Two Dice Puzzle: Part Deux
Quote: RS
If the red die is a 4 (the second one), then you're left with the matrix:
....................1-4..............
....................2-4..............
....................3-4..............
....................4-4..............
....................5-4..............
....................6-4..............
Again, only 1 of 6 ways to get a 4-4.
Although, I'm not entirely sure what Wizard means in his last post, or if he mistyped something.
This not true. When the EXPOSED die is a red 4, the 4-4 cell on this grid should get only half the weighting.
Quote: WizardHere is my interpretation of the Dalex64 question:
I maintain the answer is 1/11.
1/11 is the correct answer to that interpretation.
1/6 is the correct answer if the peeker does not randomly pick between the two dice if both are fours. So, if the peeker ALWAYS exposes the red four.
Quote: PeeMcGee1/6 is the correct answer if the peeker does not randomly pick between the two dice if both are fours. So, if the peeker ALWAYS exposes the red four.
I agree, however RS evidently did not interpret the question that way, and I think Dalex meant that the peeker would choose the die to expose randomly if both were a four.
Quote: RSIf the red die is a 4, there is a 1/6 chance the green die is a 4.
If the green die is a 4, there is a 1/6 chance the red die is a 4.
If either die (ie: color unknown) is a 4, then there is a 1/11 chance the other die (or both dice) is/are a 4.
I verified that with a simulation.
I believe in the variation where the dealer will select a red 4 when given the choice of a red 4 and a green 4, whenever a green 4 is displayed the chance that both dice are a 4 is 0, because the only time a green 4 will be selected is when the red die is NOT a 4.
So I can see now that this version of the problem won't work in the way that I wanted it to work.
The result is that two independent events become conditionally dependent (negatively dependent) given that at least one of them occurs. Symbolically:
if 0 < P(A) < 1 and 0 < P(B) < 1 (each event may or may not occur),and P(A|B) = P(A) (they are independent),then P(A|B,C) < P(A|C) where C = AuB (i.e. A or B) (A is less likely to occur given that A or B occurs and given that B occurs, than if simply A or B occurs).
In words, given two independent events, if you only consider outcomes where at least one occurs, then they become negatively dependent.
http://en.wikipedia.org/wiki/Berkson%27s_paradox
Quote: PeeMcGee1/11 is the correct answer to that interpretation.
1/6 is the correct answer if the peeker does not randomly pick between the two dice if both are fours. So, if the peeker ALWAYS exposes the red four.
So does the exposer show a green-4 when he must, by virtue of no red-4 existing to even be exposed? If yes, there are still 11 exposure events that could be revealed.
Quote: WizardI agree, however RS evidently did not interpret the question that way, and I think Dalex meant that the peeker would choose the die to expose randomly if both were a four.Quote: PeeMcGee1/6 is the correct answer if the peeker does not randomly pick between the two dice if both are fours. So, if the peeker ALWAYS exposes the red four.
I don't think exposer bias has any impact on the 1/11 probability. Unless the bias is profound to the point of failing to expose a green-4 when it's the only 4 available ... ever ... upon occasion.
Quote: WizardThis not true. When the EXPOSED die is a red 4, the 4-4 cell on this grid should get only half the weighting.
IF you're only looking for an exposed red-4, then the 4-4 should hold its equal weighting of 1/6. ie: A green-4 and a red-non-4 does not meet the criteria.
IF you're looking for "at least one 4", then I agree, it's 1/11....at least I think. Brain a little fuzzy right now.
When both dice are 4, one of them is chosen randomly to show you.
It was my belief that knowing the color would give you more useful information, but it seems that it does not.
Quote: indignant99So does the exposer show a green-4 when he must, by virtue of no red-4 existing to even be exposed? If yes, there are still 11 exposure events that could be revealed.
He can. But then the probability that the red die is a four is 0.
Quote: PeeMcGeeQuote: indignant99So does the exposer show a green-4 when he must, by virtue of no red-4 existing to even be exposed? If yes, there are still 11 exposure events that could be revealed.
He can. But then the probability that the red die is a four is 0.
Yeah! On the 5 eventualities of Green-4 without Red-4.
But those Five eventualities are still exposed, along with the Six that do contain a Red-4.
11 exposures... only 1 is a pair of Fours.
It doesn't make a damn bit of difference, the exposer's bias, when he's exposing either/or die upon pair-of-fours.
Quote: indignant99Quote: PeeMcGeeQuote: indignant99So does the exposer show a green-4 when he must, by virtue of no red-4 existing to even be exposed? If yes, there are still 11 exposure events that could be revealed.
He can. But then the probability that the red die is a four is 0.
Yeah! On the 5 eventualities of Green-4 without Red-4.
But those Five eventualities are still exposed, along with the Six that do contain a Red-4.
11 exposures... only 1 is a pair of Fours.
It doesn't make a damn bit of difference, the exposer's bias, when he's exposing either/or die upon pair-of-fours.
Yea, there are 6 exposures that are red. One of those is a pair. So 1/6.
The other 5 exposures are green. None are pairs. So 0.
So, correct, there are 11 exposures, but only 6 with an opportunity for a pair.
Quote: PeeMcGeeQuote: indignant99Quote: PeeMcGeeQuote: indignant99So does the exposer show a green-4 when he must, by virtue of no red-4 existing to even be exposed? If yes, there are still 11 exposure events that could be revealed.
He can. But then the probability that the red die is a four is 0.
Yeah! On the 5 eventualities of Green-4 without Red-4.
But those Five eventualities are still exposed, along with the Six that do contain a Red-4.
11 exposures... only 1 is a pair of Fours.
It doesn't make a damn bit of difference, the exposer's bias, when he's exposing either/or die upon pair-of-fours.
Yea, there are 6 exposures that are red. One of those is a pair. So 1/6.
The other 5 exposures are green. None are pairs. So 0.
So, correct, there are 11 exposures, but only 6 with an opportunity for a pair.
Okay... 11 exposures. How do you get 1-in-6, even with exposer's red-bias?
Quote: indignant99Quote: PeeMcGeeQuote: indignant99Quote: PeeMcGeeQuote: indignant99So does the exposer show a green-4 when he must, by virtue of no red-4 existing to even be exposed? If yes, there are still 11 exposure events that could be revealed.
He can. But then the probability that the red die is a four is 0.
Yeah! On the 5 eventualities of Green-4 without Red-4.
But those Five eventualities are still exposed, along with the Six that do contain a Red-4.
11 exposures... only 1 is a pair of Fours.
It doesn't make a damn bit of difference, the exposer's bias, when he's exposing either/or die upon pair-of-fours.
Yea, there are 6 exposures that are red. One of those is a pair. So 1/6.
The other 5 exposures are green. None are pairs. So 0.
So, correct, there are 11 exposures, but only 6 with an opportunity for a pair.
Okay... 11 exposures. How do you get 1-in-6, even with exposer's red-bias?
1 out of 11 exposures will be a pair of fours. But when you see the color of the die you gain additional information. If you see red, then it’s 1/6. If you see green, then it’s 0.
Quote: indignant99Okay... 11 exposures. How do you get 1-in-6, even with exposer's red-bias?
To clarify, here is the latest variant to the "two-dice problem":
Quote:A red and a green die are rolled. A peeker will check for at least on four. If there is exactly one four, then he will expose it. If both are a four, he will expose the red die. Question -- When a red four is exposed, what is the probability the green die is also a four?
Using Bayesian analysis:
Pr(both dice a four given red four exposed) = Pr(both dice a four)/Pr(red four exposed) = (1/36) / (1/6) = 1/6.
Quote: PeeMcGee1 out of 11 exposures will be a pair of fours. But when you see the color of the die you gain additional information. If you see red, then it’s 1/6. If you see green, then it’s 0.
But I, as the assessor of probability, DO NOT KNOW the exposer's bias.
I see 11 exposures, none the wiser that the 5 green-4's are losers.
Nor, that the 6 red-4's enjoy a 1-in-6 probability of showing pair.
I am stuck with 11 exposures, only one of which is pair-of-fours.
Quote: indignant99But I, as the assessor of probability, DO NOT KNOW the exposer's bias.
I see 11 exposures, none the wiser that the 5 green-4's are losers.
Nor, that the 6 red-4's enjoy a 1-in-6 probability of showing pair.
I am stuck with 11 exposures, only one of which is pair-of-fours.
If you do not know the exposer’s bias, then you’re simply back at the original question with 1 in 11 probability…correct.
It’s just like the Monty Hall problem—if you know the host will always reveal the goat, then you should switch.
Quote: PeeMcGeeIt’s just like the Monty Hall problem—if you know the host will always reveal the goat, then you should switch.
You're absolutely right. If there is a peeker or host in any probability question, then his behavior should be clearly spelled out or expect thousands of posts arguing about the interpretation of the question.
BTW, I'm still waiting for anybody in the 1/6 camp to come to Vegas with their money and prove me wrong.
Quote: WizardBTW, I'm still waiting for anybody in the 1/6 camp to come to Vegas with their money and prove me wrong.
I think there is only about four or five of them and and all they can do is bad mouth the experts.
And of course none of them will put their money where their mouths are.
Quote:If both are a four, he will expose the red die.
I missed this. If I know the Exposer's iron-clad Exposure Rules (or indeed I am the Exposer), then I know:
- An exposed green-4 cannot lead to a pair-of-fours. (Haunting question: How would anyone rule in/out this eventuality - exposed green-4 - before any exposure took place?)
- An exposed red-4 enjoys a 1-in-6 probability of grabbing/snapping onto the other 4, indeed a green-4, to achieve a Pair-of-Fours. (Haunting question: How would anyone rule in/out this eventuality - exposed red-4 - before any exposure took place?)
Quote: IbeatyouracesI think there is only about four or five of them and and all they can do is bad mouth the experts.
Of those supporting Alan on his own forum, there are some who don't even fall into the 1/6 camp ( E.g. Singer) They just want to use it as a springboard for a bit of posturing. According to Rob, the contagion has now moved on to another forum where Alan is going to become the target of some mockery and badmouthing of his own. Oh. hum. it'll keep him amused. Who would have thought that ancient post of the original dice question could stir up such a hornets' nest.
Quote:And of course none of them will put their money where their mouths are.
Well I've profited from a side bet ;o)
I still have a pension fund to invest if the 1/6thers are confident. The weather in Blighty is sunny and dry.
Quote: OnceDearOf those supporting Alan on his own forum, there are some who don't even fall into the 1/6 camp ( E.g. Singer) They just want to use it as a springboard for a bit of posturing. According to Rob, the contagion has now moved on to another forum where Alan is going to become the target of some mockery and badmouthing of his own. Oh. hum. it'll keep him amused. Who would have thought that ancient post of the original dice question could stir up such a hornets' nest.
Well I've profited from a side bet ;o)
I still have a pension fund to invest if the 1/6thers are confident. The weather in Blighty is sunny and dry.
I get the feeling that the only reason Singer's ban was lifted was to help troll the forum there. I've said it before and I'll say it again. He has absolutely ZERO CREDIBILITY in the gaming/gambling world and anyone who even listens to him needs a psychiatrist. Why anyone gives him a place to spew his trash is beyond me. And honestly, it makes Alan's credibility go even lower.
Good that you've made money on this!
Quote: IbeatyouracesI think there is only about four or five of them and and all they can do is bad mouth the experts.
And of course none of them will put their money where their mouths are.
Edit: Alan is still waiting on a video. Of course it won't matter because he'll dismiss that too.
Quote: IbeatyouracesEdit: Alan is still waiting on a video. Of course it won't matter because he'll dismiss that too.
I actually have a script for that video. Meets all his criteria.
Only two dice featured. No dice get lifted or spun around.
Only one throw of the dice, in total.
At least one of the dice is a two.
Not one, but both dice are 'Set' in their landing position.
And still the probability is 1/11.
But this video will be a tickets only, paid viewing. If he wants content for his site, it will cost him $100.
If he prefers to be un-enlightened, that's fine with me.
Quote: OnceDearI actually have a script for that video. Meets all his criteria.
Only two dice featured. No dice get lifted or spun around.
Only one throw of the dice, in total.
At least one of the dice is a two.
Not one, but both dice are 'Set' in their landing position.
And still the probability is 1/11.
But this video will be a tickets only, paid viewing. If he wants content for his site, it will cost him $100.
If he prefers to be un-enlightened, that's fine with me.
They keep thinking it's a once die problem. It's not! People, there are 12 TOTAL faces. Only one face gets eliminated. Not the whole Damn die! There are 11 faces left. The damn answer is 1 in 11. How GD hard is that to understand???
Quote: IbeatyouracesHow GD hard is that to understand???
Easy to understand. Impossible to persuade a Donkey. (Any donkey, no names mentioned)
I'm still happy to take bets in my 'Educate The Donkey Derby'
Set the scene...
A Donkey (Any donkey),
Two Furry Dice,
A highly numerate and literate array of trainers (teachers, not shoes),
Unlimited training resources including but not limited to...
Whiteboard,
Big screen computer,
Video equipment,
Carrots,
Sticks,
45 revolver.
You name it, every training aid conceivable.
Objective for the trainers is to teach the Donkey how to roll the two furry dice (Not just one).
Objective for the Donkey is to fail.
My money is on Donkey.
Quote: RSI'm still trying to figure out why indignant is feeding the trolls over there .
Many of us made the same mistake.
Quote: RSI'm still trying to figure out why indignant is feeding the trolls over there (not Alan, he's not a troll, since he's a member here) on Alan's forum, with his pictures and whatnot....
I like the comment about the diagrams looking like electrical ones.
BTW Alan, nice royal! More proof stop losses/win goals don't help. $67 for $5 worth of food though, not worth it :-)
Quote: IbeatyouracesI like the comment about the diagrams looking like electrical ones.
BTW Alan, nice royal! More proof stop losses/win goals don't help. $67 for $5 worth of food though, not worth it :-)
And he's still down a few $k for the year?
#StopLossesDontWork
Quote: RSI'm still trying to figure out why indignant is feeding the trolls over there...
There is truly only one troll over there and we all know who it is.
You can't eliminate the whole die because you don't know which one it is. Get it through your head. I'm not a math guy and I can clearly see this. Also, craps dealers, or most any dealers don't know squat about odds and strategy. They'd be the last to ask.
If you can't understand conditional probability, keep being the losing gambler you'll always be.
Quote: IbeatyouracesThey keep thinking it's a once die problem. It's not! People, there are 12 TOTAL faces. Only one face gets eliminated. Not the whole Damn die! There are 11 faces left. The damn answer is 1 in 11. How GD hard is that to understand???
Want proof Alan is a flat-out LIAR? Look.
Quote:I have to laugh... after quoting Ibeatyouraces here he removed this post from the WOV forum (his typo remains):
They keep thinking it's a once die problem. It's not! People, there are 12 TOTAL faces. Only one face gets eliminated. Not the whole Damn die! There are 11 faces left. The damn answer is 1 in 11. How GD hard is that to understand???
Do ya think he realized how ridiculous it was?
Here was my original response:
Quote Originally Posted by Alan Mendelson View Post
Would someone please clue them in that when you have one face showing on a die that the other five faces on that same die cannot show or be used? To quote the post on the WOV forum, "how GD hard is that to understand???"
I guess it is: "(Alan) simply can't get past a die having six sides." -- Michael Shackleford May 12, 2015
Alan Mendelson
www.AlanBestBuys.com
"(Alan) simply can't get past a die having six sides." -- Michael Shackleford May 12, 2015
Quote: indignant99Want proof Alan is a flat-out LIAR? ... In case Fraud Mendelson expunges his own post, ...
Personal insult -- third offense -- 14 days.
What's wrong with calling someone out if they are in fact a liar/fraud?
IMO calling someone a liar VS a fraud is different.Quote: RSI've never understood this "personal insult" offense/rule.
What's wrong with calling someone out if they are in fact a liar/fraud?PS: I'm not saying Alan is a liar/fraud nor am I saying he isn't.
I don't think calling someone out on a lie should be suspension worthy. Probably best not to call them a liar. Perhaps say something like... IMO that's a lie.
Quote: RSI've never understood this "personal insult" offense/rule.
What's wrong with calling someone out if they are in fact a liar/fraud?
My position is that if it sounds like an insult, then I throw the flag. It is not required of the administrators to conduct an investigation with every perceived insult to determine whether it is an opinion or fact. We pretty much assume it is always an opinion.
It is especially not incumbent on the administrators to investigate other sites involved in the perceived insult.
Quote: WizardMy position is that if it sounds like an insult, then I throw the flag. It is not required of the administrators to conduct an investigation with every perceived insult to determine whether it is an opinion or fact. We pretty much assume it is always an opinion.
It is especially not incumbent on the administrators to investigate other sites involved in the perceived insult.
Yeah, I'm very not-fond of cross-forum arguments, myself. This thread has gone 74 pages now, and Alan stopped posting in it, what, 20-some pages back? Ugghhh...
Quote: beachbumbabsYeah, I'm very not-fond of cross-forum arguments, myself. This thread has gone 74 pages now, and Alan stopped posting in it, what, 20-some pages back? Ugghhh...
It would be nice to discuss this problem without all the "Ugghhh..." I will try again, but it's too bad when people summarily call it nonsense, and then make up their own interpretations for it in the process.
I have found Alan's forum not nearly so "argumentative". He doesn't "have anything to prove" over there, especially not mathematically.
"It ain't what you don't know that gets you into trouble. It's what you know for sure that just ain't so."
- Mark Twain
Quote: KerkebetIt would be nice to discuss this problem without all the "Ugghhh..." I will try again, but it's too bad when people summarily call it nonsense, and then make up their own interpretations for it in the process.
I have found Alan's forum not nearly so "argumentative". He doesn't "have anything to prove" over there, especially not mathematically.
"It ain't what you don't know that gets you into trouble. It's what you know for sure that just ain't so."
- Mark Twain
The "Ugghhh" was in reference to Alan having left this forum to argue with itself over what he's saying over there and forum members bringing back selective bits to discuss or continue arguing over. Subject matter is almost irrelevant. If the part of the discussion he was involved in here has ended for him, why keep bringing him up? From the viewpoint of THIS forum, it's undesirable at best. Same thing when people bring arguments here they're having on a dice forum, baccarat forum, or other gambling forum. We don't need to host outside bits of conversation. There's no value in discussing it, or retaining a thread that isn't self-contained.
Quote: WizardWhat happened to all the camp 1/6 members? Is there a Flat Earth Society meeting keeping them busy?
I'm not in the flat earth society but do believe the earth is relatively flat. I feel insulted.
And with Alan you never will. THAT'S THE FRIKKIN POINT!!!!! (sorry for yelling.)Quote: WizardI still can't escape the feeling we never had proper closure on this. It would have been nice to have at least won a lunch from a 1/6 Society member.
Quote: pewAnd with Alan you never will. THAT'S THE FRIKKIN POINT!!!!! (sorry for yelling.)
On the contrary. On Alan's own forum, he has conceded the wager, though of course he never conceded the argument. He also put $50 value on that lunch. By my reckoning, if he comes to Vegas without offering an unconditional $50 or $50 lunch to Mike, then he would become a welcher.
And that is all absolutely regardless of arguments or wagers about dice.
Quote: Alan on his own forumThis is not a rolling of two dice problem. It is separate from the bet with the Wizard. Frankly, I concede the bet because the way the Wizard and I agreed to our own bet was a limited number of rolls.
Only the bolding is mine.Quote: Alan on his own forumThat won't resolve anything because anything can happen when two dice are rolled. I know that and you know that. In fact, since the bet with the Wizard that we decided on has a very limited number of rolls -- and only a $50 lunch was at stake -- I've already conceded defeat.
