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#101
10-06-2016, 06:22 PM
 SigMan Guest Join Date: Aug 2015 Location: Texas Posts: 672
If you roll two dice for one die to show a six won't that be odds of 1 in 3?
#102
10-06-2016, 06:27 PM
 x-ray vision Charter Member Join Date: Oct 2002 Location: N.J. Posts: 5,125
Quote:
 Originally Posted by octopus Yeah but in a chart of events where a die is being a revealed and a die is being concealed there are 12 permutations which I listed above will do so again. Revealed Red result Green Result Red 6 1 Red 6 2 Red 6 3 Red 6 4 Red 6 5 Red 6 6 Green 1 6 Green 2 6 Green 3 6 Green 4 6 Green 5 6 Green 6 6 Out of the 12 possibilities of a particular die being revealed and a particular die being concealed there are only 2 that have two 6s. 2/12 = 1/6. If you do this as a real life experiment with two people. You will see that 1/6 of the time when at least one six is rolled that the concealed die is a 6. Do it with real dice and two people.
You're counting 6, 6 twice, which I initially thought had to be done. See the following:

http://www.math.hawaii.edu/~ramsey/P...y/TwoDice.html

As borschevsky is alluding to in the question he asked you, you will get a 1 and a 6 twice as often as a 6 and a 6.
#103
10-06-2016, 06:44 PM
 Andy L Member Join Date: Oct 2000 Posts: 4,744
Quote:
 Originally Posted by Chronos I'll join in with the "problem is ambiguous" crowd, here. There are multiple possible experiments that are all consistent with the problem statement, and which have different answers. Here are two examples: 1: I roll a pair of dice. I then choose one die at random, look at the number on the die, and announce "At least one die is a ___", where the blank is the number I looked at. On this particular iteration of the experiment, the die I looked at was a 6, and I announce as such. What is the probability that the other die is a 6? Answer: The probability that the other die is a 6 is 1 in 6. 2: I roll a pair of dice. I then examine both dice, and if it is the case that at least one of them shows a 6, I announce "At least one die is a 6", and otherwise I say nothing. On this particular iteration of the experiment, I saw at least one 6, and so I announce as such. What is the probability that both dice are 6? Answer: The probability that both dice are 6 is 1 in 11.
I think you're right that the answer you get depends very much on how the situation is framed.

Lay out 36 pairs of dice in a grid showing all 36 possible results.

Of the 6 pairs in which the die on the right side of the pair is six, the left die will also be 6 once (one chance in 6).

Of the 6 pairs in which the die on the left side of the pair is six, the right die will also be 6 once (one chance in 6).

Of the 11 pairs in which at least one of the dice in the pair is 6, the other die will be 6 in one case (1 in 11).

Last edited by Andy L; 10-06-2016 at 06:46 PM.
#104
10-06-2016, 06:49 PM
 octopus Guest Join Date: Apr 2015 Posts: 5,792
Quote:
 Originally Posted by Thudlow Boink When the dice are thrown, there are 11 possible outcomes, all equally likely. In the case of (6, 6), there are two possible ways that one of the dice could be revealed, but that doesn't make that outcome more likely to have come up in the first place.
Take two real dice and see if you come up with 1/11.

Ok, I just rolled two dice for a good chunk of time. I discarded all rolls that did not include a 6 and I stopped after 20 double sixes. I had 132 rolls that included at least one 6.

Here are the ratios for the rolls.

4 22 5.5
5 26 5.2
6 29 4.833333333
7 30 4.285714286
8 35 4.375
9 41 4.555555556
10 85 8.5
11 86 7.818181818
12 96 8
13 98 7.538461538
14 99 7.071428571
15 111 7.4
16 112 7
17 117 6.882352941
18 118 6.555555556
19 129 6.789473684
20 132 6.6
#105
10-06-2016, 06:55 PM
 Thudlow Boink Charter Member Join Date: May 2000 Location: Lincoln, IL Posts: 24,089
I'm afraid you'll have to explain to me what those numbers represent.
#106
10-06-2016, 06:56 PM
 x-ray vision Charter Member Join Date: Oct 2002 Location: N.J. Posts: 5,125
octopus, then you think with a roll of two dice, you have an equal chance of rolling two sixes as you do rolling a 1 and a 6?

Do you agree that the probability of rolling a 1 and a 6 is 1/18?
#107
10-06-2016, 07:15 PM
 octopus Guest Join Date: Apr 2015 Posts: 5,792
Quote:
 Originally Posted by x-ray vision octopus, then you think with a roll of two dice, you have an equal chance of rolling two sixes as you do rolling a 1 and a 6? Do you agree that the probability of rolling a 1 and a 6 is 1/18?
Of course it is. You have the chance of rolling the pair 1,6 or 6,1 assuming you can tell the rolls apart with some mechanism. Now the chance of the ordered pair 1,6 or 6,1 is actually 1/36. But notice that a 1,6 is the same chance as a 6,6 if you can discern the dice.

The sum of 1 and 6 is 7 and that is a 1/6 chance of occurring. So what are we looking for sums or particular probabilities of particular events. In that example he is summing events. If 1,6 is considered exactly equivalent to 6,1 than yes those 2 have twice the probability of 6,6.

Let's make the math easier and use the coin example. Throw out all double tails. Use a penny and a nickel.

What are the outcomes

Penny Nickel

H T
T H
H H

How many possible ways can I show one head? I can't show one head with double tails so that isn't even in the denominator. It's not counted. Correct?

So how many ways can I show you a head?

A) I can show you Penny's head if Penny comes up head and Nickel comes up tails. Correct? That is one count.

B) I can show you Penny's head if Penny comes up head Nickel comes up head. That is two.

C) I can show you Nickel's head if Nickel comes up head and Penny comes up tails. That is three.

D) I can show you Nickel's head if Nickel comes up head and Penny comes up head. That is four.

Ah - ha ha ha.

Four ways to show you a head in which at least one head was flipped. These are four permutations of a state in which one coin is revealed and one is concealed with the constraint that there will only be a revelation if at least one is a head. Correct?

Now how many of those permutations satisfy the second case where the concealed coin is also a head? Two events B and D. 2/4 is 1/2 which is what you'd expect with a fair coin.
#108
10-06-2016, 07:17 PM
 x-ray vision Charter Member Join Date: Oct 2002 Location: N.J. Posts: 5,125
Quote:
 Originally Posted by octopus Of course it is.
#109
10-06-2016, 07:20 PM
 jtur88 Guest Join Date: Aug 2011 Location: Cebu, Philippines Posts: 13,282
Quote:
 Originally Posted by Skammer ETA: Thudlow Boink is also correct. If I said "the red die is a six. What are the odds that the green die is a six?" the answer is 1/6. But that's not the question.
Yes it is, it's just more detailed. If the red die is a 6, the odds of a green 6 are 1/6. If the green die is a 6, the odds of a red 6 is 1/6. That covers all possibilities, and in every case, it is 1/6. It doesn't mater if the dice are colored -- it's 1/6 in every possible case.
#110
10-06-2016, 07:22 PM
 x-ray vision Charter Member Join Date: Oct 2002 Location: N.J. Posts: 5,125
Quote:
 Originally Posted by octopus If 1,6 is considered exactly equivalent to 6,1 than yes those 2 have twice the probability of 6,6.
Then why are you claiming this?:

Quote:
 Red 6 1 Red 6 2 Red 6 3 Red 6 4 Red 6 5 Red 6 6 Green 1 6 Green 2 6 Green 3 6 Green 4 6 Green 5 6 Green 6 6
#111
10-06-2016, 07:23 PM
 LSLGuy Charter Member Join Date: Sep 2003 Location: Southeast Florida USA Posts: 20,677
What I'm enjoying now are the folks writing simulations and doing manual dice throwing trials.

The ambiguity is in the problem statement. Once you decide on your preferred interpretation of the problem statement then your result (1/11 or 1/6) is foreordained. And can be proven with a few short lines of small words.

So any effort spent to simulate is foolish. It's merely proving the trivial conclusion you already (perhaps unwittingly) assumed.

Last edited by LSLGuy; 10-06-2016 at 07:23 PM.
#112
10-06-2016, 07:25 PM
 octopus Guest Join Date: Apr 2015 Posts: 5,792
Quote:
 Originally Posted by x-ray vision Cool. Now please Google probability of rolling snake eyes.
Snake eyes is 1/36.

1,1 is one ordered pair out of 36.

Simple. I'm not the one confusing what the actual constraints of the question are.

Shoot let's say we only count the probability of the die roll after we rolled a 6. Instead of revealing a 6 and asking we are only counting 6s after we rolled a 6. What's the probability of that? Why would the probability of the die change? It only has 6 possible states regardless of what was rolled before.
#113
10-06-2016, 07:30 PM
 x-ray vision Charter Member Join Date: Oct 2002 Location: N.J. Posts: 5,125
Quote:
 Originally Posted by octopus Snake eyes is 1/36.
I'm confused. You agree the probabilty of rolling a 1 and a 6 is 1/18, and that rolling a 6 and a 6 is 1/36, yet you count 6,6 twice here:

Quote:
 Red 6 1 Red 6 2 Red 6 3 Red 6 4 Red 6 5 Red 6 6 Green 1 6 Green 2 6 Green 3 6 Green 4 6 Green 5 6 Green 6 6
#114
10-06-2016, 07:32 PM
 x-ray vision Charter Member Join Date: Oct 2002 Location: N.J. Posts: 5,125
Do you agree with the following?:

http://www.math.hawaii.edu/~ramsey/P...y/TwoDice.html

There are 36 equal rolls of two dice.
#115
10-06-2016, 07:33 PM
 Musicat Charter Member Join Date: Oct 1999 Location: Sturgeon Bay, WI USA Posts: 19,903
Quote:
 Originally Posted by Chronos Answer: The probability that both dice are 6 is 1 in 11.
That's not the OP's problem. It didn't say anything about BOTH dice, but what would happen IF one condition was already met. That condition is no longer a factor, as we are not considering multiple conditions, nor conditional conditions.

Quote:
 Originally Posted by Thudlow Boink If nothing that happened before matters, then what determines which die is the "other die," the one that you are shown?
It doesn't matter which is the "other die," since this is not a conditional situation. All conditions (no 6's found, etc.) are eliminated. All that matters is the odds of ONE toss of ONE die, with the possibilities of 1..6.

And just in case, running my program for 10,000,000 iterations confirms a 1:6 ratio, within 3 decimal places. I doubt if running it longer will change much, so it's the premise that counts here.

I stand by my post #89, more than ever. Anyone want 10 million data points to look at?
#116
10-06-2016, 07:36 PM
 x-ray vision Charter Member Join Date: Oct 2002 Location: N.J. Posts: 5,125
Quote:
 Originally Posted by octopus Shoot let's say we only count the probability of the die roll after we rolled a 6. Instead of revealing a 6 and asking we are only counting 6s after we rolled a 6. What's the probability of that? Why would the probability of the die change? It only has 6 possible states regardless of what was rolled before.
Different problems.

Quote:
 Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls? Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys? Gardner initially gave the answers 1/2 and 1/3, respectively, but later acknowledged that the second question was ambiguous.[3] Its answer could be 1/2, depending on how you found out that one child was a boy. The ambiguity, depending on the exact wording and possible assumptions, was confirmed by Bar-Hillel and Falk,[4] and Nickerson.[5]
#117
10-06-2016, 07:38 PM
 TroutMan Guest Join Date: Sep 2008 Location: Portland, OR Posts: 3,649
Quote:
 Originally Posted by LSLGuy What I'm enjoying now are the folks writing simulations and doing manual dice throwing trials. The ambiguity is in the problem statement. Once you decide on your preferred interpretation of the problem statement then your result (1/11 or 1/6) is foreordained. And can be proven with a few short lines of small words. So any effort spent to simulate is foolish. It's merely proving the trivial conclusion you already (perhaps unwittingly) assumed.
You'd like to think so. But some are still arguing about questions unrelated to the ambiguity of the problem statement, like red vs green dice and ordered pairs. For those, a simulation can be handy to demonstrate why 1/11 is correct according to one interpretation.
#118
10-06-2016, 07:41 PM
 LSLGuy Charter Member Join Date: Sep 2003 Location: Southeast Florida USA Posts: 20,677
Quote:
 Originally Posted by Musicat That's not the OP's problem. It didn't say anything about BOTH dice, but what would happen IF one condition was already met. That condition is no longer a factor, as we are not considering multiple conditions, nor conditional conditions. It doesn't matter which is the "other die," since this is not a conditional situation. All conditions (no 6's found, etc.) are eliminated. All that matters is the odds of ONE toss of ONE die, with the possibilities of 1..6. And just in case, running my program for 10,000,000 iterations confirms a 1:6 ratio, within 3 decimal places. I doubt if running it longer will change much, so it's the premise that counts here. I stand by my post #89, more than ever. Anyone want 10 million data points to look at?
Once again you've assumed one interpretation of the problem. And then spent effort tautologically proving your own assumption. Yes, Virginia, the odds of a 1:6 outcome really are 1:6.

FWIW, I prefer your interpretation of the problem too. But it's just that: one interpretation of the subtly (and deliberately) ambiguous sentences in the original problem statement.

The fact the multiple choice answers have choices which are correct for each interpretation and for the common confusion of [is R6-G6 different from G6-R6?] pretty well proves the problem was carefully deliberately designed to produce these incompatible lines of thinking.

Last edited by LSLGuy; 10-06-2016 at 07:46 PM.
#119
10-06-2016, 07:45 PM
 TroutMan Guest Join Date: Sep 2008 Location: Portland, OR Posts: 3,649
Quote:
 Originally Posted by Musicat That's not the OP's problem. It didn't say anything about BOTH dice, but what would happen IF one condition was already met. That condition is no longer a factor, as we are not considering multiple conditions, nor conditional conditions...
You're falling into the same trap I did. Given your assumptions, your solution is exactly right and 1/11 is the correct answer. But the assumption is one I didn't even realize I was making, that six was picked as the "magic" number before any roles, and therefore any roles that didn't contain a six would be discarded. The other possibility was best explained to me by Chronos in post 68. You aren't actually told that six was selected before any roles; it could just be a statement of what one of the dice showed. In that case, the fact that one of the dice is a particular number doesn't add any new information to the problem, and the other die has a 1/6 chance of being that number.
#120
10-06-2016, 07:46 PM
 octopus Guest Join Date: Apr 2015 Posts: 5,792
Quote:
 Originally Posted by x-ray vision Then why are you claiming this?:
Because what is being revealed and what is being concealed are different states. Look at it closely. There are 11 different pairs.

1,6
2,6
3,6
4,6
5,6
6,6
6,1
6,2
6,3
6,4
6,5

That's 11 things right? 6,6 is 1/11 correct? That is out of this set of ordered pairs 6/6 occurs 1 time out of 11. Makes sense right?

But we have 11 ordered pairs that could exist. The other 25 are discarded. Each of these 11 have at least one 6. If you reveal the 6 you see what the other number could be. But look! There are two different 6s in the 6,6 example that could be revealed. The left or the right, the first or second, the red or green, etc. So there are actually 12 different sixes that could be revealed. And out of that 12 only 2 of the revealed have another 6 concealed and that is reveal red 6 conceal green 6 and reveal green 6 conceal red 6. That's why the denominator is not 11. Because we aren't dealing with a set of 11 ordered pairs. We are dealing with revealing a particular 6 from a set of 11 ordered pairs.

But the question isn't what is the probability of one of these to occur. The question is what is the probability if I am shown a 6 that the other number is a 6. Well with colored dice there are 12 combinations that can occur because what matters is what is the state of the dice. Are they revealed, concealed? What number is on them. So if I can reveal a green 6 with 6 equally likely permutations on a red die that's 6 events. If I can reveal a red 6 with 6 equally likely permutations on a green die that's 6 different events. Yes or no? Is a revealing a green 6 the same as revealing a red 6? No. isn't.

If I show you a green 6 what are states the red die can be in? 1,2,3,4,5,6. Correct? Why would one number be favored over another. If I show you a red 6 what are the states the green die can be in? 1,2,3,4,5,6. Correct? Why should 6 be a lessor probability than the 1,2,3,4 or 5? There's no logical reason to treat 6 differently.
#121
10-06-2016, 07:47 PM
 jtur88 Guest Join Date: Aug 2011 Location: Cebu, Philippines Posts: 13,282
Instead of rolling the dice simultaneously, do it sequentially. Keep throwing the red one until it is a 6. Then rolll the green one. What are the odds that the green one will be six? Just superimpose the two events in an overelap time frame..

In what way would the results be skewed by the dice not being thrown simultaneously? Or by the dice being thrown in separate rooms by collaborating experimenters who then communicate the events to each other with a walkie talkie?

Last edited by jtur88; 10-06-2016 at 07:50 PM.
#122
10-06-2016, 07:48 PM
 octopus Guest Join Date: Apr 2015 Posts: 5,792
Quote:
 Originally Posted by TroutMan You're falling into the same trap I did. Given your assumptions, your solution is exactly right and 1/11 is the correct answer. But the assumption is one I didn't even realize I was making, that six was picked as the "magic" number before any roles, and therefore any roles that didn't contain a six would be discarded. The other possibility was best explained to me by Chronos in post 68. You aren't actually told that six was selected before any roles; it could just be a statement of what one of the dice showed. In that case, the fact that one of the dice is a particular number doesn't add any new information to the problem, and the other die has a 1/6 chance of being that number.
It says a 6 is revealed. That means a six was rolled. It's sort of a given.
#123
10-06-2016, 07:49 PM
 octopus Guest Join Date: Apr 2015 Posts: 5,792
Quote:
 Originally Posted by jtur88 Instead of rolling the dice simultaneously, do it sequentially. Keep throwing the red one until it is a 6. Then rolll the green one. What are the odds that the green one will be six? In what way would the results be skewed by the dice not being thrown simultaneously? Or by the dice being thrown in separate rooms by collaborating experimenters?
All those are 1/6.
#124
10-06-2016, 08:11 PM
 x-ray vision Charter Member Join Date: Oct 2002 Location: N.J. Posts: 5,125
Quote:
 Originally Posted by jtur88 Instead of rolling the dice simultaneously, do it sequentially. Keep throwing the red one until it is a 6. Then rolll the green one. What are the odds that the green one will be six? Just superimpose the two events in an overelap time frame.. In what way would the results be skewed by the dice not being thrown simultaneously? Or by the dice being thrown in separate rooms by collaborating experimenters who then communicate the events to each other with a walkie talkie?
Again, the difference is similar to it is in the boy/girl problem:

Quote:
 Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls? Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?
#125
10-06-2016, 08:13 PM
 x-ray vision Charter Member Join Date: Oct 2002 Location: N.J. Posts: 5,125
Different answers for each boy/girl problem. Also different answers depending on how this problem is phrased (see post 68).
#126
10-06-2016, 09:08 PM
 octopus Guest Join Date: Apr 2015 Posts: 5,792
Quote:
 Originally Posted by x-ray vision Different answers for each boy/girl problem. Also different answers depending on how this problem is phrased (see post 68).
(A)"Mr. Smith says: 'I have two children and at least one of them is a boy.' Given this information, what is the probability that the other child is a boy?"

(B)"Mr. Smith says: 'I have two children and it is not the case that they are both girls.' Given this information, what is the probability that both children are boys?"

Child 1 is C1. Child 2 is C2.

Here's a table of what the children can be.

C1 C2

B B
B G
G B
G G

Now we have to exclude G-G right? So we are left with

C1 C2

B B
B G
G B

You have 6 possible known, unknown pairs

Child 1 being a boy, child 2 boy
Child 1 being a boy, child 2 girl
Child 1 being a girl, child 2 boy.

Child 2 being a boy, child 1 boy
Child 2 being a girl, child 1 boy
Child 2 being a boy, child 1 girl.

2 of these 6 are the other child being a boy if the known child is a boy. So the answer is 1/3.

Example 2 is the same as example 1 just written differently. Still get 1/3. I honestly don't see what's hard about that.

Last edited by octopus; 10-06-2016 at 09:09 PM.
#127
10-06-2016, 09:51 PM
 Chronos Charter Member Moderator Join Date: Jan 2000 Location: The Land of Cleves Posts: 73,079
Quote:
 Quoth jtur88: Instead of rolling the dice simultaneously, do it sequentially. Keep throwing the red one until it is a 6. Then rolll the green one. What are the odds that the green one will be six? Just superimpose the two events in an overelap time frame.. In what way would the results be skewed by the dice not being thrown simultaneously? Or by the dice being thrown in separate rooms by collaborating experimenters who then communicate the events to each other with a walkie talkie?
There are possible results for the simultaneous throws that are not possible for your situation, such as a red 2 and a green 6.

octopus, the two statements you give of the boy-girl problem are equivalent, but those aren't the same two statements that x-ray vision just gave.
#128
10-06-2016, 09:53 PM
 x-ray vision Charter Member Join Date: Oct 2002 Location: N.J. Posts: 5,125
Quote:
 Originally Posted by Chronos octopus, the two statements you give of the boy-girl problem are equivalent, but those aren't the same two statements that x-ray vision just gave.
That's why I gave up.
#129
10-06-2016, 09:54 PM
 DrDeth Charter Member Join Date: Mar 2001 Location: San Jose Posts: 34,053
Quote:
 Originally Posted by TriPolar From the OP: "Two fair dice are rolled together" So one die isn't always a six. When one die is a six it's 1/11.
"at least one of the dice is a 6", thus one die is always a six.

Look, the question is so poorly worded it is unanswerable.
#130
10-06-2016, 10:35 PM
 Skammer Charter Member Join Date: Aug 2002 Location: Music City USA Posts: 14,041
Quote:
 Originally Posted by Musicat That's not the OP's problem. It didn't say anything about BOTH dice, but what would happen IF one condition was already met. That condition is no longer a factor, as we are not considering multiple conditions, nor conditional conditions.
Quote:
 Originally Posted by jtur88 Instead of rolling the dice simultaneously, do it sequentially. Keep throwing the red one until it is a 6. Then rolll the green one. What are the odds that the green one will be six? Just superimpose the two events in an overelap time frame..
Speaking as the OP, I'll say no, you are both misunderstanding the (admittedly ambiguously worded) problem I posed. Although I did not come up with the wording myself.

Two dice are rolled. The implication is that ANY of the 36 possible combinations may have turned up. Not, you keep rolling until you have a six. Why even bother rolling two dice if you only count the results when one of them lands on a specific face? That's just like rolling one die in the first place and is a trivial question.

So, you have one of 36 possible combinations. Then the omniscient viewer says, "Look, you rolled at least one six." At that point, what is the chance your other die is a six?

It's 1/11.
#131
10-06-2016, 10:36 PM
 octopus Guest Join Date: Apr 2015 Posts: 5,792
Quote:
 Originally Posted by Chronos There are possible results for the simultaneous throws that are not possible for your situation, such as a red 2 and a green 6. octopus, the two statements you give of the boy-girl problem are equivalent, but those aren't the same two statements that x-ray vision just gave.
Yeah, that's true but I was reading the wiki article that he linked and was addressing one of the examples given as a paradox.

With the examples x-ray vision gave the answers would be.

A) Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls?

Possibility table. Older child is OC. Younger child is YC.

OC YC

G B
G G

2 permutations to satisfy first condition of OC being a girl. 1 of the permutation satisfies the condition of 2 girls. So:

1/2

B) Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?

1/3

Same table as my previous post.

Last edited by octopus; 10-06-2016 at 10:36 PM.
#132
10-06-2016, 10:37 PM
 x-ray vision Charter Member Join Date: Oct 2002 Location: N.J. Posts: 5,125
Quote:
 Originally Posted by octopus Yeah, that's true but I was reading the wiki article that he linked and was addressing one of the examples given as a paradox.
Then maybe you should start a new thread?
#133
10-06-2016, 10:42 PM
 octopus Guest Join Date: Apr 2015 Posts: 5,792
Quote:
 Originally Posted by x-ray vision Then maybe you should start a new thread?
What's hard about calculating number of events that meet criteria A divided by number of events that meet criteria B?

Plus I answered your questions. Which since they are different than the OP's shouldn't they be in their own thread as well?
#134
10-06-2016, 10:48 PM
 x-ray vision Charter Member Join Date: Oct 2002 Location: N.J. Posts: 5,125
Quote:
 Originally Posted by octopus What's hard about calculating number of events that meet criteria A divided by number of events that meet criteria B?
No one here is discussing how hard something is.

Quote:
 Plus I answered your questions. Which since they are different than the OP's shouldn't they be in their own thread as well?
My questions are not different. They are regarding your confusion with answering the OP's question depending how the OP's question is interpreted. The answer is 1/11 in the way he just phrased it in post 130.
#135
10-06-2016, 10:58 PM
 octopus Guest Join Date: Apr 2015 Posts: 5,792
Quote:
 Originally Posted by x-ray vision No one here is discussing how hard something is. My questions are not different. They are regarding your confusion with answering the OP's question depending how the OP's question is interpreted. The answer is 1/11 in the way he just phrased it in post 130.
How do you interpret the questions you asked other than the obvious.

Older child being a girl and figuring out the probability that the 2nd is as well isn't ambiguous at all.

Being shown that one die is a 6 after a pair of dice have been rolled is ambiguous in what way precisely? If I flip a coin and roll a die and I show you heads on the coin how does that change what the die rolled? It's still a 1/6 chance on the die if you get shown a head, an ace of spades, a Draw 4 card, or another 6.

Do that with a coin. Flip one and roll a die. What's the probability if heads comes up you roll a 6? How's it different with 2 different dice?

This is a repeatable experiment.

Last edited by octopus; 10-06-2016 at 11:00 PM.
#136
10-06-2016, 11:05 PM
 septimus Guest Join Date: Dec 2009 Location: The Land of Smiles Posts: 14,891
Quote:
 Originally Posted by Skammer Here is the problem verbatim: "Two fair dice are rolled together, and you are told that 'at least one of the dice is a 6.' A 6 is removed, and you are ten shown the other die. What is the probability of the other die showing a six?"
As others agree, the problem statement is ambiguous. (Moreover, since "a six is removed," the pre-knowledge that "at least one of the dice is a 6" is redundant.)

Did the guy remove a die at random and it happened to be a six? (Then the answer is 1/6.) Did he remove any duplicated die? (The the answer is 1.) Did he remove a single six, whenever there was at least one six? (The the answer is 1/11.) If he always removes a single six when he can, but chooses the red six when both red and green are present, did he show you the color of the removed six? (Now the answers are 1/6 or 0, for red, green, resp.)

As usual, "running a simulation" makes sense only when the problem is well-defined. But if it's well-defined, why run a simulation when you can just treat 11 cases? (or even just 3 or 4 cases.)
#137
10-06-2016, 11:06 PM
 newme Guest Join Date: Dec 2009 Location: In the Woods Posts: 323
Quote:
 Originally Posted by Skammer Speaking as the OP, I'll say no, you are both misunderstanding the (admittedly ambiguously worded) problem I posed. Although I did not come up with the wording myself. Two dice are rolled. The implication is that ANY of the 36 possible combinations may have turned up. Not, you keep rolling until you have a six. Why even bother rolling two dice if you only count the results when one of them lands on a specific face? That's just like rolling one die in the first place and is a trivial question. So, you have one of 36 possible combinations. Then the omniscient viewer says, "Look, you rolled at least one six." At that point, what is the chance your other die is a six? It's 1/11.
This wording is still ambiguous. You haven't told us WHY the omniscient viewer chose to say "six". If the roll is six - one, why did he choose to say six instead of one? What would the omniscient viewer say if the roll is one - two?

The only way to get a 1/11 result is by NOT counting some of the results.
#138
10-06-2016, 11:11 PM
 octopus Guest Join Date: Apr 2015 Posts: 5,792
Quote:
 Originally Posted by septimus As others agree, the problem statement is ambiguous. (Moreover, since "a six is removed," the pre-knowledge that "at least one of the dice is a 6" is redundant.) Did the guy remove a die at random and it happened to be a six? (Then the answer is 1/6.) Did he remove any duplicated die? (The the answer is 1.) Did he remove a single six, whenever there was at least one six? (The the answer is 1/11.) If he always removes a single six when he can, but chooses the red six when both red and green are present, did he show you the color of the removed six? (Now the answers are 1/6 or 0, for red, green, resp.) As usual, "running a simulation" makes sense only when the problem is well-defined. But if it's well-defined, why run a simulation when you can just treat 11 cases? (or even just 3 or 4 cases.)

If you do the process as described. That is have one person roll two dice and show the other person a 6 whenever a 6 is present in the pair of dice you can figure out an unambiguous answer.

Obviously, you cannot show a 6 from a pair of rolled, fair dice when a 6 is not present. Where is the ambiguity? Just do this with someone else as explained and it's not hard.
#139
10-06-2016, 11:26 PM
 septimus Guest Join Date: Dec 2009 Location: The Land of Smiles Posts: 14,891
Problem statement #1:
"Two fair dice are rolled together, and you are told that 'at least one of the dice is a 6.' A 6 is removed, and you are then shown the other die. What is the probability of the other die showing a six?"
Problem statement #2:
"Two fair dice are rolled together. A 6 is removed, and you are then shown the other die. What is the probability of the other die showing a six?"
How do these problems differ from each other? Do they have the same answer?

Quote:
 Originally Posted by septimus Did the guy remove a die at random and it happened to be a six? (Then the answer is 1/6.) Did he remove any duplicated die? (The the answer is 1.) Did he remove a single six, whenever there was at least one six? (The the answer is 1/11.) If he always removes a single six when he can, but chooses the red six when both red and green are present, did he show you the color of the removed six? (Now the answers are 1/6 or 0, for red, green, resp.)
Quote:
 Originally Posted by octopus If you do the process as described. That is have one person roll two dice and show the other person a 6 whenever a 6 is present in the pair of dice you can figure out an unambiguous answer. Obviously, you cannot show a 6 from a pair of rolled, fair dice when a 6 is not present. Where is the ambiguity? Just do this with someone else as explained and it's not hard.
Reread the post you're responding to. "A six is removed" can be interpreted in multiple ways.

As for "you cannot show a 6 from a pair of rolled, fair dice when a 6 is not present" — that was my point. Read the two problems at the beginning of this post and tell us if they have the same answer.
#140
10-06-2016, 11:30 PM
 x-ray vision Charter Member Join Date: Oct 2002 Location: N.J. Posts: 5,125
Quote:
 Originally Posted by newme This wording is still ambiguous. You haven't told us WHY the omniscient viewer chose to say "six". If the roll is six - one, why did he choose to say six instead of one? What would the omniscient viewer say if the roll is one - two?
If we knew more information, such as what motivated the omniscient viewer to say that at least one six was rolled, then the probability may change. But we don't know. The question as stated in post 130 is not ambiguous.

Quote:
 The only way to get a 1/11 result is by NOT counting some of the results.
Right. We don't count any of the 36 possible rolls that don't include at least one 6.
#141
10-06-2016, 11:54 PM
 octopus Guest Join Date: Apr 2015 Posts: 5,792
Quote:
 Originally Posted by septimus Problem statement #1:"Two fair dice are rolled together, and you are told that 'at least one of the dice is a 6.' A 6 is removed, and you are then shown the other die. What is the probability of the other die showing a six?"Problem statement #2:"Two fair dice are rolled together. A 6 is removed, and you are then shown the other die. What is the probability of the other die showing a six?"How do these problems differ from each other? Do they have the same answer? Reread the post you're responding to. "A six is removed" can be interpreted in multiple ways. As for "you cannot show a 6 from a pair of rolled, fair dice when a 6 is not present" — that was my point. Read the two problems at the beginning of this post and tell us if they have the same answer.
Are you saying that when the 6 is removed it's not shown to confirm that statement A is correct? I.e. showing and removing the 6 proves that at least one 6 is rolled. That's how I interpret this. Who doesn't have dice and a friend and can't replicate the scenario?

And those people who keep saying 1/11 can't count. The person rolling, showing, and removing a 6 has 2 6's to show in the double 6 result.

I rolled dice til I got 20 sets of double 6s and I took the result of how many rolls it took that included at least one 6.

My ratio is 20/132. Which is 1/6.6. I'm going to keep rolling til I get 20 more. Brb.
#142
10-07-2016, 12:04 AM
 septimus Guest Join Date: Dec 2009 Location: The Land of Smiles Posts: 14,891
Quote:
 Originally Posted by octopus Are you saying that when the 6 is removed it's not shown to confirm that statement A is correct? ... I rolled dice til I got 20 sets of double 6s and I took the result of how many rolls it took that included at least one 6. My ratio is 20/132. Which is 1/6.6. I'm going to keep rolling til I get 20 more. Brb.
I'm not sure what you think you're arguing with me about, but your experimental result does surprise me!

Are you saying that you rolled a pair of dice many times and had 20 instances of double-six and 112 instances (132-20) of a single six? Your result is not impossible, but is surprising. On average single-sixes will outnumber double-sixes ten to one!
#143
10-07-2016, 12:21 AM
 jtur88 Guest Join Date: Aug 2011 Location: Cebu, Philippines Posts: 13,282
Quote:
 Originally Posted by Chronos There are possible results for the simultaneous throws that are not possible for your situation, such as a red 2 and a green 6. .
But the chances in simultaneous throw of green-2 and red-6 are exactly the same as red-2 and green-6, so that would not change anything, and can be ignored as a duplicating variation. Your throw of a red-2 gets absorbed as the mirrored green-2, canceling the roll of the other die which cannot be a 6 if the roll does not occur. Only if the odds are different if the 6 is green, would your argument hold sway.

Once a die of any color is a 6, the experiment proceeds with an examination of the other die, which will always be independent of the first die. Say the two dice are thrown in separate rooms, and throws continue until one of the operators announces "I have a 6". The operator in the other room examines his single die roll. What are the chances that it will be a 6? But one of the operators falls asleep, and is awakened by the first one saying "I have a 6", and suddenly rolls his die for the first time. Do the odds change?
#144
10-07-2016, 12:42 AM
 octopus Guest Join Date: Apr 2015 Posts: 5,792
Quote:
 Originally Posted by septimus I'm not sure what you think you're arguing with me about, but your experimental result does surprise me! Are you saying that you rolled a pair of dice many times and had 20 instances of double-six and 112 instances (132-20) of a single six? Your result is not impossible, but is surprising. On average single-sixes will outnumber double-sixes ten to one!
Ok! I'm back.

It takes awhile to roll that many dice.

This time my results look like this.

Double 6 At least one 6 Odds Inverse

1 9 0.111111111 9
2 13 0.153846154 6.5
3 41 0.073170732 13.66666667
4 48 0.083333333 12
5 66 0.075757576 13.2
6 80 0.075 13.33333333
7 85 0.082352941 12.14285714
8 86 0.093023256 10.75
9 95 0.094736842 10.55555556
10 97 0.103092784 9.7
11 106 0.103773585 9.636363636
12 135 0.088888889 11.25
13 149 0.087248322 11.46153846
14 150 0.093333333 10.71428571
15 154 0.097402597 10.26666667
16 181 0.08839779 11.3125
17 182 0.093406593 10.70588235
18 201 0.089552239 11.16666667
19 207 0.09178744 10.89473684
20 227 0.088105727 11.35

This looks a lot closer to 1/11.

Hmm, maybe you don't count the green and red die separately and do treat that as one event. Maybe I'm the one who can't count!

Sorry for the ugly formatting but padding spaces seem to be removed.

Last edited by octopus; 10-07-2016 at 12:45 AM.
#145
10-07-2016, 12:52 AM
 newme Guest Join Date: Dec 2009 Location: In the Woods Posts: 323
Quote:
 Originally Posted by Skammer Speaking as the OP, I'll say no, you are both misunderstanding the (admittedly ambiguously worded) problem I posed. Although I did not come up with the wording myself. Two dice are rolled. The implication is that ANY of the 36 possible combinations may have turned up. Not, you keep rolling until you have a six. Why even bother rolling two dice if you only count the results when one of them lands on a specific face? That's just like rolling one die in the first place and is a trivial question. So, you have one of 36 possible combinations. Then the omniscient viewer says, "Look, you rolled at least one six." At that point, what is the chance your other die is a six? It's 1/11.
Quote:
 Originally Posted by x-ray vision If we knew more information, such as what motivated the omniscient viewer to say that at least one six was rolled, then the probability may change. But we don't know. The question as stated in post 130 is not ambiguous. Right. We don't count any of the 36 possible rolls that don't include at least one 6.
Did you read post 130 carefully? See the bolded part?

Skammer: " Not, you keep rolling until you have a six."

x-ray: "
We don't count any of the 36 possible rolls that don't include at least one 6."

You can't have it both ways.The way I interpret Skammer in post 130 is roll two dice, and use whatever comes up. No throwing anything out.
#146
10-07-2016, 01:18 AM
 septimus Guest Join Date: Dec 2009 Location: The Land of Smiles Posts: 14,891
I use Monte Carlo simulations a lot — in problems where the number of possibilities is too numerous for exhaustive counting.

Here there are only 11 (or, depending on how you count, 36, or 4, or just 3) possibilities. Why not just list them:
6 & Not-6 ... 5 cases
Not-6 & 6 ... 5 cases
6 & 6 ........ 1 case
(Similarly, when I tell you a pair of red Kings will make a better hand than a pair of black Aces 18.553597959240882460% of the time in all-in Texas Hold-em, that's the result of enumeration, not simulation.)
#147
10-07-2016, 06:18 AM
 Colophon Guest Join Date: Sep 2002 Location: Hampshire, England Posts: 13,348
I can't believe this thread is still going. People are looking for ambiguity where there is none.

Roll two dice.

Is at least one of them a six?

P (25/36) No --> end game

P (11/36) Yes -->

There are now 11 possibilities:

6 1
6 2
6 3
6 4
6 5
6 6
1 6
2 6
3 6
4 6
5 6

Is the other die a six?

P (10/11) No

P (1/11) Yes.

#148
10-07-2016, 07:20 AM
 Chronos Charter Member Moderator Join Date: Jan 2000 Location: The Land of Cleves Posts: 73,079
Quote:
 Quoth jtur88: But the chances in simultaneous throw of green-2 and red-6 are exactly the same as red-2 and green-6, so that would not change anything, and can be ignored as a duplicating variation.
Sure, those odds are the same, but what's the duplicating variation to red-6, green-6? You can't factor out something (in this case, a factor of 2) unless it appears in all terms.
#149
10-07-2016, 07:25 AM
 septimus Guest Join Date: Dec 2009 Location: The Land of Smiles Posts: 14,891
Quote:
 Originally Posted by Colophon Answer is 1/11. Where is the ambiguity please?
"A 6 is removed." This might mean "Monty picks one of the two dice at random and shows it to you. It happens to be a six." Now the answer is 1/6.

It's similar to the ambiguity in the Monty Hall problem. "Monty opens a curtain. This exposes a goat." Did Monty deliberately avoid the curtain with the brand-new Maserati?
#150
10-07-2016, 07:45 AM
 ftg Guest Join Date: Feb 2001 Location: Not the PNW :-( Posts: 15,825
Quote:
 Originally Posted by Musicat "What is the probability of the...die showing a six?" Six possibilities, one possible outcome. Answer: 1/6 If you don't agree with me, please show me where I have defined the problem wrong, or how the first parts of this problem affect the final question.
This is such an obvious mistake! So there are 6 possibilities? That in no way shape or form means they must be equally probable!

This is the second most common mistake made in the Monty Hall problem. (The first is misstating the problem.) You see two remaining doors and think they are equally likely to have the big prize.

Look at any of the tables of eleven possible situations in this thread. Note that "the other is a 6" happens only one out of eleven times. For each of the others, they happen 2 out of eleven times.

Given that the setup leaves 11 possibilities, how on Earth do you assign 1/6 to the (6,6) case but only 1/12 to each of (3,6) and (6,3). How can that possibly be???? The chances of rolling an (a,b) and a (c,d) are the same.

You don't need to write any stupid code either. The listing of the possibilities, all equally likely, proves the answer without any further debate.

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