Two Dice Puzzle: Part Deux
Quote: RSSounds good to me. Although I'd prefer 5x36 rolls (180) to smooth out the variance a bit. My only main concern is payment -- idk how that stuff works? Like PayPal? Or can someone (wizard) be an escrow holder? (I'd prefer that over all.). And up to how much can I bet? I would flat bet the entire way through.
180 rolls is good for me. But a video crew might think it excessive.
The purpose of this exercise would be education, clarity - not mercenary ruin of some poor schmuck. So the flat bet would be $1 (one dollar), standing to win $9.
When you won $500 of my dollars ($510 in escrow), the session ends early.
Quote: indignant99180 rolls is good for me. But a video crew might think it excessive.
The purpose of this exercise would be education, clarity - not mercenary ruin of some poor schmuck. So the flat bet would be $1 (one dollar), standing to win $9.
When you won $500 of my dollars ($510 in escrow), the session ends early.
OK. I'll make it even easier for Indie.....
We already agreed that the announcer can announce either of the dice as he sees fit.
We already agreed that if it's a pair, Indi would hand over 9 units to the player.
We already agreed that if it's not a pair, the player (me or RS or whoever) would hand 1 unit to Indie.
Indie, RS, please confirm that I have that right before my next post, because I suspect that next post of mine will be a deal breaker.
Quote: OnceDearOK. I'll make it even easier for Indie.....
We already agreed that the announcer can announce either of the dice as he sees fit.
We already agreed that if it's a pair, Indi would hand over 9 units to the player.
We already agreed that if it's not a pair, the player (me or RS or whoever) would hand 1 unit to Indie.
Indie, RS, please confirm that I have that right before my next post, because I suspect that next post of mine will be a deal breaker.
I have the strange feeling that you think any pair wins for you. It does not. The only pair you win on, is a double of the announced pip-count.
Quote: indignant99I have the strange feeling that you think any pair wins for you. It does not. The only pair you win on, is a double of the announced pip-count.
Good feeling indeed.
If ANY pair is the result of the throw, how would you determine what to announce?
If it was, say a pair of deuces, what would you announce?
Quote: indignant99180 rolls is good for me. But a video crew might think it excessive.
The purpose of this exercise would be education, clarity - not mercenary ruin of some poor schmuck. So the flat bet would be $1 (one dollar), standing to win $9.
When you won $500 of my dollars ($510 in escrow), the session ends early.
OK, awesome.
Edit: Let me help you OnceDear. Indi, think about it the opposite way. First of all, doesn't matter if a die is announced or not -- the roll is either an "easy" (non-double) or a "hardway" (a double). Of the 36 ways to roll a pair of dice, 30 are easy -- you would automatically win (since both dice aren't the same). The 6 remaining rolls are hard ways. There is no need for an announcer, but even if there was an announcer, he'd only have one option. It boils down to: What's the chance of rolling a hard way? It's 1/6 - and only one pip-count can be announced. Chance of rolling an easy number? 30/36 (or 5/6) - and it doesn't matter which pip-count is announced.
Quote: OnceDearQuote: indignant99I have the strange feeling that you think any pair wins for you. It does not. The only pair you win on, is a double of the announced pip-count.
Good feeling indeed.
If ANY pair is the result of the throw, how would you determine what to announce?
If it was, say a pair of deuces, what would you announce?
C'mon Indi, this should be a Eureka moment (Sorry RS). Is there any Pair where RS and I could possibly lose?
Quote: OnceDearQuote: OnceDearQuote: indignant99I have the strange feeling that you think any pair wins for you. It does not. The only pair you win on, is a double of the announced pip-count.
Good feeling indeed.
If ANY pair is the result of the throw, how would you determine what to announce?
If it was, say a pair of deuces, what would you announce?
C'mon Indi, this should be a Eureka moment (Sorry RS). Is there any Pair where RS and I could possibly lose?
Maybe Wizard has poor vision and was banking on the fact that Wizard might announce a pip-count incorrectly. ie; Roll is a 3-3 and Wiz says "at least one 5 is showing" or whatever -- then you and I would lose, since the roll is not a double of the announced number: 5-5.
Quote: RS
Maybe Wizard has poor vision and was banking on the fact that Wizard might announce a pip-count incorrectly. ie; Roll is a 3-3 and Wiz says "at least one 5 is showing" or whatever -- then you and I would lose, since the roll is not a double of the announced number: 5-5.
I was coming round to admitting that I'm hard of hearing and that with Indi's proposed wager, I would not even care what is announced, just pop my chips onto the table and pay me fair and square. I have no need to be paying attention to the announcements as long as they are trusted
Quote: OnceDearC'mon Indi, this should be a Eureka moment (Sorry RS). Is there any Pair where RS and I could possibly lose?
It is a Eureka moment. The announcement is moot. Its only possible purpose, or value, is to tell exactly which pair resulted, when a pair indeed occurred. I cannot escape the fact that 6 pairs are possible, versus 30 outcomes which are not pairs.
Quote: indignant99It is a Eureka moment. The announcement is moot. Its only possible purpose, or value, is to tell exactly which pair resulted, when a pair indeed occurred. I cannot escape the fact that 6 pairs are possible, versus 30 outcomes which are not pairs.
Cheers Indie my friend. Have a beer on me.
Now if we could just get certain other naysayers to see the trees for the wood (And I meant it that way around) then I'd be able to get some work done.
RS has forgiven me for wising you up, I hope you'll forgive me for being unrelenting and occasionally brutal.
Assume that all dice manufactured in the world are one of four colors (red, blue, white or black). And there is an equal number of each color. And color does not affect the roll (all dice are fair).
I pick two dice at random and roll them.
At least one of the dice is a 2 and also is Red.
What is the probability that both dice are showing a 2?
Quote: PeeMcGeeIf you guys want to up the difficulty a tad; how about this variant…
Assume that all dice manufactured in the world are one of four colors (red, blue, white or black). And there is an equal number of each color. And color does not affect the roll (all dice are fair).
I pick two dice at random and roll them.
At least one of the dice is a 2 and also is Red.
What is the probability that both dice are showing a 2?
Without any evidence or proof, I'll have a stab...... And it is a guess...
1/11... Discounting fact that some dice have pictures instead of numbers
Quote: PeeMcGeeWhat is the probability that both dice are showing a 2?
My stab... Eliminate all colorings which have no red.
25% of the time, both dice are red.
75% of the time, the second die ain't red.
25% of the time, probability of second deuce is 1/11.
75% of the time, probability of second deuce is 1/6.
(.25 * 1/11) + (.75 * 1/6) = 2/88 + 11/88 = 13/88 = 14.77%
Quote: indignant99
My stab... Eliminate all colorings which have no red.
25% of the time, both dice are red.
75% of the time, the second die ain't red.
25% of the time, probability of second deuce is 1/11.
75% of the time, probability of second deuce is 1/6.
(.25 * 1/11) + (.75 * 1/6) = 2/88 + 11/88 = 13/88 = 14.77%
Hi Indie,
I'm not saying you are wrong, and happy to work this out as a team.
Where do you get
Is there something special about non-red dice?Quote:75% of the time, probability of second deuce is 1/6.
What significance at all is there in dice colour?
If we can eliminate a whole shedload of dice based on their colour, we eliminate all spot values for whatever bunch of dice they are.
Quote: OnceDearI'm not saying you are wrong, and happy to work this out as a team.
Where do you getQuote:75% of the time, probability of second deuce is 1/6.
Is there something special about non-red dice?
What significance at all is there in dice colour?
If we can eliminate a whole shedload of dice based on their colour, we eliminate all spot values for whatever bunch of dice they are.
First off, there must be one red die, perhaps two. Pee's result mandates it.
25% Red-Red
25% Red-Blue
25% Red-White
25% Red-Black 75% of the time, it's a Red plus a Non-Red.
When both are red, either could serve as the viewed/announced deuce.
We're right back to the original 1/11 probability that's been hammered into submission.
When only one is red, it must serve as the viewed/announced deuce.
The second non-red die, consequently shows {1,2,3,4,5,6} with equal probability. It absolutely cannot serve as the viewed/announced red deuce.
Quote: indignant99First off, there must be one red die, perhaps two. Pee's result mandates it.
25% Red-Red
25% Red-Blue
25% Red-White
25% Red-Black 75% of the time, it's a Red plus a Non-Red.
Agreed
I see where you are coming from.Quote:When both are red, either could serve as the viewed/announced deuce.
Quote:When only one is red, it must serve as the viewed/announced deuce.
Here we are, once again in need of more certainty about the calling rules. Does the die just happen to be red, or is that a precondition of the announcement being made. From the precedent of the original puzzle, we have sort of set a convention that it is. But, we are not certain of that. Maybe PeeMcGee will tell us that rule.
Again. I see where you are coming from and cannot disagree with you. I'll wait till we have that clarification on the announcing rules, then I'll model it and do some counting. A universal population of 16 dice should prove sufficient.Quote:The second non-red die, consequently shows {1,2,3,4,5,6} with equal probability. It absolutely cannot serve as the viewed/announced red deuce.
So Far, I'm persuaded by Indie that it is 13/88 Subject to further clarification of the rules.
I award a second pint to Indie for what seems to be a correct answer, and a third pint for explaining his reasoning in terms that I can immediately understand.
Dammit, I'll never get any work done at this rate.
$:o)
Quote: OnceDearHere we are, once again in need of more certainty about the calling rules. Does the die just happen to be red, or is that a precondition of the announcement being made. From the precedent of the original puzzle, we have sort of set a convention that it is. But, we are not certain of that. Maybe PeeMcGee will tell us that rule.
No, it's not a precondition.
It's a flat-out factual reporting of the event, after the fact, of what actually happened.
(Well, a partial reporting, anyway.)
What could be left to interpretation, that I'm missing?
Quote: WizardOnce you guys negotiate the details I'm willing to cooperate if asked. I may ask to be able to make a video of the challenge.
Hi Mike
Indie and RS and I worked out our differences of opinion. So those bets are off. All very amicable.
Please keep us informed of the Alan M wager, if it ever comes to anything.
I'm sure that one day, you'll stand as escrow in some friendly wager.
Do you care to comment on PeeMcGee's question PeeMcGee's question
$:o)
Quote: OnceDear...A universal population of 16 dice should prove sufficient...
Probably not...
In total, there would be (16 * 15) / 2 = 120 ways to pick a pair of dice.
There would only be 6 ways (each color) to get a pair of same-colored dice.
There would be (120 - 24) = 96 ways to pick a pair of dice not matching in color.
6 ways <--- matching Red pair
48 ways <--- mis-matching, but including a Red (16 with each other color).
6 ways <--- matching Blue pair
48 ways <--- mis-matching, but including a Blue (16 with each other color).
6 ways <--- matching White pair
48 ways <--- mis-matching, but including a White (16 with each other color).
6 ways <--- matching Black pair
48 ways <--- mis-matching, but including a Black (16 with each other color).
Aggregating...
24 ways <--- Matched-Color pairs (6 red, 6 blue, 6 white, 6 black)
96 ways <--- Mismatched-Color pairs.
But focusing down to interest in Red (pair of them, or just one)...
8-to-1 against randomly selecting a Red-pair versus a Red-mismatch.
This would really restrict the chances of having Red Pair (& the 1/11 dual deuces).
The universal population of colored dice greatly affects the prob of getting matched colors. We kinda need unlimited (urp) populations of each color. Or at least the guarantee that each of the (two) die-color selections, enjoys a 1/4 probability of happening.
I wish to discard my old attempt, and take a new attempt.
| Prob | Color Pairing | Why |
|---|---|---|
| 1/16 | Red-Red | 1/4 times 1/4 |
| 2/16 | Red-Blue | 1/4 times 1/4, doubled |
| 2/16 | Red-White | 1/4 times 1/4, doubled |
| 2/16 | Red-Black | 1/4 times 1/4, doubled |
| 1/16 | Blue-Blue | 1/4 times 1/4 |
| 2/16 | Blue-Black | 1/4 times 1/4, doubled |
| 2/16 | Blue-White | 1/4 times 1/4, doubled |
| 1/16 | Black-Black | 1/4 times 1/4 |
| 2/16 | Black-White | 1/4 times 1/4, doubled |
| 1/16 | White-White | 1/4 times 1/4 |
The observed outcome definitely had a Red die.
1/7th of the time, the dice were Red-Red. 6/7ths of the time, the dice were Red-Other.
(1/7 * 1/11) + (6/7 * 1/6) = 1/77 + 11/77 = 12/77 = 15.5844%
Quote: RSNeed to know how many dice there are and if we're doing the precondition requirement or the other "just say something about the roll".
Hi RS,
I'm estimating a universal dice population of somewhere between 8 and infinity. If modeling it with, say 8, gives a tiny difference to modeling with 32 then we'll know if infinity is needed. If not, well that would be great.
At first, I assumed that the redness of the die was incidental, which meant all colouration was also incidental. But this is PeeMcGee asking and from what I've read of his, he really knows his stuff about dice probability. I doubt he would do anything sneaky with us.
Based on that character analysis and on the consensus we reached on the first puzzle, then I assume that we are only considering a sample space where 'red duece' is the mandatory call and nothing else ever leads to an announcement. In that case, I tend to agree with Indie's initial 13/88 answer and his reasoning.
But, we don't know what we don't know.
We do know that this would do a certain members head in $:o)
Quote: indignant99Upon consideration of the probabilities of the initial die colors...
I wish to discard my old attempt, and take a new attempt.
Hi Indie,
I still prefer your initial answer. I too am still considering this. Your reconsidered attempt leads me to suspect that we need a universal dice population approaching infinity and that modelling with tiny finite populations like 16 just won't cut it.
Quote: OnceDear...tiny finite populations like 16 just won't cut it.
Agreed. I just idealized the probabilities to 1/4 (four colours, each equally likely).
Quote: indignant99Agreed. I just idealized the probabilities to 1/4 (four colours, each equally likely).
You could do this in real life with four "dice dispensers."
(The Red dispenser, the Blue, the White, the Black.)
Ensure a random {1 thru 4} selector (RNG ?), for each of the two dispensations.
Quote: indignant99You could do this in real life with four "dice dispensers."
(The Red dispenser, the Blue, the White, the Black.)
Ensure a random {1 thru 4} selector (RNG ?), for each of the two dispensations.
Good plan.
Using RNG in Excel, drawing 100000 dice pairs from my infinite stock dispenser, I get a pair of reds 1 in 7 of times of when at least 1 die is a red.
Who'd have thought it?
Actually a bit like the 1/11 thing: Start with 2/8 combo and strike out the duplicate double to give 1/7
So do we go with your (1/7) * (1/11) + (6/7) * (1/6)
$:o)
Quote: OnceDear...So do we go with your (1/7) * (1/11) + (6/7) * (1/6)
Well, I think so. Until any of us has a further epiphany.
Quote: OnceDearHi Indie,
I'm not saying you are wrong, and happy to work this out as a team.
Where do you get Is there something special about non-red dice?
What significance at all is there in dice colour?
If we can eliminate a whole shedload of dice based on their colour, we eliminate all spot values for whatever bunch of dice they are.
heh heh heh....shedload...heh heh heh. I really like that, especially in an English accent. Can I use that? Thanks.
Quote: indignant99...Until any of us has a further epiphany.
Good grief. I'm still working on the original problem.
A penetrating analysis is better than a bunch of fuzzy spin-offs.
Quote: indignant99Agreed. I just idealized the probabilities to 1/4 (four colours, each equally likely).
This was the intent. Assume an infinite dice population.
Quote:Here we are, once again in need of more certainty about the calling rules. Does the die just happen to be red, or is that a precondition of the announcement being made. From the precedent of the original puzzle, we have sort of set a convention that it is. But, we are not certain of that. Maybe PeeMcGee will tell us that rule.
The question was intended to read much like the original--I grab two dice, roll them, peek at them and truthfully announce that at least one die is a 2 which is also red.
And much like the original problem, there is ambiguity (and maybe this highlights the ambiguity some). Since that was hashed out already, I’ll make the question less ambiguous...
Assume this is the same setup as the Wizard/Alan’s wager.
- We grab two dice, at random from a big infinite box of dice (so, each with ¼ chance of being any one of four colors).
- Roll them until we get at least one red die showing a two.
- What is the probability the other die is a two?
Quote: WizardNice! I think this is the first WoV "video response." Let me try to draw some more attention to it below.
Before anybody takes this post out of context, the probability IS 1 in 11.
Wizard before you respond to the post, would you tell me if you think I accurately illustrated the question?
Quote: AlanMendelsonQuote: WizardNice! I think this is the first WoV "video response." Let me try to draw some more attention to it below.
Before anybody takes this post out of context, the probability IS 1 in 11.
Wizard before you respond to the post, would you tell me if you think I accurately illustrated the question?
You looked at the non-deuce, showed there were 6 sides, thus concluded it was 1/6.
So no you did not.
Quote: RSQuote: AlanMendelsonQuote: WizardNice! I think this is the first WoV "video response." Let me try to draw some more attention to it below.
Before anybody takes this post out of context, the probability IS 1 in 11.
Wizard before you respond to the post, would you tell me if you think I accurately illustrated the question?
You looked at the non-deuce, showed there were 6 sides, thus concluded it was 1/6.
So no you did not.
Remember the original question? I will refresh your memory:
"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?"
Since that was the question, why wouldn't I look at the non-deuce die? The question says "at least one of the dice is a 2," so it's only natural to consider what are the odds of a second die to have a 2. Even if I picked up the die showing a 2 the answer would still be 1/6 because it's just a two-dice question. Remember "at least one of the dice is a 2." That information tells you to consider the chance of a 2 on the second die. And it's six sided, right?
Quote: AlanMendelson
Remember the original question? I will refresh your memory:
"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?"
Since that was the question, why wouldn't I look at the non-deuce die? The question says "at least one of the dice is a 2," so it's only natural to consider what are the odds of a second die to have a 2. Even if I picked up the die showing a 2 the answer would still be 1/6 because it's just a two-dice question. Remember "at least one of the dice is a 2." That information tells you to consider the chance of a 2 on the second die. And it's six sided, right?
And you asked the question, "...if you think I accurately illustrated the question." Which you did not do.
Take 2 coins, flip them at the same time. If at least one coin lands on HEADS, what is the chance the other coin is a heads? You would say the chance is 50%, based on your illustration in the video. Unfortunately (for you) that is not the correct answer. The correct answer is 1/3 (33.33%).
Or maybe you're doing a monty-hall game show problem. Using the same logic/illustration from the dice-toss-video, you would conclude the chance of winning the car is 50% and the chance of winning the goat is 50% - and changing doors does not matter. But in reality, if you stay with the original door you have a 1/3 chance of a car and 2/3 chance of the goat -- and by switching, you have a 2/3 chance of winning the car and 1/3 chance of winning the goat.
I hope I am never in a situation where my job application is denied by someone because they use the same poor logic you use.
IF you really want to "illustrate" the scenario in a video, I suggest you roll the dice many times and note the frequency of 2-2 versus 2-X.
Actually, I think my video really does portray the problem.
And I am certainly not going to be flipping coins to solve this.
But why don't you write a script (narration) for how you think a video demonstration of the problem would be? Write it here, and I will record your script using two dice.
Thanks.
When one of the dice is a two, remove it
Record how often the one remaining die under the cup is a two, and how often it is not a two.
I think it will take a long time to do it that way, though, and would make for a boring video.
Quote: Dalex64Roll two dice under a cup
When one of the dice is a two, remove it
Record how often the one remaining die under the cup is a two, and how often it is not a two.
I think it will take a long time to do it that way, though, and would make for a boring video.
But that's not the original question either. It's not a question of how often the other die is a two, or how often it is not a two. The question is what is the probability. Thank you for recognizing it is a ONE DIE problem. And that one die has six sides.
Please kill me for coming back to this thread again. Arghhh. How about discussing the fine points of SPS?Quote: AlanMendelsonBut that's not the original question either. It's not a question of how often the other die is a two, or how often it is not a two. The question is what is the probability. Thank you for recognizing it is a ONE DIE problem. And that one die has six sides.
Quote: pewPlease kill me for coming back to this thread again. Arghhh. How about discussing the fine points of SPS?
Okay, I'll ask. What is SPS ??
Are you kidding? Your bud Singer. Ring a bell?Quote: AlanMendelsonOkay, I'll ask. What is SPS ??
Quote: AlanMendelsonWizard before you respond to the post, would you tell me if you think I accurately illustrated the question?
No, you didn't accurately depict the question. AT the 0:09 point you said, "If one of these two dice happens to show a two then our challenge is on." You never address what happens if BOTH dice are a two.
Quote: AlanMendelsonBut that's not the original question either. It's not a question of how often the other die is a two, or how often it is not a two. The question is what is the probability. Thank you for recognizing it is a ONE DIE problem. And that one die has six sides.
Sigh.
Wrong, wrong, wrong, wrong.
Your explanation is wrong. You fail to consider that there is a set up for the dice before you consider it. It is not a one die problem.
You do know what probability is, right? 1 in 6 means 1 time its a 2, and 5 times it's not a two. That's what you mean when you say 1 in 6. That is what the question is. What's the probability that dice is a two RIGHT NOW. When you pick it up.
It's not purely based on the number of sides it has. It is not a one die problem. Your interpretation is wrong.
Only way to test if you are right it is:
a) do the math
b) test empirically.
You seem to reject (a) but won't do (b).
Quote: pewAre you kidding? Your bud Singer. Ring a bell?
Well, if you read my website you would know that I do not agree with Rob Singer's special play strategy. I play conventional video poker strategy as taught by Bob Dancer. In fact, I attended his lectures and used his software. I also use the book written by John Grochowski which is also conventional play. I presented Singer's articles and videos on my site because so many people have an interest in what he says and when his own website was taken down I offered my site as a place for him to show his stuff.
I also have a section on my site for the articles of Professor Nelson Rose that he provides to me.
I also have a section on my site for the Las Vegas Convention and Visitors bureau so they can promote their monthly events calendar.
And there is a section on my site for the Vegas Realtors group.
But getting back to Singer, I make it clear his stuff is controversial and I don't agree with it myself. You can read that here: http://alanbestbuys.com/id194.html in the bold, italic font.
Quote: WizardNo, you didn't accurately depict the question. AT the 0:09 point you said, "If one of these two dice happens to show a two then our challenge is on." You never address what happens if BOTH dice are a two.
Replying to Wiz, not Alan, so please don't suspend me Face :o)
Let's explore some other ways that the depiction was wrong.
1. There was no friend/partner peeking at the dice.
2. Alan was never invited to peek at the dice, but he did.
3. Alan was never invited to touch or f*** with the dice in any way.
4. Alan was never asked to say anything about probability of 'the other die', 'the second die', 'the remaining die' or any 'just one single die' anywhere in the universe.
5. Alan does not anywhere attempt to answer the ACTUAL question and give us 'What is the probability that both dice are showing a 2?"
6. Looking at one throw of the dice means nothing.
Alan could only faithfully depict the original question if :-
1. He rolled the dice a very very large number of times and actually measured the probability 'Of a pair of twos' by observation. Rough and ready couple of hundred times would soon show that it's closer to 1/11 than 1/6.
2. He rolled the dice a number of times which represented all possible and equally likely outcomes and then actually measured the probability 'Of a pair of twos'
I would be interested in what answers Alan would expect to come up with from each of those experiments.
It would seem that because Alan wants to answer his own different question, and maybe because he cannot rustle up a friend or partner, he thinks that he can bypass the whole proof concept by just saying, here's a 6 sided die.
Maybe time for a REALLY REALLY easy question, just to test Alan's grasp of the English Language and of the meaning of probability.
"You place 1 six sided die in a cup, shake it and slam it down. You friend peeks and says 'It definitely isn't showing a one and it definitely isn't showing a three and it definitely isn't showing a four and it definitely isn't showing a five. It might be showing a two and it might be showing a six. What is the probability that it is showing a two?'
Answer with proof would be appreciated. If the answer that Alan comes up with is 1/2 Then I'd venture '"don't be daft, because everyone can see it's got 6 faces". That would be a pretty perfect analogy to the nonsense answer that he gives in his proof.
Quote: WizardNo, you didn't accurately depict the question. AT the 0:09 point you said, "If one of these two dice happens to show a two then our challenge is on." You never address what happens if BOTH dice are a two.
Actually, I think I did. Doesn't "if one of these two dice happens to show a two then our challenge is on" mean the same thing as:
======================
"Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."
What is the probability that both dice are showing a 2?
======================
You also raise the question "you never address what happens if BOTH dice are a two." Well, it doesn't matter. If both dice are a 2 then it is also a true statement that "at least one of the dice is a 2." And even if both dice show a 2, then in a 2-dice problem the answer still involves only 1 die.
And let me jump ahead: it doesn't matter which die shows a 2 or if both dice show a 2 in a two dice problem.
If die A shows the 2 then it's a question of what shows on the 6 sides of die B. If die B shows the 2 then it's a question of what shows on the six sides of die A.
And again, if both dice show a 2 then it's still a question of either die A or die B both of which have six sides.
And, to state the obvious: if the judge or partner peeked and didn't see "at least one of the dice is a 2" the "challenge" or "question" is no longer valid and the dice must be rolled again.
So, I think my video clearly illustrates the question, and explains the answer to the way the question was originally worded.
I want to add one more thing in case it somehow was missed: 1/11 is the correct answer to a different question. That question might be something like this: how many different combinations of two dice, with at least a 2 showing on at least one die, would show 2-2?" And here the answer would be 1/11 because there is only one 2-2 and 11 combinations of 2 dice showing at least one 2.
I've said it from the beginning: you have to interpret the language of the question. The language is specific.
Thanks.
Quote: AlanMendelsonActually, I think I did. Doesn't "if one of these two dice happens to show a two then our challenge is on" mean the same thing as:
Alan, I think you're getting hung up on the peeking question. Would you agree that the question you're trying to answer is equivalent to the following?
Step 1: Throw two dice on the table in front of you.
Step 2: Write down the number of 2s you see. Either 0, 1, or 2. "0" is equivalent to "this throw didn't count because there were no 2s." Either 1 or 2 is equivalent to "there is at least one 2".
Step 3: After a sufficient number of throws, compare the number of 2s to the number of 1s. The ratio should be roughly 1:10.
Do you believe the question you're trying to answer is or is not equivalent to those steps?
In the alternate, have you attempted to answer the question using Bayes' theorem? The answer is very quickly derived using that method. (So is the answer to the other question involving colored dice, FYI.)
For a restatement of how Bayes' applies here, consider these questions:
What is the probability that both dice are showing a 2 if the partner peeks under the cup and truthfully says "neither die is a 2"? It's zero, right?
We know that the overall probability that two dice show 2-2 is 1/36. (1/6 * 1/6). So you have to be able to derive that overall probability regardless of how you break up the problem.
If you split up the problem into "cases where there are no 2s" and "cases where there are at least one 2", and the probability of a pair of 2s is zero under the first case, then what does the probability of a pair of 2s have to be in the second case in order to make the overall probability equal 1/36? If you think it's 1/6, write it out and see if the math works.
