How many times have you quoted from WOV and posted it here and vice versa?Quote:
Originally Posted by Alan Mendelson
LolQuote:
Originally Posted by the troll
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How many times have you quoted from WOV and posted it here and vice versa?Quote:
Originally Posted by Alan Mendelson
LolQuote:
Originally Posted by the troll
Any why are they doing that? To show you what could happen on other throws. No one is saying the dice are being changed on any given thrownQuote:
Originally Posted by Alan Mendelson
They are simply providing you what you asked for. You wouldn't accept simple and obvious explanations. You demanded a video. So, they provided you with one. Their other option, rather than showing you the possible results by switching the die, was to create a video hours long that eventually produced the results by chance. I doubt very much you would have had the patience to watch it. It sound to me like they tried to make it as easy as possible for you.Quote:
Originally Posted by Alan Mendelson
I understand, but that does not change the question that was asked. You were told it was a trick question long ago but continued skim over our comments and argue against all of us that tried to help you understand.Quote:
Originally Posted by Alan Mendelson
If you think I'm being a jerk, you are right. It is what you earned for not taking the time to read and understand the comments of those trying to help you.
omg...Quote:
Originally Posted by Alan Mendelson
Please try and understand this:
When you say "it is 1/6 because the other die has 6 faces" you are talking probability. Probability however is not defined after the roll. At that point all you have in front of you is the outcome, the manifestation of a probability for a throw of pair of dice. One particular face for each die. No probabilities are legal at that point because you have a fact you are staring at. Odds for the faces you're looking at to be the faces you're looking at - P=1. Fact.
You measure probability upfront/derive it from all possible outcomes. How do you know the odds for a 2 to come out on a single die is 1/6? Because you've defined/measured/concluded that the 2 face is only one and there are 6 faces total. You need to know what the odds are in order to conclude that what just happened
(the actual roll) was 1 in x to happen. You don't come after the event and say "well, now I only see 6 faces for this second die, so I'm going to say it is 1/6".
You are answering a totally different question here.
Kewl, you seem to be grasping some of the problems with the trickiness of the question as written. In the question, you start with the presentation of an event, which may or may not be one of multiple events, and which can easily be read as having been a past event. Then you have a question about probability, which would seem to require future event(s), even though it has not been made clear that the description that starts the paragraph is of future or multiple events. Many, if not most, people would read the opening sentences as describing a single event in the past.
No, there is no need to be written about dice combinations at all. The question is perfectly valid as it is. There is one ambiguity, but it was cleared up by you and others (whether the peeker says something if there is no deuce).Quote:
Originally Posted by Alan Mendelson
Ok, answer this question then:
"What is the probability of a fair throw of two standard dice showing 2-2 given that at least one of the dice shows 2?"
Still 1/6 in your head, right?
So there you go. The wording is not the issue, cause you'd answer a perfectly worded question, no humans, no peekers, the same way.
To summarize the 325 threads:
YAWN!!!!!!!!!!!
Here is another way to try and point why we need to consider 11 dice combinations.
Say we play a heads up texas hold'em deep stacks. We have pocket aces and after five-bet action pre-flop called by our opponent, the flop comes 2d-Ks-2h.
Given the pre-flop action we can safely eliminate any deuce in our opponent hand and are now wondering what are the odds we are beat at this point.
Well, the only hand that beats us is pocket kings. How many ways are there that give him a kings full?
There are three kings left in the deck, so there are three different ways he can be dealt KK. Kh Kd, Kh Kc, Kd Kc. So, we need to account three ways out of 1326 total ways to calculate the probability we're beat.
And how many ways are there in a fair throw of dice to roll 2-x? Eleven,right?
So we throw the dice and when the peeker announces "there is at least one deuce under the cup" we know that one of those 11 combos has just happened.
Given that information,the odds for exactly 2-2 to be the combo under the cup are 1 out of eleven.
It makes absolutely no sense whatsoever to try and determine probability at this point, saying: "But the "other die" has 6 faces,right?". The fact that the "other die" or any die has 6 faces is relevant when we determine the probability before we throw a pair of dice. After the roll we can not possibly examine just one die and make any conclusions.
I'm afraid this will go on as long as it's allowed to go on regnis. I learned long ago that they will not stop until they can impose their beliefs & will upon anyone who disagrees, no matter how clear and valid the disagreeing facts. It's more than a sickness to them--it defines their lives. When you look back and realize these math people spent all that time as young loners, it only makes sense that they'd stay stubborn until the bitter end, where they go out kicking and crying for their mommies (or mummies for mydear :)).Quote:
Originally Posted by regnis
Let me wrap it up this way: you have one die showing a 2. The other die has a 1/6 chance of also being a 2. That's it.
Even if one die is red and one is blue??
Even if the wind is out of the northwest?
Even in a month with an "R" in it??
Even if they're Schrodinger's dice?
Quote:
Originally Posted by regnis
Quote:
Originally Posted by regnis
Even if the Wizard has a pretty picture of the 36 combinations that two dice can show. And even if our friend Arc wants to take credit for being the first to say it was a trick or tricky or confusing question.Quote:
Originally Posted by regnis
You can wrap it anyway you want, not that I care.Quote:
Originally Posted by Alan Mendelson
If you bet on it though, you'll lose money and that's what makes it real and proves you wrong.
This is what makes these guys so special. First, he DEFINITELY cares. Next, his assertion that if you bet on it means you'll lose, is not bound by any of the parameters that makes a nerd a nerd. How many rolls? At 9:1 vs. 11:1, who's to say after 100 rolls the neurotic math geek doesn't lose?Quote:
Originally Posted by kewl
You see kewl, it's all in the wording. Even the theory.
"It's not whether it was a 'good' bet; it's whether you win or lose". OBVIOUSLY!
If I have to draw a perfect scheme with dots, commas ,conditions etc every time I want to say something on a forum, just to satisfy trolls like you and language nerds, I'd deprive myself from all the fun.Quote:
Originally Posted by Rob.Singer
Whoever wants to understand will get the point easily. Whoever wants to troll and whine about commas will do it anyway.
Have fun.
I'd rather have the best of both worlds.Quote:
Originally Posted by Rob.Singer
In all seriousness, I don't care how anyone else plays. The only time I ever want someone to lose us when they are occupying a seat at a table game that we need.
Same goes for the dice problem. The way Alan reads it, he says it's 1/6, so be it.
Evidently language is less important than math. Who knew? I learn something new every day. Thanks, WoV.
All Alan really needs to know is that the bet with the Wizard is a sucker bet and should be avoided!
We've got to remember .... Alan is the same person that thinks if you set a stop loss playing VP, you will do better the next time you play. Some folks are immune to logic no matter how simple it is.
You can lead a horse to water but you can't make him drink. Whatever floats his boat.
No, that is not what I said. Geesh.Quote:
Originally Posted by arcimede$
Yes, you did. You just didn't realize that was the logical result of your approach.Quote:
Originally Posted by Alan Mendelson
My own variation on this is .... you can lead a man to knowledge but you can't make him think.Quote:
Originally Posted by jbjb
Arci has come out of his shell with a couple royals and these new WOV math geeks that think like him.
Maybe it's time for me to have another million hand VP simulation done that comes nowhere near the expected results.
That might push him over the edge.
In the real world red, it's completely opposite. I spent my working career keeping my engineers & accountants from talking to our customers without a Business Development Manager present, and even then I always had them rehearse thoroughly. The math people meant well, but most of the time they would either spill the beans or confuse the customers by coming in from an entirely different angle than what was strategized for. They were never THAT on-board with how and why the customers needed to hear something this way instead of that.Quote:
Originally Posted by redietz
This may be my favorite quote of all time. My ex-girlfriend, who is a fine editor, let me know what a comma-overusing, tense-misusing, participle-dangling mess I am, and now, somehow, I don't feel so bad. "Language nerd" sounds so cool.Quote:
Originally Posted by kewl
I think Kewl meant "deprive myself OF all the fun, by the way.
lol, maybe I should've made it clear some time earlier in the thread, that I'm not American, nor British, nor am I from a country where English is used massively.Quote:
Originally Posted by redietz
I've studied English basics as obligatory foreign language in high school for three seasons and that was it.
For someone who's native language is very different from English I'm pretty satisfied with my level - it gets my point across like 90% of the time w/o the need to rephrase or correct too much.
If I knew you guys here, are so strict on phrasing I'd have used dictionaries and grammar guides for every post just to make sure you won't get confused by my sentence structural genius :)
peace
Does this forum not show end quotation marks?Quote:
Originally Posted by redietz
You've gotta pay extra for those. This is America.
P.S. Kewl, my apologies. If not your native language, it can be tough. The ex-girlfriend does a lot of dissertation editing for international students, so if you ever need a stellar English edit for an academic paper or resume, PM me and I'll put you in touch.
But on my Best Buys TV Show I can get it for you wholesale !!Quote:
Originally Posted by redietz
Indignant from the WOV forum, since you are still reading this forum I want you to know that I am still reading your nonsense on the WOV forum.
Here is one of your recent posts -- and are you really unable to see that you are using multiple rolls of two dice to come up with the 1/11 answer? Look, here's what you posted (in bold):
You can bet your sweet *** my video will start with both dice showing DEUCE :
The first half of the video will keep the right-hand side die (blue) frozen as a Frozen Blue Deuce. Then the left-hand side die (red), starting as a 2, will rotate to all the other red faces: 1, 3, 4, 5, 6.
The second half of the video will keep the left-hand side die (red) frozen as a Frozen Red Deuce. Then the right-hand side die (blue), starting as a 2, will rotate to all the other blue faces: 1, 3, 4, 5, 6.
There will be no "Mexican Jumping Deuces."
What you and the others in the 1/11 camp fail to acknowledge is that you can't figure the odds TWICE the way you want to in your video -- as described above. Yes, when you figure the odds TWICE by saying first one die is the 2, and then the other die is the 2, you do get the answer 1/11.
But when you roll the dice ONCE and only once the answer is 1/6.
Allow me to use your "first roll" which you describe as the first half of your video:
The first half of the video will keep the right-hand side die (blue) frozen as a Frozen Blue Deuce. Then the left-hand side die (red), starting as a 2, will rotate to all the other red faces: 1, 3, 4, 5, 6.
There is the answer to the question-- in that ONE roll.
There is no need to ask the question a second time -- and to come up with the 1/11 answer you have to ask the question a second time AND COMBINE THE TWO ANSWERS.
Remember: there are only 6 sides on a die. The Wizard knows.
Alan Alan Alan. You have to quit comparing "setting a die on 2" to "rolling both dice and waiting until AT LEAST one die is a 2."
In your case, it's one in six. In the latter, it's one in eleven.
No No No whatever your real name is. The original question is the same thing as setting one of two dice as a 2 and then examining the six sides on the other die. And I know you don't see it that way. It's just mind boggling why you don't?Quote:
Originally Posted by jbjb
No it's not the same and nowhere close.Quote:
Originally Posted by Alan Mendelson
You're letting the "cup" & "peeker" interfere with the correct thinking. Those two words is what makes this a trick question.
When you set the die to a two, you eliminate the possibility of the other five numbers showing up. When you roll it, you don't eliminate them and that's where you get mixed up.
So tell us how rolling both in the open is different from doing it under a cup?
The Wizard's can't read; Alan's can't compute. Put them together, and you're left with people who can do neither but in an amusing manner. Television reporters, and gamblers who only talk about it.
Oh my! The bet will never happen. Good.
Setting the die as a two is the same thing as being told at least one of the two dice under the cup is a two. Why is that so difficult for you to understand?
There are only two friggin' dice. Just use your imagination. If you're told at least one of them is a two pick a die and make it two. It doesn't change. That leaves the other die in the two dice problem to be your variable. That other friggin' die has six sides.
No one said you had to combine the six sides on both dice and there is no reason to combine the sides on both dice WHEN ONE OF THEM is a 2. And if BOTH happen to be a 2, the solution to the problem is still the same: you pick ONE of the dice and count how many friggin' sides it has.
Geesh.
Alan,
In your video, you slammed two dice under a cup. I can't recall the exact outcome, but there was a deuce and you held that die and pointed to it. It was fixed (in your mind) as a deuce.
You then proceeded to rotate the other die in your hand, presumably to show the 6 faces. Correct so far? What does that actually mean? What has having 6 faces got to do with probability?
Why did you rotate that dice? Was it to maybe show the equally probable ways that it could have landed?
But it had landed! It either was a deuce or it wasn't. We can agree on that can't we?
If it was a deuce, then the other 5 faces could not flip face up? If it wasn't a deuce, then again it wasn't going to magically fip around to become a deuce.
Probability of being a deuce while you were looking at it was either 100% or zero, depending on where it had landed.
The only opportunity that it had to land on anything different would be to roll again. We all know that both dice get rolled at the same time.
Wouldn't that roll give both dice a new chance to land however they wanted to. Thus giving the possibility that next time they would land you might have the other die being the one with the deuce?
Nothing flips, no sleight of hand.
Now... JbJb asks an interesting question. Peeker or no peeker, are the odds that the wizard offered favourable to him or to the player? That's what THIS thread is about.
If your web site promotes 'Best Buys' then are you doing a pretty rubbish job if you don't recommend and promote a deal where Wizard is offering to pay $9 for something which has $6 of value? With unlimited purchasing money available from Wiz or from me.
No-one has considered it a worthy deal yet?
I can "compute." And the answer 1/11 is the correct answer to A DIFFERENT QUESTION.Quote:
Originally Posted by OneHitWonder
Unfortunately those who think the answer is 1/11 to THIS QUESTION don't understand what they are reading.
The Wizard's can't read the 1/11; Alan's can't compute the 1/6.
Both are right, but for the wrong reasons.
I am starting to think that you guys really have a problem with simple reasoning.Quote:
Originally Posted by OnceDear
I set aside the die that landed on a 2 because the original problem told me at least one die was a two. That die now becomes the "constant" in our problem. That leaves the other die. I picked up the other die and rotated it to show it had only six sides -- and not eleven sides.
There was no reason to pick up the die with the two because that die was our "constant."
And even if both dice showed a two, you would use the same procedure. Set one of the two dice showing a 2 as your "constant" and use the second die as your variable.
This is not high math and probability -- this is common sense.
You forgot to acknowledge or answer this bit Alan.
If your web site promotes 'Best Buys' then are you doing a pretty rubbish job if you don't recommend and promote a deal where Wizard is offering to pay $9 for something which has $6 of value? With unlimited purchasing money available from Wiz or from me.
No-one has considered it a worthy deal yet?[/QUOTE]
[/QUOTE]Quote:
Originally Posted by OnceDear
He doesn't promote false advertising dear. WoVers have proven never to walk the walk.
OnceDear I can answer that: I don't promote gambling. I don't object to sharing information about gambling but I would never suggest that anyone make a bet. Nor do I recommend stocks -- and never did in all the years I was a business news reporter at CBS or the local stations I worked at. Nor did I ever recommend that anyone invest in gold, though I reported on the gold market for decades and I was the reporter who broke the news that the Hunt Brothers failed to meet their margin call on silver which caused the metals market to tumble after its meteoric rise in the late 1970s.
You also forgot to answer or acknowledge this bit. It's a nice easy Yes or No. No need to explain or qualify your answer.
Remember from post 1Quote:
Now... JbJb asks an interesting question. Peeker or no peeker, are the odds that the wizard offered favourable to him or to the player? That's what THIS thread is about.
Quote:
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.
Here's an answer for you that you won't like: I think it's a good bet for both sides.
OK. I'll bite.Quote:
Originally Posted by Alan Mendelson
What the hell is that supposed to mean?
I did not say 'GOOD' I said 'FAVOURABLE'.
So you answered a different question once again. And if you say they mean the same thing, I'll call you out.
Then why aren't you betting the hard 4 at craps all day long? Less ways to lose there and get paid the same.Quote:
Originally Posted by Alan Mendelson
Edit that. I just looked at a table and 2-2 pays 7:1. Still better than the 6:1 odds you think it is.
I can answer your questions:
Jbjb I don't bet the hardways and when I do I bet 6 and 8 for one unit only if I have 5x odds.
OnceDear: both sides have a good bet.
Which side do you believe has the favourable bet?Quote:
Originally Posted by Alan Mendelson
To be explicit, which party do you believe, the odds favour? It's a simple enough question. It cannot be both. You cannot be meally mouthed about it and bluster won't cut it.
Unless of course you think 8:1 or 9:1 are equally neutral odds and maybe mean the same as 5:1, 10:1 and 35:1?
Or don't you know yet?
These people know nothing about AP gambling. Typical civilian wanna be's. Have fun!
First of all OnceDear you have to present the exact bet for analysis. The bet I had with the Wizard is not the bet described on this thread. Check the WOV forum for my bet.
Secondly there is a big difference between 9-to-1 and 8-to-1.
Let's start with your own paraphrasing of that bet as per post 5 of THIS THREADQuote:
Originally Posted by Alan Mendelson
Quote:
Originally Posted by Alan Mendelson
Indeed. But since wizard is a generous guy lets work with 9-to-1 as the offer.Quote:
Originally Posted by Alan Mendelson
This hurts my feelings. Probably hurts arci's feelings, too. Or, to paraphrase Captain Kirk to Picard, "We were making money while you were in diapers."Quote:
Originally Posted by jbjb
Alan's never gonna figure it out. Soon he's gonna start saying, "Of course TWO fifths of a pizza is bigger than ONE half of a pizza.....TWO IS BIGGER THAN ONE!"
I want everyone to know that I appreciate the learned contributions of the WOV forum members who have joined us. They add value to the discussions with maturity, insight and experience. No one can accuse them of spamming a message board. They are keeping the high standards of the WOV forum here.
Would one of you who still is on the WOV forum please tell "Indignant" (really a good handle he picked) that he will have to choose which set of results he wants to use when he rolls two dice: either the horizontal results or the vertical results from his chart. You cannot use both to come up with an answer for this problem.
As I mentioned before, the chart can be used to answer a different question.
"Indignant" has responded to my post on the WOV forum. (I am sure he is boring the hell out of everyone there.) He wrote: "Isn't the whole G.D. friggin' chart - all 36 outcomes - in play when rolling two dice?"
Indignant, good buddy, the whole friggin' chart with all 36 outcomes does not apply to the question. One die has already been identified as a 2, which leaves the other die with its 6 faces to be considered.
Granted, this is not a great example of probability theory, but it is common sense.
Indignant99 from the WOV forum has joined our forum but it appears he has not posted yet. He did respond to my post above on the WOV forum and here is what he posted (in bold):
Isn't the whole G.D. friggin' chart - all 36 outcomes - in play when rolling two dice?
Yes, the result itself lies on either the horizontal, or the vertical, or lo-and-behold the intersection. But none of the ELEVEN rolls are ruled out.
But before the cup is removed, and the dice revealed, is the result gonna be one of the horizontal ones? Or one of the vertical ones? Huh, genius?
(The answer is: Yes, it's gonna be either one of the horizontal, or vertical, ones.)
Before cup removal, but after "deuce" announcement, exactly which restricted zone - horizontal vs. vertical - have we been incarcerated in?
The peeker/announcer did not inform "it's vertical," nor "it's horizontal."
What he actually informed, was "it's in the criss-cross."
After the cup is removed to reveal the dice, we do not enter into some restricted fantasy zone (horizontal versus vertical). The outcome is done. It is what it is.
Unfortunately, in the real world you cannot use both the vertical and the horizontal charts. That's good for theory, but in the real world with two physical dice choose either the vertical or the horizontal.
And in the real world, dice do not appear on charts with multiple options for what appears on their faces.
I'm afraid Alan is effectively trolling his own forum at this point. He has to have realized by now he is wrong and his "interpretation" is wrong. He hasn't answered what his interpretation of the question has to do with answering the more simpler question:
"What is the probability of a fair throw of two standard dice showing 2-2 given that at least one of the dice shows 2?"
and by now I just assume he isn't admitting he's wrong, because it will make him feel ashamed or something.
kewl are you really asking this question again?
Have you answered it yet?
I have 6 men wearing 6 different colored shirts. Across the street are 6 women wearing 6 different colored shirts. Each group has one red shirt. If I select the guy with the red shirt from group 1, what are the odds of blindly picking the matching shirt from group 2.
Hint--it ain't 1 of 11.
Really? You guys pride yourselves with good comprehension/writing skills, yet you post an analogue which is completely not what the question asks?Quote:
Originally Posted by regnis
Hint--it ain't asking what are the odds of blindly picking the matching shirt from group 2.
It asks if you blindly pick a person from each group and at least one of them is wearing a red shirt, what are the odds they are both wearing red shirts. Don't you see the difference?
Gee..
regnis has it correct. That is the question. kewl, you are answering a different question with your 1/11 response.
How come? Where did you read this? Where does it say you get to choose a die? You assume that the statement "at least one die is a 2" gives you the right to set one of the dice and ask a question about the other. Your assumption is flawed. And you are looking at this at very wrong angle. An elementary one, a child will do better than this.Quote:
Originally Posted by Alan Mendelson
That is why the proposed bet will lose player's money. That's a fact(given sufficient enough trials, for trolls like RS).
And any amount of denying can't change this fact. The intrinsic probability of a fair throw of two standard dice showing 2-2 given that at least one of the dice shows 2 is 1/11. It is intrinsic probability. Not a game of words, not a trick.
And you continue to fail to understand that once a die shows a 2, it can't change to accommodate the other 5 faces. A 2 is a 2 is a 2. Now ----what are the odds the other die is a 2.Quote:
Originally Posted by kewl
My shirts weren't bleeding madras (that shows some age-huh)---so once it was red it was always red. What are the odds that I pick a red shirt from the other group.
You are blindly picking that first red shirt. You could have picked blue or yellow. Once you did, what are the odds of blindly picking the matching shirt.
Back to the dice---the friend peeked and said one is a 3. What are the odds the other is a 5? Still 1 of 6.
. The answer to that one is 1/6Quote:
Originally Posted by regnis
That's a different question - one that assumes the peeker will announce the value of a random die even if there is no 2 under the cup. The answer to that one is 1/6.
This is not the question asked here though.
It's what they do folks. When they can't understand or accept that the rest of the world sees something in the question that they don't see, some of them turn to frustration and vulgarity, some shout out insults because it makes them feel better, and some just get plain hurt by it period. Why else do you think they frequent the most liberal....most theoretical gaming forum ever?
How is it different if the peeker said one is a 3 instead of a 2. The question is still what is on the other die. That is always going to be 1 of 6.Quote:
Originally Posted by kewl
kewl you are simply making the same mistake that the Wizard and all the others made: you ignored the condition presented in the question. This is why I asked the Wizard and everyone else to do a video so we could see using real, physical dice how you interpret the question and the condition -- and you all made same mistake: you wouldn't accept that you can't change the value of a die.
Go back to the Wizard's video and what does he do? First he shows a die as a 2, and then he goes through the math mumbo-jumbo of changing the value of that die.
Of course you guys make the claim that we don't know which of the two dice is the 2. But it doesn't matter when you have a problem involving only two dice. Of course, you can't see that.
You ask: Where does it say you get to choose a die? I am going to ask you a related question: where does it say you can first have one die showing a 2 and and get the results, and then switch the 2 to another die and get different results and then add the two sets of results?
I am going to ask you to use your common sense for a moment: take two dice and if I tell you one of those two dice is showing a 2 pick one of the dice and set it as a 2. With one die known to be a 2, how many options are there on the other die in this two dice puzzle?
Again, what you are doing is finding the odds with one die being a 2, and then repeating the procedure for finding the odds when the other die is a 2, and then you are adding them together to get 1/11.
Let me rephrase.Quote:
Originally Posted by regnis
If the question was:
You roll a pair of dice and no matter what the outcome is the peeker will announce the value of one of the dice (at random). What are the odds that there will be a: pair, a (announced value) and 1, a (announced value) and 2, a (announced value) and 3...whatever),
then the answer will be 1/6.
And if the question was (in your case with 5 and 3):
You roll a pair of dice. The peeker makes announcement when at least one 3 or one 5 is under the cup. What are the odds that there are exactly 5 and 3 under the cup?
then the answer is 1/11.
Now, compare the second question with the one asked here. Keep in mind that there was discussion about whether the peeker announces a value every time or only if there is a 2 under the cup and it was agreed upon, and Alan concurred, that the case in discussion is the one where the peeker only announces if there is a 2. Otherwise he says nothing.
"What is the probability of a fair throw of two standard dice showing 2-2 given that at least one of the dice shows 2?"Quote:
Originally Posted by Alan Mendelson
Your answer?
If you asked me that at a craps table I would say 1/36 because we know that in craps the chances of throwing 2-2 are 1/36.Quote:
Originally Posted by kewl
If you asked me this: two dice have been rolled and one die came up showing a 2. What are the chances that the other die is showing a 2? I would answer 1/6.
Now to be specific about your question: "What is the probability of a fair throw of two standard dice showing 2-2 given that at least one of the dice shows 2?" As soon as you told me one die is a 2, there is a 1/6 chance that the other die is a 2.
I say we put an end to this exercise in futility and wait for a video. From Alan.
He does exactly like in the proposed bet in this thread's post#1. Throws dice, see if there is a two and counts the number of times where there is 2-x and 2-2.
Does it lots of times. Then see what happens with the ratio.
The ratio will tell us all, truthfully, what are the true odds when we throw a pair of dice and at least one of them shows 2, that they are both 2.
That's it. No more mumbo-jumbo.