Two Dice Puzzle
April 20th, 2015 at 7:21:04 AM
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Quote: Ibeatyouraces... Call it the "Doubles before Singles" bet and pay it 8:1...
Let's exercise some care here.
- If it's Double-Deuces before Single-Deuce, I can't play.
- If it's Double-Anything before Non-Double, I must play table-max.
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 20th, 2015 at 7:22:51 AM
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Quote: indignant99Let's exercise some care here.
- If it's Double-Deuces before Single-Deuce, I can't play.
- If it's Double-Anything before Non-Double, I must play table-max.
It's a work in progress. :-)
It'll be in the center of the table within a few years.
DUHHIIIIIIIII HEARD THAT!
April 20th, 2015 at 7:58:43 AM
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Quote: IbeatyouracesJust to be clear.
Roll BOTH DICE and only count the times a 2 shows up, 1/11 will be 2-2
And that's exactly what the OP question is asking. You are required to roll both dices and then looking at the outcome and determine if one 2 shows, whats the probability that both are 2's.
Some people are incapable of interpreting or thinking through this correctly and went to make a statement that the OP question is equivalent to "setting one dice as 2 and rolling the second dice only".
The only ambiguity in the original question is whether the peeker says something each time if there is no two showing (e.g." there is at least one 6 showing, what is the probability both are 6's", when 6x is the outcome) or not. That was covered in MangoJ's answer.
I think someone tried to bring some traffic to their site from this board and was trolling/misunderstanding the simple question all the way through.They even didn't disputed the no two scenario, but simply were incapable to distinguish "at least one of the two freshly rolled dice is showing a 2" from "one dice is set to two,we now roll only the other dice" despite numerous demonstrations. This sort of people do not deserve any attention and/or explanations as they are wearing blinkers.
Don't beat yourself up over past mistakes, you are going to f*** up again in the future, quite possibly in the most spectacular fashion, why worry about yesterday's f*** up's when you have tomorrow's f*** up's to look forward to?
You are a f*** up, and f***** up is part of your growth process, embrace the process.
April 20th, 2015 at 8:10:39 AM
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Quote: rawtuff...This sort of people do not deserve any attention and/or explanations as they are wearing blinkers.
Agreed and why I'd never join there. You know the old saying about a horse and water.
DUHHIIIIIIIII HEARD THAT!
April 20th, 2015 at 8:15:14 AM
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One more thing. Why in the world would they think we'd give them the wrong answer???
DUHHIIIIIIIII HEARD THAT!
April 20th, 2015 at 8:22:22 AM
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Quote: IbeatyouracesOne more thing. Why in the world would they think we'd give them the wrong answer???
Well, not for fraudulent/deceitful purposes - since they have no money at risk to us. But, because, obviously, we're all dullards, and they're brilliant. Dunning-Kruger effect.
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 20th, 2015 at 10:33:08 AM
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Quote: IbeatyouracesIf this bet paid 8:1, what would the house edge be? 20%ish?
Still would like to know the answer to this.
DUHHIIIIIIIII HEARD THAT!
April 20th, 2015 at 10:38:47 AM
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Quote: IbeatyouracesStill would like to know the answer to this.
Win 8 lose 10. +2/11 = ~ 18.18% edge on resolution. Or 2/36 = ~ 5.55% per roll.
For the banker, it's a bit different, since banker is wagering $800 to win $100 10/11's of the time and lose $800 1/11 of the time. So ( +1000 -800 ) / (11 * 800) = 2.27% (on resolution).
April 20th, 2015 at 10:41:23 AM
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Quote: RSWin 8 lose 10. +2/11 = ~ 18.18% edge on resolution. Or 2/36 = ~ 5.55% per roll.
Well hell, give em 9:1 then! Probably still a sucker bet.
DUHHIIIIIIIII HEARD THAT!
April 20th, 2015 at 10:59:50 AM
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Quote: IbeatyouracesStill would like to know the answer to this.
Hi IBY.
I'm a bit tired to do math, but estimate it with my spreadsheet Wizard Bet Sim (AKA AlansFolly.xlsx
If edge = (action+profit)/action, then simulation over 50,000 rolls seems to give 18.2% near as dammit.
I'd need to remind myself if that's how it's calculated, so don't take it as gospel.
It is actually straightforward enough to Calculate HE from first principals. To be honest, I cannot be bothered right now.
Layout might get splattered pasting it here, if so, I'll mop it up later
Count of Pairs of Deuces 1395
Count of "At Least One Deuce" Showing 15346
Count of I pay you £8 1395
Count of You pay me £1 ( You don't pay when a pair) 13951
Total That is awarded to You at £8 Per Pair of Deuces £11160
Total That is awarded to Me at £1 per "A Deuce is Showing But Not A Pair" £13951
My Profit £2791
Ratio of "Deuce" to "Pair" 11.0007
Count of DieA=2 (as a sanity Check) 8407
Count of DieB=2 (as a sanity check) 8334
Count of Simulations 50,000
1-(Action-house profit)/Action = 1-(15346-2791)/15346 = 18.19%: I'll happily fix that if my HE calculation is wrong.
I subsequently checked and believe that is 'house edge on resolution' and i should have had pushes in the denominator for convention.
So HE=1-(Action including pushes-house profit)/Action including pushes = 1-(50000-2791)/50000 = 5.582%: I'll happily fix that if my HE calculation is wrong.
Psalm 25:16
Turn to me and be gracious to me, for I am lonely and afflicted.
Proverbs 18:2
A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
April 20th, 2015 at 11:03:08 AM
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RS got it. With this edge, you guys can't afford not to get Alan and his crew betting on this!
If I were in California, I'd quickly drain his bankroll on it. Minimum 1000 dice rolls.
There's nothing like taking money from loud mouths who think they're smarter than the math.
If I were in California, I'd quickly drain his bankroll on it. Minimum 1000 dice rolls.
There's nothing like taking money from loud mouths who think they're smarter than the math.
DUHHIIIIIIIII HEARD THAT!
April 20th, 2015 at 12:53:21 PM
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Let's use Alan's example of one each Settled and Spinner dice.
- Probability the Settled Die shows Ace? ___1/6
- Probability the Settled Die shows Deuce? _1/6
- Probability the Settled Die shows Trey? __1/6
- Probability the Settled Die shows Four? __1/6
- Probability the Settled Die shows Five? __1/6
- Probability the Settled Die shows Six? ___1/6
- When the Settled Die shows Ace...
- Probability the Spinner lands Ace? ___1/6
- Probability the Spinner lands Deuce? _1/6 <-- A single Deuce
- Probability the Spinner lands Trey? __1/6
- Probability the Spinner lands Four? __1/6
- Probability the Spinner lands Five? __1/6
- Probability the Spinner lands Six? ___1/6
-
Overall, 1/36 for each outcome
{1-1, 1-2, 1-3, 1-4, 1-5, 1-6}.
- When the Settled Die shows Deuce...
- Probability the Spinner lands Ace? ___1/6 <-- A single Deuce
- Probability the Spinner lands Deuce? _1/6 <-- A double Deuce
- Probability the Spinner lands Trey? ___1/6 <-- A single Deuce
- Probability the Spinner lands Four? ___1/6 <-- A single Deuce
- Probability the Spinner lands Five? ___1/6 <-- A single Deuce
- Probability the Spinner lands Six? ____1/6 <-- A single Deuce
-
Overall, 1/36 for each outcome
{2-1, 2-2, 2-3, 2-4, 2-5, 2-6}.
-
When the Settled Die shows Trey...
- Probability the Spinner lands Ace? ___1/6
- Probability the Spinner lands Deuce? _1/6 <-- A single Deuce
- Probability the Spinner lands Trey? __1/6
- Probability the Spinner lands Four? __1/6
- Probability the Spinner lands Five? __1/6
- Probability the Spinner lands Six? ___1/6
-
Overall, 1/36 for each outcome
{3-1, 3-2, 3-3, 3-4, 3-5, 3-6}.
-
When the Settled Die shows Four...
- Probability the Spinner lands Ace? ___1/6
- Probability the Spinner lands Deuce? _1/6 <-- A single Deuce
- Probability the Spinner lands Trey? __1/6
- Probability the Spinner lands Four? __1/6
- Probability the Spinner lands Five? __1/6
- Probability the Spinner lands Six? ___1/6
-
Overall, 1/36 for each outcome
{4-1, 4-2, 4-3, 4-4, 4-5, 4-6}.
-
When the Settled Die shows Five...
- Probability the Spinner lands Ace? ___1/6
- Probability the Spinner lands Deuce? _1/6 <-- A single Deuce
- Probability the Spinner lands Trey? __1/6
- Probability the Spinner lands Four? __1/6
- Probability the Spinner lands Five? __1/6
- Probability the Spinner lands Six? ___1/6
-
Overall, 1/36 for each outcome
{5-1, 5-2, 5-3, 5-4, 5-5, 5-6}.
-
When the Settled Die shows Six...
- Probability the Spinner lands Ace? ___1/6
- Probability the Spinner lands Deuce? _1/6 <-- A single Deuce
- Probability the Spinner lands Trey? __1/6
- Probability the Spinner lands Four? __1/6
- Probability the Spinner lands Five? __1/6
- Probability the Spinner lands Six? ___1/6
-
Overall, 1/36 for each outcome
{6-1, 6-2, 6-3, 6-4, 6-5, 6-6}.
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 20th, 2015 at 1:18:13 PM
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How many "hop" bets are on a craps table?
How many of those are double-deuce?
How many of those are single-deuce? They pay 1/2 as much. Why?
How many of those are double-deuce?
How many of those are single-deuce? They pay 1/2 as much. Why?
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 20th, 2015 at 3:15:49 PM
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For Alan...
How many dice-rolls are possible with 2 dice? (Craps table or not, doesn't matter.)
^ How many of these will have a deuce (or pair of deuces) in them?
_ _ _ _ ^ (Does the spy get to announce "deuce" for all of them?)
_ ^ How many of these have two deuces in them?
How many dice-rolls are possible with 2 dice? (Craps table or not, doesn't matter.)
^ How many of these will have a deuce (or pair of deuces) in them?
_ _ _ _ ^ (Does the spy get to announce "deuce" for all of them?)
_ ^ How many of these have two deuces in them?
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 20th, 2015 at 3:27:37 PM
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FFS Indignant, Alan has failed to answer questions that are much much simpler than these.
He will not answer your questions.
He is incapable of doing so because it is irrelevant.
And Yes. I realise that that assertion is absurd.
He will not answer your questions.
He is incapable of doing so because it is irrelevant.
And Yes. I realise that that assertion is absurd.
Psalm 25:16
Turn to me and be gracious to me, for I am lonely and afflicted.
Proverbs 18:2
A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
April 20th, 2015 at 3:40:53 PM
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There is only one true die.
April 20th, 2015 at 3:49:23 PM
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Over on his forum, it looks like he's slowly thawing. I won't join over there, because I will not divulge my e-mail address, nor any other info, to him.
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 20th, 2015 at 3:54:09 PM
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Indignant, forever the optimist matey :o)
I was an optimist too. Now I'm disapointed that I had to break a personal rule. I repeated myself three times with one simple question. I should have realised that I was wasting my breath and put my time to better use.
I was an optimist too. Now I'm disapointed that I had to break a personal rule. I repeated myself three times with one simple question. I should have realised that I was wasting my breath and put my time to better use.
Psalm 25:16
Turn to me and be gracious to me, for I am lonely and afflicted.
Proverbs 18:2
A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
April 20th, 2015 at 4:41:25 PM
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I can't fathom how hard it it's to understand that you can roll 1-2, 2-1, 2-2, 2-3, 3-2, 2-4, 4-2, 2-5, 5-2, 2-6, and 6-2. That's ELEVEN possible combos, only which ONE CAN BE 2-2. Just pathetic!!
On each roll. AT LEAST ONE DIE IS A 2!!!
On each roll. AT LEAST ONE DIE IS A 2!!!
DUHHIIIIIIIII HEARD THAT!
April 20th, 2015 at 4:48:55 PM
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One more thing. I'm far from being a "math boy" and even I can understand this.
DUHHIIIIIIIII HEARD THAT!
April 20th, 2015 at 6:49:17 PM
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I realize that there probably isn't anyone left reading who doesn't believe at this point, but I came up with an explaination using Calculatus Eliminatus, in other words, the process of elimination.
I'll start with the original question:
Now, little bits at a time.
I think everyone agrees on that.
This is our starting condition. Two dice, shaken and rolled, now unmoving, and hidden from view.
Potentially, the first point of contention. I think right here, before you peek, there is an equal 1 in 36 chance that any one of the two dice arrangements could have landed under the cup. namely:
1-1
1-2 2-1
1-3 2-2 3-1
1-4 2-3 3-2 4-1
1-5 2-4 3-3 4-2 5-1
1-6 2-5 3-4 4-3 5-2 6-1
2-6 3-5 4-4 5-3 6-2
3-6 4-5 5-4 6-3
4-6 5-5 6-4
5-6 6-5
6-6
that arrangement should look familiar to craps players. 36 ways the dice can land, with equal probability.
If you don't think all 36 of those can be under the cup before we peek, under the terms of the question, please explain why not.
this is our main point of contention:
At this point, I eliminate all of the possibilities that do not have a 2. They do not meet the condition that "at least one of the dice is a two"
I marked the remaning possibilities in bold. All of the remaining die rolls fulfill the condition that "at least one of the dice is a 2"
With the informaiton given in the question, I don't have any more information that I can use to eliminate any more of the possibilities.
Of the 11 possibilities that are left, only one of them is 2-2, so the probability that both dice show a 2 is 1 in 11.
To get to 1/6, you need to eliminate 5 more of the possible rolls. Which ones do you eliminate, and why?
The people who say the answer is one in 6 say that under the cup, there must be only: 2-1, 2-2, 2-3, 2-4, 2-5, 2-6.
Or perhaps 1-2, 2-2, 3-2, 4-2, 5-2, 6-2. I'm not sure which.
I'd like to know how you eliminate the remaining possibilities that meet both the condition that at least one of the dice is a two, and both dice started shaken and hidden from view under a cup.
I'll listen to explainations as to why all 36 possibilities couldn't be under the cup before you peek, too. Please don't forget the starting condition that "You shake the dice, and slam the cup down onto the table, hiding the result."
As an aside, I think another way to say that "at least one of the dice is a 2" is "this roll of the dice has resolved the bet" that is to say, the bet where you lose on one two and win on double twos.
I'll start with the original question:
Quote: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?
Now, little bits at a time.
Quote:You have two 6-sided dice in a cup.
I think everyone agrees on that.
Quote:You shake the dice, and slam the cup down onto the table, hiding the result.
This is our starting condition. Two dice, shaken and rolled, now unmoving, and hidden from view.
Potentially, the first point of contention. I think right here, before you peek, there is an equal 1 in 36 chance that any one of the two dice arrangements could have landed under the cup. namely:
1-1
1-2 2-1
1-3 2-2 3-1
1-4 2-3 3-2 4-1
1-5 2-4 3-3 4-2 5-1
1-6 2-5 3-4 4-3 5-2 6-1
2-6 3-5 4-4 5-3 6-2
3-6 4-5 5-4 6-3
4-6 5-5 6-4
5-6 6-5
6-6
that arrangement should look familiar to craps players. 36 ways the dice can land, with equal probability.
If you don't think all 36 of those can be under the cup before we peek, under the terms of the question, please explain why not.
Quote:Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."
this is our main point of contention:
At this point, I eliminate all of the possibilities that do not have a 2. They do not meet the condition that "at least one of the dice is a two"
I marked the remaning possibilities in bold. All of the remaining die rolls fulfill the condition that "at least one of the dice is a 2"
With the informaiton given in the question, I don't have any more information that I can use to eliminate any more of the possibilities.
Quote:What is the probability that both dice are showing a 2?
Of the 11 possibilities that are left, only one of them is 2-2, so the probability that both dice show a 2 is 1 in 11.
To get to 1/6, you need to eliminate 5 more of the possible rolls. Which ones do you eliminate, and why?
The people who say the answer is one in 6 say that under the cup, there must be only: 2-1, 2-2, 2-3, 2-4, 2-5, 2-6.
Or perhaps 1-2, 2-2, 3-2, 4-2, 5-2, 6-2. I'm not sure which.
I'd like to know how you eliminate the remaining possibilities that meet both the condition that at least one of the dice is a two, and both dice started shaken and hidden from view under a cup.
I'll listen to explainations as to why all 36 possibilities couldn't be under the cup before you peek, too. Please don't forget the starting condition that "You shake the dice, and slam the cup down onto the table, hiding the result."
As an aside, I think another way to say that "at least one of the dice is a 2" is "this roll of the dice has resolved the bet" that is to say, the bet where you lose on one two and win on double twos.
April 20th, 2015 at 8:06:12 PM
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Let's lower the temperature, lest the suspensions start flying.
Again, I invite anybody not in the 1/11 camp to come to MonekyFest with plenty of money. I'll bring the dice.
Again, I invite anybody not in the 1/11 camp to come to MonekyFest with plenty of money. I'll bring the dice.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
April 20th, 2015 at 9:07:24 PM
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Quote: WizardLet's lower the temperature, lest the suspensions start flying.
Again, I invite anybody not in the 1/11 camp to come to MonekyFest with plenty of money. I'll bring the dice.
Wizard I might have some customers for you. But please describe the process of the bet exactly. For example, will two dice be rolled simultaneously? Is the pay 8-to-1 or 8-for-1, and other info. Thanks.
April 20th, 2015 at 9:14:05 PM
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Quote: Wizard... Again, I invite anybody not in the 1/11 camp to come to MonekyFest with plenty of money. I'll bring the dice.
Should you answer Alan, be a sport and offer 9:1 (that's 9-to-1, not 9-for-1).
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 20th, 2015 at 9:14:19 PM
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Two dice simultaneously and 8 to 1.
Do you seriously think we'd bet this at a disadvantage?????
Do you seriously think we'd bet this at a disadvantage?????
DUHHIIIIIIIII HEARD THAT!
April 20th, 2015 at 9:20:58 PM
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Two dice simultaneously means 2-2 is a 1/36 event. And you are paying 8-to-1 ? I doubt there will be any takers.
April 20th, 2015 at 9:23:05 PM
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Quote: AlanMendelsonTwo dice simultaneously means 2-2 is a 1/36 event. And you are paying 8-to-1 ? I doubt there will be any takers.
Any non 2 combo does not count. Only combos with a 2 in them.
DUHHIIIIIIIII HEARD THAT!
April 20th, 2015 at 9:24:54 PM
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Alan, it's not going to be a one-roll proposition. The felt already has that... Craps-2 Hard four "on the hop." Oops.
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 20th, 2015 at 9:32:39 PM
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Quote: IbeatyouracesAny non 2 combo does not count. Only combos with a 2 in them.
And so there are ten ways to lose but only one way to win? The one way to win is a 1/36 event and it pays 8-to-1. No thanks.
April 20th, 2015 at 9:35:00 PM
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Quote: AlanMendelsonAnd so there are ten ways to lose but only one way to win? The one way to win is a 1/36 event and it pays 8-to-1. No thanks.
Again, roll anything that doesn't contain a two, and it's a push. You'd only lose by rolling 2-1, 2-3, 2-4, 2-5, or 2-6. I'm sure it'll be set up with a cup and a peeker too.
Do you now concede it's 1/11?
DUHHIIIIIIIII HEARD THAT!
April 20th, 2015 at 10:01:39 PM
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It should be very clear now, that whether you roll the dice in the open or in a cup and slam the cup upside down, the odds of 2-2 showing when at least one is a 2 is 1 in 11. Not 1 in 6. Only a fool would take the 8 to 1 bet no matter which way you rolled them.
DUHHIIIIIIIII HEARD THAT!
April 20th, 2015 at 10:03:01 PM
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Quote: AlanMendelsonAnd so there are ten ways to lose but only one way to win? The one way to win is a 1/36 event and it pays 8-to-1. No thanks.
The "one way to win" isn't a 1/36 event. 25 of those events DO NOT LOSE, they push.
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 20th, 2015 at 10:04:16 PM
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Quote: indignant99The "one way to win" isn't a 1/36 event. 25 of those events DO NOT LOSE, they push.
Yellow journalism?
DUHHIIIIIIIII HEARD THAT!
April 20th, 2015 at 10:12:27 PM
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Huh? Explain.Quote: IbeatyouracesYellow journalism?
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 20th, 2015 at 10:22:52 PM
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Quote: indignant99The "one way to win" isn't a 1/36 event. 25 of those events DO NOT LOSE, they push.
The pushes don't matter. There is still only one way to win and 10 ways to lose. And the win is paid 8-to-1. You can debate this with the not-so-smart members of my forum.
April 20th, 2015 at 10:24:24 PM
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Quote: IbeatyouracesYellow journalism?
Yes please explain? Did you get this wrong, too?
April 20th, 2015 at 10:35:51 PM
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Quote: AlanMendelsonThe pushes don't matter. There is still only one way to win and 10 ways to lose. And the win is paid 8-to-1. You can debate this with the not-so-smart members of my forum.
Funny you say they don't matter when rolled in the open. Do they when unseen under the cup? Because even under the cup, there are 10 ways to lose and only 1 to win. All of which show "at least one 2"
DUHHIIIIIIIII HEARD THAT!
April 20th, 2015 at 10:56:53 PM
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Quote: IbeatyouracesFunny you say they don't matter when rolled in the open. Do they when unseen under the cup? Because even under the cup, there are 10 ways to lose and only 1 to win. All of which show "at least one 2"
That wasn't the question either. The question was... (help me Lord, help me!!)
But what's the yellow journalism comment all about?
April 20th, 2015 at 11:20:07 PM
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Quote: AlanMendelsonThe pushes don't matter. There is still only one way to win and 10 ways to lose. And the win is paid 8-to-1. You can debate this with the not-so-smart members of my forum.
This statement can be interpreted as ALL of the members of your forum are not-so smart or SOME of the members of your forum are not-so smart. Either way nice to know what you think of your own forum members.
April 20th, 2015 at 11:29:10 PM
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Quote: rudeboyoiThis statement can be interpreted as ALL of the members of your forum are not-so smart or SOME of the members of your forum are not-so smart. Either way nice to know what you think of your own forum members.
I've been over to Alan's forum, and some members are pretty damn smart. Now I ask you, rudeboyoi, turn your magnifying glass back to THIS FORUM. Whaddaya see?
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 21st, 2015 at 12:31:49 AM
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Quote: indignant99I've been over to Alan's forum, and some members are pretty damn smart. Now I ask you, rudeboyoi, turn your magnifying glass back to THIS FORUM. Whaddaya see?
I didn't actually look at his forum. I just found it humorous that he was arguing over the wording of this problem then chose to use poor wording in a statement about his own forum which could be interpreted as EVERYONE on his forum was not-so smart.
April 21st, 2015 at 12:34:25 AM
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Quote: rudeboyoiThis statement can be interpreted as ALL of the members of your forum are not-so smart or SOME of the members of your forum are not-so smart. Either way nice to know what you think of your own forum members.
No. Too many of the members of this forum have come over to my forum to tell us that we are not so smart. So, I concede -- we are not so smart. Someday maybe we'll be as smart as you guys over here.
But we're like Avis. We keep trying.
April 21st, 2015 at 1:52:27 AM
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I think Alan's use of the adjective "not-so-smart" was a teeny bit sarcastic.
I think rudeboyoi's interpretation of the comment was a poker-raise of sarcasm.
I think Alan matched/"saw" your raise with "you guys over here."
What I have been trying to do (not always completely politely, but not derogatorily either), is find an angle-of-approach to the Puzzle/Solution that made sense to Alan. (In no way did I try to obfuscate language, or twist logic, as accused by some of Alan's members.) It seems we're on the verge of clarity/understanding. But that won't actually happen, if Alan says, yet again, "that's not the problem as stated."
I think rudeboyoi's interpretation of the comment was a poker-raise of sarcasm.
I think Alan matched/"saw" your raise with "you guys over here."
What I have been trying to do (not always completely politely, but not derogatorily either), is find an angle-of-approach to the Puzzle/Solution that made sense to Alan. (In no way did I try to obfuscate language, or twist logic, as accused by some of Alan's members.) It seems we're on the verge of clarity/understanding. But that won't actually happen, if Alan says, yet again, "that's not the problem as stated."
Yeah, I made a mistake once. I thought I was wrong, when I actually wasn't. -Indignant
April 21st, 2015 at 1:56:05 AM
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indignant99... I never want to be accused of disappointing anyone. That's not the problem as stated.
April 21st, 2015 at 2:39:27 AM
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Quote: AlanMendelsonThe pushes don't matter. There is still only one way to win and 10 ways to lose. And the win is paid 8-to-1. You can debate this with the not-so-smart members of my forum.
I think you solved the mystery: it's 1/11.
April 21st, 2015 at 3:36:57 AM
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Without looking at spoilers:
1.) There is no restriction on the colors of the dice. Make one Red and one Green identical in all other respects.
2.) Red 2 has six possible outcomes
3.) Green 2 has 5 possible outcomes (2-2 is already counted)
4.) Total chances are 10 ways of 1 '2', and 1 way of 2-2. Total 11/36 rolls will have "at least one 2".
5.) Chance of 2-2 if "at least one die is a 2" is 1 in 11.
1.) There is no restriction on the colors of the dice. Make one Red and one Green identical in all other respects.
2.) Red 2 has six possible outcomes
3.) Green 2 has 5 possible outcomes (2-2 is already counted)
4.) Total chances are 10 ways of 1 '2', and 1 way of 2-2. Total 11/36 rolls will have "at least one 2".
5.) Chance of 2-2 if "at least one die is a 2" is 1 in 11.
Some people need to reimagine their thinking.
April 21st, 2015 at 3:49:32 AM
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Doesn't a proposed bet like bellow follow the original question EXACTLY ? -
A person puts two dices in a cup, shake and slam the cup on the table.
A second person peeks under the cup and truthfully announces if there is at least one deuce showing or not (he will later lift the cup and show to all he wasn't lying).
If "at least one deuce" is announced, the 1/6-ers will wager 1 unit on the premise that there are two deuces under the cup( a one in six probability according to them). If they're right, they get 9 units back. If they're wrong, they lose the 1 unit wagered.
Repeat.
A person puts two dices in a cup, shake and slam the cup on the table.
A second person peeks under the cup and truthfully announces if there is at least one deuce showing or not (he will later lift the cup and show to all he wasn't lying).
If "at least one deuce" is announced, the 1/6-ers will wager 1 unit on the premise that there are two deuces under the cup( a one in six probability according to them). If they're right, they get 9 units back. If they're wrong, they lose the 1 unit wagered.
Repeat.
Don't beat yourself up over past mistakes, you are going to f*** up again in the future, quite possibly in the most spectacular fashion, why worry about yesterday's f*** up's when you have tomorrow's f*** up's to look forward to?
You are a f*** up, and f***** up is part of your growth process, embrace the process.
April 21st, 2015 at 6:03:34 AM
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Quote: AlanMendelsonThe one way to win is a 1/36 event and it pays 8-to-1. No thanks.
Then how about offering me only 20 to 1?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
April 21st, 2015 at 8:14:33 AM
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Quote: AlanMendelsonThe pushes don't matter. There is still only one way to win and 10 ways to lose. And the win is paid 8-to-1. You can debate this with the not-so-smart members of my forum.
To swing me to your side Alan explain to me how having the cup remain on the dice and peeking changes the odds to 6-1 when leaving the cup off and looking the odds are 10-1 as you have just told us in this post.
Be careful when you follow the masses, the M is sometimes silent.
April 21st, 2015 at 1:42:58 PM
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Quote: rawtuffDoesn't a proposed bet like bellow follow the original question EXACTLY ? -
A person puts two dices in a cup, shake and slam the cup on the table.
A second person peeks under the cup and truthfully announces if there is at least one deuce showing or not (he will later lift the cup and show to all he wasn't lying).
If "at least one deuce" is announced, the 1/6-ers will wager 1 unit on the premise that there are two deuces under the cup( a one in six probability according to them). If they're right, they get 9 units back. If they're wrong, they lose the 1 unit wagered.
Repeat.
Yes, I think this bet does follow the original question and the claim that the correct answer is 1/11.
Would anyone like to offer this bet?
