OnceDear
Joined: Jun 1, 2014
• Posts: 6728
May 13th, 2022 at 3:18:33 PM permalink
Proof that 1 in 6 is a perfectly valid answer in spite of Wizard insisting otherwise

First, I restate the EXACT wording of the first post that raised this ugly mess
Quote: Dween

You have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."

What is the probability that both dice are showing a 2?

And now I stake my case completely for the potential validity of 1/6

Our calling rules are indeterminate!

I'm arguing that nowhere in the OP did it say that he was CONFIRMING twos. Wizard assumed that without good reason. He denies the possibility that he doesn't call a number at every roll, also without good reason.

Let's say he saw both dice but always called one of the dies values. Choosing randomly between the two dice seen.
That is not ruled out in the OP and is a perfectly reasonable assumption. He's not obligated to declare twos as that is not in the original premise. YES he sees both dice.

Here are all the 72 possibilities, equally weighted.

First 36 possibilities considering that he calls the first value to catch his eye, maybe being the one nearest to him

1,1 "At least one of the dice is a one!
1,2 "At least one of the dice is a one!
1,3 "At least one of the dice is a one!
1,4 "At least one of the dice is a one!
1,5 "At least one of the dice is a one!
1,6 "At least one of the dice is a one!
2,1 "At least one of the dice is a two!!
2,2 "At least one of the dice is a two!!
2,3 "At least one of the dice is a two!!
2,4 "At least one of the dice is a two!!
2,5 "At least one of the dice is a two!!
2,6 "At least one of the dice is a two!!
3,1 "At least one of the dice is a three!!
3,2 "At least one of the dice is a three!!
3,3 "At least one of the dice is a three!!
3,4 "At least one of the dice is a three!!
3,5 "At least one of the dice is a three!!
3,6 "At least one of the dice is a three!!
4,1 "At least one of the dice is a four!!
4,2 "At least one of the dice is a four!!
4,3 "At least one of the dice is a four!!
4,4 "At least one of the dice is a four!!
4,5 "At least one of the dice is a four!!
4,6 "At least one of the dice is a four!!
5,1 "At least one of the dice is a five!!
5,2 "At least one of the dice is a five!!
5,3 "At least one of the dice is a five!!
5,4 "At least one of the dice is a five!!
5,5 "At least one of the dice is a five!!
5,6 "At least one of the dice is a five!!
6,1 "At least one of the dice is a six!!
6,2 "At least one of the dice is a six!!
6,3 "At least one of the dice is a six!!
6,4 "At least one of the dice is a six!!
6,5 "At least one of the dice is a six!!
6,6 "At least one of the dice is a six!!

Now another 36 possibilities where he calls the second die to catch his eye, maybe the one furthest from him.
1,1 "At least one of the dice is a one!
1,2 "At least one of the dice is a two!
1,3 "At least one of the dice is a three!
1,4 "At least one of the dice is a four!
1,5 "At least one of the dice is a five!
1,6 "At least one of the dice is a six!
2,1 "At least one of the dice is a one!
2,2 "At least one of the dice is a two!
2,3 "At least one of the dice is a three!
2,4 "At least one of the dice is a four!
2,5 "At least one of the dice is a five!
2,6 "At least one of the dice is a six!
3,1 "At least one of the dice is a one!
3,2 "At least one of the dice is a two!
3,3 "At least one of the dice is a three!
3,4 "At least one of the dice is a four!
3,5 "At least one of the dice is a five!
3,6 "At least one of the dice is a six!
4,1 "At least one of the dice is a one!
4,2 "At least one of the dice is a two!
4,3 "At least one of the dice is a three!
4,4 "At least one of the dice is a four!
4,5 "At least one of the dice is a five!
4,6 "At least one of the dice is a six!
5,1 "At least one of the dice is a one!
5,2 "At least one of the dice is a two!
5,3 "At least one of the dice is a three!
5,4 "At least one of the dice is a four!
5,5 "At least one of the dice is a five!
5,6 "At least one of the dice is a six!
6,1 "At least one of the dice is a one!
6,2 "At least one of the dice is a two!
6,3 "At least one of the dice is a three!
6,4 "At least one of the dice is a four!
6,5 "At least one of the dice is a five!
6,6 "At least one of the dice is a six!
Of the 12 times that he calls "At least one of the dice is a two", they are both twos on two occasions.
Thus if he calls "At least one of the dice is a two" then the probability of both dice being a two is ONE in SIX
QED

That he peeks at both dice is irrelevant.
The calling rules MATTER and they are unstated.

Now..... If your friend had a declared hatred of using the word Two, after a nasty incident where 2 anvils fell on his 2 feet, but was honor bound to truthfully declare in the form "At least one of the dice is a X" and at one roll, he declared "At least one of the dice is a Two", what is the probability that the other dice is a two?
100% certain. Unless both dice were two, he would have called the other number: He only calls a two when he absolutely can't avoid it
Beware. The earth is NOT flat. Hit and run is not a winning strategy: Pressing into trends IS not a winning strategy: Progressives are not a winning strategy: Don't Buy It! .Don't even take it for free.
unJon
Joined: Jul 1, 2018
• Posts: 4082
May 13th, 2022 at 3:27:02 PM permalink
Quote: OnceDear

Proof that 1 in 6 is a perfectly valid answer in spite of Wizard insisting otherwise

First, I restate the EXACT wording of the first post that raised this ugly mess
Quote: Dween

You have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."

What is the probability that both dice are showing a 2?

And now I stake my case completely for the potential validity of 1/6

Our calling rules are indeterminate!

I'm arguing that nowhere in the OP did it say that he was CONFIRMING twos. Wizard assumed that without good reason. He denies the possibility that he doesn't call a number at every roll, also without good reason.

Let's say he saw both dice but always called one of the dies values. Choosing randomly between the two dice seen.
That is not ruled out in the OP and is a perfectly reasonable assumption. He's not obligated to declare twos as that is not in the original premise. YES he sees both dice.

Here are all the 72 possibilities, equally weighted.

First 36 possibilities considering that he calls the first value to catch his eye, maybe being the one nearest to him

1,1 "At least one of the dice is a one!
1,2 "At least one of the dice is a one!
1,3 "At least one of the dice is a one!
1,4 "At least one of the dice is a one!
1,5 "At least one of the dice is a one!
1,6 "At least one of the dice is a one!
2,1 "At least one of the dice is a two!!
2,2 "At least one of the dice is a two!!
2,3 "At least one of the dice is a two!!
2,4 "At least one of the dice is a two!!
2,5 "At least one of the dice is a two!!
2,6 "At least one of the dice is a two!!
3,1 "At least one of the dice is a three!!
3,2 "At least one of the dice is a three!!
3,3 "At least one of the dice is a three!!
3,4 "At least one of the dice is a three!!
3,5 "At least one of the dice is a three!!
3,6 "At least one of the dice is a three!!
4,1 "At least one of the dice is a four!!
4,2 "At least one of the dice is a four!!
4,3 "At least one of the dice is a four!!
4,4 "At least one of the dice is a four!!
4,5 "At least one of the dice is a four!!
4,6 "At least one of the dice is a four!!
5,1 "At least one of the dice is a five!!
5,2 "At least one of the dice is a five!!
5,3 "At least one of the dice is a five!!
5,4 "At least one of the dice is a five!!
5,5 "At least one of the dice is a five!!
5,6 "At least one of the dice is a five!!
6,1 "At least one of the dice is a six!!
6,2 "At least one of the dice is a six!!
6,3 "At least one of the dice is a six!!
6,4 "At least one of the dice is a six!!
6,5 "At least one of the dice is a six!!
6,6 "At least one of the dice is a six!!

Now another 36 possibilities where he calls the second die to catch his eye, maybe the one furthest from him.
1,1 "At least one of the dice is a one!
1,2 "At least one of the dice is a two!
1,3 "At least one of the dice is a three!
1,4 "At least one of the dice is a four!
1,5 "At least one of the dice is a five!
1,6 "At least one of the dice is a six!
2,1 "At least one of the dice is a one!
2,2 "At least one of the dice is a two!
2,3 "At least one of the dice is a three!
2,4 "At least one of the dice is a four!
2,5 "At least one of the dice is a five!
2,6 "At least one of the dice is a six!
3,1 "At least one of the dice is a one!
3,2 "At least one of the dice is a two!
3,3 "At least one of the dice is a three!
3,4 "At least one of the dice is a four!
3,5 "At least one of the dice is a five!
3,6 "At least one of the dice is a six!
4,1 "At least one of the dice is a one!
4,2 "At least one of the dice is a two!
4,3 "At least one of the dice is a three!
4,4 "At least one of the dice is a four!
4,5 "At least one of the dice is a five!
4,6 "At least one of the dice is a six!
5,1 "At least one of the dice is a one!
5,2 "At least one of the dice is a two!
5,3 "At least one of the dice is a three!
5,4 "At least one of the dice is a four!
5,5 "At least one of the dice is a five!
5,6 "At least one of the dice is a six!
6,1 "At least one of the dice is a one!
6,2 "At least one of the dice is a two!
6,3 "At least one of the dice is a three!
6,4 "At least one of the dice is a four!
6,5 "At least one of the dice is a five!
6,6 "At least one of the dice is a six!
Of the 12 times that he calls "At least one of the dice is a two", they are both twos on two occasions.
Thus if he calls "At least one of the dice is a two" then the probability of both dice being a two is ONE in SIX
QED

That he peeks at both dice is irrelevant.
The calling rules MATTER and they are unstated.

Now..... If your friend had a declared hatred of using the word Two, after a nasty incident where 2 anvils fell on his 2 feet, but was honor bound to truthfully declare in the form "At least one of the dice is a X" and at one roll, he declared "At least one of the dice is a Two", what is the probability that the other dice is a two?
100% certain. Unless both dice were two, he would have called the other number: He only calls a two when he absolutely can't avoid it

You’re wasting time. 1 in anything is a valid answer to the original wording.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
OnceDear
Joined: Jun 1, 2014
• Posts: 6728
May 13th, 2022 at 3:32:06 PM permalink
Quote: unJon

Quote: OnceDear

Proof that 1 in 6 is a perfectly valid answer in spite of Wizard insisting otherwise

First, I restate the EXACT wording of the first post that raised this ugly mess
Quote: Dween

You have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."

What is the probability that both dice are showing a 2?

And now I stake my case completely for the potential validity of 1/6

Our calling rules are indeterminate!

I'm arguing that nowhere in the OP did it say that he was CONFIRMING twos. Wizard assumed that without good reason. He denies the possibility that he doesn't call a number at every roll, also without good reason.

Let's say he saw both dice but always called one of the dies values. Choosing randomly between the two dice seen.
That is not ruled out in the OP and is a perfectly reasonable assumption. He's not obligated to declare twos as that is not in the original premise. YES he sees both dice.

Here are all the 72 possibilities, equally weighted.

First 36 possibilities considering that he calls the first value to catch his eye, maybe being the one nearest to him

1,1 "At least one of the dice is a one!
1,2 "At least one of the dice is a one!
1,3 "At least one of the dice is a one!
1,4 "At least one of the dice is a one!
1,5 "At least one of the dice is a one!
1,6 "At least one of the dice is a one!
2,1 "At least one of the dice is a two!!
2,2 "At least one of the dice is a two!!
2,3 "At least one of the dice is a two!!
2,4 "At least one of the dice is a two!!
2,5 "At least one of the dice is a two!!
2,6 "At least one of the dice is a two!!
3,1 "At least one of the dice is a three!!
3,2 "At least one of the dice is a three!!
3,3 "At least one of the dice is a three!!
3,4 "At least one of the dice is a three!!
3,5 "At least one of the dice is a three!!
3,6 "At least one of the dice is a three!!
4,1 "At least one of the dice is a four!!
4,2 "At least one of the dice is a four!!
4,3 "At least one of the dice is a four!!
4,4 "At least one of the dice is a four!!
4,5 "At least one of the dice is a four!!
4,6 "At least one of the dice is a four!!
5,1 "At least one of the dice is a five!!
5,2 "At least one of the dice is a five!!
5,3 "At least one of the dice is a five!!
5,4 "At least one of the dice is a five!!
5,5 "At least one of the dice is a five!!
5,6 "At least one of the dice is a five!!
6,1 "At least one of the dice is a six!!
6,2 "At least one of the dice is a six!!
6,3 "At least one of the dice is a six!!
6,4 "At least one of the dice is a six!!
6,5 "At least one of the dice is a six!!
6,6 "At least one of the dice is a six!!

Now another 36 possibilities where he calls the second die to catch his eye, maybe the one furthest from him.
1,1 "At least one of the dice is a one!
1,2 "At least one of the dice is a two!
1,3 "At least one of the dice is a three!
1,4 "At least one of the dice is a four!
1,5 "At least one of the dice is a five!
1,6 "At least one of the dice is a six!
2,1 "At least one of the dice is a one!
2,2 "At least one of the dice is a two!
2,3 "At least one of the dice is a three!
2,4 "At least one of the dice is a four!
2,5 "At least one of the dice is a five!
2,6 "At least one of the dice is a six!
3,1 "At least one of the dice is a one!
3,2 "At least one of the dice is a two!
3,3 "At least one of the dice is a three!
3,4 "At least one of the dice is a four!
3,5 "At least one of the dice is a five!
3,6 "At least one of the dice is a six!
4,1 "At least one of the dice is a one!
4,2 "At least one of the dice is a two!
4,3 "At least one of the dice is a three!
4,4 "At least one of the dice is a four!
4,5 "At least one of the dice is a five!
4,6 "At least one of the dice is a six!
5,1 "At least one of the dice is a one!
5,2 "At least one of the dice is a two!
5,3 "At least one of the dice is a three!
5,4 "At least one of the dice is a four!
5,5 "At least one of the dice is a five!
5,6 "At least one of the dice is a six!
6,1 "At least one of the dice is a one!
6,2 "At least one of the dice is a two!
6,3 "At least one of the dice is a three!
6,4 "At least one of the dice is a four!
6,5 "At least one of the dice is a five!
6,6 "At least one of the dice is a six!
Of the 12 times that he calls "At least one of the dice is a two", they are both twos on two occasions.
Thus if he calls "At least one of the dice is a two" then the probability of both dice being a two is ONE in SIX
QED

That he peeks at both dice is irrelevant.
The calling rules MATTER and they are unstated.

Now..... If your friend had a declared hatred of using the word Two, after a nasty incident where 2 anvils fell on his 2 feet, but was honor bound to truthfully declare in the form "At least one of the dice is a X" and at one roll, he declared "At least one of the dice is a Two", what is the probability that the other dice is a two?
100% certain. Unless both dice were two, he would have called the other number: He only calls a two when he absolutely can't avoid it

You’re wasting time. 1 in anything is a valid answer to the original wording.

How very true.
Funny that the OP stated...9 years ago...
Quote: Dween

I'm interested to see how people solve this one.

But he never once re entered that thread!
Beware. The earth is NOT flat. Hit and run is not a winning strategy: Pressing into trends IS not a winning strategy: Progressives are not a winning strategy: Don't Buy It! .Don't even take it for free.
Dieter
Joined: Jul 23, 2014
• Posts: 4312
May 13th, 2022 at 4:45:16 PM permalink
Quote: OnceDear

BUT You assume that the peeker is confirming twos. Exactly the mistake Wizard makes.
The peeker is calling the value of one of the dice. We don't know which one. We don't know how he decides which one. We don''t know what he declares if no two is rolled. See my reply to Gordon.

This is exactly what I came to as I mulled it over for the last 3 or 4 hours.

Is there something special about twos in this context?

Until I know that, I am making the assumption that a dice value is randomly revealed.

I am greatly enjoying the opportunity to analyze my assumptions.
May the cards fall in your favor.
OnceDear
Joined: Jun 1, 2014
• Posts: 6728
May 13th, 2022 at 5:11:12 PM permalink
Quote: Dieter

Quote: OnceDear

BUT You assume that the peeker is confirming twos. Exactly the mistake Wizard makes.
The peeker is calling the value of one of the dice. We don't know which one. We don't know how he decides which one. We don''t know what he declares if no two is rolled. See my reply to Gordon.

This is exactly what I came to as I mulled it over for the last 3 or 4 hours.

Is there something special about twos in this context?

Until I know that, I am making the assumption that a dice value is randomly revealed.

I am greatly enjoying the opportunity to analyze my assumptions.

Have fun Dieter. You weren't here in 2013. This threat was the source of a lot of argument and a lot of expended mental effort. It brought out analogies, simplifications, alternative interpretations and even real life wagers.
And there are still two people that just don't get it.
Beware. The earth is NOT flat. Hit and run is not a winning strategy: Pressing into trends IS not a winning strategy: Progressives are not a winning strategy: Don't Buy It! .Don't even take it for free.
gordonm888
Joined: Feb 18, 2015
• Posts: 4409
Thanks for this post from:
May 13th, 2022 at 5:12:13 PM permalink
Quote: OnceDear

Quote: unJon

Quote: billryan

I appreciate the explanation and accept it, but don't fully grasp it.

Potentially an easier way to think about it:...

So fair odds to the two dice problem are 1 / (1+10) = 1/11.

BUT You assume that the peeker is confirming twos. Exactly the mistake Wizard makes.
The peeker is calling the value of one of the dice. We don't know which one. We don't know how he decides which one. We don''t know what he declares if no two is rolled. See my reply to Gordon.

The peeker might have said "one of the dice is a 5." There is nothing special about a two. Then the odds would have been 1/11 that the other die is a 5.

This 2-dice puzzle challenges you to understand the additional information that has been conveyed to you and its effect on your ability to assign probabilities to the unknown outcomes. I am teaching senior-level high-school math at home and the textbooks constantly reinforce the need to understand
1. what the known and unknown variables are in a problem statement
2. What parameter you are asked to calculate or define
3. Writing an equation that expresses the requested parameter in terms of the known and unknown variables

Approximately half the people posting on this thread seem to have never been taught the above approach in their education and don't seem to understand how to interpret a word problem. The problem statement isn't War and Peace, you don't need to analyze motivations of the "friend," you are simply asked to straightforwardly interpret new information from your friend (at face value.)

For some reason, some otherwise fairly intelligent people seem determined on this thread to provoke others by pretending to be obtuse. I understand that BillRyan has strengths other than math -that's okay, I'm cool with that. But for OnceDear to constantly denounce people on other threads who don't understand "Expected Value" and then to play at being obtuse about dice probabilities on this thread is very discouraging. No wonder Wizard lost his cool with Alan and the other flat-earthers.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
unJon
Joined: Jul 1, 2018
• Posts: 4082
Thanks for this post from:
May 13th, 2022 at 5:27:48 PM permalink
Quote: gordonm888

Quote: OnceDear

Quote: unJon

Quote: billryan

I appreciate the explanation and accept it, but don't fully grasp it.

Potentially an easier way to think about it:...

So fair odds to the two dice problem are 1 / (1+10) = 1/11.

BUT You assume that the peeker is confirming twos. Exactly the mistake Wizard makes.
The peeker is calling the value of one of the dice. We don't know which one. We don't know how he decides which one. We don''t know what he declares if no two is rolled. See my reply to Gordon.

The peeker might have said "one of the dice is a 5." There is nothing special about a two. Then the odds would have been 1/11 that the other die is a 5.

This 2-dice puzzle challenges you to understand the additional information that has been conveyed to you and its effect on your ability to assign probabilities to the unknown outcomes. I am teaching senior-level high-school math at home and the textbooks constantly reinforce the need to understand
1. what the known and unknown variables are in a problem statement
2. What parameter you are asked to calculate or define
3. Writing an equation that expresses the requested parameter in terms of the known and unknown variables

Approximately half the people posting on this thread seem to have never been taught the above approach in their education and don't seem to understand how to interpret a word problem. The problem statement isn't War and Peace, you don't need to analyze motivations of the "friend," you are simply asked to straightforwardly interpret new information from your friend (at face value.)

For some reason, some otherwise fairly intelligent people seem determined on this thread to provoke others by pretending to be obtuse. I understand that BillRyan has strengths other than math -that's okay, I'm cool with that. But for OnceDear to constantly denounce people on other threads who don't understand "Expected Value" and then to play at being obtuse about dice probabilities on this thread is very discouraging. No wonder Wizard lost his cool with Alan and the other flat-earthers.

Gordon,

I disagree with much of what you say. The rules that the peeker follows are critically important to the problem. And those rules are not disclosed in the original formulation of the problem.

I’ll give you two examples.

1) A challenge. I’ll be the person that peeks. I will sometimes when I look at the dice make the truthful statement “there is at least one 2.” Every time I make that statement, you can make a bet that there are 2 twos showing on the dice. I will offer you 20 to 1 odds. Should you take the bet?

2) The rule the peeker follows is:

a) if there is at least one 1, say “there is at least one 1”.
b) if a) doesn’t happen and there is at least one 2, say “there is at least one 2.”

The dice are rolled and the speaker says “there is at least one 2.” Now what is the probability that there are two 2s?
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
OnceDear
Joined: Jun 1, 2014
• Posts: 6728
May 13th, 2022 at 5:45:17 PM permalink
Quote: gordonm888

The peeker might have said "one of the dice is a 5." There is nothing special about a two. Then the odds would have been 1/11 that the other die is a 5.

SO WRONG on so many levels
Gordon.....
If the peeker had seen a 5 and a 2 or a 2 and a 5, what would he have declared and why?

I assert that we do not know and we do not know.

Very similar to what I just PM'd wizard
Quote: oncedear to wizard a few minutes ago

Stop speed reading. ! Stop seeing what you want to see. You're doing an [redacted]

If the dice had been a two and a four, or a four and a two, what would the peeker have said? And why?

Beware. The earth is NOT flat. Hit and run is not a winning strategy: Pressing into trends IS not a winning strategy: Progressives are not a winning strategy: Don't Buy It! .Don't even take it for free.
unJon
Joined: Jul 1, 2018