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unJon
unJon
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djtehch34tOnceDear
May 14th, 2022 at 5:00:06 AM permalink
Iíll try once more. Even if Gordon doesnít want to hear it maybe it will make sense to others.

Itís not about whether the ďpartnerĒ is a partner or a logician or a Monty Hall type.

Itís literally a Bayes information question.

Also Wizard didnít formulate the original question.

Thereís a real simple restatement of the question that leads to 1/11. And Iím very sure thatís how the OP meant the question to be taken. And itís 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.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Wizard
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Wizard
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OnceDeargordonm888
May 14th, 2022 at 5:22:33 AM permalink
Hi 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 wording

You 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.
It's not whether you win or lose; it's whether or not you had a good bet.
unJon
unJon
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Thanks for this post from:
OnceDeargordonm888
May 14th, 2022 at 5:27:59 AM permalink
Quote: Wizard

Hi 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 wording

You 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.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
OnceDear
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OnceDear 
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May 14th, 2022 at 5:38:02 AM permalink
Quote: Wizard

Hi 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 wording

You 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)
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
unJon
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May 14th, 2022 at 5:45:04 AM permalink
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

Trimmed.

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.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Dieter
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Dieter
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gordonm888OnceDear
May 14th, 2022 at 5:46:03 AM permalink
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
May the cards fall in your favor.
Torghatten
Torghatten
Joined: Feb 3, 2012
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May 14th, 2022 at 5:48:22 AM permalink
Lets imagine a Craps table offers a "double before single bet" It pays 9:1

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.
AlanMendelson
AlanMendelson
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May 14th, 2022 at 11:22:23 AM permalink
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
Last edited by: AlanMendelson on May 14, 2022
Dieter
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Dieter
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May 14th, 2022 at 12:16:47 PM permalink
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.
May the cards fall in your favor.
AlanMendelson
AlanMendelson
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May 14th, 2022 at 12:41:40 PM permalink
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



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?

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