Two dice puzzle -- Part Trois

Quote: Dieter

Quote: AlanMendelson

If I know there are only two dice and at least one die is a 2, there are only six sides on the remaining die. So the answer is 1/6.
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Except... since you don't know which dice is a 2, there are 11 other faces in play, not 6.

If you reveal one dice, you then know which dice is a 2, and the revealed non-top sides are out of play, returning it to 1/6.

The discrepancy comes from thinking about the probabilities of one dice vs a pair of dice.
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Bolded added.

That’s not right.

If the person says at least one die is a two, and then removes one of the die and shows that two to you, it’s still 1/11.

ETA: it’s a little like the Monty Hall problem in that way.

Quote: AlanMendelson

…Why are you making it difficult?
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That says it all! You’ve used this “argument” before. Tell me, do you stomp your feet, as well, when you right this? I have a hard time picturing anything else.

Quote: unJon

Quote: Dieter

Quote: AlanMendelson

If I know there are only two dice and at least one die is a 2, there are only six sides on the remaining die. So the answer is 1/6.
link to original post

Except... since you don't know which dice is a 2, there are 11 other faces in play, not 6.

If you reveal one dice, you then know which dice is a 2, and the revealed non-top sides are out of play, returning it to 1/6.

The discrepancy comes from thinking about the probabilities of one dice vs a pair of dice.
link to original post

Bolded added.

That’s not right.

If the person says at least one die is a two, and then removes one of the die and shows that two to you, it’s still 1/11.

ETA: it’s a little like the Monty Hall problem in that way.
link to original post

I dont need to know what's on 12 faces or 11 faces. I just need to know that one die is showing a 2. So the other die has six options.

And since there are only 2 dice I dont care which die is a 2, because the answer is still 1/6 on the remaining die.

Alan I would love you to try the experiment at home. Roll two dice. Track how many rolls have zero 2s, one 2 and two 2s and report back.

What you should find is that:

69.5% have zero 2s
27.8% have one 2
2.8% have two 2s

And when you look at the ratio of rolls that have two 2 vs “at least one” 2 you’ll see that it’s 1/11.

Quote: unJon

Quote: Dieter

Quote: AlanMendelson

If I know there are only two dice and at least one die is a 2, there are only six sides on the remaining die. So the answer is 1/6.
link to original post

Except... since you don't know which dice is a 2, there are 11 other faces in play, not 6.

If you reveal one dice, you then know which dice is a 2, and the revealed non-top sides are out of play, returning it to 1/6.

The discrepancy comes from thinking about the probabilities of one dice vs a pair of dice.
link to original post

Bolded added.

That’s not right.

If the person says at least one die is a two, and then removes one of the die and shows that two to you, it’s still 1/11.

ETA: it’s a little like the Monty Hall problem in that way.
link to original post

OK, I can be wrong.

If my understanding is wrong, I invite explanation.

https://wizardofvegas.com/forum/questions-and-answers/math/21845-two-dice-puzzle-part-deux/#post452706

Part Deux problems 1 and 3 each involve a pair of dice.

In Part Deux problem 2, there is a single dice involved.

Quote: Dieter

If my understanding is wrong, I invite explanation.

https://wizardofvegas.com/forum/questions-and-answers/math/21845-two-dice-puzzle-part-deux/#post452706

Part Deux problems 1 and 3 each involve a pair of dice.

In Part Deux problem 2, there is a single dice involved.
link to original post

I’ll give it the college try.

There’s two ways that problem 2 might go.

Way 1:

Roll a die. If it’s not a 2, roll the same die again. If you get a 2, roll a second die under a cup.

Chance of 2/2 is 1/6.

Way 2:

Roll two dice under a cup. If no 2s then reroll. If at least one 2 then remove a die with a two to show the world that 2, and keep the second die hidden.

Chance of 2/2 is 1/11.

Follow?

Way 2 is like the Monty Hall problem. The dealer will always pick a die with a 2 to show, no matter which die it is.

Way 1 is different. The first die roll is always shown.

Quote: camapl

Quote: AlanMendelson

…Why are you making it difficult?
link to original post

That says it all! You’ve used this “argument” before. Tell me, do you stomp your feet, as well, when you right this? I have a hard time picturing anything else.
link to original post

I stomp my feet when I' hear about 11 faces.

I dont need to know 11 faces. I need to know if I die has a 2.

Quote: unJon

Quote: Dieter

If my understanding is wrong, I invite explanation.

https://wizardofvegas.com/forum/questions-and-answers/math/21845-two-dice-puzzle-part-deux/#post452706

Part Deux problems 1 and 3 each involve a pair of dice.

In Part Deux problem 2, there is a single dice involved.
link to original post

I’ll give it the college try.

There’s two ways that problem 2 might go.

Way 1:

Roll a die. If it’s not a 2, roll the same die again. If you get a 2, roll a second die under a cup.

Chance of 2/2 is 1/6.

Way 2:

Roll two dice under a cup. If no 2s then reroll. If at least one 2 then remove a die with a two to show the world that 2, and keep the second die hidden.

Chance of 2/2 is 1/11.

Follow?

Way 2 is like the Monty Hall problem. The dealer will always pick a die with a 2 to show, no matter which die it is.

Way 1 is different. The first die roll is always shown.
link to original post

You're giving unnecessary steps. You've already been told that one die is showing a 2.

Why dont you pick up 2 dice.

Put one die in your pocket.

With the die not in your pocket set it to 2. This is where the problem starts. You've been told one die is a 2.

Now ask yourself... if I want to have 2-2 what does the die in my pocket have to show? Let's count the possibilities...

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

Quote: AlanMendelson

Hi Wizard. No apology is needed.

I'm sorry to say this but my position hasnt changed even with the new quote.

If I know there are only two dice and at least one die is a 2, there are only six sides on the remaining die. So the answer is 1/6.

Unless you're allowing the two dice to continue to shake, rattle and roll? In which case 2/2 is 1/36 once the dice settle.

Sorry but I cant budge on this.

I have suggested that a different question might be answered by 1/11.

But when you have two dice and the result of one die is known, there are only six options on the second dice.

Best,
Your thick and stupid friend, Alan
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THAT, sir, is exactly how one answers the question wrongly or answers the wrong question altogether.

Alan, You are just so totally wrong..... But more than that, you are stubborn. If you cannot understand this now, after nine years of every conceivable proof, I feel confident that you never will. Your brain's programming on this is flawed but hard wired.

Your [insert appropriate, pejorative adjective] answer to this question, and especially this clarified version of it, defines you.

There are a few proverbs that apply to this situation, but to quote them would be insulting to you.