Two Dice Puzzle: Part Deux
May 17th, 2015 at 5:46:54 PM
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Quote: indignant99Nobody can. Not even himself.
I think he's actually disputing the meaning of the word "or"...
"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
May 18th, 2015 at 12:23:31 AM
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Hopefully they keep this discussion over on his board now. Such silliness.
DUHHIIIIIIIII HEARD THAT!
May 18th, 2015 at 6:55:44 AM
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I've been meaning to do this for a while, so here is a simulator that anyone can run in their web browser.
copy and paste the code into a file with a .html extension and double-click on it and it should run in your web browser.
I made some style and algorithm choices that aren't the prettiest or most efficient, but should be easier to understand.
copy and paste the code into a file with a .html extension and double-click on it and it should run in your web browser.
<!DOCTYPE html>
<html>
<head>
<script>
// number of times we will evalulate the situation where at least one of the dice is a two
var numTrials = 11000;
function simulator() {
// counters
var at_least_one_die_is_a_two = 0;
var both_dice_are_two = 0;
var stop_now = false;
while (false == stop_now) {
// roll two dice
var d1 = Math.floor((Math.random() * 6) + 1);
var d2 = Math.floor((Math.random() * 6) + 1);
// count when at least one die is a two
if ( 2 == d1 || 2 == d2) {
at_least_one_die_is_a_two = at_least_one_die_is_a_two + 1;
}
// count when both dice are two
if ( 2 == d1 && 2 == d2) {
both_dice_are_two = both_dice_are_two + 1;
}
// evaluate stop condition, set stop flag
if (at_least_one_die_is_a_two >= numTrials) {
stop_now = true;
}
}
r = document.getElementById("results");
r.innerHTML = both_dice_are_two + "/" + at_least_one_die_is_a_two +
" = " + both_dice_are_two/at_least_one_die_is_a_two;
}
window.onload=simulator
</script>
</head>
<body>
<div id="results"></div>
</body>
</html>
I made some style and algorithm choices that aren't the prettiest or most efficient, but should be easier to understand.
May 18th, 2015 at 8:26:44 AM
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What are the odds? I just made a video of myself making a program of this very problem. Enjoy.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
May 18th, 2015 at 9:20:54 AM
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Ahahahahahaha. I got called a "penny loser" by Rob Singer over there. I paid over $126,000 in taxes in 2012 alone jacka$$!! Where are your returns you keep beating around the bush about?? You sir, have NEVER won a dime in a casino.
If he knew how I really play in a casino, he'd never play vp again in his life, if he ever has, which I doubt.
End of rant, carry on.
If he knew how I really play in a casino, he'd never play vp again in his life, if he ever has, which I doubt.
End of rant, carry on.
DUHHIIIIIIIII HEARD THAT!
May 18th, 2015 at 9:51:37 AM
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I've kinda ignored this thread since something like page 7. But since I noticed that the Wiz made the most recent post, I figured I'd check it out.
Note: For the record, in this thread I have always been in the 1/11 camp. In the other thread, I was briefly in the 1/6 camp.
I love/hate Mike's comment in the above video, "I doubt this will convince anyone..." Love it because it made me laugh. Hate it because ya gotta have more faith in yourself Mike!
I saw the other video as well. I have only one comment about the 36 pairs of dice. It may have had a better impact if you had 36 with dice, and put all the white dice on one side of the pair. Similarly, the first video may also have had a better impact if you had used white and colored dice.
Also, for shits & giggles, I entered Dalex64's code and ran it in my browser. I was shocked to see how fast it ran. I had to add 4 zeros to get a measurable duration. 38.4 seconds on my several years old MacBook Pro.
Note: For the record, in this thread I have always been in the 1/11 camp. In the other thread, I was briefly in the 1/6 camp.
I love/hate Mike's comment in the above video, "I doubt this will convince anyone..." Love it because it made me laugh. Hate it because ya gotta have more faith in yourself Mike!
I saw the other video as well. I have only one comment about the 36 pairs of dice. It may have had a better impact if you had 36 with dice, and put all the white dice on one side of the pair. Similarly, the first video may also have had a better impact if you had used white and colored dice.
Also, for shits & giggles, I entered Dalex64's code and ran it in my browser. I was shocked to see how fast it ran. I had to add 4 zeros to get a measurable duration. 38.4 seconds on my several years old MacBook Pro.
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? 😁
May 18th, 2015 at 9:58:38 AM
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Quote: DJTeddyBear...I love/hate Mike's comment in the above video, "I doubt this will convince anyone..." Love it because it made me laugh. Hate it because ya gotta have more faith in yourself Mike!...
Of course it won't convince the other side. They love being argumentative. And it gives guys like Singer a chance to attack the experts only because his good buddy Mendelson disagrees with Mike and almost everyone else also.
And the colored shirt analogy by Reg or Red, I don't remember which one, is completely wrong also. Those supposed English comprehension experts over there pretty much can't comprehend $h!t!
DUHHIIIIIIIII HEARD THAT!
May 18th, 2015 at 10:15:08 AM
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Quote: MathExtremistI think he's actually disputing the meaning of the word "or"...Quote: indignant99
Nobody can. Not even himself.
No, I simply stated that the 1/11 is a 100% matter of 'and' rather than 'or'. I went on to confirm this with some other thoughts also which apparently no one wanted to hear so far as a genuine attempt was made to understand any of it. I hope it's actually w/i the grasp of the majority of the persons here, because I'd hate to think that people would actually give up a real mind for the AP silliness of blackjack, etc, to "fit in" on such a waning internet forum. Scary boo hoo. If the tacit leadership such as teliot, the Wizard, Face, and BBB, want to continue to encourage or condone more of this silliness as in the thread https://wizardofvegas.com/forum/off-topic/general/22060-a-new-blog-on-the-mathematics-of-blackjack/#post458259 , that's their prerogative. More "customers", perhaps. But scary boo hoo, if you ask me. I'd actually rather work at something real about gaming and gambling, hence which has the potential for other applications along the way.
You have to and/add the groups of dice which have one 2, two 2's, and up, to arrive at the chance of n 2's in n dice. For n=2, of two 2's in two dice. In other words, the sum of ten "one-2 outcomes" plus one "two-2 outcomes" is eleven. So that there's a one in eleven chance of two 2's in two dice given the sum of those distinctly possible rolls of those groupings of the dice.
For n=3 dice, there are the groupings of "one-2 outcomes", "two-2 outcomes", and "three-2 outcomes". You would have to add - not or - those, and take the inverse or "1 over" to get the answer then. A matter of and; not or. Of 216 distinctly possible rolls for n=3: 125 have no 2; 75 have one 2; 15 have two 2's; and 1 has three 2's. Adding, 75 + 15 + 1 = 91. Inverting, we have 1/91 chance of three 3's given the groupings of one AND more 2's up to three (for three dice).
For the chance of two 2's with two dice for the groupings of "one-2 outcomes" OR "two-2 outcomes", then the answer would be 0/10 OR 1/1 respectively.
Add on: What is far more interesting is how various exercises such as this combine to confirm particular results at the deep levels of math and physics.
Nonsense is a very hard thing to keep up. Just ask the Wizard and company.
May 18th, 2015 at 10:23:09 AM
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Quote: DJTeddyBearAlso, for shits & giggles, I entered Dalex64's code and ran it in my browser. I was shocked to see how fast it ran. I had to add 4 zeros to get a measurable duration. 38.4 seconds on my several years old MacBook Pro.
I spend a few moments a week online for that.
Nonsense is a very hard thing to keep up. Just ask the Wizard and company.
May 18th, 2015 at 10:47:50 AM
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Quote: DJTeddyBearHate it because ya gotta have more faith in yourself Mike!
It isn't so much a lack of faith in myself. As I've said many times, the more ridiculous a belief is, the more tenaciously it tends to be held.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
May 18th, 2015 at 10:58:43 AM
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Quote: WizardIt isn't so much a lack of faith in myself. As I've said many times, the more ridiculous a belief is, the more tenaciously it tends to be held.
A belief based on tenacious assertion rather than clarity?
Nonsense is a very hard thing to keep up. Just ask the Wizard and company.
May 18th, 2015 at 11:23:38 AM
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Quote: KerkebetFor the chance of two 2's with two dice for the groupings of "one-2 outcomes" OR "two-2 outcomes", then the answer would be 0/10 OR 1/1 respectively.
And then you'll study Boolean algebra and disabuse yourself of these semantic gymnastics.
Quote:But scary boo hoo, if you ask me. I'd actually rather work at something real about gaming and gambling, hence which has the potential for other applications along the way.
If you think you can find a profitable gambling-related application for "doing math wrong," you're welcome to try. Casino coffers are full of cash from your intellectual forebears, having lost too much of their stake, and nevertheless incredulous that their "non-traditional" interpretation of the odds didn't pan out where it matters, on the felt.
I believe the purpose of Mike's websites over the years has been in part to prevent that from happening. But you can't make the horse drink.
"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
May 18th, 2015 at 2:17:45 PM
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Quote: KerkebetA belief based on tenacious assertion rather than clarity?
See, bass-ackwards again. ASSERTION BASED ON TENACIOUSLY-HELD BELIEF.
You dare to mention "clarity?" You are the most "un-clarity" poster in this colossal thread.
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
May 18th, 2015 at 5:53:17 PM
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Quote Regnis on Alan's forum:
"I have 6 men wearing 6 different colored shirts. Across the street are 6 women wearing 6 different colored shirts. Each group has one red shirt. If I select the guy with the red shirt from group 1, what are the odds of blindly picking the matching shirt from group 2.
Hint--it ain't 1 of 11."
That's not the original question nor was it how it is worded. Let's try his analogy with the proper wording:
I have 6 men wearing 6 different colored shirts. Across the street are 6 women wearing 6 different colored shirts. Each group has one red shirt. If I select SOMEONE with the red shirt from ANY GROUP, what are the odds of blindly picking the matching shirt from the REST OF THE PEOPLE.
Hint--it IS 1 of 11.
Since you don't know where I plucked the red shirt from, you can't discount the 11 remaining.
"I have 6 men wearing 6 different colored shirts. Across the street are 6 women wearing 6 different colored shirts. Each group has one red shirt. If I select the guy with the red shirt from group 1, what are the odds of blindly picking the matching shirt from group 2.
Hint--it ain't 1 of 11."
That's not the original question nor was it how it is worded. Let's try his analogy with the proper wording:
I have 6 men wearing 6 different colored shirts. Across the street are 6 women wearing 6 different colored shirts. Each group has one red shirt. If I select SOMEONE with the red shirt from ANY GROUP, what are the odds of blindly picking the matching shirt from the REST OF THE PEOPLE.
Hint--it IS 1 of 11.
Since you don't know where I plucked the red shirt from, you can't discount the 11 remaining.
DUHHIIIIIIIII HEARD THAT!
May 18th, 2015 at 6:58:19 PM
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What is the probability that anybody in the 1/6 camp ever took a course in probability or statistics?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
May 18th, 2015 at 6:59:58 PM
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Quote: WizardWhat is the probability that anybody in the 1/6 camp ever took a course in probability or statistics?
Definitely not 1 in 11!
I never did. And I still know better.
DUHHIIIIIIIII HEARD THAT!
May 18th, 2015 at 7:24:52 PM
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Quote: IbeatyouracesLet's try his analogy with the proper wording:
I have 6 men wearing 6 different colored shirts. Across the street are 6 women wearing 6 different colored shirts. Each group has one red shirt. If I select SOMEONE with the red shirt from ANY GROUP, what are the odds of blindly picking the matching shirt from the REST OF THE PEOPLE.
Hint--it IS 1 of 11.
Since you don't know where I plucked the red shirt from, you can't discount the 11 remaining.
That's a really good analogy. It's intuitively obvious when you phrase it that way.
It falls down a bit when you consider that you can't pick two men or two women, but that's easily sidestepped when considering that you didn't say whether you picked a man or woman to start.
"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
May 18th, 2015 at 7:33:34 PM
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Quote: MathExtremistThat's a really good analogy. It's intuitively obvious when you phrase it that way.
It falls down a bit when you consider that you can't pick two men or two women, but that's easily sidestepped when considering that you didn't say whether you picked a man or woman to start.
Exactly. And they are the supposed English comprehension experts.
There is a complete difference between knowing exactly which die has the 2 or just knowing with one could at least be a 2.
DUHHIIIIIIIII HEARD THAT!
May 19th, 2015 at 6:37:27 AM
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I just asked Alan a question. If he ever deigns to answer the question, we might be nearer to figuring out why he disagrees with us.
Quote: Me to AlanI wonder.
When you read the question, do you perceive
'at least one of the dice is ALWAYS a deuce'
or do you perceive it as
'ALWAYS, at least one of the dice is a deuce'
The difference between those two phrases is massive. I hope that you would agree that those phrases have different meaning?
I read it as the latter.
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.
May 20th, 2015 at 12:13:06 PM
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Quote: MathExtremistAnd then you'll study Boolean algebra and disabuse yourself of these semantic gymnastics.
I'm not the one of many here who thinks that simulation proves something. The problem here is not with the numbers, but how the question as written is interpreted. In engineering terms, simulations are normally done after the manual research and set of a project. And hence, are the least resource intensive, usually at around 5%. The only persons I can think of who would do this first are AP's. But, look at the results when we "jump to conclusions". More on this later on.
I'm also not the one of many here who has to show that the question about a specific roll of two dice as written involves an inclusive or. I was maligned over at Alan's forum by Indignant99 for claiming that there's no inclusive or in the question; but he/she still, after continuing with many other similarly distracting replies at Alan's, hasn't indicated the inclusive or. Please note http://www.mathwords.com/i/inclusive_or.htm . Oh, and pardon me for actually getting into Boolean algebra instead of silly simulations of simple calculations of summary mathematical interpretations.
Quote: MathExtremistIf you think you can find a profitable gambling-related application for "doing math wrong," you're welcome to try.
To let you in on a not so little secret, the only profitably gambling is on the casino side of the equation. It's correctly called a business or industry. They make relatively sure that any players who profit by it essentially waste their lives in the process. Certainly, no one here who has claimed to meaningfully profit has shown anything for it by way of tact or technique, or any actual proof. Teliot posted up the Dragon baccarat counting system a while back, after others on other forums more or less did the same, but like John May's tie system it's not worth someone's time or money to play it as a career to be quickly (or not) told to move on. Or in teliot's words a while back - to mirror Mike's of don't let the door slam on your way out - GTFOOH. If you want to sit alone and count cards every day of the year without any real benefits and security, a family, job training, growth, etc, then fine. What do I care. I guess you'll "enjoy" also living in a basement apartment forever alone and on internet forums such as this seems intended to be. Tacitly agree with everything Mike says. He's the one who ought to care more about his readers' well being than their agreeing with everything he says.
Quote: MathExtremistCasino coffers are full of cash from your intellectual forebears, having lost too much of their stake, and nevertheless incredulous that their "non-traditional" interpretation of the odds didn't pan out where it matters, on the felt.
If you want to sit there and count cards every day of the year without real benefits and security, a family, etc, then I guess you'll enjoy living in a basement apartment alone on a forum such as this...
And continue to assert idiotic conjectures such as to my own "intellectual forebears". Amateurish, to say the least.
Quote: MathExtremistI believe the purpose of Mike's websites over the years has been in part to prevent that from happening.
If Mike doesn't want to be accused of tenaciously holding onto also the ridiculous belief that he's somehow a "Wizard of Vegas" in some way, then I suggest that he closely look at the other possible interpretations presented of this exercise instead of tenaciously simulating his own interpretation to prove it's the right interpretation. (Of course he's going to get the same 1/11 he got by manual calculation.) At least try to understand from where the dissenters come if you don't want to do the real work yourself. Parse the thing out, either mathematically or in plain English. Teliot likes to go into such unnecessary detail if it can be said simply, but where is he now when the language doesn't seem to agree with the math?
The 1/11 answer doesn't rule out the 1/6 answer because it's merely the 1/6 answer written another way. Paying 10 to 1 for two 2's is the SAME as twice paying 5 to 1 for one two. It comes down to taking the rolls together, or taking the rolls separately. But, the 1/6 answer taken alone in the correct manner does not allow for also the 1/11 answer.
Quote: MathExtremistBut you can't make the horse drink.
Why is this even a matter of making someone do something. You've been here too long ME. If people are shown how to develop a thirst for knowledge, then they will want or choose to drink all by themselves.
Suffice it say, Mike isn't a math genius by any stretch of the imagination. He works for a casino coupon forum to endorse an array on online casinos. I would call him a "rental scientist".
If you and the others still want to argue our points in a calm and rational manner, then please continue. I'm not going anywhere. If we have to take the question as written to a real math board for confirmation, then so be it.
Nonsense is a very hard thing to keep up. Just ask the Wizard and company.
May 20th, 2015 at 4:47:07 PM
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Is algebra.com a real math board?
http://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.434512.html
Similar question - after rolling two dice, if it is known that one of them is a 4, what are the odds that the other one is a 5? Similar answer - 2/11. There is that pesky 11 again.
Xkcd has a pretty serious math forum, you might try there.
http://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.434512.html
Similar question - after rolling two dice, if it is known that one of them is a 4, what are the odds that the other one is a 5? Similar answer - 2/11. There is that pesky 11 again.
Xkcd has a pretty serious math forum, you might try there.
May 21st, 2015 at 9:59:14 AM
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Quote: Dalex64Is algebra.com a real math board?
http://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.434512.html
Similar question - after rolling two dice, if it is known that one of them is a 4, what are the odds that the other one is a 5? Similar answer - 2/11. There is that pesky 11 again.
Xkcd has a pretty serious math forum, you might try there.
How is that a similar question? The same denominator as the Wizard's means nothing. You can't be serious. Anyway, as they say, "Close only counts in horseshoes."
I'd rather someone answer my simpler question, to get us off on the right foot here first.
Quote:Where is the "inclusive or" in the question as written?
The 'or' may be assumed inclusive, but that isn't necessarily always possible. There will be a 2 in one row or the other (of the chart) but not in both on a specific roll; or there will be two 2's but not both (one 2 and two 2's).
Still waiting for anyone to answer this question.
Nonsense is a very hard thing to keep up. Just ask the Wizard and company.
May 21st, 2015 at 10:08:22 AM
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Quote: KerkebetSuffice it say, Mike isn't a math genius by any stretch of the imagination. He works for a casino coupon forum to endorse an array on online casinos. I would call him a "rental scientist".
Personal insult -- seven-day suspension (second offense).
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
May 21st, 2015 at 10:22:20 AM
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Quote: WizardPersonal insult -- seven-day suspension (second offense).
Well, if I'm not in red yet, I will assume I'm being allowed a reply.
You failed the Mensa test, Mike. (I passed it, by the way, at 156. I was a teenager in high school.)
Nonsense is a very hard thing to keep up. Just ask the Wizard and company.
May 21st, 2015 at 10:30:43 AM
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Quote: KerkebetWhere is the "inclusive or" in the question as written?
The 'or' may be assumed inclusive, but that isn't necessarily always possible. There will be a 2 in one row or the other (of the chart) but not in both on a specific roll; or there will be two 2's but not both (one 2 and two 2's).
"At least one" is equivalent to "one or more". The question as written states that you are informed that the roll of two dice contains "at least one 2."
A roll of two dice containing any of the previously-identified ten combinations having exactly one 2 meets that criterion. So does the single combination of 2-2. The total number of combinations having "at least one 2" is 11. But you appear to be suggesting that because a pair of the dice cannot simultaneously contain "exactly one 2" and "exactly two 2s" that, somehow, the question is poorly written and/or unsolvable. That's nonsense.
The question as written is equivalent to the question "what is the probability that a fair roll of two standard dice contains two 2s conditioned upon the roll containing at least one 2," which is in turn equivalent to the expression p(two 2s | at least one 2) and that expression has already been solved:
https://wizardofvegas.com/forum/questions-and-answers/math/21845-two-dice-puzzle-part-deux/26/#post454379
"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
May 21st, 2015 at 10:32:36 AM
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Quote: Kerkebet(I passed it, by the way, at 156. I was a teenager in high school.)
They must not have asked any probability questions on it, fortunately for you.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
May 21st, 2015 at 11:13:54 AM
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The 4,5 problem and the 2,2 problem follow the same method to find the solution.
There are 11 possible results "if it is known that one of the dice is a 4". (a little vague, since you might interpret that to mean that ONLY one of the dice is a 4)
The explanation then highlights in red the 4,5 and 5,4 possibilities, to indicate that when one of the dice is a 4 that the chance that the other die is a 5 is 2/11.
if ONLY one of the dice can be a four, you eliminate 4,4 and get 2/10 or 1/5 for the answer.
There are 11 possible results "if at least one of the dice is a two" "if it is known that one of the dice is a two" or more specifically "if it is known that at least one of the dice is a two" of those 11 possible results, one of them is 2,2.
There are 11 possible results "if it is known that one of the dice is a 4". (a little vague, since you might interpret that to mean that ONLY one of the dice is a 4)
The explanation then highlights in red the 4,5 and 5,4 possibilities, to indicate that when one of the dice is a 4 that the chance that the other die is a 5 is 2/11.
if ONLY one of the dice can be a four, you eliminate 4,4 and get 2/10 or 1/5 for the answer.
There are 11 possible results "if at least one of the dice is a two" "if it is known that one of the dice is a two" or more specifically "if it is known that at least one of the dice is a two" of those 11 possible results, one of them is 2,2.
May 21st, 2015 at 12:12:05 PM
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Quote: WizardPersonal insult -- seven-day suspension (second offense).
I believe that if you insult the owner/moderator, it should be a nuke!
DUHHIIIIIIIII HEARD THAT!
May 21st, 2015 at 1:51:10 PM
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Quote: KerkebetHow is that a similar question? The same denominator as the Wizard's means nothing. You can't be serious.
It is not just a similar question. It is the exact same question just ask for the outcome of 4,5 instead of 4,4.
4,4 1/11
4,1 2/11
4,2 2/11
4,3 2/11
4,5 2/11
4,6 2/11
Total 11/11
Kerkbet
If you agree that the 4,5 is correct at 2/11 as the Math Website says but say that the 4,4 is 1/6 then
4,4 1/6
4,1 2/11
4,2 2/11
4,3 2/11
4,5 2/11
4,6 2/11
Total 71/66 >1
And you know there is usually something wrong in a calulation prob when the total sum of all probs exceeds 1.
May 21st, 2015 at 2:36:32 PM
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Aright, finally I realize why I don't understand anything this guy says. It's simply beyond me.Quote: KerkebetWell, if I'm not in red yet, I will assume I'm being allowed a reply.
You failed the Mensa test, Mike. (I passed it, by the way, at 156. I was a teenager in high school.)
May 21st, 2015 at 4:00:41 PM
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Same problem, on wikibooks, but with 6's instead of 2's
http://en.m.wikibooks.org/wiki/Probability/Conditional_Probability
Same answer, 1/11.
Same explainations that have been given here before.
There are a ton of these out there. This one asks what the probabity is of getting a sum of ten given that one of the dice is a 5.
http://mathcentral.uregina.ca/QQ/database/QQ.09.00/wallace1.html
It explains why 1/3 is the wrong answer, and explains how 1/3 could be the right answer if you were given slightly different information.
http://en.m.wikibooks.org/wiki/Probability/Conditional_Probability
Same answer, 1/11.
Same explainations that have been given here before.
There are a ton of these out there. This one asks what the probabity is of getting a sum of ten given that one of the dice is a 5.
http://mathcentral.uregina.ca/QQ/database/QQ.09.00/wallace1.html
It explains why 1/3 is the wrong answer, and explains how 1/3 could be the right answer if you were given slightly different information.
May 22nd, 2015 at 12:39:19 PM
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Here's a different conditional probability problem.
You have three bags. One bag has two red chips, one bag has two blue chips, one bag has one red chip and one blue chip.
You reach into a bag and pull out a red chip. What are the odds that the chip still in the bag is red?
Here is a fun variation of the two dice problem.
You need: a red die, a green die, a colorblind person, and a person with normal color vision.
The two colored dice are shaken and slammed on the table, under a cup. A previously unmentioned third party reaches under the cup, and without changing the value of either die, pulls a 4 out from under the cup and places it in plain view.
The two previously mentioned people are asked what the odds are that both dice are four.
The right answer for the colorblind person is 1/11
The right answer for the person with normal color vision is 1/6
One more -
We know that when you roll two dice, and one of them is an N, that they will both be an N one time in 11
We also know that when you roll two dice, you will get doubles 6 times out of 36, or 1 time in 6
So, how can the odds that the other die is the same 1 time in 11, and the chances that you roll doubles 1 time in 6, at the same time?
You have three bags. One bag has two red chips, one bag has two blue chips, one bag has one red chip and one blue chip.
You reach into a bag and pull out a red chip. What are the odds that the chip still in the bag is red?
Here is a fun variation of the two dice problem.
You need: a red die, a green die, a colorblind person, and a person with normal color vision.
The two colored dice are shaken and slammed on the table, under a cup. A previously unmentioned third party reaches under the cup, and without changing the value of either die, pulls a 4 out from under the cup and places it in plain view.
The two previously mentioned people are asked what the odds are that both dice are four.
The right answer for the colorblind person is 1/11
The right answer for the person with normal color vision is 1/6
One more -
We know that when you roll two dice, and one of them is an N, that they will both be an N one time in 11
We also know that when you roll two dice, you will get doubles 6 times out of 36, or 1 time in 6
So, how can the odds that the other die is the same 1 time in 11, and the chances that you roll doubles 1 time in 6, at the same time?
May 22nd, 2015 at 12:55:35 PM
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Quote: Dalex64You need: a red die, a green die, a colorblind person, and a person with normal color vision.
The two colored dice are shaken and slammed on the table, under a cup. A previously unmentioned third party reaches under the cup, and without changing the value of either die, pulls a 4 out from under the cup and places it in plain view.
The two previously mentioned people are asked what the odds are that both dice are four.
The right answer for the colorblind person is 1/11
The right answer for the person with normal color vision is 1/6
Are you sure you're telling this right? Can the colorblind person see the numbers on both dice? Can you tell us anything about the behavior of the third party? As with the Monty Hall problem, it sometimes is important.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
May 22nd, 2015 at 7:16:27 PM
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I am trying to say that both people know that there is at least one four, the colorblind person can not distinguish which four it is, but the other person can.
The third party is the dealer role, is fair and truthful, and only pulls out a 4 when there is at least one 4 under the cup, and does not change the results of the dice when he moves one of them into view.
The colorblind person can see the numbers on the dice, he just can't tell which one is red and which one is green - they look the same, for the purposes of this puzzle anyway.
The third party is the dealer role, is fair and truthful, and only pulls out a 4 when there is at least one 4 under the cup, and does not change the results of the dice when he moves one of them into view.
The colorblind person can see the numbers on the dice, he just can't tell which one is red and which one is green - they look the same, for the purposes of this puzzle anyway.
May 22nd, 2015 at 10:04:07 PM
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You can pass Mensa tests all day long. It's meaningless if you're working at Walmart, living in your parents basement looking for the next best blow up doll. Remember the time when we got that extra case of tampons (Good times, good times).Quote: KerkebetWell, if I'm not in red yet, I will assume I'm being allowed a reply.
You failed the Mensa test, Mike. (I passed it, by the way, at 156. I was a teenager in high school.)
This guy passed at 156 as well.
http://webdesignzseo.com/funny/wp-content/uploads/2011/06/The-Most-Nerdy-Guy-Nerdiest-Nerd-300x227.jpg
♪♪Now you swear and kick and beg us That you're not a gamblin' man Then you find you're back in Vegas With a handle in your hand♪♪ Your black cards can make you money So you hide them when you're able In the land of casinos and money You must put them on the table♪♪ You go back Jack do it again roulette wheels turinin' 'round and 'round♪♪ You go back Jack do it again♪♪
May 24th, 2015 at 3:53:15 PM
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Quote: Dalex64You need: a red die, a green die, a colorblind person, and a person with normal color vision.
The two colored dice are shaken and slammed on the table, under a cup. A previously unmentioned third party reaches under the cup, and without changing the value of either die, pulls a 4 out from under the cup and places it in plain view.
The two previously mentioned people are asked what the odds are that both dice are four.
The right answer for the colorblind person is 1/11
The right answer for the person with normal color vision is 1/6
Quote: Dalex64I am trying to say that both people know that there is at least one four, the colorblind person can not distinguish which four it is, but the other person can.
The third party is the dealer role, is fair and truthful, and only pulls out a 4 when there is at least one 4 under the cup, and does not change the results of the dice when he moves one of them into view.
The colorblind person can see the numbers on the dice, he just can't tell which one is red and which one is green - they look the same, for the purposes of this puzzle anyway.
Is the number 4 specifically chosen before the dice are shaken?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
May 24th, 2015 at 4:26:19 PM
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I hadn't thought about that. It is supposed to be just like the "2" dice puzzle, with the added wrinkle of instead of just being told that at least one of the dice is a 4, you are also shown that one of the dice is a 4.
It is my belief that saying at least one of the dice is a 4 and then showing one of the fours as proof doesn't change the odds of both dice being a 4, if you can't tell which die was revealed.
The colorblind person can't tell the difference, so he can't eliminate any of the 11 possibilities. The other person can see the color of the die, though, so he can eliminate 5 of the possibilities.
It is my belief that saying at least one of the dice is a 4 and then showing one of the fours as proof doesn't change the odds of both dice being a 4, if you can't tell which die was revealed.
The colorblind person can't tell the difference, so he can't eliminate any of the 11 possibilities. The other person can see the color of the die, though, so he can eliminate 5 of the possibilities.
May 24th, 2015 at 6:12:53 PM
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It is critical to explain the behavior of the host to give a correct answer. That is why there is a huge dispute about the Monty Hall problem every times it comes up, because it is almost always asked badly, so people arrive at different answers according to the behavior they assume for Monty.
In this case, I think it is important to clarify:
1. How did the host arrive at asking about four specifically, as opposed to another number?
2. How did he choose the one die?
In this case, I think it is important to clarify:
1. How did the host arrive at asking about four specifically, as opposed to another number?
2. How did he choose the one die?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
May 24th, 2015 at 8:49:42 PM
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I picked the number 4 arbitrarily, to differentiate it in discussion from the problem with double twos.
4 is chosen before the dice are rolled. If there is no 4, the dice would have to be rolled again.
If there is only one four, then that is the die revealed. If there are two fours, he randomly picks one to expose.
4 is chosen before the dice are rolled. If there is no 4, the dice would have to be rolled again.
If there is only one four, then that is the die revealed. If there are two fours, he randomly picks one to expose.
May 24th, 2015 at 10:28:03 PM
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Quote: Dalex64I picked the number 4 arbitrarily, to differentiate it in discussion from the problem with double twos.
4 is chosen before the dice are rolled. If there is no 4, the dice would have to be rolled again.
If there is only one four, then that is the die revealed. If there are two fours, he randomly picks one to expose.
Thank you. So, why would the answer be 1/6 for the colored-vision observer?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
May 25th, 2015 at 5:33:17 AM
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He can differentiate the two dice. He can tell that, for instance, if the red die is revealed, the red die is 4 - eliminating the possibilities of a red 12356 leaving the 6 possibilities of a red 4 and a green 123456
May 25th, 2015 at 7:36:56 AM
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Quote: Dalex64He can differentiate the two dice. He can tell that, for instance, if the red die is revealed, the red die is 4 - eliminating the possibilities of a red 12356 leaving the 6 possibilities of a red 4 and a green 123456
I disagree. The way you worded the question, I maintain the answer is 1 in 11. Would the original question have a different answer if the dice were different colors?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
May 25th, 2015 at 10:44:18 AM
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I think the original question would have a different answer if the dice were different colors AND you were told or shown which die was a 2.
In the original question, with colored dice and no reveal, either
The red die is 2 and the green die is 13456,
The green die is 2 and the red die is 13456
Both red and green are blue.
In the new version, if you see a red 4 you have
The red die is 4 and the green die is 12356
Both the red die and the green die are 4
Or when you see a green 4
The green die is 4 and the red die is 12356
Both dice are 4
This version of the puzzle imposes a new condition which splits the puzzle into two -
Two dice are rolled and at least the red die is a 4. With these conditions, the green die will be 4 one time out of 6.
Two dice are rolled and at least the green die is a 4. With these conditions, the red die will be 4 one time out of 6.
My intention was to express these two conditional statements, and add a person who could not differentiate the color condition, leaving him with only the same information as the original puzzle, that at least one of the dice is a 2 (or 4) but he doesn't know which one.
If we can fix the wording to make what I am trying to say more clear, that would be great.
In the original question, with colored dice and no reveal, either
The red die is 2 and the green die is 13456,
The green die is 2 and the red die is 13456
Both red and green are blue.
In the new version, if you see a red 4 you have
The red die is 4 and the green die is 12356
Both the red die and the green die are 4
Or when you see a green 4
The green die is 4 and the red die is 12356
Both dice are 4
This version of the puzzle imposes a new condition which splits the puzzle into two -
Two dice are rolled and at least the red die is a 4. With these conditions, the green die will be 4 one time out of 6.
Two dice are rolled and at least the green die is a 4. With these conditions, the red die will be 4 one time out of 6.
My intention was to express these two conditional statements, and add a person who could not differentiate the color condition, leaving him with only the same information as the original puzzle, that at least one of the dice is a 2 (or 4) but he doesn't know which one.
If we can fix the wording to make what I am trying to say more clear, that would be great.
May 25th, 2015 at 4:58:51 PM
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Quote: Dalex64If we can fix the wording to make what I am trying to say more clear, that would be great.
Consider this wording: "There are two dice, a red and a green one. They are rolled until the red die is a four. At that roll, what is the probability the green die is also a four?" The answer to that problem is 1 in 6.
However, I still can't think of a way to work the color-blindness element into it and have different answers.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
May 25th, 2015 at 5:39:08 PM
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They are rolled until either or both dice are four, not just the red die.
So either a red 4 or a green 4 will be shown.
Red die 4 shown means green 4 1/6
Green die 4 shown means red 4 1/6
Can't tell red from green means other die 4 1/11
I can try to make some time over the next few days to write a simulation.
The jist of it is:
If there is a 4 add one to master denominator
If there is a red 4 add one to red denominator
If there is a green 4 add one to green denominator
If both dice are 4 add one to numerator
Repeat many times, then look at n/md, n/rd, n/gd
N/md is the colorblind person's view on the problem, the other two are for a person with normal color vision.
So either a red 4 or a green 4 will be shown.
Red die 4 shown means green 4 1/6
Green die 4 shown means red 4 1/6
Can't tell red from green means other die 4 1/11
I can try to make some time over the next few days to write a simulation.
The jist of it is:
If there is a 4 add one to master denominator
If there is a red 4 add one to red denominator
If there is a green 4 add one to green denominator
If both dice are 4 add one to numerator
Repeat many times, then look at n/md, n/rd, n/gd
N/md is the colorblind person's view on the problem, the other two are for a person with normal color vision.
May 25th, 2015 at 7:56:25 PM
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Quote: Dalex64They are rolled until either or both dice are four, not just the red die.
So either a red 4 or a green 4 will be shown.
Red die 4 shown means green 4 1/6
Green die 4 shown means red 4 1/6
Can't tell red from green means other die 4 1/11
I disagree. If both are a four, and the exposed die is chosen randomly, then I maintain the probability the other die is a 4 is 1/11.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
May 26th, 2015 at 4:17:01 AM
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If the red die is a 4, there is a 1/6 chance the green die is a 4.
If the green die is a 4, there is a 1/6 chance the red die is a 4.
If either die (ie: color unknown) is a 4, then there is a 1/11 chance the other die (or both dice) is/are a 4.
Essentially, you're taking the matrix:
G-R [Green-Red]
1-1, 1-2, 1-3, 1-4, 1-5, 1-6
2-1, 2-2, 2-3, 2-4, 2-5, 2-6
3-1, 3-2, 3-3, 3-4, 3-5, 3-6
4-1, 4-2, 4-3, 4-4, 4-5, 4-6
5-1, 5-2, 5-3, 5-4, 5-5, 5-6
6-1, 6-2, 6-3, 6-4, 6-5, 6-6
And removing everything that doesn't have a 4 in it, so you get a cross looking shape:
....................1-4..............
....................2-4..............
....................3-4..............
4-1, 4-2, 4-3, 4-4, 4-5, 4-6
....................5-4..............
....................6-4..............
If the green die is a 4 (the first one), then you're left with the matrix:
.......................................
.......................................
.......................................
4-1, 4-2, 4-3, 4-4, 4-5, 4-6
.......................................
.......................................
Where there are only 1 of 6 ways to get a 4-4.
If the red die is a 4 (the second one), then you're left with the matrix:
....................1-4..............
....................2-4..............
....................3-4..............
....................4-4..............
....................5-4..............
....................6-4..............
Again, only 1 of 6 ways to get a 4-4.
Although, I'm not entirely sure what Wizard means in his last post, or if he mistyped something.
If the green die is a 4, there is a 1/6 chance the red die is a 4.
If either die (ie: color unknown) is a 4, then there is a 1/11 chance the other die (or both dice) is/are a 4.
Essentially, you're taking the matrix:
G-R [Green-Red]
1-1, 1-2, 1-3, 1-4, 1-5, 1-6
2-1, 2-2, 2-3, 2-4, 2-5, 2-6
3-1, 3-2, 3-3, 3-4, 3-5, 3-6
4-1, 4-2, 4-3, 4-4, 4-5, 4-6
5-1, 5-2, 5-3, 5-4, 5-5, 5-6
6-1, 6-2, 6-3, 6-4, 6-5, 6-6
And removing everything that doesn't have a 4 in it, so you get a cross looking shape:
....................1-4..............
....................2-4..............
....................3-4..............
4-1, 4-2, 4-3, 4-4, 4-5, 4-6
....................5-4..............
....................6-4..............
If the green die is a 4 (the first one), then you're left with the matrix:
.......................................
.......................................
.......................................
4-1, 4-2, 4-3, 4-4, 4-5, 4-6
.......................................
.......................................
Where there are only 1 of 6 ways to get a 4-4.
If the red die is a 4 (the second one), then you're left with the matrix:
....................1-4..............
....................2-4..............
....................3-4..............
....................4-4..............
....................5-4..............
....................6-4..............
Again, only 1 of 6 ways to get a 4-4.
Although, I'm not entirely sure what Wizard means in his last post, or if he mistyped something.
May 26th, 2015 at 4:38:41 AM
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Yes, that is what I was trying to say and meant to say.
May 26th, 2015 at 7:21:50 AM
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Here is my interpretation of the Dalex64 question:
I maintain the answer is 1/11.
Quote:Two dice, a red one and a green one are rolled. In addition a coin is flipped. A peeker shall check both dice for a four. If there is exactly one four, then he will expose it. If there are two fours, then if the coin is heads he will expose the red die, if it is tails, then he exposes the green die. He continues doing this until a red four is exposed. When the red four is exposed, what is the probability the green die is also a four?
I maintain the answer is 1/11.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
May 26th, 2015 at 7:38:01 AM
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You guys ars driving the guys on Alan's board nuts. They have four or five threads about this question.
DUHHIIIIIIIII HEARD THAT!
