reason these people need to keep two dice in play is due to their incessant requirement to make complicated math problems out of the simplest of events.
It is all so very similar to the Monty Hall problem. If that question clearly identified that all 3 doors were in play when the question about odds was asked, of course the answer would be 2 in 3. But the way it was asked appeared to be "what are the odds NOW" which naturally would be 1 in 2. As my articles said, it is all in how one interprets the wordsmithing. Simple, more sensible people would prefer to take the shortest, surest route. Those forever in conflict-mode who enjoy appearing like the smartest in the room, will always take the lengthier, more complicated route.
Here's another example of their stubborn thinking that cannot consider ONE die but must consider only two dice answers. This comes form "Dalex64":
I realize that there probably isn't anyone left reading who doesn't believe at this point, but I came up with an explaination using Calculatus Eliminatus, in other words, the process of elimination.
I'll start with the original question:
Quote:
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?
Now, little bits at a time.
Quote:
You have two 6-sided dice in a cup.
I think everyone agrees on that.
Quote:
You shake the dice, and slam the cup down onto the table, hiding the result.
This is our starting condition. Two dice, shaken and rolled, now unmoving, and hidden from view.
Potentially, the first point of contention. I think right here, before you peek, there is an equal 1 in 36 chance that any one of the two dice arrangements could have landed under the cup. namely:
1-1
1-2 2-1
1-3 2-2 3-1
1-4 2-3 3-2 4-1
1-5 2-4 3-3 4-2 5-1
1-6 2-5 3-4 4-3 5-2 6-1
2-6 3-5 4-4 5-3 6-2
3-6 4-5 5-4 6-3
4-6 5-5 6-4
5-6 6-5
6-6
that arrangement should look familiar to craps players. 36 ways the dice can land, with equal probability.
If you don't think all 36 of those can be under the cup before we peek, under the terms of the question, please explain why not.
Quote:
Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."
this is our main point of contention:
At this point, I eliminate all of the possibilities that do not have a 2. They do not meet the condition that "at least one of the dice is a two"
I marked the remaning possibilities in bold. All of the remaining die rolls fulfill the condition that "at least one of the dice is a 2"
With the informaiton given in the question, I don't have any more information that I can use to eliminate any more of the possibilities.
Quote:
What is the probability that both dice are showing a 2?
Of the 11 possibilities that are left, only one of them is 2-2, so the probability that both dice show a 2 is 1 in 11.
To get to 1/6, you need to eliminate 5 more of the possible rolls. Which ones do you eliminate, and why?
The people who say the answer is one in 6 say that under the cup, there must be only: 2-1, 2-2, 2-3, 2-4, 2-5, 2-6.
Or perhaps 1-2, 2-2, 3-2, 4-2, 5-2, 6-2. I'm not sure which.
I'd like to know how you eliminate the remaining possibilities that meet both the condition that at least one of the dice is a two, and both dice started shaken and hidden from view under a cup.
I'll listen to explainations as to why all 36 possibilities couldn't be under the cup before you peek, too. Please don't forget the starting condition that "You shake the dice, and slam the cup down onto the table, hiding the result."
As an aside, I think another way to say that "at least one of the dice is a 2" is "this roll of the dice has resolved the bet" that is to say, the bet where you lose on one two and win on double twos.
Again, their only answer seems to come from the idea that the answer must come from a two dice combination of numbers. They don't even consider the idea that the answer can come from a single die with six faces.
Is this something taught in school? That the easy, simple answer can't be right?
I think this is the most important part of that post on WOV:
I'll listen to explainations as to why all 36 possibilities couldn't be under the cup
Well how about this one: there is one other die with only six faces on it, 1 thru 6?
first you tell me there are only 6 numbers on one die. Now they say there's no santa claus. I am a broken man.
Wizard is now inviting anyone not in the 1 in 11 camp to his MonekyFest, and to bring "plenty of money" while he brings the dice.
AGAIN, WoVers....I will GLADLY show up. PLEASE tell him! All he need do is follow the parameters of the original question: set one of the dice--either one, doesn't matter--to a 2; then roll the other die. 1 in 6 will yield a 2nd 2. And I will happily send him back to begging the pseudo-mensas for "mo money".
Count me in too Rob. Can't let you have all the fun,
If the Wizard conducts his bet the way Rob describes and is offering 8 for 1 OR 8 to 1 pays for every hard-4 (2-2) I am also in. But I am afraid that is not what's going to happen. They will want two dice rolled simultaneously like in an everyday game of craps -- and in that case 2-2 occurs 1/36 times. And in that case the Wizard is only paying 8 to 1 or 8 for 1 for a 1/36 event.
I vowed to stay out of this any further, but I don't find the language or question ambiguous at all. One die is a 2--that is clearly stated. Therefore, the question is what are the odds that the other is a 2. Answer-1 of 6. As a licensed arbiter for many years, I am required to look at the simplest and plain meaning of a contract or other document in resolving a dispute. In this case, it is so clear as to not even constitute a dispute. I must take the stated fact--that one die is a 2--in its simplest form. that leaves only the other die in question.
Math, excel spreadsheets, all the other hypothecations are meaningless to the simple question that was presented. If you want to answer the question of what are the odds of rolling two dice and one or more of them being a 2, then go ahead and do the math and create your fancy spreadsheets. But that is totally unnecessary and unreasonable in this context.
Actually--just roll one die as the other is a 2. Then pay 8-1 on every 2 on the die that is rolled. I'm in all day.
Which one is a 2 though?
The one with 2 dots on it.
That's funny regnis....but also so pathetic that a collection of geniuses just can't comprehend that.Quote:
Originally Posted by regnis
Singer's a moron, fraud, and has no credibility so he's ignored.
The bet is this. Roll 2-2 get paid 8:1. Roll 2-1, 2-3, 2-4, 2-5 or 2-6 and lose 1:1. Roll any other number and it's a push so re-roll. Minimum 500 rolls.
So there are ten ways to lose and you are only paying 8-to-1 ??Quote:
Originally Posted by jbjb
I asked for details on the WOV site and got this response:
Two dice simultaneously and 8 to 1.
Do you seriously think we'd bet this at a disadvantage?????
Once again, I wouldn't take the bet. So jbjb I don't think you will have to ignore Rob because I doubt he will show up.
Then I take it back about the geniuses, esp. since I'm a moron :) Better to be an AP forced to beg members for cash I suppose. Oh where did I go wrong!Quote:
Originally Posted by jbjb
This whole thread is a disaster, but at least it brought several new members to the forum.
So bravo, Alan, I guess...
No Dan. The disaster is that people don't understand how to read and understand a simple premise and question.
There is nothing wrong with their math. It's just that it is the wrong math for the question that was asked.
Other than them creating a complicated issue out of a very simple one--and I suppose it was because of how smart :) they like to appear--the only thing really accomplished was them getting away with insulting you (a WoV member) on their forum, which if you insulted any of the geniuses over there the way they insulted you, you'd have gotten an immediate suspension from one of their goofy administrators.
I have a question for Dan since I'm not a live poker player. In that video they put up--as a way of belittling you out of frustration with Alan I guess--why would Helmuth stay in with such a lousy hand as suited Q2? Was it because he had such a big stack that he really didn't care if he lost, or does he regularly get away with bluffing like that with basically nothing? We saw the odds of you winning with the pair of 3's, and it was a far superior hand (although neither of you could have even come close to guessing those odds since no one but the TV knows what cards are out of play at that point).
Well, looks like this discussion is pretty much done. Hope the entire population of 4 members here enjoy jerking each other off. Farewell.
Would it make it easier to understand if we imagined the 2 dice were of different colors, Red and Blue?Quote:
Originally Posted by Alan Mendelson
R/B
2 1
2 2
2 3
2 4
2 5
2 6
1 2
2 2
3 2
4 2
5 2
6 2
Sure when we know the red dice is 2 the odds are 1/6. When we know the blue dice is 2 the odds are 1/6.
Because you are rolling BOTH dice and only know "1 of them is a 2", don't we consider ALL possible outcomes where at least one of them is a 2, but don't know which one?
There are 11 unique outcomes with 1 winning outcome of 2,2.
I see several folks are trying to save face by changing the rules of the game. It was never stated that any particular dice was a 2. At least we now know you guys actually understand you would lose under the initial rules. Otherwise, why would you be trying to change them.
Okay, I'm coming to the rescue (or I am not coming to the rescue). If your friend would be so kind as to take the die with the two showing out of the cup and give it away to someone in need of a single die, what is the chance that the poor die still in the cup shows a 2?
100,000,000,000,000,000 in 1,100,000,000,000,000,000 ?
1,000,000,000 to 11,000,000,000?
Frank--in the game I ran in 8th grade behind the school ball fields, there probably wouldn't have even been another 2 on that second die. House edge--my dice.Quote:
Originally Posted by FrankScoblete
When you write experimental conditions, the onus is on the writer to state any and all conditions, using multiple, overlapping descriptions if necessary, so as to avoid all confusion, especially since many researchers reading the descriptions are not reading them in their native language. The onus for understanding something like this is on the writer. Any failure of interpretation is the fault of the writer, given a reasonable audience. Now the WoV folks may think this is an audience of less than average intelligence. That's probably not a correct interpretation, given the general success of many (non-anonymous) posters here.
I understand, according to Rob, that the WoV folks are berating the members of this forum. That's certainly their right. Having spent much of my life hanging out with math professors and people who teach statistics, I can tell you that this forum has its math deficiencies on occasion. This particular silly argument, however, is about a brief statement purposefully written in an unclear fashion so as to create the look of a paradox. An editor at any academic journal would have required an expansion of the statement and a rewrite.
I understand the WoV members have an annual get together. I think it appropriate that Rob and I attend so that we can be properly impressed by the personal attributes of the membership and report on those attributes to the outside world. So when and where is the annual shindig?
Sorry red but I disagree. I thought the description was very accurate and concise. Yes, since this situation is very subtle there will always be misunderstanding by those who are not trained to look for these kind of subtleties. The problem is designed specifically to force people to think outside the box. If a person missed it, it is not a big deal. Lots of people will miss it. The key is to learn from the problem and be able to handle future problems a little better. To deny the correct answer is actually a sign of a closed mind or massive ego.
You shouldn't be so hard on yourself Arci.
Correct, The question couldn't be clearer. The dice are thrown together.
You're missing the point, Arci, and actually making my point. In fact, in an ironic way, what you just did in your interpretation of what I said is directly analogous to those people who "misread" the problem. Here's what I mean:
Concise and clear are not the issue. Something can be concise and clear -- to an expert in a field or someone looking from the writer's perspective -- and still be bad writing because it's not understandable to the majority of the readers. If, as you said, many if not most people would get it wrong, then it has failed as a piece of writing to the majority of the readers. That means it has failed as a piece of writing, period. You can't say, well, this is clear to some or a minority of readers, so it is clear. That's not really the way one should write. One writes for an audience. If the WoV folks are writing for "their" audience, great. But in this case, they are not.
Now here's the thing that ticks me off. This was clearly designed to be just what you described -- a piece of writing, a description, that would partition the reading audience into two factions. There would be (probably a minority) of readers who would "get it" and say 1 in 11. There would (probably) be a majority of readers who would not "get it" and say 1 in 6. So the entire thing is an exercise in partitioning an audience for no good reason so some are "clever" and some are not.
Clear and concise, in this case, doesn't mean "understood." It is the writer's responsibility to be understood. Designing something to be clear and concise in an obvious and superficial sense, but still be something easily read two ways or misinterpreted, is purposeful bad writing. It also has an aristocratic, condescending context to it that makes me a little ill.
Arci, I know you're a math guy. I hope I've written this well enough that you understand what I'm saying here. I actually really want you to understand a writer's angle on this, so let me know if this isn't clear.
And Rob, I get your feeling that the WoV folks have an unwarranted superiority complex.
What in the question is unclear or difficult to understand pertaining to both dice being thrown at the same time?
As a wordsmith yourself how would you change it so that more folks would get it?
As you can likely tell by now red, I got over WoV's obnoxious "superiority complex" first when I successfully called them about playing my strategy, then when their beloved leader 's AP gambling mantra failed him enuf to be humbled before the masses. And what have they done in their visit here? Not much more than exhibit agitated insulting and namecalling--something they are not allowed to do over there, & just what you'd expect from the genius crowd when they aren't getting their way or able to change minds.
Red's depiction was right on, and all arci's doing is trying to show he belongs more over there than over here. But he seriously missed the OP's question being purposely provocative & vague. As soon as it was defined in relationship to the infamous "two-coin problem" it was a dead giveaway that only a mother would miss. In short, arci slipped up and is now taking a logic-beat down from red for doing so.
What is unclear is that it's specific wording can be taken in two ways. Do you approach it straight on by understanding that since there's one die already showing a 2, that we determine the odds of just the other die showing a 2? Or, do you back into the solution by trying to solve the problem using the two original dice? The short route or the long route? Easy vs. more complex. Those who choose to get to the answer quickly vs. those who enjoy analyzing statistics for all they're worth. In essence, there never really was a totally right or wrong answer, thus the heated discussions.Quote:
Originally Posted by quahaug
The OP could have given specific & clear parameters about whether one or both dice were still in play, but then why even bother to post such a boring question?
Quahaug, you're missing the point. If the majority of people misread or interpret the material differently, the material is flawed. It doesn't matter how I would try to fix it or how anyone would try to fix it. It's not my job to fix it. It's up to the writer to (1) explain the process so most people reading it would get it, (2) go out of his/her way to not bifurcate the audience, and (3) write nothing to make the writer look smart and the reader stupid. Conciseness in this case probably works against being understood.
I'd probably scrap the description as is and rewrite the thing step by step, expanding and explaining as I went. It should be written so the answer is obvious, but then that takes the whole cute trickiness aspect out of the thing. Without the trickiness, there's no "aha" aspect to it, and nobody would care about it.
It's written to provide an "aha" moment. The only problem is, stuff that tries to do that is self-absorbed and almost by definition bad communication.
There is no way around the fact that something that would be interpreted different ways by large chunks of a reading audience is not written appropriately. It has failed.
When Singer's mug is put into OSN or Griffin, then maybe, just maybe he'll get and ounce of credibility. Until then, anything he says has no merit, anywhere.
As for the OQ, that's "Original Question" for you all. Yes, it's the wording of the question that's the trick.
1/36 times 2-2 will show up
1/11 times 2-2 will show up when you EXCLUDE any non two numbers
1/6 times 2-2 will show up when setting one die on 2 and rolling the second separately.
I don't think the majority do misread it. Clearly the dice are thrown at the same time. How anyone can get one die is a two, then the other die is thrown is beyond my understanding. It would be like having a die with only twos, but there is no way to infer that either. What part of the question is unclear?
One of my concerns here is that a blatant failure in writing is somehow being considered less onerous than a blatant failure in math. If "the wording of the question is the trick," then the question should have been, obviously, worded differently. That is a failure of writing, a failure in communication, and that failure is worse because it's (1) self-serving vis-a-vis the writer and (2) dividing the audience into camps for no reason. Those are major failures just so some folks can feel clever.
If I can phrase 1+1=2 in such a clever way that people misinterpret it, I'm not being clever. I'm being a bad writer.
Quahaug, my simple response is -- why don't you rewrite the problem so Alan can easily understand what you're saying? Alan knows basic math. Alan can read. Alan is a world-class journalist. Just write it for him. You don't get bonus points for using the original question. There's no need for the original question.
Red, I get what you are saying, but in this case the problem is meant to make a person think. It is not meant to lead them directly to the correct answer. When you look at it from this perspective I think you will see that the writer achieved his goals. In this case there is nothing wrong with getting the wrong answer. The problem was meant to be tricky.
What bugs me is the number of people unwilling to admit they were wrong and still insisting they got it right. If I hadn't read the first few posts before thinking it through I might have gotten it wrong too. Big deal.
First you're done then you're not....get your stories straight :)Quote:
Originally Posted by jbjb
What I find interesting from these pseudo-mensas is their inability to comprehend basis mechanics, like when two dice are rolled and one of them is identified as being a 2, why they always have to pretend only one of the dice was rolled in order for the answer to be 1 in 6.
Nerves, I guess.
Alan. If you toss a pair of FAIR dice with your eyes closed, wait 60 seconds for both dice to settle, then you look and see that ONE of them is a two. What are the odds of the OTHER die being a two? I think the original was clearer but we'll see.
Q, you know what he'll say because your wording eliminated the first die for the answer to be arrived at. Just as the OP's question did, albeit in somewhat more of a tricky fashion.Quote:
Originally Posted by quahaug
Or maybe you could ask Alan how many ways a FAIR pair of dice can show when ONE and ONLY ONE shows a two? That does seem kind of simple but it doesn't make a very good riddle.
1/6. Sorry to disappoint you but given the question you wrote it can only be 1/6. Any die with six different faces only gives you 1/6.Quote:
Originally Posted by quahaug
You want a question with the 1/11 answer? You ask: how many combinations of two dice with at least one die showing a 2 can show 2-2? There's your fucking question. Clear and concise. I can't post that on the WOV forum but one of their English majors can edit it.
Cool, you finally get it.
Hopefully you and the others finally get it. There was only one correct answer to the original question and that was 1/6 because in the original question you are only talking about ONE die.Quote:
Originally Posted by quahaug
The way I wrote the question, you have to consider the different combinations of dice with a 2 showing.
Good stuff.
Hopefully we're now done with the WoV visitors' insults, namecalling and they're annoying stomping up & down like spoiled little brats, since there can be no doubt they feel they've been taken out to the woodshed over this question....more than once for that matter.
What Arci said about the purpose of the question being "to make people think" relates to what Alan said initially about the whole enterprise reminding him of the fake language questions on his law exam. I experienced something similar when I took the GREs after they had first introduced a logic section. I thought I would do best on it, but I actually scored worst on it. I kept reading the questions, thinking "I know what they're asking, but I would have written it this way." The unnecessary complexity of the logic section's writing, in an effort to test your abilities with purposefully tricked-up obfuscation, was a helluva distraction. They dropped the logic section shortly afterwards.
I just think the whole WoV presentation, argument, and aftermath (what Rob's reported) are adolescent. But who knows? Maybe a lot of the WoV forum people are real adolescents. I don't know. Maybe teachers or professors are underrepresented. Maybe they don't do much actual writing. Maybe this is how they get their jollies, by being anonymously clever. The whole thing was silly, and had nothing to do with Las Vegas or gambling.
Over on the Wizard's forum, one of the members proposed a different bet. I think this bet follows the assetion of the 1/11ers that the answer to the original question is 1/11:
Quote: rawtuff
Doesn't a proposed bet like bellow follow the original question EXACTLY ? -
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.
They're all the same question.Quote:
Originally Posted by Alan Mendelson
First of all, any no. from 1-6 can be shown because no one knows what number is on the peaked-at die. We'll accept that it's a 2 so as not to get the WoVers whining any more.
There is now a 1 in 6 chance that the other die has a 2. Anyone who claims the other die's chances are 1 in 11 is smoking bad weed.
Where'd the 9-1 come from?
Rob, once again I posted on the WOV site:
I will repeat: Does anyone want to bank THIS BET:
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.
The 9 to 1 came from Rawtuff. My "guess" is he thought that was a fair, or middle-number, between the two sides.
And just a moment ago "RS" said he would bank. See his post copied here and my response:
Quote: RS
Like I already said, yes. And to avoid confusion, that's 9-for-1 not 9-to-1.
Please post where and when so that I can provide the details on my forum -- and to the members on this forum. And please include what size bets you will accept.
And here is the link:
http://wizardofvegas.com/forum/quest...32/#post450084
Erm No. BOTH dice are peeked at. We agreed that several hundred posts ago, before you rode to Alan's rescue.... But a bit after you misquoted the original question on your comments page.Quote:
Originally Posted by Rob.Singer
Quote:
8. Can we agree that from the meaning of the question, that we can reasonably assume that the 'partner' saw both dice? It's implied, but central to the question. YES, BUT IT DOESN'T MATTER FROM MY POINT OF VIEW. JUST SEEING ONE DIE IS ALL THE INFO I NEED.
Quote:
"This is the problem. The ORIGINAL QUESTION deals with TWO DICE under a cup with at least one of them showing a 2. YOU MUST address the problem using a pair of dice with at least ONE DIE showing a 2. If you don't do this you are altering the question."?
Agree or disagree? I AGREE AND I STAND BY THAT. IN ORDER TO ANSWER THE QUESTION OF 1/6 (OR IN YOUR CASE 1/11) YOU MUST CONSIDER BOTH DICE.
OnceDear the position of everyone who believes the answer is 1/6 (and I think I can speak for them all) is that we just need to know what one die shows.
Now, over on the WOV site you said that the "new bet" outlined above still offers an advantage to the banker. Are you willing to bank that bet as written by Rawtuff??
Regarding your quote from Rob, I think you took that out of context and not related to the original debate. I think Rob is now talking about the new bet offer.
Just so everyone knows, I consulted with Jesus Christ on this. He too is a simple man, and understands the question to be what are the chances the other die is a 2. As in, taking the known 2 out of the equation right then and there. You know, the simple way.
Of course, there's two problems with this: first, WoVers want it to be all about knowing 1/2 the answer without actually identifying which die HAS that info, just so they can throw the monkey wrench into the problem solving. Then, of course, the mensas don't believe in Jesus Christ, because being "men of pure science" they haven't actually ever met him. But they do believe in the collection of other freaks over there, because as anonymous posters, they "said so".
So if we have two peekers, Your peeker and the wizards peeker, and your peeker only looks at one die, and he doesn't say 'At Least one of the dice is a deuce, do you want the bet to be off? Or do you still want the bet to be on if the wizard's peeker says 'at least one of the dice is a deuce'?Quote:
Originally Posted by Alan Mendelson
It makes a hell of a difference. There would be a lot more action and a lot less pushes if you take the action when the wizard's peeker looked. Totally different odds would apply and we can guess that the odds would be 1/6 on your subset of the bets and 1/11 in the full set of the wizards bets. Or do you assert (again) that it makes no difference?
I'm willing to bank this bet (rawtuff on WOV posted and I quoted on page 31). I'd have a max (idk yet). Let me know if you're interested in wagering $100's...or $1,000's and I'll (most likely) be able to get the money together.
Once dear, betting.& challenges with your crowd over there does not have a credible history. If they get backed into the corner that shocks them into realizing that they actually HAVE to make the bet with real cash even by offering go put the escrow right into the wizard's hands for safe keeping, the geniuses then ask their opponents to pay for their travel and other expenses. I however, would trust ANYTHING evenbob says about this.Quote:
Originally Posted by OnceDear
We don't really support EvenBob....he's not one of our own. He just trolls the forum.
Let me just make this clear. Rawtuff wrote on the Wizard's forum a "bet" which I think follows the original question that evoked two answers: 1/6 or 1/11. This bet would, in my opinion, monetize the disagreement. I am asking if there are any "takers" meaning who wants to take the bet and who wants to bank the bet. If someone wants to bank the bet, I want them to clearly state where and when and what limits. I don't care if you want to bet dollars or donuts or even pennies -- the amount is between you and the other participants. I just want the "bet" clearly defined here following Rawtuff's proposal.
Once again, here is Rawtuff's proposal:
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.
Actually, I worry that EvenBob is off his game. He's not contributed to this trollfest at all. I reckon he's keeping his powder dry. Wise old Troll. When he does bob in he will milk this for 5,000 posts minimum $:o)Quote:
Originally Posted by RS__
Hey Alan, could you answer my last question please for the elimination of doubt. Cheers.
Wish I had some dough but I've been hit hard by the Obama depression. All in on 1 in 11. Singers history on bets not withstanding, this bet will never happen. Singer knows the real answer. He's just stroking Allan for the time being
Alan needs Singer.
Agreed.Quote:
Originally Posted by quahaug
Would you like a sidebet Q that Rob and Alan blame 'The wizards crowd' for putting obstacles in the way?
We already know that "It's a push if 'At least one dice is a deuce' is not declared" is a whole different ballgame to "The bet is on if 'At least one dice is a deuce' Is declared."
And actually it is a whole different ballgame unless the bettor is forced to bet EVERY TIME BOTH DICE ARE EXAMINED and at EVERY POSSIBLE OPPORTUNITY 'At least one dice is a deuce' is declared'
By insisting on only betting when only one of the dice is examined, Alan skews the odds to the 1/6 that they truly are. We always knew that.
Heck It's even been agreed a zillion times that the odds are 1/6 if only one die is peeked at.
Please post it again.Quote:
Originally Posted by OnceDear
At this point I am asking all of you to stop the personal attacks and naming people who do not participate here.
For the sake of clarity I would like to discuss the "Rawtuff bet" which I think monetizes the original question and dispute.
Sure thing Alan.Quote:
Originally Posted by Alan Mendelson
So if we have two peekers, Your peeker and the wizards peeker, and your peeker only looks at one die (because it really doesnt matter), and he doesn't say 'At Least one of the dice is a deuce, do you want the bet to be off? Or do you still want the bet to be on if the wizard's peeker says 'at least one of the dice is a deuce'?
It makes a hell of a difference. There would be a lot more action and a lot less pushes if you take the action when the wizard's peeker looked. Totally different odds would apply and we can guess that the odds would be 1/6 on your subset of the bets and 1/11 in the full set of the wizards bets. Or do you assert (again) that it makes no difference?
OnceDear I would like to follow Rawtuff's bet because the cup will be lifted to show that there wasn't a lie. You only need one person to do this and I don't think it matters who that person is. If the Wizard wants to shake up the dice in the cup and slam the cup down and peek himself it would be fine with me and it should be fine with everyone because after the bets are made the cup will be removed showing the dice. Don't you agree?
This is why I asked on the WOV forum if anyone would bank Rawtuff's bet, and to spell out exactly the conditions, details, etc.
I am not discussing any bet besides Rawtuff's bet. As I said, I think Rawtuff's bet monetizes the original question.
Allen I think you are being clear. I don't know where the oncedear obfuscation comes from. Alan stated that at least ONE die is a two. You would have to look at both to say at least one die is a two. I think?
I briefly ducked into WoV to see what was up. Folks, it sounds like a collection of gambling wannabes, probably young gambling wannabes, who have some math backgrounds but likely not three doctorates between them. They seem too pretentious to be actual scholars, so I wouldn't take them too seriously. They're like a friendly math club. And probably young (did I say that?).
My humble opinion is that they couldn't make a profit in LV with Dillinger's schlong and Steve Wynn's bankroll. They're just looking to cherry pick some poor fellow gamblers and spit cider in their ear (and they may be too young to get the reference).
Just give them a wide berth and don't take them too seriously. Cider's great if you drink it.
Evenbob is awesome. You can just picture him picking his toes as he posts his special little paragraphs of wisdom.
I'm not understanding this challenge, seeing that multiple personalities keep inputting different parameters. If the peeker sees "at least one 2" will that die be eliminated from under the cup? If so and with those quoted payouts, I'm in for whatever amount wizard wants to put up.
If the die with a 2 is not identified or eliminated, ie., if it remains "random" then the 9-1 odds are something only a big phony storeteller like mickeycrimm would take.