A roll of the dice

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Re: A roll of the dice

Post by Gord » Sun Jan 27, 2013 3:08 pm

Rob Lister wrote:Gord is vindicated

Nifty! :thumbsup:

Thanks for that. Can't do those sorts of things for myself anymore. I wish I'd kept up with my computer programming courses, but they literally drove me crazy.
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Re: A roll of the dice

Post by Vanguard » Sun Jan 27, 2013 6:07 pm

*facepalm* :roll:

If you don't understand the logic behind the math, you can't write a program or build a model. Rob and John, you wrote programs to do the exact same wrong thing that Gord did. That doesn't make him right.

Your computer simulations come up with the number of times out of 36 that at least one die will show a 1 and separately how many times out of 36 you get snakeyes. Then you tell it to arbitrarily have the computer set them up as a probability solution in an improper manner and label it as the result you want to find out. But that's not how it works. The number on the right is the total number of possibilities. Both of the numbers are the results, not total possibilities.

Once we have been told that at least one of the die is a 1, there are two distinct possibilities. Either die A is a 1 or die B is a 1. It doesn't matter that both can be a one yet, because we only know that one die says a 1.

If die A is a 1, then we remove all of the rows on the probability table that aren't 1. It looks like this:
1,1 1,2 1,3 1,4 1,5 1,6

If die B is a 1, then we remove all of the columns on the probability table that aren't 1. It look like this:
1,1
2,1
3,1
4,1
5,1
6,1

Knowing that 'at least one die says 1' means that either table can exist but both cannot exist simultaneously as the die that is a 1, whichever die it is, will not change to another number once the other die is revealed. It doesn't matter if you know whether die A or die B is a 1.

If die A is the known 1, there is only a 1/6 chance that die B is a 1. If die B is the known 1, there is only a 1/6 chance that die A is a 1.
Your model overlaps table A and table B in a way that isn't consistent with reality and so gives you an incorrect answer.

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Re: A roll of the dice

Post by Rob Lister » Sun Jan 27, 2013 6:30 pm

Vanguard wrote:*facepalm* :roll:

If you don't understand the logic behind the math, you can't write a program or build a model. Rob and John, you wrote programs to do the exact same wrong thing that Gord did. That doesn't make him right.


Don't know about John, but I used Gord's model almost exactly

If his pseudo code model represents the conditions--and one MUST presume it does since they are his conditions--then what he has been saying is correct, ergo 11 is the answer.

There's no wiggle room here. There's no room for debate.

You can refuse to like what he's saying, but it stands on its own. You can argue that what he wrote longhand does not match what he wrote in pseudo code, but that's a different argument.

The proof is in there.

eta: in my translation of his pseudo code I made certain small assumptions. Compare for yourself.
eta2: in my presentation of my translation, there is a missing quote and crlf that has been fixed.
eta3: reasons for strikes at later post.
Last edited by Rob Lister on Sun Jan 27, 2013 7:54 pm, edited 2 times in total.
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Re: A roll of the dice

Post by John Jackson » Sun Jan 27, 2013 6:43 pm

Vanguard wrote:*facepalm* :roll:

If you don't understand the logic behind the math, you can't write a program or build a model. Rob and John, you wrote programs to do the exact same wrong thing that Gord did. That doesn't make him right.

Or you could just be wrong :!:

I'd be interested in seeing your program or algorithm to see how you would tackle it.

Vanguard wrote:Your computer simulations come up with the number of times out of 36 that at least one die will show a 1

Which is what we need to know.

Vanguard wrote:and separately how many times out of 36 you get snakeyes.

Again, this is what we need to know.

Vanguard wrote:Then you tell it to arbitrarily have the computer set them up as a probability solution in an improper manner and label it as the result you want to find out.

There's nothing arbitrary about it. We're trying to answer the question: what is the probability that we've thrown a [1,1] given that at least one of the dice is a 1.

Vanguard wrote:Once we have been told that at least one of the die is a 1, there are two distinct possibilities. Either die A is a 1 or die B is a 1.

[.....]

If die A is the known 1, there is only a 1/6 chance that die B is a 1. If die B is the known 1, there is only a 1/6 chance that die A is a 1.

The important bits are highlighted in red.

We never know whether die A or die B is a 1. It is not known. This is different from the situation where we know which die is a 1 (I included this extra information in my program which clearly shows that if we know which die is a 1, the answer to whether the other one is also a 1 is 1 in 6.

Knowing that at least one die is a 1 is different from knowing which die is a 1. That's the crucial insight you seem to be missing.

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Re: A roll of the dice

Post by SweetPea » Sun Jan 27, 2013 6:51 pm

Here's an offbeat question:

To help understand where we commonly go wrong, is the "why" a useful question ? For example, in the Monty Hall Problem, does taking the element of human actions and relationships into account, help solve the puzzle - and further, can it help correct thinking ?

viewtopic.php?f=33&t=19488
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Re: A roll of the dice

Post by Rob Lister » Sun Jan 27, 2013 7:30 pm

Late entry.

Gord, you're not vindicated.

If I change line 60 to do 10,000 rolls, I get a substantially different answer.
That should not happen.


Code: Select all

5 counter1 = 0
6 counter2 = 0
7 counter3 = 0
10  counter1 = counter1 + 1
20                       'Roll two [6 sided] dice.
die1 = int((rnd * 5)+1)
die2 = int((rnd * 5)+1)

30 If die1 > 1 and die2 > 1 goto 10
40  counter2 = counter2 + 1
50  if die1=1 and die2=1 then counter3=counter3+1
60  if counter1<10000 goto 10
70  print "Total number of rolls: ";counter1
80  print "Rolls where at least one die shows a 1:  ";counter2
90  print "Rolls where both dice show 1s: ";counter3
100  print "Chances of both dice showing 1s when it is known that"
     print "at least one die shows a 1 is  ";counter2/counter3


Image

We approach 9
(which makes since given the code. but at least it's not 1/6)

Gord, you are wrong.

:D
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Re: A roll of the dice

Post by SweetPea » Sun Jan 27, 2013 7:59 pm

Rob Lister wrote:Late entry.

Gord, you're not vindicated.

If I change line 60 to do 10,000 rolls, I get a substantially different answer.
That should not happen.


Code: Select all

5 counter1 = 0
6 counter2 = 0
7 counter3 = 0
10  counter1 = counter1 + 1
20                       'Roll two [6 sided] dice.
die1 = int((rnd * 5)+1)
die2 = int((rnd * 5)+1)

30 If die1 > 1 and die2 > 1 goto 10
40  counter2 = counter2 + 1
50  if die1=1 and die2=1 then counter3=counter3+1
60  if counter1<10000 goto 10
70  print "Total number of rolls: ";counter1
80  print "Rolls where at least one die shows a 1:  ";counter2
90  print "Rolls where both dice show 1s: ";counter3
100  print "Chances of both dice showing 1s when it is known that"
     print "at least one die shows a 1 is  ";counter2/counter3


Image

We approach 9
(which makes since given the code. but at least it's not 1/6)

Gord, you are wrong.

:D
but what exactly is he wrong about other than the answer?
How do the Deniers get so lucky?
http://www.skepticforum.com/viewtopic.php?f=16&t=24129" onclick="window.open(this.href);return false;

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Re: A roll of the dice

Post by John Jackson » Sun Jan 27, 2013 8:12 pm

Rob Lister wrote:We approach 9
(which makes since given the code. but at least it's not 1/6)

Gord, you are wrong.

:D


Your line 100: print "Chances of both dice showing 1s when it is known that" print "at least one die shows a 1 is ";counter2/counter3

Should be counter3/counter2 (*100 to give a percentage answer).

The correct percentage is ~9% which is the same as 1 in 11.

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Re: A roll of the dice

Post by Rob Lister » Sun Jan 27, 2013 8:30 pm

John Jackson wrote:
Rob Lister wrote:We approach 9
(which makes since given the code. but at least it's not 1/6)

Gord, you are wrong.

:D


Your line 100: print "Chances of both dice showing 1s when it is known that" print "at least one die shows a 1 is ";counter2/counter3

Should be counter3/counter2 (*100 to give a percentage answer).

The correct percentage is ~9% which is the same as 1 in 11.


Not my line. His line. Let him tell me.
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Re: A roll of the dice

Post by Rob Lister » Sun Jan 27, 2013 8:33 pm

SweetPea wrote:but what exactly is he wrong about other than the answer?


I don't know. I'm playing straight man here.
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Re: A roll of the dice

Post by John Jackson » Sun Jan 27, 2013 8:43 pm

Rob Lister wrote:
SweetPea wrote:but what exactly is he wrong about other than the answer?


I don't know. I'm playing straight man here.

There's nothing actually wrong with the calculation. I expect the slightly lower than expected chance is simply due to variation or a non-too-good random number generator.

I would think that if you run the program several times, the answer will move back towards the expected value of 11 (or 9%)

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Re: A roll of the dice

Post by Vanguard » Sun Jan 27, 2013 9:26 pm

How many times do I have to say it?

Your model is wrong. You are getting the result you want because you aren't thinking about the problem right. Forget the computer.

Here is the model you need to use:

Set one die on the table with 1 facing up. Roll the 2nd die 50 times. Count how many times the two dice both have 1s versus the number of rolls.
Next turn the die you rolled to a 1 and set it on the table. Roll the die that was previously the 1 50 times. Count the same way as the first time.

Do you get around 17 snakeyes or 9?

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Re: A roll of the dice

Post by SweetPea » Sun Jan 27, 2013 9:31 pm

Vanguard wrote:How many times do I have to say it?

Your model is wrong. You are getting the result you want because you aren't thinking about the problem right. Forget the computer.

Here is the model you need to use:

Set one die on the table with 1 facing up. Roll the 2nd die 50 times. Count how many times the two dice both have 1s versus the number of rolls.
Next turn the die you rolled to a 1 and set it on the table. Roll the die that was previously the 1 50 times. Count the same way as the first time.

Do you get around 17 snakeyes or 9?
Why do you set one die on the table ?
How do the Deniers get so lucky?
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Re: A roll of the dice

Post by Vanguard » Sun Jan 27, 2013 9:36 pm

To represent that s die is already known to be a 1.

Gord's question happens after the dice are rolled, not before. The flaw in his model is that he inclued the possibility that the dice we know to be a 1 still needs to be rolled. It does not.

The chances that a 1 will be rolled on 2d6 is 11/36. The chances that a 1 was rolled in Gord's set up is 100%. He seems to think that it's only 1/6 despite the fact that he is already looking at a die that came up a 1.
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Re: A roll of the dice

Post by SweetPea » Sun Jan 27, 2013 9:40 pm

Vanguard wrote:To represent that s die is already known to be a 1.
in the roll already performed.

Gord's question happens after the dice are rolled, not before. The flaw in his model is that he inclued the possibility that the dice we know to be a 1 still needs to be rolled. It does not.
then neither does the other die need to be rolled. It's been rolled already, too.
Last edited by SweetPea on Sun Jan 27, 2013 9:44 pm, edited 1 time in total.
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Re: A roll of the dice

Post by Vanguard » Sun Jan 27, 2013 9:43 pm

Except that we know the result of one die already. We are only looking for probability thst the 2nd die is a 1.

On that die there are 6 sides. One side says 1. The probability is 1/6.

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Re: A roll of the dice

Post by SweetPea » Sun Jan 27, 2013 9:47 pm

Vanguard wrote:Except that we know the result of one die already.
in that roll

We are only looking for probability thst the 2nd die is a 1.
I think that is the place where a decision is commonly being made
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Re: A roll of the dice

Post by Vanguard » Sun Jan 27, 2013 9:53 pm

Only the result of that singular roll is being considered in the question.

Yes, that is where you all are getting it wrong. Since we know the result of one die already, the only factor is the probability of the 1 coming up on the 2nd die. The die that is known cannot change the number it shows. It is independent.

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Re: A roll of the dice

Post by SweetPea » Sun Jan 27, 2013 10:01 pm

Vanguard wrote:Only the result of that singular roll is being considered in the question.

Yes, that is where you all are getting it wrong. Since we know the result of one die already, the only factor is the probability of the 1 coming up on the 2nd die. The die that is known cannot change the number it shows. It is independent.
That's why it's like the MHP. Maybe a "reverse" MHP.
How do the Deniers get so lucky?
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Re: A roll of the dice

Post by SweetPea » Sun Jan 27, 2013 10:11 pm

Vanguard wrote:Only the result of that singular roll is being considered in the question.

Yes, that is where you all are getting it wrong. Since we know the result of one die already, the only factor is the probability of the 1 coming up on the 2nd die. The die that is known cannot change the number it shows. It is independent.
Would your method then involve "erasing" all results containing non 1 results formerly considered obtainable by what we call "die A"?
How do the Deniers get so lucky?
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Re: A roll of the dice

Post by Vanguard » Sun Jan 27, 2013 10:18 pm

Yeah, that's why I said it was nothing like the MHP. That is an example of Dependent Probability, like rolling dice in a sequence. This is an example of Independent Probability.

On the probability table, I would remove or erase either 5 entire columns or 5 entire rows to represent the impossibility of one of the die being anything but a 1. Gord didn'remove enough on his modified table and so included the possibility that the known die could be anything other than a 1. The 2d6 table doesn't apply for this question, only the 1d6.

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Re: A roll of the dice

Post by SweetPea » Sun Jan 27, 2013 10:20 pm

I'm just trying to think of how to experiment with your approach.
Mark die "A", roll the pair of dice a number of times and subtract all the results that have die "A" result other than 1?

I see you've already started to answer about those things as I was typing.
How do the Deniers get so lucky?
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Re: A roll of the dice

Post by Gord » Mon Jan 28, 2013 12:17 am

Rob Lister wrote:Don't know about John, but I used Gord's model almost exactly

I keep saying my math is poor. :P

counter1 is the total number of rolls.
counter2 is the total number of rolls where at least one 1 was rolled.
counter3 is the total number or rolls where two 1s were rolled.

In my code, I tried (but failed! by reversing the two values) to make the result equal to "the total number or rolls where two 1s were rolled" / "the total number of rolls where at least one 1 was rolled" -- as in, the outcome should have looked like "30/330" rather than simply "11". I don't know how to program anymore, but I seemed to remember it was something like the way I presented it:

print "Chances of both dice showing 1s when it is known that at least one die shows a 1: "; counter2; "/"; counter3

:pardon: Sorry for my poor programming skills, too.
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Re: A roll of the dice

Post by SweetPea » Mon Jan 28, 2013 12:23 am

Vanguard, Am I correct to say that by using your approach this is where Gord made a wrong decision?
When I told you at least one of them showed a 1, you have to discard all the combinations that don't contain at least one 1. It turns out, there are 25 of them

in that he should also have discarded combinations that do not show 1 on die "A"?
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Re: A roll of the dice

Post by SweetPea » Mon Jan 28, 2013 1:14 am

Highly orchestrated attacks by well-funded Dice Denialists
How do the Deniers get so lucky?
http://www.skepticforum.com/viewtopic.php?f=16&t=24129" onclick="window.open(this.href);return false;

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Re: A roll of the dice

Post by SweetPea » Mon Jan 28, 2013 2:17 am

Gord, I believe the Vanguard reasoning can be said like this:
You chose to discard all combinations that do not show at least one 1.
That was based on the information that at least one die DID show a 1
If we name the dice Red/Blue and you can choose to name the one actually rolled and showing 1, as Blue, the other die is Red and has the unknown result.
In your square when you are discarding impossible combination results, all combinations that do not have at least one 1, are taken away.
You should have also taken away any combination with ONLY one die showing 1 if it's not the Blue one, because the Blue one DID show a1
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Re: A roll of the dice

Post by John Jackson » Mon Jan 28, 2013 2:20 am

Vanguard wrote:Here is the model you need to use:

Set one die on the table with 1 facing up. Roll the 2nd die 50 times. Count how many times the two dice both have 1s versus the number of rolls.

This just confirms that it's you who hasn't grasped the nature of the problem.

Here is the model you need to use:

Throw two dice. Check them both to see whether at least one of them is a 1. If so, increase your tally by one. Check to see whether both dice are a 1. If so increase that tally by one.

Do it enough times to get a decent sample and then work out the number of [1,1]s you get when at least one of them was initially a 1.

Instead of arguing, do the experiment :!:

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Re: A roll of the dice

Post by John Jackson » Mon Jan 28, 2013 2:29 am

SweetPea wrote:If we name the dice Red/Blue and you can choose to name the one actually rolled and showing 1, as Blue, the other die is Red and has the unknown result.

We don't get to know the result of one throw before another one is made though. Nor do we get to look at one die to see whether it is a 1 before calculating the chance of the second one being a 1 also.

Two dice are thrown and all we know is that at least one of them is a 1. There are eleven ways that this can occur and one of them is [1,1] hence it will happen one time in eleven under these conditions (which is the conditions stated in the original problem).

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Re: A roll of the dice

Post by Vanguard » Mon Jan 28, 2013 2:43 am

If I perform your experiment I will get the wrong answer because your experiment doesn't fit the question.
Your experiment is determining the probability of getting a 1 when neither of the dice are known. Gord screwed up by changing the question by making one of the die known. This changes everything.


SweatPea has the hang of it.

So, tell me the answer to this. Assuming Gord isn't lying, what is the probability of at least one die being a 1 out of the two dice he rolled?

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Re: A roll of the dice

Post by John Jackson » Mon Jan 28, 2013 2:53 am

Here's the problem as stated in the opening post:

Gord wrote:Here's one for your statistical analysis:

I'm sitting at a table. You walk in and hand me a pair of dice. Each die has six sides, and you are certain that they are fair -- in other words, they aren't loaded at all, and they would be acceptable to anyone who wanted to roll fairly.

You sit down across the table from me. Then I roll both dice behind a screen where you can't see them.

"Did you roll at least one 1?" you ask me.

I look at the dice. "Yes," I reply. "At least one of these two dice shows a 1 on top."

Now here's the question: What are the odds that I rolled two 1s? :mrgreen:
(Red highlighting mine.)

The question is unequivocal.

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Re: A roll of the dice

Post by Vanguard » Mon Jan 28, 2013 2:59 am

So you agree that the probability of there being a 1 on at least one die in the given scenario is 100%?

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Re: A roll of the dice

Post by SweetPea » Mon Jan 28, 2013 3:08 am

Vanguard wrote:So you agree that the probability of there being a 1 on at least one die in the given scenario is 100%?
Answering an unequivocal question with another ( I mean that in a good way). Dice Denialism is where the action is. Thirty years is too long to wait with Climate.
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Re: A roll of the dice

Post by SweetPea » Mon Jan 28, 2013 3:13 am

John Jackson wrote:
SweetPea wrote:If we name the dice Red/Blue and you can choose to name the one actually rolled and showing 1, as Blue, the other die is Red and has the unknown result.

... Nor do we get to look at one die to see whether it is a 1 before calculating the chance of the second one being a 1 also.
We can name it though, even though we do not know it's physical location. It's got a 1 on top. So you can tell later.
How do the Deniers get so lucky?
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Re: A roll of the dice

Post by Vanguard » Mon Jan 28, 2013 3:23 am

Well, I need to walk them through the solution because they don't get it when I describe it in a broader sense. Their logic is flawed here so I can't take for granted that they understand any of the concepts involved.

I'm halfway expecting one of them to tell me the probability of there being a 1 present on at least one of the dice in question to be 11/36, despite the fact that we know there is one for certain. Their computer model makes this assumption, actually, though they may not realize it. Essentially, I'm plumbing the depths of their derptitude.

The question requires an answer. What is the probability that at least one die says a 1 in the given scenario?

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Re: A roll of the dice

Post by SweetPea » Mon Jan 28, 2013 3:29 am

Vanguard wrote:Well, I need to walk them through the solution because they don't get it when I describe it in a broader sense. Their logic is flawed here so I can't take for granted that they understand any of the concepts involved.

I'm halfway expecting one of them to tell me the probability of there being a 1 present on at least one of the dice in question to be 11/36, despite the fact that we know there is one for certain. Their computer model makes this assumption, actually, though they may not realize it.
In Climate Science the data is always wrong when there is disagreement.
This is much faster

The question requires an answer. What is the probability that at least one die says a 1 in the given scenario?
How do the Deniers get so lucky?
http://www.skepticforum.com/viewtopic.php?f=16&t=24129" onclick="window.open(this.href);return false;

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Re: A roll of the dice

Post by SweetPea » Mon Jan 28, 2013 3:46 am

Is this another hide the die cline.?
How do the Deniers get so lucky?
http://www.skepticforum.com/viewtopic.php?f=16&t=24129" onclick="window.open(this.href);return false;

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dicey connections

Post by SweetPea » Mon Jan 28, 2013 3:59 am

That's why humans have the advantage. We can name things.
No wonder James Hansen is all about dice now.

, [ytube]http://www.youtube.com/watch?v=TX2KyF0p-xU[/ytube]
How do the Deniers get so lucky?
http://www.skepticforum.com/viewtopic.php?f=16&t=24129" onclick="window.open(this.href);return false;

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Re: A roll of the dice

Post by John Jackson » Mon Jan 28, 2013 5:26 am

Vanguard wrote:Well, I need to walk them through the solution because they don't get it when I describe it in a broader sense. Their logic is flawed here so I can't take for granted that they understand any of the concepts involved.

The more smug and condescending you get the sillier you're going to look when (or if) you ever manage to conceptualize the problem correctly. :!:

Vanguard wrote:I'm halfway expecting one of them to tell me the probability of there being a 1 present on at least one of the dice in question to be 11/36

Why? No one has said this. We know from the question set that at least one of the dice is a 1. The important step is to understand that there are eleven possible ways that at least one of them can be a 1.

Vanguard wrote:The question requires an answer. What is the probability that at least one die says a 1 in the given scenario?

100% - it's in the question.

OK, what's the next step?

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Re: A roll of the dice

Post by SweetPea » Mon Jan 28, 2013 6:08 am

Vanguard wrote:How many times do I have to say it?

Your model is wrong. You are getting the result you want because you aren't thinking about the problem right. Forget the computer.

Here is the model you need to use:

Set one die on the table with 1 facing up. Roll the 2nd die 50 times. Count how many times the two dice both have 1s versus the number of rolls.
why fitty times not sitty?.60 rolls each makes 20 out of 120 rolls which is nice. Then you don't have to bother
How do the Deniers get so lucky?
http://www.skepticforum.com/viewtopic.php?f=16&t=24129" onclick="window.open(this.href);return false;

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Right turn at Albuquerque

Post by SweetPea » Mon Jan 28, 2013 6:27 am

How about a twist on things?
Suppose Gord rolls the dice behind blinds and only reveals to himself one die at a time. If he opens the first door ( Blue) and finds a 1, he reports finding a 1 and does not open the Red door.

Does that scenario help out?
edited to fix
1/After 120 rolls, he's opened the blue door and found a 1, (120/6) times, or .20 times.
2/After 120 rolls he's had to open the red door 100 times.
3/He's found a 1 behind the red door (100/6) times in 120 rolls, but those 1s don't matter because they're not snake-eyes..
4/He's found a1 behind neither door (100 -100/6) times in 120 rolls.

The problem then is the red doors Gord did not open because he found a 1 behind the blue door.
5/Out of those 20 times that Gord found a 1 behind the blue door in 120 rolls, he should have expected to find a 1 behind the red door how many times? (20/6) times.

The problem becomes how many of the 120 rolls did our Gord have to deal with ?
6/Because he only has to deal with rolls that have a 1, there were (120 rolls - rolls with neither door holding a 1) = 120 - (100 - 100/6) rolls for Gord.

This way at least the die behaves naturally!
How do the Deniers get so lucky?
http://www.skepticforum.com/viewtopic.php?f=16&t=24129" onclick="window.open(this.href);return false;