Two dice puzzle -- Part Trois
Quote: billryanRemind me why we keep Indiana around again
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Gives the bookmakers something to do next weekend?
Plus there are all those Amish craftsmen manufacturing recreational vehicles and travel trailers.
Quote: AlanMendelsonAgain... everyone tells me I must consider two dice but no one tells me why?
Everyone says the dice cup problem/question is not the same as a die spinning on a craps table but no one tells me why?
Again I'm going to say the question was poorly worded and was a setup to create controversy.
Unless someone tells me WHY two dice must be considered and the problem is not like a spinning die, I'm sticking with my answer.
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In short - because that's how the reality in this Universe we're all in works.
Many people here have explained why, including my attempt earlier which also tries to explain why in layman terms.
"BOTH dice work together to produce eleven outcomes with at least one 2 showing. And EITHER die can show a non-two in those eleven outcomes thus evading the 2-2 combo for longer than 1/6 of the time, namely only one in eleven times will they both show a 2.
...
You just need to realize there is a combined probability between BOTH dice for producing the "at least one 2" combos AND the exactly 2-2 combo.
"
You just seem to lack the willing to consciously read through these and put the mental effort in understanding it.
Quote: billryanQuote: billryanAllen,
Please answer each question, step by step. Maybe then you can see why you are mistaken
1) There are 36 combinations you can roll with two dice Yes or No
2) Of those 36 combinations, there are eleven possible combinations that contain a Six. Are we in agreement?
3) After they are rolled and we know there is at least one six, the twenty-five other combinations are eliminated because we know there is a six. Do you agree to this?
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Why won't Allen answer these simple questions?
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Because Alan was busy.
1. Yes
2. Yes
3. Please restate this question. I don't understand.
Quote: billryanQuote: AlanMendelsonQuote: billryanQuote: AlanMendelsonQuote: billryanI suppose the author did a bad job as he only managed to fool one person.
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I'll put money on this: I'm not the only one who says 1/6.
How much?
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How many of us looked at the evidence and still insisted it was 1 in 6. How is it that someone like myself can look at the evidence, see how I was mistaken and use the actual evidence to come up to the correct answer while you keep saying there is no evidence and it doesn't matter anyway.
When you look at the various charts that show there are 11 possible combinations and only one is a pair, what is your reaction?
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Tell you what...
I'll go on my TV show with the question, solicit postcards with an answer as part of a drawing.
How many 1/6 votes do I need?
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Need for what? We already know most American adults can't pass fifth grade math.
How about this.
Do a show and ask people to write in. Then get anyone who understands math to come on your show, walk everyone thru the proper way to answer the question. Then see how many still agree with you. I've found there are always one or two in every crowd.
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"The proper way to answer the question"? LOL
Quote: rawtuffQuote: AlanMendelsonAgain... everyone tells me I must consider two dice but no one tells me why?
Everyone says the dice cup problem/question is not the same as a die spinning on a craps table but no one tells me why?
Again I'm going to say the question was poorly worded and was a setup to create controversy.
Unless someone tells me WHY two dice must be considered and the problem is not like a spinning die, I'm sticking with my answer.
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In short - because that's how the reality in this Universe we're all in works.
Many people here have explained why, including my attempt earlier which also tries to explain why in layman terms.
"BOTH dice work together to produce eleven outcomes with at least one 2 showing. And EITHER die can show a non-two in those eleven outcomes thus evading the 2-2 combo for longer than 1/6 of the time, namely only one in eleven times will they both show a 2.
...
You just need to realize there is a combined probability between BOTH dice for producing the "at least one 2" combos AND the exactly 2-2 combo.
"
You just seem to lack the willing to consciously read through these and put the mental effort in understanding it.
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Why do we need eleven outcomes?
And to get 11 outcomes you need to count one die twice.
Do you count your deductions twice when you fill out your taxes?
Quote: AlanMendelsonQuote: billryanQuote: billryanAllen,
Please answer each question, step by step. Maybe then you can see why you are mistaken
1) There are 36 combinations you can roll with two dice Yes or No
2) Of those 36 combinations, there are eleven possible combinations that contain a Six. Are we in agreement?
3) After they are rolled and we know there is at least one six, the twenty-five other combinations are eliminated because we know there is a six. Do you agree to this?
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Why won't Allen answer these simple questions?
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Because Alan was busy.
1. Yes
2. Yes
3. Please restate this question. I don't understand.
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Sure. If we look at the 36 combinations, there are 11 combinations that have a six, and 25 possible combinations that don't have a six.
Once we know there is a six, that means it must be one of the eleven combinations that have at least one. We can eliminate any combination that does not have a six. For example, you cant have 1-1 because we have a six. So we have eliminated twenty-five combinations that don't have a 6 in it. Are you following?
Quote: AlanMendelson
Why do we need eleven outcomes?
And to get 11 outcomes you need to count one die twice.
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Alan,
My teaching hat is off. My green pointy mod hat is on.
If you're trying to understand, you're discussing.
If you are not trying to understand, you are trolling.
You have several outs, one of which is to drop the subject.
Plenty of very smart people have given you explanations of why considering only one dice is inappropriate in some situations.
You have apparently decided not to test your assumptions with dice and cash money.
I suggest you consider the functional differences, if any, between the situation you envision where one dice may be excluded and the cash money test that has been proposed.
Quote: Dieter
Alan,
My teaching hat is off. My green pointy mod hat is on.
If you're trying to understand, you're discussing.
If you are not trying to understand, you are trolling.
Interesting
I sense that at least one of the moderators is losing patience. What's the probability that...
Quote: DieterI suggest you consider the functional differences, if any, between the situation you envision where one dice may be excluded and the cash money test that has been proposed.
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But Dieter still can't resist feeding him with excuses to fail again.
Quote: billryanQuote: AlanMendelsonQuote: billryanQuote: billryanAllen,
Please answer each question, step by step. Maybe then you can see why you are mistaken
1) There are 36 combinations you can roll with two dice Yes or No
2) Of those 36 combinations, there are eleven possible combinations that contain a Six. Are we in agreement?
3) After they are rolled and we know there is at least one six, the twenty-five other combinations are eliminated because we know there is a six. Do you agree to this?
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Why won't Allen answer these simple questions?
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Because Alan was busy.
1. Yes
2. Yes
3. Please restate this question. I don't understand.
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Sure. If we look at the 36 combinations, there are 11 combinations that have a six, and 25 possible combinations that don't have a six.
Once we know there is a six, that means it must be one of the eleven combinations that have at least one. We can eliminate any combination that does not have a six. For example, you cant have 1-1 because we have a six. So we have eliminated twenty-five combinations that don't have a 6 in it. Are you following?
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Bill, give it up. I had to try giving up three times before it took. That’s more times than I had to try before I gave up smoking. The fact that Alan (a) won’t roll dice to test himself, or (b) won’t make a wager he should think is way +EV or (c) avoids answering questions when it gets too close to the bone like yours . . .
. . . should tell you everything you need to know. Be billryan, not Billy goat gruff.
Unjon- I borrowed your illustrated chart to show a few non-believers on another forum and it is the best example I have come across.
Quote: unJon
Bill, give it up. I had to try giving up three times before it took. That’s more times than I had to try before I gave up smoking. The fact that Alan (a) won’t roll dice to test himself, or (b) won’t make a wager he should think is way +EV or (c) avoids answering questions when it gets too close to the bone like yours . . .
. . . should tell you everything you need to know. Be billryan, not Billy goat gruff.
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Here is a customer service announcement.
Are you addicted to trying to persuade someone of something?
Is it still fun?
When the fun stops, Stop.
Link to unsubscribe and hide this thread: https://wizardofvegas.com/block-thread/37156/
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And if you change your mind... https://wizardofvegas.com/preferences
Link to waste your time on and free your mind https://www.websudoku.com/?level=4
Stay Safe
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Quote: OnceDear
Link to ignore any posts by OnceDear: https://wizardofvegas.com/block/oncedear/
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Completeness demands:
https://wizardofvegas.com/block/dieter/
Another timewasting link:
Proceed with caution: This rabbit hole runs deep.
https://wikipedia.org/wiki/Special:Random
Quote: OnceDearQuote: unJon
Bill, give it up. I had to try giving up three times before it took. That’s more times than I had to try before I gave up smoking. The fact that Alan (a) won’t roll dice to test himself, or (b) won’t make a wager he should think is way +EV or (c) avoids answering questions when it gets too close to the bone like yours . . .
. . . should tell you everything you need to know. Be billryan, not Billy goat gruff.
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Here is a customer service announcement.
Are you addicted to trying to persuade someone of something?
Is it still fun?
When the fun stops, Stop.
Link to unsubscribe and hide this thread: https://wizardofvegas.com/block-thread/37156/
Link to ignore any posts by Alan Mendelson: https://wizardofvegas.com/block/alanmendelson/
Link to ignore any posts by OnceDear: https://wizardofvegas.com/block/oncedear/
Link to ignore any posts by BillRyan: https://wizardofvegas.com/block/billryan/
Link to ignore any posts by Unjon: https://wizardofvegas.com/block/unjon/
And if you change your mind... https://wizardofvegas.com/preferences
Link to waste your time on and free your mind https://www.websudoku.com/?level=4
Stay Safe
Stay Sane.
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The problem with that is some posters like to quote who they are responding to so even though you block the poster, you still see their posts as they get "quoted". I find it's easier to just scroll past a person's post than to use the ignore. It's no big deal as I only have put one person on ignore, only to see how ineffective it is. I wish I had an alternative to offer.
Be Pure.
Be Vigilant
Be Nice.
Given any of six faces on any die, 1/6 on another die gives you matching faces.
Words like "at least one die" don't change that and only make you look for complex answers.
Rolling dice doesn't answer the original question which is simply -- without the extra words-- if you've got a 2 on one die what are the chances a 2 will show on another die.
This entire problem reminds me of the trick test we got in 6th grade:
Our teacher handed us a one page test paper. The instruction at the top read: read all instructions before answering. 20 questions followed including math and writing questions. And immediately everyone in the class started doing the math and writing answers... only to come to #20 which said "write your name on the top and leave the rest of the page blank."
I have nothing else to say.
Quote: AlanMendelsonWhy do we need eleven outcomes?
Because that is how one assess the probability of two dice rolling at least one 2 and the probability of having both dice being a 2 on that roll.
It seems the root of the misunderstanding is your ignorance on the basics of what probability is and how is it assessed.
I feel no one can help you with that and you need to read on those basics and try and realize how correctly calculating probabilities isn't some abstract math mumbo jumbo but the most accurate approach to not only gambling related matter but also to most essential everyday situations.
Quote: AlanMendelsonAnd to get 11 outcomes you need to count one die twice.
No, you don't.
What if you roll two dice one at time? You roll the first one and no matter what you then roll the second one and observe what happens in that (combined) roll of two dice.
You roll a 2 with the first die. Then the chances of rolling a 2 with the second die is 1/6 and consequently having a 2-2 rolled out within the both rolls is now 1/6.
But what if you roll a non-two with the first die (5/6 of the time)? The chance of rolling a 2 with the second die is again 1/6, but your chance of rolling a 2-2 out of the both rolls is now suddenly 0% and you would've still observed "at least one 2 is showing" out of the combined two dice roll should the second die landed on a 2 (basically what Gordon said in his last excellent post, but seems someone didn't bother reading or comprehend it). Whatever will we do?
Will we still boldly proclaim that rolling two dice and having one showing a 2 gives us a 1/6 chance of both of them showing a 2, cause a die has six faces? What gives?
The above is not rhetorical, you need to read and think about posts trying to answer your puzzlement and not just gloss over and dismiss'em.
Quote: rawtuffBut let us consider a single die roll at a time for a moment in the same context to satisfy the insisting ones.
What if you roll two dice one at time? You roll the first one and no matter what you then roll the second one and observe what happens in that (combined) roll of two dice.
You roll a 2 with the first die. Then the chances of rolling a 2 with the second die is 1/6 and consequently having a 2-2 rolled out within the both rolls is now 1/6.
But what if you roll a non-two with the first die (5/6 of the time)? The chance of rolling a 2 with the second die is again 1/6, but your chance of rolling a 2-2 out of the both rolls is now suddenly 0% and you would've still observed "at least one 2 is showing" out of the combined two dice roll should the second die landed on a 2 (basically what Gordon said in his last excellent post, but seems someone didn't bother reading or comprehend it). Whatever will we do?
Will we still boldly proclaim that rolling two dice and having one showing a 2 gives us a 1/6 chance of both of them showing a 2, cause a die has six faces? What gives?
The above is not rhetorical, you need to read and think about posts trying to answer your puzzlement and not just gloss over and dismiss'em.
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This has nothing to do with rolling one die or two dice. This is knowing that when you have any number showing on one six sided die, another die has a 1/6 chance of matching.
It's that simple, written in an awkward way to mislead you.
8:1 is the last offer I heard, in case anyone thinks 5:1 is fair odds and wants to get in.
Quote: DieterI calmly suggest that the discussion be paused until after the dice are thrown.
8:1 is the last offer I heard, in case anyone thinks 5:1 is fair odds and wants to get in.
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Can I bet with the Wizard? He's expected to win. But his win has nothing to do with "the question."
ANY number?Quote: AlanMendelsonThis has nothing to do with rolling one die or two dice. This is knowing that when you have any number showing on one six sided die, another die has a 1/6 chance of matching.
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Yes. With ANY number, you have 1/6 chance of matching.
But that’s not what the original problem stated. Not even close.
Quote: DJTeddyBearANY number?Quote: AlanMendelsonThis has nothing to do with rolling one die or two dice. This is knowing that when you have any number showing on one six sided die, another die has a 1/6 chance of matching.
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Yes. With ANY number, you have 1/6 chance of matching.
But that’s not what the original problem stated. Not even close.
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Any number.
1 gives you 1/6 chance of matching
2 gives you 1/6 chance of matching
3 gives you 1/6 chance of matching
4 gives you 1/6 chance of matching
5 gives you 1/6 chance of matching
6 gives you 1/6 chance of matching
Seriously? I mean seriously?Quote: AlanMendelsonAny number.Quote: DJTeddyBearANY number?Quote: AlanMendelsonThis has nothing to do with rolling one die or two dice. This is knowing that when you have any number showing on one six sided die, another die has a 1/6 chance of matching.
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Yes. With ANY number, you have 1/6 chance of matching.
But that’s not what the original problem stated. Not even close.
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1 gives you 1/6 chance of matching
2 gives you 1/6 chance of matching
3 gives you 1/6 chance of matching
4 gives you 1/6 chance of matching
5 gives you 1/6 chance of matching
6 gives you 1/6 chance of matching
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You gave me proof for the part of your post that I was agreeing to, and ignored the part of my post that says that it has nothing to do with the original problem???
Quote: DJTeddyBearSeriously? I mean seriously?Quote: AlanMendelsonAny number.Quote: DJTeddyBearANY number?Quote: AlanMendelsonThis has nothing to do with rolling one die or two dice. This is knowing that when you have any number showing on one six sided die, another die has a 1/6 chance of matching.
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Yes. With ANY number, you have 1/6 chance of matching.
But that’s not what the original problem stated. Not even close.
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1 gives you 1/6 chance of matching
2 gives you 1/6 chance of matching
3 gives you 1/6 chance of matching
4 gives you 1/6 chance of matching
5 gives you 1/6 chance of matching
6 gives you 1/6 chance of matching
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You gave me proof for the part of your post that I was agreeing to, and ignored the part of my post that says that it has nothing to do with the original problem???
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I've explained how I "read" the original problem.
You can stop trying to teach me. My mind is made up. Move on to another subject. Why don't you bring up the 18 yos?
Quote: billryanThe mind is like a parachute. Deadweight and useless until its owner chooses to open it.
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If you're going to insult me, be a man about it and put my name in your post.
Quote: AlanMendelsonQuote: DieterI calmly suggest that the discussion be paused until after the dice are thrown.
8:1 is the last offer I heard, in case anyone thinks 5:1 is fair odds and wants to get in.
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Can I bet with the Wizard? He's expected to win. But his win has nothing to do with "the question."
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https://wizardofvegas.com/forum/questions-and-answers/all-other/37156-two-dice-puzzle-part-trois/3/#post850487
June 8th.
Quote: billryanWhen someone insists for weeks on end that something is this but wants to bet that it is that, I have to pause and think WTF.
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Does that depend on which meaning you ascribe to "bet with"?
A man is charged with murder for killing three people in a traffic accident. He is charged with murder because he allegedly was on drugs at the time of the crash.
The defendant was in another fatal car crash three years earlier. That crash also killed three people and the defendant was found to be on drugs. The defendant was convicted of manslaughter in that crash.
You are told this because you are in the jury pool.
The prosecutor, wishing to find a fair jury, asks you:
1. Can you judge the defendant without prejudice even though he's had a prior conviction for the first fatal crash while on drugs? That is to say, the first crash would have no impression on you about possible guilt in the second crash.
2. Can you judge the defendant fairly, and can you say the first crash had no bearing on his guilt in the second crash?
3. But if the defendant is found guilty in the second crash can you then use the guilty verdict from the first crash to impose a much harsher sentence even though the verdict from the first crash did not influence your verdict in the second crash?
Also I don’t understand how the prejudicial information about the first conviction made it to the jury during voire dire, when the predicate is supposed to be that it’s not relevant to determining guilt.
1. Yes
2. Yes
3. Unknown. In most states, the punishment is decided by the judge and laws dictate that repeat offenders get harsher punishments.
So what does this gave to do with you ignoring evidence that dictates the answer in the two dice question is 1/11?
Quote: AlanMendelson
-snip-
I have nothing else to say.
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Yesterday you posted this and then stop posting. I thought you had reached an endpoint on this subject! i am sorry to see you resume your posting.
Alan, you are now approaching 90 posts on this thread, most in about the past 6 days. Of course, you are free to have your own opinion and to express it, but for several days you have been repeating yourself over and over and not really engaging any of the new comments that others are making. Please consider that if you have nothing new to say on this subject then it is time to move on to different subject on a different thread.
I am trying to be as polite as possible.
All you have to do is play the game in real-life to see that it has to be 1 out of 6.
That's all you gotta do, lol
Quote: TravisRI heard there was only one guy saying the answer was 1 out of 6 and I'm here to support that one guy because the answer *is* 1 out of 6.
All you have to do is play the game in real-life to see that it has to be 1 out of 6.
That's all you gotta do, lol
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Welcome to the forum.
There may be an opportunity for interesting odds coming up soon. Please play responsibly.
If the guy peeks under his cup 60 times and each time announces a number that he has do you really think he isn't gonna be rolling doubles 1/6th of the time? Or do you think that about 10 of those 60 times he's going to have doubles?
so every roll comes up once.
Our opponent randomly selects one of the dice to announce.
36 rolls so he's going to announce each number 6 times if he's doing it randomly
all non-doubles come up twice......2-5 comes up twice..... one time he annonces a 2 the other time he announces a 5
So after 36 trials, he's announced he has a 2 six times
one of those six times is the double 2's
the other 5 is half of the remaining 10 rolls that have a 2
1 out of 6..... like it should be because doubles come up 1 out of 6 rolls
Actually yes!Quote: TravisRThere are other fun ways to show it's 1 out of 6..... like..... what happens if you didn't quite hear what the guy mumbled about his dice? Would the answer be 1 out of 11 then ?..... nope.....
Here's that problem, spelled out.
Your friend rolls two dice, hiding the outcome. He peeks under the cup to see what was rolled. He says, truthfully, "At least one of the dice is a..." And at that exact moment a loud crash comes from the other room... You didn't hear what he said!. He then asks you, "What is the probability BOTH dice are a.... " CRASH! Another loud crash comes from the room! You again didn't hear him finish his sentence!
However, you heard enough that you can answer his question... which is 1/11. (Since it doesn't matter what number he says, of course.)
Welcome to the forum. (TravisR and I are members a backgammon forum.)
I'll play this game every time and every time I will bet that the guy has doubles for 1:10 odds , and 1 out of 6 times I'm going to collect
Its that simple
Quote: TravisRLet's play this game 36 times in our head distributing everything evenly.....
so every roll comes up once.
Our opponent randomly selects one of the dice to announce.
Stop right there. In the original problem, that's not what was done. That changes everything.
If he DOES do that, the answer is 1/6.
If he selects the RED die, for example, and announces the number, the answer is 1/6.
If he selects the LEFT die, for example, the answer is 1/6.
If he selects the die that has stopped spinning, the answer is 1/6.
But none of that was presented in the original question. (After rolling the dice, he peeks under the cup and says, truthfully, "At least one of the dice is a 2." Question: What is the probability both dice show a 2?)
Quote: TravisRBut the easiest proof is to simply offer to play this game with someone who want to pay out 10:1 odds
I'll play this game every time and every time I will bet that the guy has doubles for 1:10 odds , and 1 out of 6 times I'm going to collect
Its that simple
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That’s a different game and would indeed be 1/6. In the problem under discussion though the game isn’t played 25 out of 36 rolls, because the caller only calls it when he sees at least one 2.
Our friends keep on talking about how you need to know how one particular die landed. You dont.
You dont need to roll any dice.
You only need to know there are six different numbers on a die. And any other die also has six numbers. And one face out of six faces on each die will make a pair.
It's so simple even a fourth grader who plays Monopoly can figure it out.
Quote: TravisRI heard there was only one guy saying the answer was 1 out of 6 and I'm here to support that one guy because the answer *is* 1 out of 6.
All you have to do is play the game in real-life to see that it has to be 1 out of 6.
That's all you gotta do, lol
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All you have to do is take up the challenge then.
Quote: billryanAll you have to do is take up the challenge then.
Are you offering the challenge?
Quote: coachbellyQuote: billryanAll you have to do is take up the challenge then.
Are you offering the challenge?
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Absolutely, although Mike has first dibs. If Mike passes, I'll gladly play the house. I may no longer be bound by math, but I know a great opportunity when I see it.
"I just rolled 100 dice and at least 99 of them are 2's"
what's the odds that all 100 dice are 2's?
Quote: billryanQuote: TravisRI heard there was only one guy saying the answer was 1 out of 6 and I'm here to support that one guy because the answer *is* 1 out of 6.
All you have to do is play the game in real-life to see that it has to be 1 out of 6.
That's all you gotta do, lol
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All you have to do is take up the challenge then.
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Wrong. The challenge is a DIFFERENT problem.
The challenge does not represent the original question. It is another misinterpretation of ENGLISH.
I would bet along with the Wizard on the challenge.
SO he anounces a 2 as the other die is still spinning thru the air...... what's the odds he rolled a double 2?
