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.
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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.Quote:
Originally Posted by Alan Mendelson
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."Quote:
Originally Posted by Alan Mendelson
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.Quote:
Originally Posted by 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.
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" .....Quote:
Originally Posted by Alan Mendelson
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?Quote:
Originally Posted by arcimede$
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.
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.
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.Quote:
Originally Posted by RS__
Regarding my lunch bet with the Wizard I'd love to do it.
You say it's the wrong interpretation because you don't want it to be. In simple English it is the correct interpretation.Quote:
Originally Posted by arcimede$
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.
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.Quote:
Originally Posted by redietz
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.
Agreed...........Quote:
Originally Posted by Alan Mendelson
Bolded: that's my point.Quote:
Originally Posted by Alan Mendelson
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?
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.Quote:
Originally Posted by RS__
Nope. "At least" has a specific meaning. You want to interpret it to mean one specific die. That is not what it means.Quote:
Originally Posted by Alan Mendelson
Your answering a different question. Your assuming one specific die is a 2, that is not what the question asks.Quote:
Originally Posted by Alan Mendelson
Odds are based on multiple dice throws. That is what odds refers to.Quote:
Originally Posted by Alan Mendelson
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.Quote:
Originally Posted by 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.
And your nonsense continues...Quote:
Originally Posted by Alan Mendelson
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)
If I roll the dice 66 times and in each roll, at least one of the dice is a 2....how many times can I expect both dice to be a 2? PS: I'm not setting one die to a 2 or anything like that -- just giving them a typical roll where both dice are random.
If you answered 11....just try it out for yourself.
Yes it should be 11 or 1/6 times but you are rolling real physical dice and you could get zero or 66 of the "hard 4" and we all know that.Quote:
Originally Posted by RS__
More importantly: JUSTIFY how the Wizard can "double count" the die showing a 2 when answering the Original Question.
I don't care what "could" happen. We all know what could happen. I'm asking what is expected. Ok. So your answer is 1/6 (11/66) for this case, yes?Quote:
Originally Posted by Alan Mendelson
That's where you're wrong (again). The "die showing a 2" is NOT being double-counted. Actually, it is being removed, giving us 1/11 NOT 2/12 (or 1/6).Quote:
Originally Posted by Alan Mendelson
This is like the many similar examples/scenarios I've posted....none of which you've even bothered to think about. Let's try this again:
Let's say I have 20 cans on my table. Each can has a 50% chance of being sprite (green) or 50% chance of being coke (red). I ask you to pick a can at random. Before you pick a can, I know you have a 50% chance at picking a red can. Still with me?
Now, let's say before you join me, you're in another room. I remove 3 of the red cans from the table. So now there are 17 cans on the table. If you pick a can at random, what's the chance it'll be red?
Remember -- a can has a 50% chance of being red and 50% chance of being green. This is true of cans in general. However, I have now removed 3 red cans from the table. The chance of you picking a red can, at random, is no longer 50%.
If you don't believe me, you can test this out. You don't need cans, you can use something else. Something you have a lot of.
The reason why this is similar to the 2-dice problem is because the can I'm removing is directly related to the can I want you to pick (or rather, the can that I'm figuring the probability of). In the same way, with the dice problem, I am removing a 2, then trying to figure the probability of the other die being a 2.
Now, let's look at another similar but not identical scenario.
Let's say all the same stuff is true about the cans. But this time, I still remove the 3 red cans. You come in the room, and I ask you "If you pick a can at random, and you pick either green or red [at random], what is the chance/probability you will choose a can of the same color you've picked?" In this situation, it is [oddly enough] -- 50% chance. This is because the color I'm "searching for" is not directly related to the color I have removed. 50% of the time (you pick color red) your chances of choosing a red can decrease, and the other 50% of the time (when you pick color green) your chances of choosing a green can increase.
Let's say I roll 2 dice, you can't see them. I only will "count" a roll if at least one of the dice is a 2 [and you don't know this]. I ask you "pick a random number between 1 and 6". You have a 1/6 chance of being correct [the number you pick is the value of the other die]. In the can problem, if you chose red, you'd have less than 50% chance of being right. And if you chose green, you'd have a greater than 50% chance of being correct. But the sum is still 50%. This is the same for the dice problem. If you choose a 2, you have less than 1/6 chance of being correct. If you choose a non-2, you'd have greater than 1/6 in being correct.
I am sure that you made an error here -- a mistype, or you misspoke. Because when you remove one die from a two dice problem there are only six faces remaining on the die that remains.Quote:
Originally Posted by RS__
I didn't have to get an advanced degree for this deduction. I just looked at a die that I keep in my top desk drawer.
I love all of your little examples of math problems but the question is specific. I don't really care about cans, or shirts, or even your fancy graphs and charts and "expectations." Because this is a specific question and a specific problem.
Again, I am going to ask you to tell me why you have to "double count" the die with a 2?? Answer.
I already responded, Alan.
It is clear that you are either trolling or a real big fart.
Teaching someone smart is easy. Teaching someone dumb is hard. Teaching someone ignorant is impossible.
And it continues....The math people who cannot impose their will of probability & theory on those who choose to view things and events in all their glorious reality, always resort to vulgarity & name calling because they can't get their way.
That's why wizard's forum constantly uses lengthy threads on subjects heavily laced with theory: it's the SAFE way to bring something up--anonymously and theoretically. If most of them actually went to casinos and played a hundred bucks, they'd pee their pants.
RS___ will be taking a week off for his contribution above.
I have my doubts that RS is in a teaching profession.
I had to commit the crime of showing the BLUE DEUCE twice.Quote:
Originally Posted by indignant99
I had to commit the crime of showing the RED DEUCE twice, too.
http://www.imageurlhost.com/images/1...gnzf29mv6m.jpg
My last post had a graphic. After editing my graphic, the original link (had worked), is now pointing into "the void."
http://www.imageurlhost.com/images/b...ypqlbd0wx7.jpg
I like the "or" graphics. 1/6.
I have tried to improve the upper half of my graphic.
Retain the bottom half - Alan prefers the bottom half.
http://www.imageurlhost.com/images/3...k7wr171qob.jpg
Looks like television circuit schematics?Quote:
Originally Posted by Alan Mendelson
Quote:
Originally Posted by indignant99
Geniuses teach themselves. They have a mind of their own.Quote:
Originally Posted by RS__
When I was in high school, I had to go to a school to get a FCC First Class RadioTelephone Operators License so I could work at radio stations and do double-duty baby-sitting the transmitter and being the DJ or news reporter or sports reporter. I worked my way through high school and college doing that. At school we had to study schematics and memorize charts. I had no idea what I was looking at. But I passed the tests at the FCC on the first try. First the third class, then the second class, and then the first class license. Then one day I am working at a radio station and the transmitter failed and I had to change the "main plate" said the Chief Engineer in a phone call. And I said "what's that?" He talked me through it over the phone. I still have no idea what a main plate is. By the way, that was in 1970.Quote:
Originally Posted by OneHitWonder
Quote:
Originally Posted by OneHitWonder
Gentlemen, thank you. Your comments have helped.Quote:
Originally Posted by Alan Mendelson
I understand where I am prone displaying unclear graphics.
I am trying my best to be clear. (I'm failing.)
Welcome your critique, on a fresh graphic.
http://www.imageurlhost.com/images/8...zleldjdyfa.jpg
Now eliminate the other options on the die that already showed a 2, as unlike a leopard, it does not change its spots.
1 of 6
I don't think I showed extra faces on the die that showed a 2.Quote:
Originally Posted by regnis
Please put a big X or circle slash on "the options" to eliminate.
Alan, now your friend mickeycrimm is bringing up the two dice problem on vpFree, and badmouthing you in the process. For some reason he chose to only give the 1in11 explanation while ignoring why you concluded it was 1in6. He was his usual drunken self of course, and the first responder spotted that right off the bat. But don't forget, this is an "AP" who regularly pulls in $596.97 every day from those well-known loosely goosey Montana machines! (And of course, he also lives with his mommy.....)
I don't read VP Free nor do I care what he writes there.
So the other day on another thread Alan posted a picture of a sequential royal that he hit. He was dealt 4 pieces and drew the fifth. Now let me phrase this for the 1-11's:
I had a deck of cards and dropped four cards. They were the AKQJ of hearts. If I now flip over or select or draw one more card, what are the odds of drawing the royal?
Different situation, regnis ... there aren't two different A of hearts (or K or Q or J or T)..
Not so sure arc---there was one deuce remaining on the other die. Here there is one card remaining for the royal.
Good point regnis because PROBABILITY says the card answer is 1/47 so why can't the answer for dice be 1/6 ??
Yes-unless these cards have the ability to change their faces like the dice that the 1-11's are using. Because I'm waiting for them to argue that we don't know which of the 52 cards was the Ace and therefore it could be any of the 52 x some multiplier or some convoluted answer.
P.S.--1 of 48 because I only dropped 4 cards--there are 48 remaining.
You're right -- you dropped four cards.Quote:
Originally Posted by regnis
And I am still waiting to see a video proving the answer is 1/11 without changing the value of the die that shows a 2. The Wizard on his forum says he is waiting for the 1/6ers to show up with money. I'm waiting for his video.
Because it's a different situation as I explained above. Duh.Quote:
Originally Posted by Alan Mendelson
That's still not how you compute probability. You will never get what you want because it doesn't apply to the question being asked.Quote:
Originally Posted by Alan Mendelson
Arc, you are hopeless. Do you really think this is a question of probability -- or simply a question involving looking at two dice?
For you to use your probability answer you have to do funny things with two dice -- like rotate them to show different numbers. In a court of law that would be called tampering with the evidence.
Both. Here is the question ...Quote:
Originally Posted by Alan Mendelson
"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?"
How difficult is it to see that both are in the original question? I've bolded them to make it obvious
No, you don't do anything to the two dice. The "probability" asks what are the odds across an infinite number of independent throws. You can estimate what this might be by throwing the dice yourself a large number of times. Recording 100-200 throws with a 2 showing will get you close.Quote:
Originally Posted by Alan Mendelson
Why do you keep denying what the question asks? Simple, you are emotionally invested in your incorrect answer.
Arc I am going to defer to redietz who responded to your question in previous posts. He addressed the question and definition of "probability" and the wording of the question.
Frankly -- you can maintain that this involves multiple throws of two dice but I will never see it that way.
By the way, don't ask for multiple chances at a craps table either -- won't happen.
Yup, and I showed that his interpretation was silly.Quote:
Originally Posted by Alan Mendelson
No one is asking for "multiple chances" when they asked about "probability". For example, what is the probability you will be dealt 3 of a kind from a randomly shuffled deck of 52 cards? No one is asking about one hand being dealt. When "probability" is inserted into the question it directly implies that we are looking at a set number of events. Not a single event.Quote:
Originally Posted by Alan Mendelson
Hence to compute the "probability" you look at the number of distinct situations of 3oak and compare that to the total number of possible deals.
http://en.wikipedia.org/wiki/Classic...of_probability
Same applies to the dice question. You look at the total number of distinct situations of a pair of 2s (1) compared to the total number of possibilities that contain at least one 2 (11)
I am glad you wrote this:
Because the dice in a cup question is a single event.Quote:
Originally Posted by arcimede$
No, Alan, the dice cup question ... "What is the probability that both dice are showing a 2?" .... is a probability question. It is no different than my question about 3oak. You are intentionally ignoring the real question because you KNOW that it leads to the correct answer of 1-11.
You cannot compute probability from a single event. Go read the reference I provided you.
Arc here is the original question again from the Wizard's site:
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?
There is no way that I can consider this to be more than a one roll, one time event.
There is no way that I can't "freeze" one die on its face showing a 2 and then consider the six faces of the second die.
Now if in your mind it is something different and you get to change the value showing on a die to accommodate your 1/11 answer, then be my guest.
Sorry Alan, your denial of the definition of "probability", which I provided you above, is silly. There is no such thing as probability (other than 1 or 0) for "one time events"
Here's the same type of question ... You are dealt 5 cards on a VP machine. A friend covers the screen so you can't see the cards. He tells you that the hand contains at least a one deuce. What is the probability you have at least two deuces?
Note the question does not specify which deuce is dealt. It could be hearts, diamonds, clubs or spades. Hence all possible combinations of a pair of deuces is possible. Your approach would limit it to just cases when the deuce is in hearts (or any one of the other suits) and all pairs would have to contain the heart deuce. Of course, that is obviously not what the question asks.
You are making this exact mistake with the dice.
Arci, why do you think the asking for "probability" should override the initial description of a single event? This is not an equation. This is a piece of writing, and one word at the end does not override a bunch of words in the beginning just because you say so. You are not the arbiter of logic, how this piece of writing works, how it should be interpreted, and so on. You have neither the authority nor the credentials to do that just because you understand basic probability theory.
Rarely have I gone head to head with arci over something. In this case, I feel better about my position partly because it's the humble position. It requires no self-designated expertise regarding what an author meant, nor how something should be read or interpreted. Arci assumes a lot here, and even acknowledges that most people may not see it his way. Well, if a piece of writing leads to 1/6 as an answer for most people, and 1/6 is wrong, that is not the readers' fault. Arci, go look up the author of the question and give him/her hell, because that is where the fault lies. It's the writer's responsibility if he leads readers astray.
And actually, yes, there is a definition of "probability" that can refer to single events. IT'S THE PRIMARY DEFINITION, for God's sake. Now everybody please go to their actual dictionary and look it up.
P.S. Arci, you've been semi-obnoxious and I do not know why, but let me get in an academic dig here. If you're using Wikipedia as your reference in a serious discussion, you're in trouble. And you know this as well as I do.
There's several reasons for his obnoxious attitude today, but let's just pick the easy, less obtrusive one: another gambling binge that just didn't work out so well.
This question is analogous to setting one die to a deuce and then asking what is the probability of rolling a deuce with the other die.Quote:
Originally Posted by regnis
It is NOT analogous to the question where we roll two dice until at least one of them is a two—then asking what is the probability both are deuces.
For that analogy to work, you would have to ask:
"I dropped five cards until at least four of the cards are part of a hearts Royal Flush. What is the probability that the fifth card completes the Royal?"
The answer to that question is not 1/47.
It's 1/235.
The single event is just describing the situation (just like I did in my VP example). It is not asking a question. How can one ask a question without some descriptive information? Look at where the question mark appears. The question is obvious. It has nothing to do with me. Is it a little tricky? Sure, but that is only if you don't take the time to parse it carefully.Quote:
Originally Posted by redietz
Use any other source you choose. They will all say the same thing for an obvious math question.Quote:
Originally Posted by redietz
Wow. You 1/11ers have to twist words just like rotating dice to prove your point.Quote:
Originally Posted by arcimede$
I love this. With a 52 card deck the answer is 1/235.Quote:
Originally Posted by Zedd
(By the way, regnis pointed out the answer is 1/48 since only four cards were dropped, not 5.)
People look stuff up on the internet instead of thinking it through from first principles themselves.Quote:
Originally Posted by redietz
I think that the Wizard's "autobiography" is in there too. Funny, how it comes off when you do it for, by, and of, yourself.
Yes, the correct answer to his question is 1/48.Quote:
Originally Posted by Alan Mendelson
I proceed to make a similar mistake with my question—the answer should be 1/236.
And yea, that might seem ridiculous to you. But it is true nevertheless. That’s why this dice/coin/child/card/whatever problem is referred as a veridical paradox.
But if you can’t grasp the 1/11 for the dice problem, I wouldn’t go near this card/Royal Flush variant.
Why is the "dice problem" as written here any sort of a paradox? Is it listed as written in Wikipedia under veridical paradox? How so?Quote:
Originally Posted by Zedd
Again, the problem as written is ambiguous. The problem as written to produce the 1/11 answer is a veridical paradox because the 1/11 answer is very counterintuitive, yet it’s true.Quote:
Originally Posted by OneHitWonder
No twisting at all. Just silly denial on your part. BTW, why didn't you respond to the VP example? Maybe because it makes your position ridiculous?Quote:
Originally Posted by Alan Mendelson
Do you mean this question?Quote:
Originally Posted by arcimede$
Sorry. I ignored it.Quote:
Originally Posted by arcimede$
Of course you did. It destroys any possibility you are right.Quote:
Originally Posted by Alan Mendelson
No, I ignored it because I wanted to.
I didn't ask you for your definition of veridical paradox. I asked about how the problem as written is a paradox of any sort. Ie, what is the counter intuitive other aspect here which isn't part of working the problem one way? The reason that I asked for even a Wikipedia certification from you.Quote:
Originally Posted by Zedd
Saying that "1/6 seems to be the answer because one die is a 2", would merely be not considering the other column or row of the dice-chart given the proper question with the 1/11 chance answer. No paradox to that.
Clearly, you aren't "up to" discussing this. There is a reason that the Wizard failed the Mensa test. And, that no one with degrees in pure math endorses him, let alone the casinos of the companies for which he works, or has worked. Sorry, but Jacobson and Collins aren't independent entities. They have gambling books to sell, or similar web-pages of their own; as well as perennial involvement with the Wizard's.
Indeed, there is a bit of a paradox to the dice questions in general, but you are nowhere near this yet(?). Please don't re-post the Wikipedia conditional lingo on that which AceTwo posted up at the Wizard's. Nobody responded to it there either.
______________________________________________
And again, no, the problem as written is not ambiguous in your way, or even ambiguous enough. You have to be intelligent enough to work with what you are given, and realize what information is necessary to which answer.
Even were it ambiguous, one would have a bad bet that the first roll just happened to be of 2, and just what the "peeker" was looking for.
Me too. People with the right answers don't have to "change the channel".Quote:
Originally Posted by Alan Mendelson
It’s not me giving it the 'paradox' label. I have a strong background in mathematics, so the question is not counterintuitive to me—and is not a paradox in my opinion. But for most people, the question is counterintuitive and therefore it is considered a paradox.Quote:
Originally Posted by OneHitWonder
If you google “Boy Girl Problem” (which is the same problem as this dice question), nearly all the top searches refer to it as a paradox.
But I am unsure why we are debating whether this is a paradox or not…why does it even matter?
And I’m sorry, but your third and fourth paragraphs read as gibberish to me. I have no clue who those people are. And what does any of that have to do with the problem at hand?
The original question is open to multiple interpretations—that’s the definition of being ambiguous. You seem more interested in arguing semantics than mathematics.
if the dice were thrown one at a time the odds would be 1/11 against 2 2 idk if the peeking is the same as them being thrown one at a time but once the first dice is on 2 then its a 1/6. so once both dice are sitting there and one is on 2 the chance of the other die being on 2 is 1/6.
What happens when the first die lands on 6? Chance of getting a 2-2 is 0. But you can still get "at least one die is a 2". Or did you not think that far ahead?Quote:
Originally Posted by champ724
Yup, it's called denial. Common human condition when faced with uncomfortable information.Quote:
Originally Posted by Alan Mendelson
No, there is no longer the "or more is a 2" part about the specific roll in question then. Here, the roll of 6-X.Quote:
Originally Posted by RS__
You must ask for a roll which is an element of the set of rolls which contain the number 2. The only specific roll which can satisfy "at least one die is a 2" is when only the first die is seen, and it's a 2. This allows for the roll to satisfy "at least one die is a 2". The other die could still be another or the more 2.
Quote:
Originally Posted by Zedd
You brought the word paradox into this!Quote:
Originally Posted by Zedd
I asked you a couple of specific questions. No answer.
The reason Wikipedia isn't considered any sort of a scientific standard. Lol.Quote:
Originally Posted by Zedd
I don't know in which branch of mathematics you believe you have any background, but arguing semantics is fundamental to all.Quote:
Originally Posted by Zedd
All I said was that the problem is referred as a paradox.Quote:
Originally Posted by OneHitWonder
What were the couple of specific questions that you had asked in which I did not give an answer?Quote:
Originally Posted by OneHitWonder
Did I say anything about Wikipedia?Quote:
Originally Posted by OneHitWonder
I simply said to google the problem. Just look at the top searches—on the first page (top 9 results…and yes, one of them is Wikipedia but go ahead and ignore it if you wish). I’m seeing 7 of them with the word ‘paradox’ in the main tagline (despite leaving the word out of the search). So I was correct in saying ‘the problem is referred as a paradox’. But again, paradox or not, this has much to do about nothing.
My background is strong enough to be able to answer this simple probability problem. For someone who just wants to argue semantics…you’re not being great at it.Quote:
Originally Posted by OneHitWonder
Zedd's dead.