A roll of the dice

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Re: A roll of the dice

Postby SweetPea » Sat Feb 02, 2013 10:47 pm

Vanguard wrote:I was showing you that the color of the die doesn't matter..
Vanguard wrote:
SweetPea wrote:What are the chances that the blue die would be a die showing 1, in rolls which have results meeting the conditions { at least one die showing 1)?
50%


Even though you might be tempted to say "red/blue die " is like "six of one/half dozen of the other", so it's 50:50, that is an ill allusion.

The colour doesn't matter, but it does matter.

The colour does not matter, in the sense that either die could be named the blue die, or the "A" die, or any other signifier.
But the colour does matter, in keeping track of which die is which.

So it's a misleading simile.

Start with the correct statement that there are 11 ways to get results showing at least one 1.
Then count how many have the blue die showing 1.
The answer is 6. Out of 11.
That's not a probability of 50 %

That matters!

.
How do the Deniers get so lucky?
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Re: A roll of the dice

Postby peanuts72 » Sun Oct 20, 2013 2:19 pm

Hi, I just joined the forum to post here, because I can't get my head around why this is 11/1

I think it is clearly 6 to 1.
I'll outline my logic below,

So the two dices get rolled behind the screen and the observer cant see them, ok

now after the two dices have been rolled, they don't move so whatever state they are in now is fixed.

Clearly the odds of both of the dices coming up 1 is 36 to 1.

Now there is a question, "is at least one of the die showing a 1?"

To which the answer is given yes.

as an aside, this question gives us precisely the same information as would the questions:

Is one of the die a 1? (if both were 1, the answer would still be yes)

Is there a 1 showing on either die? (same information being deduced, answer would be yes).

Now, as the answer is yes, our knowledge of the state of the die has changed.

so say one dice is red and one dice is blue.

One of the following states definitely exists:

a)
The blue dice shows 1.

In this situation, the possible states, are 1,1 1,2 1,3 1,4 1,5 1,6

So in this situation the odds of both die showing a 1 is 6 to 1

b)
The red dice is 1

in this situation the possible states are the same as above. and the odds are 6 to 1

So in either state, the odds are 6 to 1

where am I going wrong? I cant see how it could be 11 to 1.

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Re: A roll of the dice

Postby Austin Harper » Mon Oct 21, 2013 1:16 pm

It helps to think of the question backwards.
What are the odds of each die not showing a one individually?
Spoiler:
There are 5 other options so the odds of not-one are 5/6

But you have two dice and you want to know the odds of either of them showing one.
Spoiler:
So you multiply the odds of each of them not being one and you get (5/6)*(5/6)=25/36.

That leaves the odds for either die showing a one as 11/36.
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Re: A roll of the dice

Postby Gord » Mon Oct 21, 2013 3:08 pm

peanuts72 wrote:One of the following states definitely exists:

Or better yet, each of the following states might exist and therefore cannot yet be discounted:

a)
The blue dice shows 1.

In this situation, the possible states, are 1,1 1,2 1,3 1,4 1,5 1,6

So in this situation the odds of both die showing a 1 is 6 to 1

There are those six possibilities, out of a total of 36 original possibilities.

And:
b)
The red dice is 1

in this situation the possible states are the same as above. and the odds are 6 to 1

So in either state, the odds are 6 to 1

where am I going wrong? I cant see how it could be 11 to 1.

There are those six possibilities, out of a total of 36 original possibilities.

When we group the two sets of possibilities, we discover that each set of six contains one identical possibility, so rather than having twelve distinct possibilities out of the original 36, we only have eleven -- both sets contain the possibility of having rolled 1,1 (in other words, two 1s).

It's a problem of adding together two sets:

Set 1: (1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
Set 2: (1,1), (2,1), (3,1), (4,1), (5,1), (6,1)

Most people add two sets of 6 together in their heads and get one set of 12, because they forget not to double-count identical units within the two sets. If we do that, we get a total of 12 remaining possibilities, 2 of which are the roll of double ones (1,1). If that worked mathematically, then it would indeed be 2/12 which reduces to 1/6. But no, that's an error: Because of the duplicate unit (1,1), when adding both sets together, we end up with a combined set of 11 possibilities, only one of which is (1,1):

Combined set: (1,1), (1,2), (2,1), (1,3), (3,1), (1,4), (4,1), (1,5), (5,1), (1,6, (6,1)
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Re: A roll of the dice

Postby peanuts72 » Mon Oct 21, 2013 4:51 pm

Hmm, ok thanks. I'm have trouble with the first thing you say... take it one step at a time :?

Or better yet, each of the following states might exist and therefore cannot yet be discounted:


Isn't it correct to say that as that as only one of the two possible states can exist at a time.....

state A being that blue dice is definitely 1

state B being that red dice is definitely 1

..... and the probabilities within each state are identical, then the probability of the overall problem is the same as that of either state?

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Re: A roll of the dice

Postby Austin Harper » Mon Oct 21, 2013 5:03 pm

Not exactly. You want to know the odds of either die being 1, not the probability one die alone being one or of both of them being one.
The odds of one die being one are 1/6 = 16.7%
The odds of both dies being one are (1/6)2 = 1/36 = 2.8%
It makes sense that the odds of one die or the other coming up 1 would be greater than the odds of both of them together or of each of them alone. And 10/36 = 30.6%, which is just that.
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Re: A roll of the dice

Postby peanuts72 » Mon Oct 21, 2013 5:39 pm

thanks, yes im know its right now, its is 11/1

I think I've proved it to myself as follows, (with rounded numbers)

If I was to roll a dice 10,000 times, I would expect 1,666 1s
so if i roll simultaneously, 5000 times per dice

5000 times with the first dice I would expect 833 1s

if I use the 11/1 probability, each time i roll a one on this dice, the other dice will be a one 75 times

833 + 75 = 909

the same applied to the other dice, gives a total of 1818

of course though, we have duplicated the snake eyes,

of which we should expect (5000 /36) = 138

1818 - 138 = 1680 AKA 1,666 :D

YAY, did I do that right?

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Re: A roll of the dice

Postby Gord » Tue Oct 22, 2013 9:14 am

peanuts72 wrote:Hmm, ok thanks. I'm have trouble with the first thing you say... take it one step at a time :?

Or better yet, each of the following states might exist and therefore cannot yet be discounted:


Isn't it correct to say that as that as only one of the two possible states can exist at a time.....

state A being that blue dice is definitely 1

state B being that red dice is definitely 1

..... and the probabilities within each state are identical, then the probability of the overall problem is the same as that of either state?

It's important to remember that the two states overlap when both dice are definitely 1, in which case both states exist simultaneously. That means you can't discount one state simply by assuming the other state exists.

peanuts72 wrote:thanks, yes im know its right now, its is 11/1

I think I've proved it to myself as follows, (with rounded numbers)

If I was to roll a dice 10,000 times, I would expect 1,666 1s
so if i roll simultaneously, 5000 times per dice

5000 times with the first dice I would expect 833 1s

if I use the 11/1 probability, each time i roll a one on this dice, the other dice will be a one 75 times

You can't use the 11/1 probability here, because the two dice are independent of each other; use the 1/6 probability instead. So you would expect the other die to be a one 138 times.

833 + 75 = 909

Adding the number of 1s on the first die to the number of 1s on the second die seems like an odd thing to do.

the same applied to the other dice, gives a total of 1818

of course though, we have duplicated the snake eyes,

of which we should expect (5000 /36) = 138

1818 - 138 = 1680 AKA 1,666 :D

YAY, did I do that right?

It looks wrong to me. Try it this way instead:

Instead of working with large numbers, let's work with smaller numbers. And we'll make the dice behave as if they were prefect representations of large numbers. So, we roll two dice six times each, and they come up with all 36 possible results: Each rolls 1, 2, 3, 4, 5, and 6 six times, and they combine to make every possible combination from (1,1) to (6,6). Then you get to draw up the neat chart that I love to draw up!

(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)

How many times did we roll snake eyes?

Once.

How many times did we roll at least one 1?

11

How many times did we roll no 1s at all?

25

Edit: Hang on, I'll try to draw a box around the interesting bits....

Okay, that is the worst box ever. But hopefully you can see the parts I've separated out:

- In the top left we have a box containing just one set, snake eyes.
- Down the left side we have a box containing 6 sets, all the times where the red die rolled a 1.
- Across the top we have a box containing 6 sets, all the times where the blue die rolled a 1.
- And we have a big box containing all 25 rolls where no 1s came up at all; it's the big box comprising the bottom right of the graph.

Anyhoooo.... We can see from the boxes (I hope) that there are five instances where the red die rolled a 1 and the blue die rolled something other than a 1; and there are five instances where the blue die rolled a 1 and the red die rolled something other than a 1; and there is just one instance where the red die and the blue die both rolled 1s. These represent all the times when "at least one die rolled a 1": There are a total of 11 such cases, and only in 1 of them do both dice roll 1s.
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Re: A roll of the dice

Postby djembeweaver » Mon Oct 19, 2015 4:00 pm

Wow...this has totally frazzled my head! The maths tells me that it is indeed 1/11 but the intuition that it is 1/6 is so strong that it is difficult to believe the maths...even when I've worked it through and arrived at the solution myself!

Brilliant problem!

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Re: A roll of the dice

Postby xouper » Mon Oct 19, 2015 9:16 pm

Sometimes it's interesting to get empirical results to confirm (or not) a hypothesis.

So here is some javascript code I wrote to test Gord's hypothesis:

Code: Select all

<!DOCTYPE html>
<html>
<head>
<style>
   body { background: #ffeedd; }
   table { margin: 1em; }
   td { background: white; border: 1px solid black; }
   td { text-align: center; padding: 1em; }
   td.nob { background: transparent; border: none; font-weight: bold;}
</style>
</head>

<body>
<div align="center">

<h1>An Experiment in Probabilities</h1>

<p>Each click of the button plays the game a million times.
<button onclick="playgame()"> Play </button></p>

<table>
<tr>
<td> Number of games played: </td>
<td id="ngames">000000</td>
</tr>
<tr>
<td> Number of times a one is rolled: </td>
<td id="n1">000000</td>
</tr>
<tr>
<td> Number of times snake eyes rolled: </td>
<td id="nboth">000000</td>
</tr>
<tr>
<td> Odds of getting snake eyes: </td>
<td> <span id="r1ngames">000</span> / <span id="r1nboth">000</span> = 1 in <span id="result1">000</span> </td>
</tr>
<tr>
<td> Odds of getting a one: </td>
<td> <span id="r2ngames">000</span> / <span id="r2n1">000</span> = 1 in <span id="result2">000</span> </td>
</tr>
<tr>
<td> Odds of getting snake eyes<br/>if at least one die is a one: </td>
<td> <span id="r3n1">000</span> / <span id="r3nboth">000</span> = 1 in <span id="result3">000</span> </td>
</tr>
</table>

<script>

   // define global game variables

   var maxgames = 1000000;   // number of games to play

   var die1 = 0;      // the number rolled
   var die2 = 0;      // the number rolled

   var n1 = 0.0;      // tally how many times a one is rolled
   var nboth = 0.0;      // tally how many snake eyes


//--------------------------------------------------------------------------
function getrandom(n) {
   return Math.floor((Math.random() * n) + 1);
}


//--------------------------------------------------------------------------
function display_results() {

   var odds1 = maxgames / nboth;  // odds of getting snake eyes
   var odds2 = maxgames / n1;     // odds of getting at least one one
   var odds3 = n1 / nboth;        // odds of getting snake eyes contingent on a die = one

   document.getElementById("ngames").innerHTML = maxgames;
   document.getElementById("n1").innerHTML     = n1;
   document.getElementById("nboth").innerHTML  = nboth;

   document.getElementById("r1ngames").innerHTML = maxgames;
   document.getElementById("r1nboth").innerHTML  = nboth;
   document.getElementById("result1").innerHTML  = odds1.toFixed(2);

   document.getElementById("r2ngames").innerHTML = maxgames;
   document.getElementById("r2n1").innerHTML     = n1;
   document.getElementById("result2").innerHTML  = odds2.toFixed(2);

   document.getElementById("r3n1").innerHTML     = n1;
   document.getElementById("r3nboth").innerHTML  = nboth;
   document.getElementById("result3").innerHTML  = odds3.toFixed(2);

}


//--------------------------------------------------------------------------
function reset_variables() {

   maxgames = 1000000;   // number of games to play

   die1 = 0;      // the number rolled
   die2 = 0;      // the number rolled

   n1 = 0.0;      // tally how many times a one is rolled
   nboth = 0.0;      // tally how many snake eyes

}


//--------------------------------------------------------------------------
function playgame() {

   reset_variables();

   // play the game a bunch of times

   for (var game = 0; game < maxgames; game++) {

      // step 1. Roll the dice
      die1 = getrandom(6);
      die2 = getrandom(6);

      // step 2. keep track of how many times either die is a one
      if ((die1 == 1) || (die2 == 1)) {
         n1 += 1;
      }

      // step 3. keep track of how many times snake eyes are rolled
      if ((die1 == 1) && (die2 == 1)) {
         nboth += 1;
      }

   }

   // After all games played, display the results
   display_results();

}

</script>

</div>
</body>
</html>


Cut and paste the above code to a file with the extension .htm and then open the file in your browser.

The results confirm Gord's analysis (assuming I understood the rules of the puzzle correctly and did not make any programming errors).

The probability of getting snake eyes out of all rolls is 1 in 36.
The probability of at least one die showing a one is 11 in 36 (or 1 in 3.27).
The probability of getting snake eyes contingent on a one already showing is 1 in 11.

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Re: A roll of the dice

Postby Gord » Mon Oct 19, 2015 10:05 pm

I got my empirical results by rolling actual dice multiple times. Which is odd, considering how lazy I am! But not that odd, considering how obsessive I am. My uncle's name was Odd. What're the odds of that?
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Re: A roll of the dice

Postby bobbo_the_Pragmatist » Mon Oct 19, 2015 10:24 pm

djembeweaver wrote:Wow...this has totally frazzled my head! The maths tells me that it is indeed 1/11 but the intuition that it is 1/6 is so strong that it is difficult to believe the maths...even when I've worked it through and arrived at the solution myself!

Brilliant problem!

Mathematically...probability is about future events. What screws with your head is assigning "probabilities" to events that have already happened. I know...........so close.
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Re: A roll of the dice

Postby Gord » Mon Oct 19, 2015 10:30 pm

bobbo_the_Pragmatist wrote:
djembeweaver wrote:Wow...this has totally frazzled my head! The maths tells me that it is indeed 1/11 but the intuition that it is 1/6 is so strong that it is difficult to believe the maths...even when I've worked it through and arrived at the solution myself!

Brilliant problem!

Mathematically...probability is about future events. What screws with your head is assigning "probabilities" to events that have already happened. I know...........so close.

Probability is about unknown outcomes; as long as the answer is unknown to someone, the event is still a "future event" the way you've put it, even though the actual event has already happened. That's why I can flip a coin and look at it but not show it to you, and to you the chances of it being heads or tails are 50/50, but to me the chances are pretty much 100% heads or 100% tails (I say "pretty much" because I might have seen the coin wrong, or misread it 'cause I'm an idiot or on Teh Drugs or something).
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Re: A roll of the dice

Postby xouper » Mon Oct 19, 2015 10:33 pm

bobbo_the_Pragmatist wrote:
djembeweaver wrote:Wow...this has totally frazzled my head! The maths tells me that it is indeed 1/11 but the intuition that it is 1/6 is so strong that it is difficult to believe the maths...even when I've worked it through and arrived at the solution myself!

Brilliant problem!

Mathematically...probability is about future events.

Agreed.

It is simple enough to rephrase the puzzle so that it refers to future events.

In the future you roll two dice. If one of them shows a one, what is the probability that the other is also a one?

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Re: A roll of the dice

Postby bobbo_the_Pragmatist » Mon Oct 19, 2015 11:03 pm

Ha, ha....xoouper.....you got me. I'm confused.===Let it pass. I did a few years of statistics but quit probability study after one class.

I haven't gotten any better at it either.
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Re: A roll of the dice

Postby ElectricMonk » Tue Jan 12, 2016 2:18 pm

I only just looked at this puzzle, I, too consider the answer to be 1 in 6.

Here my reasoning: we are not just facing the probabilities to two dice rolling and both of them coming up 1 - we've also added what amounts to a Monty Hall problem.
By revealing that one dice is one (analogous to showing that one door is empty), information has been injected into the system and we have to recalculate the probabilities.

And the probability of one dice showing a 1 is 1 in 6.
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Re: A roll of the dice

Postby Monster » Tue Jan 12, 2016 2:25 pm

Everyone in this thread is wrong. The correct answer is: indeterminable. Since the results are behind a screen, Gord might be lying about how many 1s are rolled and it's impossible to know.
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Re: A roll of the dice

Postby Austin Harper » Tue Jan 12, 2016 5:34 pm

That son of a bitch!
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Re: A roll of the dice

Postby Gord » Tue Jan 12, 2016 7:07 pm

ElectricMonk wrote:By revealing that one dice is one (analogous to showing that one door is empty), information has been injected into the system and we have to recalculate the probabilities.

And the probability of one dice showing a 1 is 1 in 6.

But you don't know which die has the 1 showing. Six times out of 36 it's die #1, and six times out of 36 it's die #2, but one time out of 36 is it both -- and those chances all overlap, so exclusively speaking, five times out of 36 it's ONLY die #1, five times out of 36 it's ONLY die #2, and one time out of 36 it's BOTH.

5/36 + 5/36 + 1/36 = 11/36

So you end up with 11 different possibilities once I've given you that information (which is true, because I never lie). Of those 11 possibilities, only one of them has both dice showing 1s.

1/11
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Re: A roll of the dice

Postby ElectricMonk » Tue Jan 12, 2016 8:30 pm

@Gord

I agree that that is the case if you put the question the way Xouper has done:

xouper wrote:In the future you roll two dice. If one of them shows a one, what is the probability that the other is also a one?


then yes, the answer is 1 in 11.

But that is not the OP scenario. The OP scenario, only the probabilities of one dice matters, since we have the information that one dice is one before we have to consider the other one. All five 'not 1' doors of dice no.1 have been opened, to speak with Monty Hall. Which reduces the 1 in 11 to 1 in 6.

Or let me put it another way: the probability of a throw of two dice has been replaced by a: how big are my chances to pick the right side of the second dice?
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Re: A roll of the dice

Postby xouper » Tue Jan 12, 2016 10:00 pm

ElectricMonk wrote:@Gord

I agree that that is the case if you put the question the way Xouper has done:

xouper wrote:In the future you roll two dice. If one of them shows a one, what is the probability that the other is also a one?


then yes, the answer is 1 in 11.

But that is not the OP scenario.

Not so. They are logically equivalent.

ElectricMonk wrote:The OP scenario, only the probabilities of one dice matters, ...

Not so. That is where your analysis goes astray.

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Re: A roll of the dice

Postby Gord » Wed Jan 13, 2016 2:24 am

ElectricMonk wrote:The OP scenario, only the probabilities of one dice matters, since we have the information that one dice is one before we have to consider the other one. All five 'not 1' doors of dice no.1 have been opened, to speak with Monty Hall. Which reduces the 1 in 11 to 1 in 6.

There's a 50% chance I wasn't talking about die #1.

Consider that I have two dice, one red, one blue. I roll them and tell you that at least one of them has rolled a 1. Am I describing the red one, or the blue one? You don't know, so you have to account for both possibilities.
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Re: A roll of the dice

Postby ElectricMonk » Wed Jan 13, 2016 8:01 am

But no one asks which dice is which - the distinction disappears once the extra information is revealed.

Again, I agree that the initial chance was 1:11 - it just moves after the new information comes in. Or to put it another way: forget what the chance are of rolling the dice - what is the chance of you picking the right number?

When I have to bet on what the not-revealed dice is, I can completely ignore the original throw: I only ask: how many possible results can the other dice have? For a single dice, the answer can not be 12, it's 6.

Or what would you do in a casino, given exactly the OP scenario and given odds of 1:10?
Would you bet or not?
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Re: A roll of the dice

Postby Gord » Wed Jan 13, 2016 9:10 am

No, the distinction is inherent in reality. It doesn't matter that you can interchange the dice in your mind; in reality, the dice are separate entities.

Note that odds are written as 'the chances for":"the chances against", so 1 in 11 chances would be written as 1:10 (one chance of winning, 10 chances of losing, 11 possibilities overall -- they could also be written as 1-10, but I prefer using the colon over using the hyphen). I will take it you meant that the initial odds were 1:10, and that the casino would be offering odds of 1:9.

And in that case, I would not bet on snake-eyes when given odds of 1:9, because the odds are 1:10 and therefore the betting favours the house.

This can be confirmed by experimentation. Try it for yourself. Take two dice and roll them until at least one of them shows a one (that's the initial condition required by the original problem). Then mark down whether or not the other die shows a 1 as well, or whether it doesn't. The more tries you make, the closer the outcome will shift toward 1 in 11. I've done it myself, as have other posters in this thread.
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Re: A roll of the dice

Postby ElectricMonk » Wed Jan 13, 2016 9:32 am

In my proposed bet, the house would always offer slightly worse than probabilistic odds to make a net win - thanks for correcting me on the actual way odds are expressed.

The question is: does the casino think it's 1 in 11 or 1 in 6.

It depends on whether I sit in front of the 'double-dice'-dealer with his screen and he rolls and I wait and he rolls and I wait until he proclaims: "I have at least one dice at 1 - please place your bet if you think it's snake-eyes."

In that case, yes, I have to take the combined probability (i.e. the result of the first dice impacts the probabilities of the second).

What I envision is essentially a Monty Hall setup with 21 doors (considering that 1,2 is equivalent to 2,1 if we don't distinguish the dice), and Monty (by telling it's at least 1 1) opens the 15 doors with no 1's.
But this I guess is not quite the same as the OP setup.
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Re: A roll of the dice

Postby Gord » Wed Jan 13, 2016 9:38 pm

15 doors with no 1s? There should be 25 doors with no 1s.

22 23 24 25 26
32 33 34 35 36
42 43 44 45 46
52 53 54 55 56
62 63 64 65 66


A five-by-five cube. It's only the top row and the left column where the 1s occur:

11 12 13 14 15 16
21
31
41
51
61


The cube of no 1s fits snugly in the niche:

11 12 13 14 15 16
21 22 23 24 25 26
31 32 33 34 35 36
41 42 43 44 45 46
51 52 53 54 55 56
61 62 63 64 65 66

Mmm, pretty colours.... :drool:
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Re: A roll of the dice

Postby ElectricMonk » Wed Jan 13, 2016 9:51 pm

Not if you consider 2:3 and 3:2 etc. to be equivalent.
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Re: A roll of the dice

Postby xouper » Wed Jan 13, 2016 10:39 pm

ElectricMonk wrote:Not if you consider 2:3 and 3:2 etc. to be equivalent.

In the context of this puzzle, they are not equivalent.

In combinatorics, the context determines whether they are equivalent (or not).

Another example: What are the odds of getting snake eyes on any given roll? In that question, (2,3) is not equivalent to (3,2), otherwise the answer would not be 1/36.

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Re: A roll of the dice

Postby Gord » Tue May 24, 2016 1:42 am

Oh for.... I got copied! :P

https://www.youtube.com/watch?v=lTJydnfGyyk

And people in the comments section are arguing with him. :roll:
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Re: A roll of the dice

Postby Matthew Ellard » Tue May 24, 2016 2:07 am

Gord wrote:And people in the comments section are arguing with him. :roll:


The voice over gentleman does make the sensible point about using a spreadsheet. I find that if sit down and create spreadsheets that I understand the questions and solutions, a bit better. I think it's because you have to physically consider each step of the formula you are trying to achieve.

I did think it was 1/6 when I started watching the video and can now see my error.

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Re: A roll of the dice

Postby Scott Mayers » Tue May 24, 2016 3:55 am

Gord wrote:Oh for.... I got copied! :P

https://www.youtube.com/watch?v=lTJydnfGyyk

And people in the comments section are arguing with him. :roll:


I don't want to get embroiled in another extended debate here, but while I understand this "puzzle", like I keep seeing all the time with questions of using probability, they are often intentional 'tricks' of how one words things and/or presumes some extended unspoken premise lacking sufficient clarity.

Simply for the question as asked above, you can ask whether 'trust' counts here. Otherwise, why should the person being 'tested' require closing their eyes? To assume there is ONE correct answer based on that simple set up statement and its question is no different than a magician staging a trick and expecting the audience to require solving THE PARTICULAR means of how the magician did it. Sure, we can assume he/she did it in one specific way. But it is an error to assume that only one specific way is possible without more information.

In this example, some hidden assumption is, "We trust the set up as some universal truth", which begs, "why should we require faith here? (being blind to the toss)" Do we assume we hear the dice hit the table? Do we trust two have actually been thrown? Perhaps one is already just quietly placed down with a one facing up and so only one die is tossed, right?

The "puzzle", if it intended not to trick anyone, should be more specific, such as,

Given:
1) Two die are thrown of which ANY one of them specifically indicates one dot (a '1'). In other words, each die is considered specific to solving this problem. Take one die as specifically colored green, and the other red, for instance, the red die can be tossed to have a one, or the green die can have a one, OR both.

2) Further, assume that given any one of these 'ones' is revealed to a God you trust who informs you this is the case without you looking. That is, you "trust" that some '1' actually has been tossed but you are not given which die, the red or the green, or both have this '1'.

Then, Puzzle Question: what is the probability that the other die left over will be a '1'?

Now the answer can be narrowed down as 1/11. But...

...still, the language here can be problematic merely because now it is sounding too robotic of a set up for anyone to bother caring to try the puzzle. I hate how statistics and probability are deemed (by some) as somehow a 'superior' math, when it's use can be easily abused to intimidate others into simply giving up their own skepticism to trade for some FAITH in them authoritatively. It is not the fault of the ones answering the puzzles that are necessarily 'wrong' for their perspective, but the ones proposing them who often think they've presented something with some strict unique answer according to their presentation.
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Re: A roll of the dice

Postby Poodle » Tue May 24, 2016 9:20 am

I've been through this thread a dozen times, and I STILL can't see where the goats come into it.

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Re: A roll of the dice

Postby OlegTheBatty » Tue May 24, 2016 8:35 pm

Poodle wrote:I've been through this thread a dozen times, and I STILL can't see where the goats come into it.


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Re: A roll of the dice

Postby Monster » Tue May 24, 2016 9:25 pm

I didn't watch the video yet.

"What is the probability both dice show a 1?" is the question. The previous statement about what one of the dice already show is irrelevant. If it was relevant, then I'd say the probability is 1/6. But the way the question is worded, it's regarding both dice. It's a 3% chance to get two 1s.
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Re: A roll of the dice

Postby OlegTheBatty » Tue May 24, 2016 9:46 pm

Monster wrote:I didn't watch the video yet.

"What is the probability both dice show a 1?" is the question. The previous statement about what one of the dice already show is irrelevant. If it was relevant, then I'd say the probability is 1/6. But the way the question is worded, it's regarding both dice. It's a 3% chance to get two 1s.

If the first die is a 1, there are 6 possibilities for the second die. if second die is a 1, there are 6 possibilities for the first die, for a total of twelve. One of the possibilities (where both are 1) is a duplicate in each set of 6, so there are only 11 possibilities. One of them is both dice having a 1.

One chance in 11.
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Re: A roll of the dice

Postby Gord » Wed May 25, 2016 9:09 am

Scott Mayers wrote:Simply for the question as asked above, you can ask whether 'trust' counts here.

That's true for every riddle ever told. Also for every test question ever asked on any exam, ever.

What's 3 time 9?

Twenty-seven.

Wrong! I lied! The real question was: what's 9 times 8. You said twenty-seven, so you fail maths class.

So yes, you can assume I may have been lying when I opened the question with the statement, "Here's one for your statistical analysis." But honestly, I don't see the point.
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Re: A roll of the dice

Postby Scott Mayers » Wed May 25, 2016 10:41 am

Gord wrote:
Scott Mayers wrote:Simply for the question as asked above, you can ask whether 'trust' counts here.

That's true for every riddle ever told. Also for every test question ever asked on any exam, ever.

What's 3 time 9?

Twenty-seven.

Wrong! I lied! The real question was: what's 9 times 8. You said twenty-seven, so you fail maths class.

So yes, you can assume I may have been lying when I opened the question with the statement, "Here's one for your statistical analysis." But honestly, I don't see the point.

You missed my point. Given a 'puzzle', by the nature of the one presenting it to assume some specific answer requires sufficient language to narrow down the actual problem. Most 'puzzles' are often presented with missing information or information that is FALSELY assumed as sufficient to solve the problem. Often, the one presenting it presumes some common 'cultural' assumptions of its listeners expecting some things are understood by default when they are not. I pointed to the way this presentation lacks sufficient information to draw ONE specific conclusion in this way, as an example.

"With your eyes closed, you roll a pair of dice."

When given that something is a "puzzle", it is often interpreted by the listener to require paying some more 'anal' attention to detail. For example, a parent to a child says, "come here", to which the kid doesn't move and might playfully expand his disobedience as follows: "I AM 'here' ", where he is implying obedience with respect to his own reference of "here". Obviously, the kid is being too "anal" but is making a clear point about definition. But if the kid did this following the parent demanding, "Do exactly as I say, or else", it actually rationally challenges the kid to respond 'anally' with partial justice until he learns the etiquette of the language and appropriateness to the situation.

My point here up front is that if you assert something as a type of puzzle, you have to accept others to BE 'anal' and the puzzler is at fault should they leave specific gaps in explanation.

"With your eyes closed,..." begs to a listener the relevance of this act with appropriateness to such detail that 'puzzles' ask. So, when...

"The Casino dealer announces that at least one of the rolls is a one,"

...the puzzler's 'trick' here is to make the audience 'guess' to which understanding he has in his mind. Are we allowed to 'trust' the dealer's claim that one of the rolls is actually a "one" since we are given we must initially close our eyes? Or are we to interpret this to mean that by some reason we KNOW (like with eyes wide open) that one of the dice does have a "one" recognizing also that each die is unique and required to be represented in the probability? Asserting that one die has been tossed, even to 'trust' this as true, should also imply that the toss has already been done and so the question may be asking what the odds of the second one to be a "one" now. That's why I demonstrated how to be more precise to enable others to rationally draw only one specific conclusion without a doubt to meaning. Otherwise there IS more than one answer without more information.

The set-up thus, acts like a magician who creates an illusion for the audience, believing he knows specifically WHERE his audience IS (their perspective), but errs if he didn't take caution to make sure the audience actually IS where they are. Without realizing it, the magician may have accidentally left a mirror on the stage from the last performance and accidentally enabled the audience to see the magician doing something behind his back. They'd thus have more extended perspective than the magician thinks he's demonstrated 'correctly'.

I actually gave a different example above though where the magician does the trick (correctly) but the audience is then asked to determine HOW the trick was done. One could actually posit many possible solutions to which the magician simply answers "yes" or "no". So then if one suggests ANY possible solution that IS rational from their perspective, it is odd should the magician just say, "your wrong" or "you are an idiot". Why would the audience be at fault for NOT reading into the mind of the magician to determine which possibility is correct without demanding more information?
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Re: A roll of the dice

Postby bobbo_the_Pragmatist » Wed May 25, 2016 3:18 pm

"Come Here." Implies movement.

Some puzzles are simply wrong. Especially when money is involved.
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Re: A roll of the dice

Postby psychiatry is a scam » Wed May 25, 2016 8:26 pm

basic craps odds - 36 possible combinations - only one way to role a 2 or 12
36 to 1 odds
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Re: A roll of the dice

Postby Gord » Wed May 25, 2016 8:58 pm

Scott Mayers wrote:it is odd should the magician just say, "your wrong" or "you are an idiot".

you're
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