Two Dice Puzzle
October 21st, 2013 at 10:01:16 AM
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Along 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.
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.
Pleased to meet you, name's Francis Pumphandle, but everyone calls me Pip. Cheese balls are one of my all-time favorite foods. I always seem to meet the most interesting people when I'm around them, too. In fact, cheese balls bring to mind the time I met Bob Barker. Yes, indeed -- Bob Barker, star of the most popular morning game show. He's an emcee, a host and a celebrity all rolled into one. Anyway, eight months ago, it was Tuesday the 17th, I believe, or it might have been the 18th -- no, no, it was definitely the 17th, because it was precisely one week after my Aunt Lucretia's birthday, which is the 10th. Aunt Lucretia's quite a woman -- loves to cook. She prepares a fabulous "war-shoo-off" -- that's a Chinese duck dish. I love Chinese food. I once went to a party where they served Chinese food and cheese balls. Now that was a Catch-22 situation. Catch-22 was a movie, you know. It was long -- VERY long. They say the book was better, but it was a novel, and I never finish reading those things. Of course, a lot of people don't read much nowadays; they watch television. I caught a program on PBS last night -- a very good show on chimpanzees in the media. They had a clip of Jay Fred Mug and a chimp on the Today Show, but it was Fred's chimpanzee's girlfriend that had me stumped. I couldn't remember her name, so I looked it up. Her name was Fibi B. Bibi. Anyway, as I was saying, eight months ago, Tuesday the 17th, I went downtown on a nice relaxing stroll. I love to relax. In fact, relaxing is a hobby of mine. Some people play golf, others like tennis, horseshoes, bridge, canasta, and other such fancy hobbies. Now another hobby enjoyed by many is knitting. My grandmother was a great knitter -- knitted this sweater I'm wearing. It's red, which is not my favorite color. I prefer mauve or a mustard yellow. Now, don't get me wrong, red is o.k. for ties and suspenders, but with sweaters I prefer more neutral colors. But when I'm relaxing, I don't care WHAT I wear -- long pants, bermuda shorts, t-shirts, or formal attire, you name it -- anything goes. Now, on the 17th, during my relaxing stroll, I recall wearing my herringbone jacket, my Laughlin, Nevada souvenir tie, and my charcoal grey slacks -- or was it the navy slacks? Oh, I suppose it doesn't really matter, does it? What matters is comfort. You know, I once stayed at a Comfort Inn -- warm, cozy, comfortable. I love comfort. It goes along with that pastime of mine -- relaxing. Now, for me, there's nothing more relaxing than a nice leisurely stroll like the one I took eight months ago on the 17th. It was a bright, sunny day, which of course is the optimum condition for relaxed strolling. And as I walked along, I found myself humming a haunting melody. I kept humming and humming and humming and humming. I couldn't get the tune out of my head. I racked my brain to come up with the title, but to no avail. You see, I'm not terribly musical -- and yet, I'd always wanted to play a musical instrument and be like my musical hero, Leo Sayer. But who can compete with Leo? I think I was just scared I'd fail. Well, I decided right then and there to go buy a musical instrument. So on the particular Tuesday the 17th to which I was referring, I went down to the Sixth Street Musical Emporium to buy a new tambourine, a terribly soothing instrument contrary to popular opinion. And as I was strolling along, I detected a wonderful scent in the morning air. "What could it be?" I asked myself. So I went toward that marvelous scent, distracted by its aroma from my musical mission. The odor was a mix of orchid flowers and bologna, which, of course, is one of the world's most underappreciated luncheon meats -- that and pimento loaf. I love a good pimento-loaf-and-mayo sandwich -- the more pimentos, the better. Why just the mention of pimentos makes my taste buds stand up and say "Howdy". Now there's an interesting word -- "howdy". Is it from "How are you?", or maybe "How ya doing?" "Howdy"'s one of those strange words that really HAS no origin. I like saying, "How do" more than "Howdy" -- more formal, I think -- not too flowery. But the flowery aroma of that particular Tuesday morning carried me on my fragrant quest. Now, the smell was actually less bologna and more orchid, the beautiful flower found on the island state of Hawaii. Of course, I wasn't in Hawaii, so I needed to search out the location of the nearest orchid. So, I visited every flower shop in town. Well, to make a long story short, not a SINGLE flower shop in town had ANY orchids in stock, which seemed mighty curious to me. Now, as we all know, curiosity killed the cat, but since I'm not feline, I wasn't too worried. Felines are funny creatures, don't you think? I had a cat once. It used its claws to tear my living room couch to shreds. It was a comfy couch, too -- had a sleepaway bed in it with a foam rubber mattress. Now, I bought the couch AND the mattress at Levine's department store on Third Avenue the very same afternoon of that relaxing stroll aforementioned. I also bought myself a lovely tambourine on that same shopping expedition. Anyway, I didn't want to pay extra for the delivery of the couch, so I decided to carry the couch home myself. It was quite cumbersome, and getting it through the store's revolving doors was a bit of a challenge. And just as I emerged onto the street, by accident I bumped into a well-dressed man with an orchid in his lapel. It was Bob Barker, and he was eating and bologna-and-cheese-ball sandwich. Well, it's been nice chatting with you. Bye!
-Dween!
October 21st, 2013 at 10:22:05 AM
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pr(two sixes)/pr(one or more six) = (1/36)/(11/36) = 1/11
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
October 21st, 2013 at 10:57:07 AM
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The Wizard is correct under the assumption that both dice are standard six-sided dice with 1..6 on them, instead of something like:
Last edited by: unnamed administrator on Oct 4, 2016
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice."
-- Girolamo Cardano, 1563
October 21st, 2013 at 1:39:00 PM
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... 36 combinations of two 6-sided dices, only 11 of them show at least a 2, from which one is 2-2. Hence the probability would be 1/11.
..this would only hold if the player would say nothing when no 2 were spotted.
If (before peeking) your friend is determined to say "at least one of the dice is a X" where X is one of the numbers he may spots, the probability for a pair would be different.
If the friend sees any pair (1/6), probability of "X-X" where X is the named number, is one.
If the friend sees a non-pair (5/6), probability of "X-X" is zero whatever he choses for X.
Hence, like the Monty-Hall-Problem, if your friend must always say "At least one of the dice is a X" (while not lying), probability of a pair is 1/6.
April 13th, 2015 at 1:28:17 PM
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Short answer: 1/6 = 16.667% probability.
(God damn it, Michael, we've ruled OUT a bunch of your denominators.)
I love showing the Wizard wrong!
Short answer:
One-Third of all possible rolls have a deuce (or two of them) in them.
Two-thirds of all possible rolls lack any deuce.
Out of the 12 rolls that contain a deuce (or two deuces), Two have a pair of deuces.
Two out of twelve = 1/6 = 16.667% Period, the end.
(God damn it, Michael, we've ruled OUT a bunch of your denominators.)
I love showing the Wizard wrong!
Short answer:
| Prob | Die-1 | Die-2 | Note |
|---|---|---|---|
| 1/36 | 2 | 1 | |
| 2/36 | 2 | 2 | |
| 1/36 | 2 | 3 | |
| 1/36 | 2 | 4 | |
| 1/36 | 2 | 5 | |
| 1/36 | 2 | 6 | |
| Prob | Die-1 | Die-2 | Note |
| 1/36 | 1 | 2 | |
| 0 | 2 | 2 | Already accounted for above (pair of deuces) |
| 1/36 | 3 | 2 | |
| 1/36 | 4 | 2 | |
| 1/36 | 5 | 2 | |
| 1/36 | 6 | 2 |
One-Third of all possible rolls have a deuce (or two of them) in them.
Two-thirds of all possible rolls lack any deuce.
Out of the 12 rolls that contain a deuce (or two deuces), Two have a pair of deuces.
Two out of twelve = 1/6 = 16.667% Period, the end.
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 13th, 2015 at 2:12:10 PM
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Quote: indignant99Short answer: 1/6 = 16.667% probability.
(God damn it, Michael, we've ruled OUT a bunch of your denominators.)
I love showing the Wizard wrong!
And perhaps some day, you will...but it's not today.
The probability of throwing a pair of 2s is not 1/18.
Think about it. Let's change the problem from 2s to 6s. According to you, the probability of throwing a pair of 6s is 1/18. Pretty much every craps table in Vegas offers 29-1 or better on this. Talk about an advantage play!
In your table, change "die 1" to "red die" and "die 2" to blue die. 6 out of 36 rolls will have the red die be 2, and of those 6, only one will have the blue die be 2 as well.
April 13th, 2015 at 2:22:07 PM
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In this problem, the probability is not the same as rolling two dice since you have new data. I think 1/6 may be correct, since you have truthful knowledge that at least one di is already a 2.
"Rule No.1: Never lose money. Rule No.2: Never forget rule No.1." -Warren Buffett on risk/return
April 13th, 2015 at 2:47:28 PM
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How do you repress the occurrence of "2" on the unknown die, to
1=twice
2=once
3=twice
4=twice
5=twice
6=twice
(one out of 11)? Absurd.
1=twice
2=once
3=twice
4=twice
5=twice
6=twice
(one out of 11)? Absurd.
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 13th, 2015 at 2:50:00 PM
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Quote: surrender88sIn this problem, the probability is not the same as rolling two dice since you have new data. I think 1/6 may be correct, since you have truthful knowledge that at least one di is already a 2.
What do you mean, "you have new data"?
Either you have "at least one 2", or you have zero 2s, so the probability of at least one 2 = 1 - the probability of zero 2s = 1 - 25/36 = 11/36. In other words, 11 out of the 36 rolls have one 2. How many of these 11 have two 2s?
April 13th, 2015 at 2:59:55 PM
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Quote: indignant99How do you repress the occurrence of "2" on the unknown die, to
1=twice
2=once
3=twice
4=twice
5=twice
6=twice
(one out of 11)? Absurd.
Either die can be "the unknown die".
Let one of the dice be red, and the other blue.
The columns in this table refer to the blue die, and the rows to the red die.
As you can see, 11 of the 36 rolls have at least one 2; of these, only one has two 2s.
Probability = 1 / 11.
The columns in this table refer to the blue die, and the rows to the red die.
| Red Die | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| 1 | 0 | 1 | 0 | 0 | 0 | 0 |
| 2 | 1 | 2 | 1 | 1 | 1 | 1 |
| 3 | 0 | 1 | 0 | 0 | 0 | 0 |
| 4 | 0 | 1 | 0 | 0 | 0 | 0 |
| 5 | 0 | 1 | 0 | 0 | 0 | 0 |
| 6 | 0 | 1 | 0 | 0 | 0 | 0 |
As you can see, 11 of the 36 rolls have at least one 2; of these, only one has two 2s.
Probability = 1 / 11.
April 13th, 2015 at 3:03:56 PM
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Quote: ThatDonGuyAnd perhaps some day, you will...but it's not today.
Yes, it is today. With one deuce, the other can only be 1,2,3,4,5,6. One-sixth.
Quote: ThatDonGuyThe probability of throwing a pair of 2s is not 1/18.
When one of them is already a deuce, the second deuce is 1 out of six.
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 13th, 2015 at 3:18:28 PM
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Quote: ThatDonGuyWhat do you mean, "you have new data"?
Don't damage your own reputation, Don.
The "reveal" has already narrowed the outcomes to:
2-1, 2-2, 2-3, 2-4, 2-5, 2-6... (first die)
1-2, 2-2, 3-2, 4-2, 5-2, 6-2. (second die)
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 13th, 2015 at 3:41:57 PM
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Agree with wizard.
btw for those people who have mastered the spoiler tags, the trick is
(i) use "[" and "]" putting "SPOILER" between
(ii) adds words etc to be hidden
(iii) then close using "[/" and "]" again putting "SPOILER" between.
Another method is to quote someone else and look at the resultant code.
Of the original chances, there were 36 possibles, but only 11 possible outcomes remain: 1-2 2-2 3-2 4-2 5-2 6-2 2-1 2-3 2-4 2-5 2-6. Of these only one has 2-2.
Hence chances are 1/11.
Hence chances are 1/11.
(i) use "[" and "]" putting "SPOILER" between
(ii) adds words etc to be hidden
(iii) then close using "[/" and "]" again putting "SPOILER" between.
Another method is to quote someone else and look at the resultant code.
April 13th, 2015 at 3:59:21 PM
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Count again: 2 out of twelve.
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 13th, 2015 at 4:29:15 PM
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Quote: indignant99Don't damage your own reputation, Don.
The "reveal" has already narrowed the outcomes to:
2-1, 2-2, 2-3, 2-4, 2-5, 2-6... (first die)
1-2, 2-2, 3-2, 4-2, 5-2, 6-2. (second die)
The 2-2 on the first line means "the first die is a 2, and the second die is a 2".
The 2-2 on the second line means "the first die is a 2, and the second die is a 2".
These are the same roll. Either that, or you can roll 2-2 twice out of 36 rolls, which means you can roll 6-6 twice out of 36 rolls as well, which means that 29-1 the casino offers for rolling a 12 is the best bet in the casino.
I have a feeling the problem is, you're misunderstanding what "at least one of the dice is a 2" means.
April 13th, 2015 at 5:46:34 PM
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i see this
yes, i hide the other
what you think?
trick question for sure
Mully
yes, i hide the other
what you think?
trick question for sure
Mully
I Heart Vi Hart
April 13th, 2015 at 7:39:48 PM
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I graciously concede.
These are all the outcomes that can fall out of the cup:
Now, abbreviating the table:
One out of eleven "qualified" events, yield a double deuce.
My blind brain did not see (1-2, 2-1), (3-2, 2-3), (4-2, 2-4), (5-2, 2-5), (6-2, 2-6).
These are all the outcomes that can fall out of the cup:
| Prob | Die-1 | Die-2 | Note |
|---|---|---|---|
| 1/36 | 1 | 1 | excluded |
| 1/36 | 1 | 2 | one deuce |
| 1/36 | 1 | 3 | excluded |
| 1/36 | 1 | 4 | excluded |
| 1/36 | 1 | 5 | excluded |
| 1/36 | 1 | 6 | excluded |
| 1/36 | 2 | 1 | one deuce |
| 1/36 | 2 | 2 | two deuces |
| 1/36 | 2 | 3 | one deuce |
| 1/36 | 2 | 4 | one deuce |
| 1/36 | 2 | 5 | one deuce |
| 1/36 | 2 | 6 | one deuce |
| 1/36 | 3 | 1 | excluded |
| 1/36 | 3 | 2 | one deuce |
| 1/36 | 3 | 3 | excluded |
| 1/36 | 3 | 4 | excluded |
| 1/36 | 3 | 5 | excluded |
| 1/36 | 3 | 6 | excluded |
| 1/36 | 4 | 1 | excluded |
| 1/36 | 4 | 2 | one deuce |
| 1/36 | 4 | 3 | excluded |
| 1/36 | 4 | 4 | excluded |
| 1/36 | 4 | 5 | excluded |
| 1/36 | 4 | 6 | excluded |
| 1/36 | 5 | 1 | excluded |
| 1/36 | 5 | 2 | one deuce |
| 1/36 | 5 | 3 | excluded |
| 1/36 | 5 | 4 | excluded |
| 1/36 | 5 | 5 | excluded |
| 1/36 | 5 | 6 | excluded |
| 1/36 | 6 | 1 | excluded |
| 1/36 | 6 | 2 | one deuce |
| 1/36 | 6 | 3 | excluded |
| 1/36 | 6 | 4 | excluded |
| 1/36 | 6 | 5 | excluded |
| 1/36 | 6 | 6 | excluded |
Now, abbreviating the table:
| Prob | Die-1 | Die-2 | Note |
|---|---|---|---|
| 1/36 | 1 | 2 | one deuce |
| 1/36 | 2 | 1 | one deuce |
| 1/36 | 2 | 2 | two deuces |
| 1/36 | 2 | 3 | one deuce |
| 1/36 | 2 | 4 | one deuce |
| 1/36 | 2 | 5 | one deuce |
| 1/36 | 2 | 6 | one deuce |
| 1/36 | 3 | 2 | one deuce |
| 1/36 | 4 | 2 | one deuce |
| 1/36 | 5 | 2 | one deuce |
| 1/36 | 6 | 2 | one deuce |
One out of eleven "qualified" events, yield a double deuce.
My blind brain did not see (1-2, 2-1), (3-2, 2-3), (4-2, 2-4), (5-2, 2-5), (6-2, 2-6).
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 13th, 2015 at 7:45:25 PM
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If the person says "The first die I looked at was a two. What's the odds that the second is also a two?" Then the answer is 1/6.
But the person said "At least one die is a two." You don't know if it was the first die he looked at or not. All you know is that there are eleven combinations that will produce at least one die with the two showing. And out of them, there's only one double two.
So the Wiz is right: 1/11.
But the person said "At least one die is a two." You don't know if it was the first die he looked at or not. All you know is that there are eleven combinations that will produce at least one die with the two showing. And out of them, there's only one double two.
So the Wiz is right: 1/11.
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
April 13th, 2015 at 7:52:13 PM
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Here's another way to look at it.
The person peeks, then says "At least one die is NOT a two."
What are the odds that neither is a two?
The person peeks, then says "At least one die is NOT a two."
What are the odds that neither is a two?
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
April 13th, 2015 at 8:11:14 PM
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Quote: indignant99I graciously concede.
Defending 1/6 was not entirely without merit:
- If the person looks at ONE die and sees a 2, the probability of a 2-2 pair is of course 1/6
- If the person looks at BOTH dice and sees at least one 2, the probability of a 2-2 pair is 1/11
The question, as worded, has the person looking at both dice and thus a probability of 1/11.
April 13th, 2015 at 8:34:44 PM
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12,22,32,42,52,62,21,23,24,25,26
Are the possibilities.
1/11
If it's "there are 2 dice, 1 red and 1 blue, if the blue one is a 2, what's the chance the red one is a 2?" then the answer is surely 1/6.
Are the possibilities.
1/11
If it's "there are 2 dice, 1 red and 1 blue, if the blue one is a 2, what's the chance the red one is a 2?" then the answer is surely 1/6.
April 13th, 2015 at 8:54:42 PM
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"Your partner peeks under the cup,Quote: JB-<snip>
The question, as worded,
has the person looking at both dice
and thus a probability of 1/11.
and tells you,
truthfully, "At least one of the dice is a 2."
huh?
nothing about partner actually looked at both dice
"Sally, did you look at both dice?"
Sally replies
"maybe, maybe nots" (i can be bought)
"has the person looking at both dice"
no way in California
still a trick question
Sally
I Heart Vi Hart
April 14th, 2015 at 12:52:59 AM
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Quote: mustangsallystill a trick question
I interpreted it as seeing both dice for two reasons:
1) by lifting the lid of the cup, it seems unlikely that they wouldn't see both dice
2) the question would be pointless if only one die was seen
April 14th, 2015 at 1:08:07 AM
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Quote: DJTeddyBearIf the person says "The first die I looked at was a two. What's the odds that the second is also a two?" Then the answer is 1/6.
But the person said "At least one die is a two." You don't know if it was the first die he looked at or not. All you know is that there are eleven combinations that will produce at least one die with the two showing. And out of them, there's only one double two.
So the Wiz is right: 1/11.
I admit I'm not smart when it comes to math, but this makes no sense to me.
If someone told me that at least one die was a 2, that means you have a deuce on at least one die. That leaves the question what is on the second die? And on a six sided dice with each face showing a number from 1 to 6 the answer is 1/6 is a deuce.
So it only makes sense to me that the answer can only be one out of six when asked the question.
I don't think it's a matter of looking at all of the combinations that have a deuce showing. The questioner has already answered that at least one die is showing a 2. That only leaves what are the possibilities for the second die. And that's 1/6.
So in my simple, non math mind, I think the simple answer is the correct answer: 1/6.
It is very similar to what we players go through at craps. We see one die come to rest on, for example, the 6 -- while the second die spins. And we say to ourselves, there is only a 1/6 chance of a 7-out.
April 14th, 2015 at 4:18:53 AM
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Quote: indignant99I graciously concede.
These are all the outcomes that can fall out of the cup:
Prob Die-1 Die-2 Note 1/36 1 1 excluded 1/36 1 2 one deuce 1/36 1 3 excluded 1/36 1 4 excluded 1/36 1 5 excluded 1/36 1 6 excluded 1/36 2 1 one deuce 1/36 2 2 two deuces 1/36 2 3 one deuce 1/36 2 4 one deuce 1/36 2 5 one deuce 1/36 2 6 one deuce 1/36 3 1 excluded 1/36 3 2 one deuce 1/36 3 3 excluded 1/36 3 4 excluded 1/36 3 5 excluded 1/36 3 6 excluded 1/36 4 1 excluded 1/36 4 2 one deuce 1/36 4 3 excluded 1/36 4 4 excluded 1/36 4 5 excluded 1/36 4 6 excluded 1/36 5 1 excluded 1/36 5 2 one deuce 1/36 5 3 excluded 1/36 5 4 excluded 1/36 5 5 excluded 1/36 5 6 excluded 1/36 6 1 excluded 1/36 6 2 one deuce 1/36 6 3 excluded 1/36 6 4 excluded 1/36 6 5 excluded 1/36 6 6 excluded
Now, abbreviating the table:
Prob Die-1 Die-2 Note 1/36 1 2 one deuce 1/36 2 1 one deuce 1/36 2 2 two deuces 1/36 2 3 one deuce 1/36 2 4 one deuce 1/36 2 5 one deuce 1/36 2 6 one deuce 1/36 3 2 one deuce 1/36 4 2 one deuce 1/36 5 2 one deuce 1/36 6 2 one deuce
One out of eleven "qualified" events, yield a double deuce.
My blind brain did not see (1-2, 2-1), (3-2, 2-3), (4-2, 2-4), (5-2, 2-5), (6-2, 2-6).
I absolutely withdraw my concession. With the proviso that at least one die shows a deuce, here are all ways the pair of dice can fall:
| Event # | Die-1 | Die-2 |
|---|---|---|
1 | 2 | 1 |
2 | 2 | 3 |
3 | 2 | 4 |
4 | 2 | 5 |
5 | 2 | 6 |
6 |
2 | 2 |
7 | 1 | 2 |
8 | 3 | 2 |
9 | 4 | 2 |
10 | 5 | 2 |
11 | 6 | 2 |
Although you don't know whether Die-1 or Die-2 was revealed, the incontrovertible fact is that it was either Die-1 exclusive-or Die-2.
If it was Die-1 (revealed), then only 6 outcomes are possible. Events 1, 2, 3, 4, 5, 6.
If it was Die-2 (revealed), likewise only 6 outcomes are possible. Events 6,7,8,9,10,11.
The answer is one-sixth = 1/6 = 16.667%
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 14th, 2015 at 4:25:24 AM
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Perhaps the logic will make more sense to some in a gaming context. You see a table offering to pay 8 times your wager if the dice come up four the hard way. No die showing two is a push, one die at two is a loss. Do you tap your 401K and play table max, or do you pass on the offer?
“You don’t bring a bone saw to a negotiation.” - Robert Jordan, former U.S. ambassador to Saudi Arabia
April 14th, 2015 at 6:30:31 AM
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I wrestled with the Monty Hall problem and eventually got it in my "minds eye" (people definitely see math in different ways conceptually) Now this, my head is going to fall off thinking about it.
April 14th, 2015 at 6:46:19 AM
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Quote: pew...my head is going to fall off thinking about it.
Nah. The simple answer is the right answer. One Die shows a 2.
The other die shows 1, 2, 3, 4, 5, 6. One-sixth of the times, the second die is a 2.
There has been way too much skewed thinking on this thread/puzzle.
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 14th, 2015 at 6:56:43 AM
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Um, MangoJ's response and subsequent caveat back on Page 1 sufficiently explain both the 'correct' answer and the 'other correct' answer to the original question.
BTW, why are we digging up 2 year old threads that have been asked and answered?
BTW, why are we digging up 2 year old threads that have been asked and answered?
"Dealer has 'rock'... Pay 'paper!'"
April 14th, 2015 at 7:05:56 AM
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and interpreted i agree that is what you didQuote: JBI interpreted it as seeing both dice for two reasons:
and any one can do the same
only your opinion hereQuote: JB1) by lifting the lid of the cup, it seems unlikely that they wouldn't see both dice
read the stated question as i did and C
(also testing with a cup and dice it is easy to only view one die, but not every time as the many times the dice land together and are too close to see one and not the other)
Oh, never pointless questionsQuote: JB2) the question would be pointless if only one die was seen
sad if that is what you really believe (like a religion)
that is what a trick question is all about
i guess here
you
have not been in a college stats and probs class for some time (more than 4 years)
these type of trick questions (meaning more than one meaning could be seen) have a great purpose
at least
that is what i was told in me college days
Sally
I Heart Vi Hart
April 14th, 2015 at 7:24:41 AM
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excellent observation!Quote: AlanMendelsonIt is very similar to what we players go through at craps.
We see one die come to rest on, for example, the 6 -- while the second die spins.
And we say to ourselves, there is only a 1/6 chance of a 7-out.
i see this many times too (2)
but only one die spins like a top, never both (unless i do not pay attention)
how does the shooter make that happen, where only one die spins from tossing two dice?
I ask and they never say how they do it (the shooter that is)
i wood pay to learn that skill
(i have only done this meself maybe twice, and did not know how i did it)
Alan, you have a video of what you talk about?
Mully
I Heart Vi Hart
April 14th, 2015 at 7:31:51 AM
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Below are results of ten simulations of 3600 dice rolls each:
| Trial | At least one die was a 2. | Both dice were 2. |
|---|---|---|
1 | 1119 | 98 |
2 | 1079 | 82 |
3 | 1123 | 83 |
4 | 1106 | 116 |
5 | 1110 | 89 |
6 | 1133 | 97 |
7 | 1080 | 97 |
8 | 1115 | 93 |
9 | 1068 | 100 |
10 | 1095 | 101 |
TOTAL | 11028 | 956 |
April 14th, 2015 at 7:38:42 AM
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i say you went after a different questionQuote: ChesterDogBelow are results of ten simulations of 3600 dice rolls each:
the OP question is too (2) vague
too much info missing
i could also assume the dice are rolled for a purpose to see how often they will both be the same
meaning a double
but the author of the question did not word the question correctly
i look to be right (left) too
Sally
I Heart Vi Hart
April 14th, 2015 at 8:03:52 AM
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The question is like the two envelopes in that possibilities, and the intersection thereof, may become confused with expected value.
If someone said to me that at least one die were a 2, I would say prove it. The someone would then reveal the one of the dice which is a 2. But, the someone would always have the choice of which of the two dice to reveal first (coming into the scenario). Are many other non-2 possibilities for the die not revealed (but associated with the 2 of the die revealed first) than six. The die not revealed brings more possibilities into the question because of the "choice" of which die to reveal in the event of at least a 2.
When at least one die is 2, then half the time the one die is the claimed-to-exist 2, but the other half of the time the other die is the claimed-to-exist 2? So that half of six possibilities (of the other die when one die is 2) plus half of another six possibilities (of one die when other die is 2) equals six possibilities. No.
If someone said to me that at least one die were a 2, I would say prove it. The someone would then reveal the one of the dice which is a 2. But, the someone would always have the choice of which of the two dice to reveal first (coming into the scenario). Are many other non-2 possibilities for the die not revealed (but associated with the 2 of the die revealed first) than six. The die not revealed brings more possibilities into the question because of the "choice" of which die to reveal in the event of at least a 2.
When at least one die is 2, then half the time the one die is the claimed-to-exist 2, but the other half of the time the other die is the claimed-to-exist 2? So that half of six possibilities (of the other die when one die is 2) plus half of another six possibilities (of one die when other die is 2) equals six possibilities. No.
Nonsense is a very hard thing to keep up. Just ask the Wizard and company.
April 14th, 2015 at 8:11:21 AM
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Quote: ThatDonGuyI have a feeling the problem is, you're misunderstanding what "at least one of the dice is a 2" means.
Apparently, we can't come to a consensus on what it means.
I for one read it as, "Either one of the dice is a 2 and the other isn't, or both dice are 2," which is the interpretation used in what indignant99 eventually called his "wrong" analysis.
April 14th, 2015 at 9:23:40 AM
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Quote: AlanMendelsonIt is very similar to what we players go through at craps. We see one die come to rest on, for example, the 6 -- while the second die spins. And we say to ourselves, there is only a 1/6 chance of a 7-out.
That's a different situation, and 1/6 is correct. In your example, the dice are segregated. One has settled the other is still an unknown random result. It's kinda like the prior example of a red and blue die and the person says the red die is a two.
The original question is more like a craps throw where both die are spinning and someone says "If one die settles on two, what's the odds that the other will also be a two?" Answer: 1/11.
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April 14th, 2015 at 9:50:31 AM
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Quote: DJTeddyBearThat's a different situation, and 1/6 is correct. In your example, the dice are segregated. One has settled the other is still an unknown random result. It's kinda like the prior example of a red and blue die and the person says the red die is a two.
The original question is more like a craps throw where both die are spinning and someone says "If one die settles on two, what's the odds that the other will also be a two?" Answer: 1/11.
It is best to say "if either dice...". That way the Alans of the world can try to have a chance of understanding.
Alan, remember this scenario: the first die lands on a non-2 number, and the second die is spinning. If that die lands on a 2, then the roll meets the " if either of the dice land on 2..." requirement.
April 14th, 2015 at 11:19:46 AM
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Quote: RS...If it's "there are 2 dice, 1 red and 1 blue, if the blue one is a 2, what's the chance the red one is a 2?" then the answer is surely 1/6.
And if both dice are the same color, it doesn't change the answer from 1/6.
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 14th, 2015 at 11:32:48 AM
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Quote: indignant99And if both dice are the same color, it doesn't change the answer from 1/6.
You might be right, if the original statement was, "One of the dice is a 2", instead of, "At least one of the dice is a 2."
If both dice have stopped, and both of the numbers are known, then the answer is 11/36, as you yourself discovered before tossing out the correct (in this case) solution.
If only one die has stopped, then the probability is 1/6.
However, we don't know what the probability is that both dice have stopped, so in that instance, the answer is, "It is impossible to calculate without knowing if both dice have stopped or not."
April 14th, 2015 at 11:38:28 AM
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Quote: ThatDonGuy...If both dice have stopped, and both of the numbers are known, then the answer is 11/36, as you yourself discovered before tossing out the correct (in this case) solution...
That is not correct. Please review this.
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 14th, 2015 at 12:43:39 PM
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Quote: indignant99That is not correct. Please review this.
I did review it. You show all 11 of the possible rolls with at least one 2. Each one is equally likely.
You can't say "2-2 is twice as likely as the others because either one can be the first die", as then all of the others have to be counted twice as well as, for example, 1-2 can be "first die 1, second die 2" and "second die 1, first die 2 - if it's just as likely for either die to be the first die when you roll 2-2, it should be just as likely if you roll 1-2 (or 2-1)."
Quote: indignant99Although you don't know whether Die-1 or Die-2 was revealed, the incontrovertible fact is that it was either Die-1 exclusive-or Die-2.
If it was Die-1 (revealed), then only 6 outcomes are possible. Events 1, 2, 3, 4, 5, 6.
If it was Die-2 (revealed), likewise only 6 outcomes are possible. Events 6,7,8,9,10,11.
The answer is one-sixth = 1/6 = 16.667%
Does it matter which die is revealed if the roll is 2-2?
If not, then let's assume that if you roll 2-2, Die 1 is always revealed.
The possible events if Die 1 is revealed are 1, 2, 3, 4, 5, and 6.
The possible events if Die 2 is revealed are 7, 8, 9, 10, and 11. 6 is not possible as 2-2 is always a "Die 1 revealed" situation.
Even if, on a 2-2, you reveal Die 1 with probability P and Die 2 with probability 1-P, then if Die 1 is revealed, Events 1 through 5 would be revealed with probability 1, and 6 with probability P, while if Die 2 is revealed, Events 7 through 11 would be revealed with probability 1 and Event 6 with probability 1-P. It is not "two out of 12 equally likely cases."
April 14th, 2015 at 1:11:17 PM
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Quote: RSPlease, just stop.
No. I don't mind having an army of wrong opponents.
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 14th, 2015 at 1:13:45 PM
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Get two dice, throw them a few hundred times and record:Quote: indignant99No. I don't mind having an army of wrong opponents.
A = number of rolls where one (or both) of the dice shows a 2.
B = number of rolls where both dice show 2.
Divide B/A. Is it closer to 1/6 or 1/11 ?
If you're not convinced, repeat the experiment.
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April 14th, 2015 at 2:45:06 PM
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Can we please return to the original post? This is what was written in the original post:
We know that one die is a 2. The other die has six faces, numbered 1, 2, 3, 4, 5, 6. There is a 1/6 chance that the other die is a 2.
In the original post we are told that it is known that at least one die is a 2. That limits the possibility to 1/6. It's a matter of English.
Quote: DweenYour 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?
We know that one die is a 2. The other die has six faces, numbered 1, 2, 3, 4, 5, 6. There is a 1/6 chance that the other die is a 2.
In the original post we are told that it is known that at least one die is a 2. That limits the possibility to 1/6. It's a matter of English.
April 14th, 2015 at 3:07:07 PM
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Quote: AlanMendelsonCan we please return to the original post?
We know that one die is a 2. The other die has six faces, numbered 1, 2, 3, 4, 5, 6. There is a 1/6 chance that the other die is a 2.
In the original post we are told that it is known that at least one die is a 2. That limits the possibility to 1/6. It's a matter of English.
We know that at least one die is a 2.
There are 36 distinct ways to roll two dice; each is equally likely. 25 of them do not have any 2s, so we can ignore them. That leaves 11 equally likely ways to roll two dice such that at least one die is a 2.
Of these 11 remaining equally likely ways, only 1 has two 2s.
1 out of 11. 1/11.
April 14th, 2015 at 3:53:20 PM
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Quote: teliotGet two dice...
You get 2 dice. Set one of them with a deuce on top.
Now roll the other die. You'll get equal 1/6th populations of each pip-count.
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 14th, 2015 at 3:57:57 PM
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Quote: indignant99You get 2 dice. Set one of them with a deuce on top.
Now roll the other die. You'll get equal 1/6th populations of each pip-count.
What you guys are absolutely failing to recognize, is that one die being a deuce,
has absolutely eliminated 5 of your magical 11 "possibilities."
Which are eliminated? The 1, 3, 4, 5, 6 pip-faces on the die showing deuce.
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 14th, 2015 at 4:17:15 PM
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Unfortunately, (message number two in the thread) Wizard's probability expression is wrong. It should be pr(both 2's)/pr(2 and any) = 1/6.
Or equivalently pr(both 2's)/pr(any and 2) = 1/6.
That's a damned important "OR." It ain't
pr(both 2's) / pr( [2 and any]+[any and 2]-[1 of the double-counted 2-2] ).
Or equivalently pr(both 2's)/pr(any and 2) = 1/6.
That's a damned important "OR." It ain't
pr(both 2's) / pr( [2 and any]+[any and 2]-[1 of the double-counted 2-2] ).
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 14th, 2015 at 4:23:00 PM
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Quote: indignant99What you guys are absolutely failing to recognize, is that one die being a deuce,
has absolutely eliminated 5 of your magical 11 "possibilities."
Which are eliminated? The 1, 3, 4, 5, 6 pip-faces on the die showing deuce.
That'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.
