It’s not about whether the “partner” is a partner or a logician or a Monty Hall type.

It’s literally a Bayes information question.

Also Wizard didn’t formulate the original question.

There’s a real simple restatement of the question that leads to 1/11. And I’m very sure that’s how the OP meant the question to be taken. And it’s a really interesting question.

********

You have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. You ask your partner if at least one of the dice show a 2. Your partner peeks under the cup at both dice, and tells you, truthfully, "Yes"

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

*********

See. Not hard. And the answer now is clearly 1/11.

Quote:New wordingYou have two dice in a shaker. You ask a friend to shake the dice and peek at the outcome, which he does. You ask him, "Is at least one of the dice a two?" He says, "Yes." What is the probability both dice are a two?

I would like to say I was a bit harsh on Alan, and for that, I apologize.

Quote:WizardHi Everyone. It's nice to be out of WoV jail. I spent my time arguing with my judge and jailer about this problem. Not much else to do in there. Suffice it to say, I'm ready to quit tearing apart the original wording, which was not mine, and suggest a rewording. I submit this for the consideration of the forum:

Quote:New wordingYou have two dice in a shaker. You ask a friend to shake the dice and peek at the outcome, which he does. You ask him, "Is at least one of the dice a two?" He says, "Yes." What is the probability both dice are a two?

I would like to say I was a bit harsh on Alan, and for that, I apologize.

link to original post

Welcome back!

Yes I think that wording does it. Same idea as my wording in post above.

Quote:WizardHi Everyone. It's nice to be out of WoV jail. I spent my time arguing with my judge and jailer about this problem. Not much else to do in there. Suffice it to say, I'm ready to quit tearing apart the original wording, which was not mine, and suggest a rewording. I submit this for the consideration of the forum:

Quote:New wordingYou have two dice in a shaker. You ask a friend to shake the dice and peek at the outcome, which he does. You ask him, "Is at least one of the dice a two?" He says, "Yes." What is the probability both dice are a two?

I would like to say I was a bit harsh on Alan, and for that, I apologize.

link to original post

Wizard, Welcome back. It was fun debating you in PM. Thanks for the apology to Alan.

Your paraphrased question has the unequivocal answer of 1/11, It is geared, like the Monty Hall question, to draw out the incorrect, intuitive answer of 1/6. and I'm sure there is one here who would insist that your question should be answered 1/6. He was wrong 9 years ago and he always will be. If/when he or his fellow [Insert correct word] challenge this version, I expect fur to fly.

Yours lacks the entertainment value of the original in-explicit version. Maybe we can get some fool to accept a wager paying 8 to 1 on your version. Maybe someone will take on Unjon's 20 to 1. It's about time someone monetized this.

$:o)

Trimmed.Quote:OnceDear

Yours lacks the entertainment value of the original in-explicit version. Maybe we can get some fool to accept a wager paying 8 to 1 on your version. Maybe someone will take on Unjon's 20 to 1. It's about time someone monetized this.

$:o)

link to original post

To monetize this, I propose a craps side bet called the Harder 4. The bet pays 9:1 if a hard four is rolled before a roll that contains only one 2.

Quote:OnceDear

If/when he or his fellow [Insert correct word] challenge this version, I expect fur to fly.

link to original post

Won't be this muttonhead!

My mind is now turned to fond memories of a flugelhorn player I used to know.

Welcome back, Wizard. Thanks for sparring, Gordon. -D

You bet on all numbers. Every 6th roll (on average) you win 9 units. All other rolls you lose 2 units.

So you lose 1 unit every 6th roll on.

That sounds for me like an 1/11th chance of winning.

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

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.

Quote:DieterQuote: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

Really? There are only two dice. If we know any one die is showing a 2 then the other die has six options.

Why are you making it difficult?