Two Dice Puzzle
I think the original poster set up everyone for a joke.
He used the "penny puzzle" to set everyone up and to make them think a certain way. And that was diversion from the simple question: what is the probability of a single die?
And how did he end his post? By saying he was interested in the process of answering the question.
What tipped me off is whats in his spoiler tag.
Quote: DweenAlong the vein of the Two Coin Puzzle, I wanted to put this in another thread as to not clog up the other.
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?
Based on the discussion in the Two Coin thread, I'm interested to see how people solve this one.
Quote: indignant99...
So I am a convert to the 1/11.
My epiphany? Thought experiment about whether I would wager on this game in the casino. Here was my scenario, in the casino...Self-delusion is a terrible thing. Clarity of analysis is a wondrous thing. This scenario is accurate whatever pip-count is displayed.
- 2 dice are shaken in an electronic cup.
- Electronic cup for honesty. No human deceit possible.
- Slam down the cup with the 2 dice, as yet unrevealed.
- Bottom screen display of the cup is now on top.
- Cup's display - an optical reader - now shows one and only one of the die's pip-count.
- Casino let's me bet whether there's a pair under there.
- No pair - my bet loses one unit. Pair - casino pays me 9-to-one.
- Duh. I think not.
Jeez, I might have to flip-flop again.
Sight-unseen, there's a 6/36 probability of a pair (1-1, 2-2, 3-3, 4-4, 5-5, 6-6).
Getting a revelation of one pip-count surely doesn't destroy the original probability.
Quote: indignant99
Jeez, I might have to flip-flop again.
Lol@Indignant.
Regarding what the poster said about one die being a two? Again he wrote:
"Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2.""
We have to assume that this was a true part of his post. The part of his post which is the diversion is this:
"Along the vein of the Two Coin Puzzle," and this: "Based on the discussion in the Two Coin thread, I'm interested to see how people solve this one" when simply he stated the problem this way:
there are two dice, one is a 2, and what are the odds that the other die is also a 2?
It's diversion -- to make you think in one direction and steer you away from the simple answer.
Why would he do such a thing? I don't know. But why would he write such a lengthy article in his "spoiler." Anyone who would write such a thing is out to make a point (of some sort).
Has he returned to the forum? Has he commented on his post? Why hasn't he given HIS answer? Because he wanted to poke fun at everyone with his diversion.
By the way, most "magic" or illusions are accomplished through diversion. You make the audience look for something else.
Quote: Dween... Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2." ...
For those who are still fighting for the 1/6'th route, well....math & logic is pretty tough stuff. I apologize, apparently we cannot explain this well enough for you guys (just Alan, now?) to understand.
My offer goes out to you or anyone else, who would like to bet 7:1 on a 2-2 being rolled (losing 1 for 12/32/42/52/62/12/23/24/25/26 and pushing on everything else). I don't understand why anyone wouldn't take me up on this offer, if they live in LV or will be visiting LV.
Quote: RSIndignant -- I'm glad you've discovered 1/11 as the correct answer. My hat goes off to you for being able to own up (?) to a mistake.
For those who are still fighting for the 1/6'th route, well....math & logic is pretty tough stuff. I apologize, apparently we cannot explain this well enough for you guys (just Alan, now?) to understand.
Me too. Persuade me RS. Please :o) I'm not fighting for the 1/6th route
https://wizardofvegas.com/forum/questions-and-answers/math/15508-two-dice-puzzle/2/#post448990
1. Put two dice in a cup, shake them, and turn them over onto a table.
2. Have a friend peek under the cup and tell you TRUTHFULLY what one of the dice shows. (Example: 6)
3. Then figure the odds that the other die under the cup is the same number.
4. Will you go through the math and the charts, or will you simply say: there is a one in six chance that a 6 is showing on the second die.
Let's try that simple, basic, simulation.
Now if you want to go through the lengthy process of determining the number of heads and tails combinations that there might be with two pennies, knock yourself out.
h,h
t, t
h, t
t, h
Quote: AlanMendelsonIt's a matter of trying to make the forum think they should answer the question using the same methodology as with the coin/penny puzzle.
Nah! I hadn't even read the two penny puzzle.
Like many, I was torn between 1/6 and 1/11. I could see and replicate the simulations and demonstrations that show 1/11 as being correct. But it isn't. I could see and understand the logic behind the 1/6 answer two, but that too is incorrect.
It is that both the solutions rely on different, but equally valid assumptions. Apart from the partner telling the truth about seeing a two, why the hell should we assume that he always said anything, and under what circumstances WOULD he make that true assertion?
Quote: OnceDearMe too. Persuade me RS. Please :o) I'm not fighting for the 1/6th route
https://wizardofvegas.com/forum/questions-and-answers/math/15508-two-dice-puzzle/2/#post448990
We can only go off of the information that we have. If he always says "at least one die is a 2" when both are a 2...then that changes it. Or if he says "at least one die is a 2" when the other isn't a 2...then that obviously also changes things.
But, with the given information, you can't say, "We cannot figure out the probability of a 2-2."
Of course we can figure out the probability -- we've done it multiple times. I did it in an Excel spreadsheet, someone else did too. And if you want, you can put more rules/info into it.....well ok, we can figure out those probabilities, too.
Quote: AlanMendelsonFor everyone who doesn't believe that this thread was a joke or a mockery (please read my last several posts) and believes in the validity of simulations, please do this:
1. Put two dice in a cup, shake them, and turn them over onto a table.
2. Have a friend peek under the cup and tell you TRUTHFULLY what one of the dice shows. (Example: 6)
3. Then figure the odds that the other die under the cup is the same number.
4. Will you go through the math and the charts, or will you simply say: there is a one in six chance that a 6 is showing on the second die.
Let's try that simple, basic, simulation.
Now if you want to go through the lengthy process of determining the number of heads and tails combinations that there might be with two pennies, knock yourself out.
h,h
t, t
h, t
t, h
I'll do it in Excel and post the results.
Edit: Maybe later.
Quote: AlanMendelson
2. Have a friend peek under the cup and tell you TRUTHFULLY what one of the dice shows. (Example: 6)
There's a difference between
'Have a friend peek under the cup and tell you TRUTHFULLY what one of the dice shows. (Example:6 )'
and
'Have a friend peek under the cup and tell you TRUTHFULLY if at least one dice shows a 6'
$:o)
Quote: RSQuote: OnceDearMe too. Persuade me RS. Please :o)
We can only go off of the information that we have.
Why?
Quote:And if you want, you can put more rules/info into it.....well ok, we can figure out those probabilities, too.
If it's possible and permissible for me to put in other rules and those rules do not break the rules in the original information, then what is invalid about my answer?
A bit like "A chicken crosses the road from side A to side B on a Monday morning, so which side of the road is he on on Wednesday morning"
You would assert 'Side B', but I would assert 'I dunno, what happened on Tuesday?'
$:o)
Quote: OnceDearThere's a difference between
'Have a friend peek under the cup and tell you TRUTHFULLY what one of the dice shows. (Example:6 )'
and
'Have a friend peek under the cup and tell you TRUTHFULLY if at least one dice shows a 6'
$:o)
Actually there isn't. Let me give you two examples:
Example #1 One of the dice shows a 6 and this is the truth, OnceDear. What are the odds that the other die shows a 6?
Example #2 At least one of the dice shows a 6 and this is the truth, OnceDear. What are the odds that the other die shows a 6?
So please answer: what are the odds for Example #1 and what are the odds for Example #2?
In both cases we are talkig about "the other die." And the other die has six options. And only six options.
Quote: AlanMendelsonActually there isn't. Let me give you two examples:
Example #1 One of the dice shows a 6 and this is the truth, OnceDear. What are the odds that the other die shows a 6?
Example #2 At least one of the dice shows a 6 and this is the truth, OnceDear. What are the odds that the other die shows a 6?
So please answer: what are the odds for Example #1 and what are the odds for Example #2?
In both cases we are talking about "the other die." And the other die has six options. And only six options.
Which one is the 'one' and which one is 'the other one'?
If you cannot tell the difference between "One of the dice" and "At least one of the dice" then I cannot really help you.
If 'One of the people in a crowded room is a woman' then how many women are in that room?
If 'At least one of the people in a crowded room is a woman' then how many women are in that room?
If "One of the dice is (wrongly/unconventionally) interpreted as "Exactly One of the dice..." then the other dice cannot be showing a 6 so no chance. By convention, we take " One of" to mean "At least one of"
#2 : Absolutely 1 in 11If we don't impose any unstated rules such as if the friend says "One of the dice is a 3" when 6,3 shows.
Quote: OnceDearQuote: RS
We can only go off of the information that we have.
Why?
Because we can't go off of information that we do not have.
For example, I played $10 pass line bets at the craps table with $52/$54 across, did same bet (no pressing). I counted a total of 100 dice rolls? How much should I have lost?
The proper way to figure out my expected loss would be to figure out the HE on all the bets times the amount times the number of rolls.
But, the way you are analyzing the double-deuce-dice scenario would have you analyzing my craps session as something like, "We don't know -- how many shooters were there? Did anyone have a long roll? Was there a 45 minute roll? Or was it point-7, point-7, point-7 all session?"
You could figure out my expected loss (or profit) given the other information -- if it was point-7 point-7 all night....or if there were two 30-minute shooters.
But, without knowing how the table went (if it was "cold" or "hot" that night), then we can only go off of the information that we know -- the average/expected loss of each of my bets.
Quote: OnceDear
If it's possible and permissible for me to put in other rules and those rules do not break the rules in the original information, then what is invalid about my answer?
A bit like "A chicken crosses the road from side A to side B on a Monday morning, so which side of the road is he on on Wednesday morning"
You would assert 'Side B', but I would assert 'I dunno, what happened on Tuesday?'
$:o)
Because you're adding rules/information that we don't know about.
I don't like your example. But, with the information we have, side B would be correct -- since we aren't told the chicken moved from side B.
I flip a coin 10 times, I say I expect (average) 5 heads and 5 tails. You say, "I don't know, does the coin have heads on both sides? Is the coin rigged somehow [magnetic, one side weighted, skilled coin-flipper]."
With the double-deuce-dice scenario, you could say, "If the first die that landed was a 2, then there's a 1/6 chance the second die is a 2." And, that's all good and fair.
BUT, that's not the scenario that was presented. The scenario that was presented is that one or the other die is a 2 [the deuce is unknown]. Given this bit of information, we must conclude the chance that both dice are a 2 is 1/11.
Unfortunately everyone is caught up in the "coin puzzle" and the spread sheet showing multiple combinations with two coins. But we are not looking at multiple combinations with two dice. We are looking at ONE DIE and the choice of one of its six faces.
ONCE AGAIN... the original poster set up the question about the dice with a mention of the coin puzzle which acted as a diversion and made you treat the dice problem the same way you would treat the coin problem with heads and tails. But that's not the situation with the dice.
When choosing a face from a single six sided die, it will always be 1/6. Always. It will never change.
Quote: RSQuote: OnceDearQuote: RS
We can only go off of the information that we have.
Why?
Because we can't go off of information that we do not have.
Why not? Like a magician's audience, you assume that you have all the facts.
Sorry. Craps means nothing to meQuote:
For example, I played $10 pass line bets at the craps table...
Quote:Because you're adding rules/information that we don't know about.
I don't like your example. But, with the information we have, side B would be correct -- since we aren't told the chicken moved from side B.
I flip a coin 10 times, I say I expect (average) 5 heads and 5 tails. You say, "I don't know, does the coin have heads on both sides? Is the coin rigged somehow [magnetic, one side weighted, skilled coin-flipper]."Granted, you could say something like, "If the chicken moved to side X, then he'll be on side X." or "If the coin is rigged, then we don't know" / "If the coin has heads on both sides, then you'll flip 10 heads."
With the double-deuce-dice scenario, you could say, "If the first die that landed was a 2, then there's a 1/6 chance the second die is a 2." And, that's all good and fair.
BUT, that's not the scenario that was presented. The scenario that was presented is that one or the other die is a 2 [the deuce is unknown]. Given this bit of information, we must conclude the chance that both dice are a 2 is 1/11.
Of course, if "The scenario that was presented is that one or the other die is a 2" is the truth, the whole truth and nothing but the truth, then 1/11 is fine. (counter-intuitive, but fine).
That's not the original scenario. There we had one observed event devoid of rules except those stated
Quote:You have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."
BUT it's not necessarily the whole truth.
Below is a table of events and 'partner assertions' The assertions that my partner would make consistently with her twisted mind. The original poster knows my wife and knows her to be a truthful but mischievous sod. He observes ONE roll of the dice and observes her assertion that 'At least one of the dice is a 2'. He calculates the probability at 100% based on his knowledge of all the facts. He's spot on in his calculation.
1,1 At least one of the dice is a 1
1,2 At least one of the dice is a 1
1,3 At least one of the dice is a 1
1,4 At least one of the dice is a 1
1,5 At least one of the dice is a 1
1,6 At least one of the dice is a 1
2,1 At least one of the dice is a 1
2,2 At least one of the dice is a 2
2,3 At least one of the dice is a 3
2,4 At least one of the dice is a 4
2,5 At least one of the dice is a 5
2,6 At least one of the dice is a 6
3,1 At least one of the dice is a 1
3,2 At least one of the dice is a 3
3,3 At least one of the dice is a 3
3,4 At least one of the dice is a 3
3,5 At least one of the dice is a 5
3,6 At least one of the dice is a 3
4,1 At least one of the dice is a 1
4,2 At least one of the dice is a 4
4,3 At least one of the dice is a 4
4,4 At least one of the dice is a 4
4,5 At least one of the dice is a 5
4,6 At least one of the dice is a 4
5,1 At least one of the dice is a 1
5,2 At least one of the dice is a 5
5,3 At least one of the dice is a 5
5,4 At least one of the dice is a 5
5,5 At least one of the dice is a 5
5,6 At least one of the dice is a 6
6,1 At least one of the dice is a 1
6,2 At least one of the dice is a 6
6,3 At least one of the dice is a 3
6,4 At least one of the dice is a 5
6,5 At least one of the dice is a 6
6,6 At least one of the dice is a 6
With only one observed event, we don't know the rules. We know all about the rules and probabilities of how dice will land, but we do not know the rules that my partner makes. We hazard a guess at our peril.
Quote: AlanMendelsonAs I explained to you, it doesn't matter how the question is worded because there is only one other die to be considered. That other die has only six faces. So the odds of matching the number can only be 1/6. That's all. It's that simple
....
When choosing a face from a single six sided die, it will always be 1/6. Always. It will never change.
You're wrong Alan. It's not one other die. Either of the dice can be the 'second die'.
Call the dice a and b.
If die a is the first die and it's a two then there is one chance in 6 that die b is a two: That's 1 chance in 6 of us seeing a second 2 from die b.
If die a is not the first die then die b is the first die and it is a two: That's 1 chance in 6 of us seeing a second 2 from die a.
So initially that gives us 2 chances in twelve of seeing a pair of twos.
BUT
the above implies that a=2 and b=2 can occur as well as b=2 and a=2 which is one possibile outcome and not two
so we have to discount that possibility from numerator and denominator
What we initially assumed to be 2/12 becomes 1/11
It's hellishly counter-intuitive, but you really REALLY are wrong. (assuming no other absurd rules apply)
Quote: OnceDearLol@Indignant.
Keep laughing. Because I'm going to get funnier.
Quote: OnceDearMe too...
Read on.
Quote: RSIndignant -- I'm glad you've discovered 1/11 as the correct answer...
For those who are still fighting for the 1/6'th route...
My offer goes out to you or anyone else, who would like to bet 7:1 on a 2-2 being rolled (losing 1 for 12/32/42/52/62/12/23/24/25/26 and pushing on everything else)...
I haven't discovered that 1/11 is the correct answer.
And right now, I am going to keep fighting for 1/6.
I will play your game. (Although you should offer 9-to-1. After all, the odds are 10-to-1 against me, aren't they?)
When Die-1 is showing deuce, I have a 1/6 shot at Die-2 = deuce.
When Die-2 is showing deuce, I have a 1/6 shot at Die-1 = deuce.
The probability calculation should be...
(Die-1 is a deuce, grey Die-2 deuces are dead) --> 1/(11 - 5) = 1/6 or
(Die-2 is a deuce, grey Die-1 deuces are dead) --> 1/(11 - 5) = 1/6
Isn't the sight-unseen probability for Deuce-Deuce 1/36 ?
(Die-1 is a deuce) --> Knocked out 30 other possibilities. (1/36)/(6/36) = 1/6
(Die-2 is a deuce) --> Knocked out 30 other possibilities. (1/36)/(6/36) = 1/6
Quote: indignant99
Isn't the sight-unseen probability for Deuce-Deuce 1/36 ?
Yeah. But We don't have sight unseen. One Die is a deuce. granted we don't know which, but all outcomes that don't feature a deuce need to be extracted from numerator and denominator of the calculations.
Quote: kenarmanIt has been my experience that the people that start screaming are usually the ones that are wrong.
Usually, yeah. But not this time.
Quote: OnceDearQuote: indignant99
Isn't the sight-unseen probability for Deuce-Deuce 1/36 ?
Yeah. But We don't have sight unseen. One Die is a deuce. granted we don't know which, but all outcomes that don't feature a deuce need to be extracted from numerator and denominator of the calculations.
Razzberries. Oh man, this is becoming ridiculously hilarious.
Nearly though.Quote: indignant99I haven't discovered that 1/11 is the correct answer.
Stop digging.Quote:And right now, I am going to keep fighting for 1/6.
When Die-2 is showing deuce, I have a 1/6 shot at Die-1 = deuce.
Agreed, though your image didn't show. and 'grey' did nothing for me. I can only guess. you only have that 1/6 chance When Die-2 is showing deuce Which isn't very often.
You could also say...
When Die-1 is showing deuce, I have a 1/6 shot at Die-2 = deuce. When Die-1 is showing deuce
Which would be equally valid and equally rare.
Quote:The probability calculation should be...
Ermmmm. you lose me with this. Maybe your missing picture made it clearer.
(Die-1 is a deuce, Die-2 deuces are dead) --> 1/(11 - 5) = 1/6 or
(Die-2 is a deuce, grey Die-1 deuces are dead) --> 1/(11 - 5) = 1/6
Isn't the sight-unseen probability for Deuce-Deuce 1/36 ?
(Die-1 is a deuce) --> Knocked out 30 other possibilities. (1/36)/(6/36) = 1/6
(Die-2 is a deuce) --> Knocked out 30 other possibilities. (1/36)/(6/36) = 1/6
Close but no cigar. Start from that probability of Deuce-Deuce as 1/36 which is fine.
Then throw away all outcomes where no deuces. There are 25 of those.
1/(36-25) =1/11
Quote: DweenAlong the vein of the Two Coin Puzzle, I wanted to put this in another thread as to not clog up the other.
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?
Based on the discussion in the Two Coin thread, I'm interested to see how people solve this one.
Quote: RSWe can only go off of the information that we have.
Yes, and then we must make reasonable assumptions like:
- Each of the 6 sides are equal-likely to show. You could have weighted dice or some sort of 6-sided "pyramid section" dice with different sized faces.
- Each die has each number 1-6 on it's faces. The dice could be from the board game Outburst, which have three 1's, two 2's and one 3 on their faces, IIRC.
This information is not stated explicitly in the question, but I think all reasonable people agree that they are implicit facts of the problem. However, if we don't make these assumptions (and others I haven't listed), you cannot come up with a definitive answer.
Now, what has been at hand for the last 13 pages is whether it is reasonable to assume that the partner will only reveal that one of the dice is a "2," and otherwise be silent, or if he'll blurt out any number he sees. I prefer to assume the former (like the 1/11 crowd), but I don't find it unreasonable to assume the latter (like the 1/6 crowd).
If we look back at Page 1, MangoJ provides the answer for both scenarios.
Quote: JoemanOP:
Quote: RSWe can only go off of the information that we have.
Yes, and then we must make reasonable assumptions like:...
This information is not stated explicitly in the question, but I think all reasonable people agree that they are implicit facts of the problem. However, if we don't make these assumptions (and others I haven't listed), you cannot come up with a definitive answer.
...
If we look back at Page 1, MangoJ provides the answer for both scenarios.
Nicely summed up Joe.
I agree that it is reasonable to assume that these are regular dice: No reason to suspect otherwise.
It's neither reasonable nor unreasonable to assume that the partner is only going to make a comment if a deuce shows and always going to make that comment about the visibility of a deuce, because we have only observed one event, and it so happens that something was said that time and that that something happened to be about deuces ( The MontyHall problem relies on the fact that Monty always makes a comment and there the rules were quite explicit.) It could be equally valid that I had asked 'Please tell me something about the dice that have been cast', or 'Please tell me if any deuces have landed'
There are some that are doing convoluted maths that still happen to be wrong. I'm not bothered about wrong maths, only about invalid interpretation and some degree of bullying about which interpretation is authoritative.
Start with a question:
Quote: DweenYou 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?
Come up with a hypothesis: 1/11
Show a mathematical proof:
Quote: Wizardpr(two sixes)/pr(one or more six) = (1/36)/(11/36) = 1/11
Verify it with an experiment:
Several people here have done it with excel. Here is a way to do it with perl:
#!/usr/bin/perl
use warnings;
use strict;
my $NUM_TRIALS=1000000;
my $rollHasATwo=0;
my $rollHasDoubleTwo=0;
my %individual_histogram;
my %total_histogram;
for (my $i = 0 ; $i < $NUM_TRIALS ; $i++)
{
# roll two dice
my $d1 = int(rand(6)) + 1;
my $d2 = int(rand(6)) + 1;
# this tracks the number of rolls where there was at least one two
# I.E. Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."
$rollHasATwo++ if ((2 == $d1) || (2 == $d2));
# this tracks the number of rolls where there were double twos
$rollHasDoubleTwo++ if ((2 == $d1) && (2 == $d2));
# gather statistics
$individual_histogram{$d1}++;
$individual_histogram{$d2}++;
$total_histogram{$d1+$d2}++;
}
print "histogram of individual dice distribution\n";
foreach my $i (sort {$a <=> $b} keys %individual_histogram)
{
my $total = $individual_histogram{$i};
my $pct = sprintf("%02.1f", $total/$NUM_TRIALS*100);
print "$i: $total $pct%\n";
}
print "\nhistogram of dice sum distribution\n";
foreach my $i (sort {$a <=> $b} keys %total_histogram)
{
my $total = $total_histogram{$i};
my $pct = sprintf("%02.1f", $total/$NUM_TRIALS*100);
print "$i: $total $pct%\n";
}
print "\nRolls with a two:\n";
print "$rollHasATwo " . sprintf("%02.1f", $rollHasATwo/$NUM_TRIALS*100) . "\n";
print "\nRolls with double-two\n";
print "$rollHasDoubleTwo " . sprintf("%02.1f", $rollHasDoubleTwo/$NUM_TRIALS*100) . "\n";
print "\npercent of rolls with double two when a roll had at least one two: ";
print sprintf("%02.1f", $rollHasDoubleTwo/$rollHasATwo*100);
results from a run:
$ ./twodice.pl
histogram of individual dice distribution
1: 333353 33.3%
2: 333275 33.3%
3: 333576 33.4%
4: 333479 33.3%
5: 333670 33.4%
6: 332647 33.3%
histogram of dice sum distribution
2: 27597 2.8%
3: 55697 5.6%
4: 83383 8.3%
5: 111859 11.2%
6: 138355 13.8%
7: 166366 16.6%
8: 139346 13.9%
9: 110982 11.1%
10: 82838 8.3%
11: 55935 5.6%
12: 27642 2.8%
Rolls with a two:
305641 30.6%
Rolls with double-two
27634 2.8%
percent of rolls with double two when a roll had at least one two: 9.0
All of the one in sixers are changing the question to: if you have a two, what are the odds that you will roll another two?
They are discounting from the probabilities all of the trials where there is no two at all.
I'd barely call that a proof. More 'a formula' or 'an equation'.Quote: Dalex64Here's a little math and science.
Start with a question:
Come up with a hypothesis: 1/11
Show a mathematical proof: "pr(two sixes)/pr(one or more six) = (1/36)/(11/36) = 1/11"
Quote:Verify it with an experiment:
Several people here have done it with excel. Here is a way to do it with perl:
%<snip some nice code >%
# this tracks the number of rolls where there was at least one two
# I.E. Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."
$rollHasATwo++ if ((2 == $d1) || (2 == $d2));
%<snip some nice code >%
Based on an interpretation of the rules. Not only the probability rules about how the dice will land, but also about the representativeness of the one observed statement of the participant. I contend that "The number of rolls where there was at least one two" might not equal the number of times "Your partner peeks under the cup, and tells you, truthfully, 'At least one of the dice is a 2.'"
Some are, some are not. Alan is. I'm not.Quote:All of the one in sixers are changing the question to: if you have a two, what are the odds that you will roll another two?
Quote:They are discounting from the probabilities all of the trials where there is no two at all.
Hmmmm. That's what some are failing to do. On the Wizard's hypothesis, and on RS's accepted criteria, we are supposed to discard the trials with no two at all. Rightly so, because we know for a fact that there is at least one Deuce showing.
P(Pair of Deuces) = Count(Events with pair of Deuces)/Count(Possible events with at least one Deuce) = 1/11 : Having discarded the events with no Deuces.
Quote: pewI will repeat, the toss of BOTH dice is a single event. You don't toss one until it comes up a two and then toss the other die once 1/6.
Hmmm. Pew, are you saying that it's 1/6? That doesn't fit your own argument. You toss both dice together until one or both 'comes up a two'. Actually, in the OP, the dice were cast once and 'one of them came up a two once'
Quote: Dalex64
All of the one in sixers are changing the question to: if you have a two, what are the odds that you will roll another two?
They are discounting from the probabilities all of the trials where there is no two at all.
Wow, have you got it wrong. We "one in sixers" haven't changed anything. Need I remind you what the question was? Well here it is -- from the very first post:
Quote: DweenYou 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?
The ONLY question is if you have two dice, and you are told that "at least one of the dice is a 2" then "what is the probability that both dice are showing a 2?"
THAT is the only question. And that is why it all depends on the other die. And the other die, no matter if it's die A or die B -- because it's the OTHER die -- has a 1/6 chance for showing a 2.
That's it. Plain and simple. The information has been given for a simple, easy answer. No convoluted math is needed.
Wow, have you got it wrong, Alan. No Seriously. You have it wrong.Quote: AlanMendelsonWow, have you got it wrong. We "one in sixers" haven't changed anything. Need I remind you what the question was? Well here it is -- from the very first post:
The ONLY question is if you have two dice, and you are told that "at least one of the dice is a 2" then "what is the probability that both dice are showing a 2?"
THAT is the only question. And that is why it all depends on the other die. And the other die, no matter if it's die A or die B -- because it's the OTHER die -- has a 1/6 chance for showing a 2.
That's it. Plain and simple. The information has been given for a simple, easy answer. No convoluted math is needed.
Can we agree that the peeker saw both dice before stating the observation? Can we agree that the peeker will always tell you if there is at least one 2 showing? Can we further agree that the peeker will say nothing, or maybe say 'Neither dice came up a 2' if that's the case? On those rules, and on those rules alone, it's categorically 1/11.
But then I personally don't subscribe to those rules.
$:o)
Quote: OnceDearWow, have you got it wrong, Alan. No Seriously. You have it wrong.
Can we agree that the peeker saw both dice before stating the observation?
YES. We can agree that the peeker saw both dice. But that doesn't change anything. It doesn't change the fact that if he says at least one die is a 2, then it is still a 1/6 chance that the other die is a 2.
Why are you trying to make this so complicated? It is so basic, so simple.
And where is the original poster? Probably laughing his butt off.
I am going to ask you one more time. Take two dice and set one on your table or desk showing a 2. Take the "other die" in your hand and look at it. Turn it around. Look at all the faces and count the number of faces. Then count the number of faces you have that would also be a two. Do you get one out of six? Of course you do. You do not get one out of eleven.
Quote: AlanMendelsonI am going to ask you one more time. Take two dice and set one on your table or desk showing a 2. Take the "other die" in your hand and look at it. Turn it around. Look at all the faces and count the number of faces. Then count the number of faces you have that would also be a two. Do you get one out of six? Of course you do. You do not get one out of eleven.
Wrong wrong WRONG with a capital ONG!.
Take two dice and set both on your table or desk.. How many ways can you do that? What is the probability of the dice landing in each of those possible ways ( I contend each way is 1/36). How many of those ways has two deuces showing? (I'll contend just one way) How many equally likely ways has at least one deuce showing (I'll contend 11 different ways)
For equally probable events,
Probability of a favourable event is DEFINED as (Count of favourable possibilities/ Count of ALL possibilities)
Where 'Favourable event' is defined as the event for which you are calculating the probability, whether you like it or not.
Quote: pew"If I roll two dice and one shows a "2" and the second die is a spinner, what are the chances that the "spinner" will settle on another "2"? I think 1/6. What do you think? Are there 11 sides on the spinner?" One out of eleven times it will land on a two.
No Pew. If the first Die has settled as a two, then the spinner has a 1/6 chance of also landing a two. Different scenario, because now we have defined first and second dice.
Quote: ThatDonGuyQuote: ThatDonGuyThat's if a particular die has a 2 on it. We don't know which die/dice has a 2 - only that none of the 25 rolls that do not have any 2s were rolled.
I can't explain it any better than I - and, in fact, you, before you disregarded your correct analysis - already have, so I won't bother.
I knew you'd crack...
P / 6 + (1 - P) / 11, where P is the probability that one of the two dice is still spinning.
There are two cases:
(a) Both dice have stopped, in which case the answer is 1/11 as already shown;
(b) One die has stopped, which, in this case, has to be a 2, and the probability that the spinner lands on 2 is 1/6.
However, it is impossible for the person who is not looking at the dice to determine P.
Effectively, it is the same as pulling the die with the two out from under the cup, then asking, "What is the probability the the unrevealed die is also a two?"
The answer is one-in-six because the peeker's statement assigns a value to one of the dice, and thereby assigns an order to the two dice before the question is asked. This effectively eliminates the five "reverse" options from the 1/11 answer, leaving 1 chance in 6.
Quote: AyecarumbaOn a side track, I notice that some of the arguments for 1/11 only count the occurence of the pair of two's one time. According to the reasoning that every combination of unknown dice needs to be considered (e.g., 5,2 and 2,5 are discrete since you don't know which is first or second), aren't there two occurences of each pair (e.g., "2,2") to consider, since you don't know which was the first or second?
Really? I mean, REALLY?
If you REALLY think about what you're saying, then you'll finally understand what the 1/11 crowd is saying.
Quote: DweenYou 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."
Here is how we will bet (I am not the first to make this offer in this thread):
- We will take turns rolling two dice under a cup (for fairness.)
- The roller will peek under the cup. If one of the dice is a two he will say so. He will then reveal the dice.
- Upon confirmation that there are no twos the bet is a push and we move to the next roll.
- When there is at least one two, every time there it is a pair of twos I will pay you 7 units. Every time it is not a pair of twos you will pay me one unit.
- I will play this game for up to $10 units until one of us loses $1000. I could be convinced to go higher but I don't want to bankrupt anyone.
Quote: AyecarumbaI still agree with Alan. The question is not what is the probability prior to the peek. The question is asked after the peek, and after the information that one die is showing a two.
Effectively, it is the same as pulling the die with the two out from under the cup, then asking, "What is the probability the the unrevealed die is also a two?"
That's intersting, but NO.
Look here at what possible scenarios exist if you pull out an identified Deuce.
There are actually 11 scenarios where the peeker can pull a Deuce from under, and with one of those he even gets a choice of which.
Die A Die B What's left after removing the Identified Deuce
1____1
2____1________1
3____1
4____1
5____1
6____1
1____2________1
2____2________2______Whichever Deuce you pull out here would leave one Deuce. This is the 1/11 possibility
3____2________3
4____2________4
5____2________5
6____2________6
1____3
2____3________3
3____3
4____3
5____3
6____3
1____4
2____4________4
3____4
4____4
5____4
6____4
1____5
2____5________5
3____5
4____5
5____5
6____5
1____6
2____6________6
3____6
4____6
5____6
6____6
That Leaves 11 possible scenarios for the 'Other Die'
Quote:On a side track, I notice that some of the arguments for 1/11 only count the occurence of the pair of two's one time. According to the reasoning that every combination of unknown dice needs to be considered (e.g., 5,2 and 2,5 are discrete since you don't know which is first or second), aren't there two occurences of each pair (e.g., "2,2") to consider, since you don't know which was the first or second?
There is no first and second.
Quote: jml24
- I will play this game for up to $10 units until one of us loses $1000. I could be convinced to go higher but I don't want to bankrupt anyone.
In the UK, I'll go to £20 per roll on condition that we play until 100 Deuces show. Player must come to me though.
Quote: OnceDearWrong wrong WRONG with a capital ONG!.
Take two dice and set both on your table or desk.. How many ways can you do that? What is the probability of the dice landing in each of those possible ways ( I contend each way is 1/36). How many of those ways has two deuces showing? (I'll contend just one way) How many equally likely ways has at least one deuce showing (I'll contend 11 different ways)
For equally probable events,
Probability of a favourable event is DEFINED as (Count of favourable possibilities/ Count of ALL possibilities)
Where 'Favourable event' is defined as the event for which you are calculating the probability, whether you like it or not.
Why are you doing this with two dice? We are told in the original problem that at least one die is a two. Therefore, with one die as a 2 the question becomes what are the chances that a 2 will show on the second die.
Simply take the second die -- and it doesn't matter whether you use die A or die B -- and determine the chance. It's 1 out of 6.
There is absolutely no reason IN THIS QUESTION/PROBLEM to jump through all of the hoops you are jumping through. We are given information to deal with, and that information is at least ONE DIE IS A TWO therefore the question pertains to the second die.
And since there are ONLY TWO DICE FOR THIS EXERCISE we can simply put one die to the side (with a 2 showing) and use the second die to determine those odds. Again, one out of six.
You are just making it too complicated for a simple problem.
If there were more than two dice, or if we were not specifically told that at least one die is a 2 the problem would need a more detailed method to solve.
Quote: jml24I think it is telling that none of the 1 in sixers are willing to bet money with an amazing advantage. I will repeat the offer but unfortunately I may not be in Vegas again until the end of the year. If any of the math-impaired live near Seattle I will meet you to make a bet based as closely as possible on the original question.
Here is how we will bet (I am not the first to make this offer in this thread):
- We will take turns rolling two dice under a cup (for fairness.)
- The roller will peek under the cup. If one of the dice is a two he will say so. He will then reveal the dice.
- Upon confirmation that there are no twos the bet is a push and we move to the next roll.
- When there is at least one two, every time there it is a pair of twos I will pay you 7 units. Every time it is not a pair of twos you will pay me one unit.
- I will play this game for up to $10 units until one of us loses $1000. I could be convinced to go higher but I don't want to bankrupt anyone.
Just stop it. This bet has nothing to do with the question at hand or the problem.
A bigger question and certainly a much bigger problem is why so many people here can't understand that the question involves what a second die shows after the original poster told us that at least one die is a two? This is not a question about two unknown dice. This is a question of figuring the odds with at least one die known. There is a problem here with reading comprehension.
Quote: AlanMendelsonJust stop it. This bet has nothing to do with the question at hand or the problem.
How does it not? I followed the wording of the original question exactly. The only change is that instead of asking "what is the probability that both dice are showing a two" I changed it to "I will pay you 7:1 if both dice are twos." If the probability is 1 in 6 then 5:1 would be a fair payoff and you will make a killing.
Quote: AlanMendelsonWhy are you doing this with two dice? We are told in the original problem that at least one die is a two. Therefore, with one die as a 2 the question becomes what are the chances that a 2 will show on the second die.
You are So So wrong. If you think we are jumping through hoops, you do what has been asked of you ! look at my recent post just above here Read through all of the possible outcomes. Count them. Rather than beating on about being able to set one die aside, just find a flaw, any flaw in the logic there. No need to talk of spinners or first or second. Just count the possible outcomes. You might just notice a cluster of results where die B is pulled out as a deuce.
If you do it your way and pull out either of Die A or Die B at the outset, then the answer is 1/6. BUT you need to allow both dice the opportunity of fulfilling its destiny. Thems the rules.
Quote:Simply take the second die -- and it doesn't matter whether you use die A or die B -- and determine the chance. It's 1 out of 6.
No No NO NO NO. You HAVE to account for every different possibility where one or more of a pair of dice is a deuce: Not just count the possibilities of your favourite die is a deuce.
Take me up on my wager. Please.
Quote: jml24Perhaps Alan is making a non-obvious assumption about the question. If it is stipulated that the peeker can only see one of the dice, and he sees that it is a two without looking at the second die, then the answer is 1 in 6.
Nope. From earlier where I challenged him.
Quote: AlanMendelsonYES. We can agree that the peeker saw both dice. But that doesn't change anything. It doesn't change the fact that if he says at least one die is a 2, then it is still a 1/6 chance that the other die is a 2.
Psalm 25:16 Turn to me and be gracious to me, for I am lonely and afflicted. Proverbs 18:2 A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
This is the kind of problems in basic probability courses that test the understanding of conditional probalities and dependent events.
The people who say 1/6 think that the following 2 problems are equivalent.
1. 2 Dice are rolled. At LEAST one of the dice is 2. What is the probability the both dice are 2.
2. 1 Dice is rolled. What is the probability that it is 2.
The 2 problems are NOT equivalent.
The first problem becomes equivalent to the second ONLY IF the one dice is shown to you and it is a 2 and then asked what is the probability that the other unseen dice is a 2. Then its is 1/6.
ALTERNATIVE PRESENTATION OF THE PROBLEM.
As stated BUT one dice is randomly chosen out of the cup and you are asked the probability of the specific dice being a 2.
IF you have zero information then the answer is 1/6.
With the information as stated in the problem, the specific dice chosen has probability of 6/11 of being a 2.
Not 1/6 as a random dice, Not 100% for the dice that is a 2 but not known which one, and NOT 1/2 (1/2 would be if the problem stated that one and only one dice is a 2).
The same probability of 6/11 applies to the other not chosen dice.
In independent events you would multiply the 2 probs, ie 6/11 * 6/11 = 36/121 for both dice to be 2.
But the 2 dice are not independent
The probability is then calculated as 6/11 * 1/6 = 1/11
