Two dice puzzle -- Part Trois
Quote: AlanMendelsonQuote: unJonQuote: DieterIf 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.
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Ill give it the college try.
Theres two ways that problem 2 might go.
Way 1:
Roll a die. If its 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.
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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.
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Do the experiment I proposed in this thread. When you get 1/11 and dont understand why, Ill try to explain it to you.
Do some investigative journalism through empirical experimentation!
Quote: OnceDearQuote: AlanMendelsonHi 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.
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Tsk tsk. You want to insult me but you wont pick up two dice and look at them?
You're not considering a roll of two dice. The roll is over.
You know one die has settled on two.
To have 2-2 what are the odds that remain?
Careful... only one die now.
Alan, Sir.....Quote: AlanMendelsonTsk tsk. You want to insult me but you wont pick up two dice and look at them?
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What I scream at the screen is acceptable under the rules here. If I typed it, then as a moderator I am held to 'A higher standard' and I would be suspended.
You are wrong. You are allowed to be wrong.
I would be a fool to try any further to set you right.
I would be a [Insert really offensive adjective] fool to believe that you would ever be set right.
No offence to unjon or anyone else here, but be assured that you will not get Alan to understand this.
I KNOW that Alan will remain forever wrong in his interpretation and answer here. If he wants to continue to make a [redacted pejorative] of himself, I'll let others in this thread entertain him.
My work here is complete.
Quote: AlanMendelson
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.
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You have separated the pair of dice into two individual dice; one known, one unknown.
The problem is about a pair of dice with partial information.
Quote: AlanMendelsonIf 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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It's time to put up or shup up. I'll bet whatever stakes you wish. If you think the probability is 1/6, then fair odds would pay 5 to 1. I'll offer you 8 to 1.
It's a simple question with a simple answer.
But to get this 1/11 answer you complicate it.
So again, returning to the simple approach:
It's been announced that two dice have been rolled and at least one of those two dice is showing a 2.
That die is 2.
Now, you're asked what are the odds that the second die is also showing a 2?
You pause for a moment and realize that the second die has faces marked 1, 2, 3, 4, 5, 6. So to have both dice showing 2-2 it would be a 1/6 chance.
No one needs to concern themselves will 11 faces on two dice.
It's a one die with six faces question.
Why complicate it?
Let me ask this: why count the five other faces of the die showing a 2? That die is no longer moving. It's showing a 2.
Quote: WizardQuote: AlanMendelsonIf 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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It's time to put up or shup up. I'll bet whatever stakes you wish. If you think the probability is 1/6, then fair odds would pay 5 to 1. I'll offer you 8 to 1.
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Put up or shut up about what?
The odds of rolling 2-2 is 1/36.
The odds of rolling a 2 on one die is 1/6.
The odds of throwing two dice on a table, with one die immediately stopping on 2 and the second die spinning for ten seconds, and then stopping on 2 is also 1/6.
Where is there a result with odds of 1/11 throwing two dice?
Show me those odds on any casino bet.
Thanks.
Quote: AlanMendelsonDieter, how much more information do you need? We're talking about two six sided dice... and one of them has settled on a number that has been announced.
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In the problem, you do not know which of the two dice shows a 2.
In your rearrangement of the problem, you do know whice dice shows a 2.
That matters.
You've either been offered an amazing opportunity to win Wizard's money or an amazing opportunity to grow Wizard's bankroll. I suggest at least 40 units, and a clear statement of the rules of contest.
Only bet money you can afford to lose. Best of luck.
Quote: DieterQuote: AlanMendelsonDieter, how much more information do you need? We're talking about two six sided dice... and one of them has settled on a number that has been announced.
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In the problem, you do not know which of the two dice shows a 2.
In your rearrangement of the problem, you do know whice dice shows a 2.
That matters.
You've either been offered an amazing opportunity to win Wizard's money or an amazing opportunity to grow Wizard's bankroll. I suggest at least 40 units, and a clear statement of the rules of contest.
Only bet money you can afford to lose. Best of luck.
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Why is everyone saying you need to know which of the two dice is showing a 2?
It makes no difference in a two dice problem.
Tell me why it makes a difference. Please.
And what is the bet?
I originally thought the answer was 1/6.
I had to be shown the math to see that it's 1/11.
But when I read (Alan's?) comment that it has to do with the wording, I rattled that around for a couple days.
As a result I came up with two scenarios.
Scenario 1: The dice are shaken and rolled but hidden. One person peeks and states that at least one die is a 2.
Scenario 2: The dice are shaken and rolled but hidden. One person asks the other to verify that at least one die is a 2.
In each scenario, what are the odds that both dice are 2?
In Scenario 1 the odds are 1/6.
In Scenario 2 the odds are 1/11.
Why are there different results?
In Scenario 1, there are no ground rules that the person had to look only for 2s. I.E. No rule saying that he must say there are no 2s as opposed to saying there is at least 1 of whatever value he happens to see. In this case, he sees a die which happens to be a 2 and reports that there's at least one 2. Therefore, there's a 1/6 chance the other die matches what he happens to see.
In Scenario 2, the odds are high (25/36) that the person will be forced to report that he sees no 2s. But that's the only option. 25 times there are no 2s and 11 times there is at least one 2. And we know that of those 11 times, there is only one combination of a pair of 2s. Therefore, if he's not saying there aren't any 2s, there's a 1/11 chance of it being a pair of 2s.
