Two Dice Puzzle
Quote: AceTwoIt is remarkable that a simple probability problem which is very well stated (no ambiguties) and an obvious answer of 1/11 has a thread of 15 pages with a few people arguing that it is 1/6.
To be fair, I think we are pretty much down to one person insisting on 1/6 : Three cheers for Alan, who will kick himself when he sees the light.
Now you've blown it. Alan will say you are simply overcomplicating thingsQuote:This is the kind of problems in basic probability courses that test the understanding of conditional probabilities and dependent events.
The people who say 1/6 think that the following 2 problems are equivalent.
1. 2 Dice are rolled. At LEAST one of the dice is 2. What is the probability the both dice are 2.
2. 1 Dice is rolled. What is the probability that it is 2.
The 2 problems are NOT equivalent.
Yep. Now you have definitely made it too complicated for the sixthers $:o)Quote:
The first problem becomes equivalent to the second ONLY IF the one dice is shown to you and it is a 2 and then asked what is the probability that the other unseen dice is a 2. Then its is 1/6.
ALTERNATIVE PRESENTATION OF THE PROBLEM.
As stated BUT one dice is randomly chosen out of the cup and you are asked the probability of the specific dice being a 2.
IF you have zero information then the answer is 1/6.
With the information as stated in the problem, the specific dice chosen has probability of 6/11 of being a 2.
Not 1/6 as a random dice, Not 100% for the dice that is a 2 but not known which one, and NOT 1/2 (1/2 would be if the problem stated that one and only one dice is a 2).
The same probability of 6/11 applies to the other not chosen dice.
In independent events you would multiply the 2 probs, ie 6/11 * 6/11 = 36/121 for both dice to be 2.
But the 2 dice are not independent
The probability is then calculated as 6/11 * 1/6 = 1/11
Make two columns: A and B
Roll two dice.
If you don't see a two, do nothing.
If you see a two, put a mark in column A
If you see double twos, ALSO put a mark in column B
Repeat 1000 times
The ratio of marks in column B to Column A shold be close to 1/11. You might expect it to be 1/6
Now, that will take too long, so we can do the same thing with two-sided dice, i.e. coins
Throw the two coins
If you see two tails, do nothing
If you see a head, put a mark in column A
If you see two heads, ALSO put a mark in column B
You won't have to repeat this one so many times, pretty quickly a ratio of 1 to 3 (NOT 1 to 2) will be apparent.
Want to use dice for the same ratio?
Roll two dice
If you see all even numbers, do nothing
If you see at least one odd number, mark col A
If you see two odd numbers, also mark col B
This will also approach 1/3 pretty quickly.
If the math isn't good enough, the simulation isn't good enough, maybe you'll take the time for "seeing is believing" by playing this game.
Not gonna happen. I love Alan but he's too stubtuse to ever concede, give in, or get it.
Quote: pewstubtuse
Thank you for this word. Ain't English grand.
Quote:Now if you want to think that way, that's fine. But if you were working for me I'd fire you. In fact, if I knew you even thought that way and ignored the basic information in the question, I wouldn't even hire you.
Why......
As a junior technician, many many moons ago, I had a senior manager get furious at me when I showed him how a simple electrical circuit consisting of a few lights and switches could not possibly work. I stood my ground at great peril and eventually built it for him to PROVE that he was wrong. No apology for his apoplectic fury. No thanks for trying to save time or money. The scars remain.
Being blind to an articulately argued and reasoned explanation is unfortunate. Being deaf to the truth and not listening to the truth are slightly different.
Back to the OP.... the 1/11 is counter intuitive, even I was thrown when it was later suggested that we extract the deuce as it was discovered. But the simplest, and I mean THE SIMPLEST list of outcomes made it crystal clear what was wrong with Alan's and Aycarumba's model of the question.
Grasp the nettle Alan. Work through all of the proofs and show us in one tiny way where they are wrong and you are right. Better still do the honourable thing and just accept that you don't understand.
I'm hoping to get it in Miriam WebsterQuote: mrcleanThank you for this word. Ain't English grand.
Quote: mrcleanThank you for this word. Ain't English grand.
An anagram for USS Butte,
Quote: mrcleanThank you for this word. Ain't English grand.
I like that word too. It doesn't matter that it isn't considered a "real" word. We recognize what it means. Getting the message across is really all that's important about language and the more concise the better.
A made up word thread would be fun.
When the dice are thrown on the table and the first one comes to rest on a 4 don't you say to yourself as you watch the other die "no 3, no 3"?
Of course, you will say that is not relevant to the original post. But let's once again examine the question in the original post:
Quote: DweenYou 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, you tell me why looking at just ONE DIE would not be the right way to answer THIS QUESTION ??
Let me give you a hint:
THE QUESTION: Dice in a cup are shaken and come to rest under a cup on the table.
REAL CRAPS: Dice are thrown on a table and come to rest.
THE QUESTION: The friend looks and says that at least one of the dice is a 2.
REAL CRAPS: One of the dice is visible and shows a 2. But the second die is behind a stack of player's chips. The stickman calls out "well one of them is a two, Rob (a base dealer) you call it!"
THE QUESTION: What is the probability that both dice are showing a 2?
REAL CRAPS: The die behind the stack of chips has only six sides. The probability is that 1/6 is another 2, and 1/6 is a 5 for a 7-out.
How do you get 1/11 ???
Quote: AlanMendelsonHow do you get 1/11 ???
The unseen die may be the single(?) 2 as well as the seen one, which would bring five more non-2/2 possibilities in the seen one whatever it is now. "At least one 2" connects the two scenarios into one. See the diagram on pg 3 (?).
Quote: DweenAlong the vein of the Two Coin Puzzle, I wanted to put this in another thread as to not clog up the other.
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?
Based on the discussion in the Two Coin thread, I'm interested to see how people solve this one.
Is there any merit to this approach:
Prior to peeking, or even after peeking the question is posed:
Q: "If at least one of the dice is a two, what is the probability that both dice are showing a two?"
A: 1/11 - As stated and proven in many previous posts.
However, that is not the way the question was posed. There is no ambiguity about the value of one of the dice. It is a two. Under these unchanging conditions, the question is then posed, "One die is a two, what is the probability that the other is also a two?" There is no "If".
If we played the wagering game according the original post, we would do it as follows:
1) Take one die out and place it aside with a two showing. You set it this way because this was the final condition in the original post, just prior to posing the pair probability problem. The question was not asked before the value of one die was already established; the die with the two is not a random variable, it is fixed.
2) Shake the other die in the cup.
3) Reveal. Theoretically, one time in six it will be a two.
The key to me is what facts are available when the question is posed.
Answer: 11 ways.
There are 11 ways to roll one or two twos on two dice.
Of the 11 possible ways that you can possibly roll at least one two, only one of those ways is 2-2.
On the craps table, if you throw two dice, one of them lands on a two where you can see it, and one of them lands out of sight - the odds of the one out of sight being a two is 1/11
Because guess what - there are 11 combinations of dice that have one or two twos in them, and you just rolled one of those 11. Only one of the 11 is 2-2, so the odds of fhe hidden die being a 2 is 1/11.
Why don't you try it out?
Quote: KerkebetThe unseen die may be the single(?) 2 as well as the seen one, which would bring five more non-2/2 possibilities in the seen one whatever it is now. "At least one 2" connects the two scenarios into one. See the diagram on pg 3 (?).
ridiculous response.
The friend saw at least one die with a 2. And the dealer saw at least one die with a 2.
The final outcome now rests with the other die under the cup, or the die behind the stack of chips on the table.
Both the die under the cup, and the die behind the stack of chips has six sides.
Quote: jml24...none of the 1 in sixers are willing to bet money...
You missed my message... I AM WILLING, EVEN EAGER.
Quote: AyecarumbaIs there any merit to this approach:
Prior to peeking, or even after peeking the question is posed:
Q: "If at least one of the dice is a two, what is the probability that both dice are showing a two?"
A: 1/11 - As stated and proven in many previous posts.
However, that is not the way the question was posed. There is no ambiguity about the value of one of the dice. It is a two. Under these unchanging conditions, the question is then posed, "One die is a two, what is the probability that the other is also a two?" There is no "If".
If we played the wagering game according the original post, we would do it as follows:
1) Take one die out and place it aside with a two showing. You set it this way because this was the final condition in the original post, just prior to posing the pair probability problem. The question was not asked before the value of one die was already established; the die with the two is not a random variable, it is fixed.
2) Shake the other die in the cup.
3) Reveal. Theoretically, one time in six it will be a two.
The key to me is what facts are available when the question is posed.
That is changing the rules.
To me the only ambiguous part, because it isn't explicitly stated, is what you do when neither die is a two.
I say you roll the dice again, together, in the same cup, at the same time, until at least one of them is a two. Only then do you ask the question.
Quote: Dalex64
On the craps table, if you throw two dice, one of them lands on a two where you can see it, and one of them lands out of sight - the odds of the one out of sight being a two is 1/11
Are you serious?
Hey buddy, there are only six sides on a die, and the six sides are numbered: 1, 2, 3, 4, 5, 6. That die that's out of sight can be ONE of those SIX numbers.
Get real.
Quote: indignant99You missed my message... I AM WILLING, EVEN EAGER.
If you are going to bet, you better agree on the rules first. What do you do when no two is rolled?
You don't even need a cup to hide the dice for this game. Everyone can see, so no question of fairness.
Just throw them down the table. How many times, when you roll at least one two, do you get doubles?
I'm trying to figure out how to monitize this bet. Do you bet something somehow on each toss, or do you just bet that the number of doubles thrown will be over or under some number relative tothe number of times that any two was thrown?
Unfortunately I won't be in Vegas any time soon.
Quote: AlanMendelsonAre you serious?
Hey buddy, there are only six sides on a die, and the six sides are numbered: 1, 2, 3, 4, 5, 6. That die that's out of sight can be ONE of those SIX numbers.
Get real.
Open your mind and think.
Try it with two coins.
This is very easy to see.
Seeing is believing.
Quote: Dalex64Open your mind and think.
Try it with two coins.
This is very easy to see.
Seeing is believing.
Oh no. This is OVER. And it's all because of this ridiculous statement:
Quote: Dalex64
On the craps table, if you throw two dice, one of them lands on a two where you can see it, and one of them lands out of sight - the odds of the one out of sight being a two is 1/11
And here is a copy of my original response, so everyone can see this in one place:
Quote: AlanMendelsonAre you serious?
Hey buddy, there are only six sides on a die, and the six sides are numbered: 1, 2, 3, 4, 5, 6. That die that's out of sight can be ONE of those SIX numbers.
Get real.
Quote: IbeatyouracesYou guys are arguing with a guy who's only trusted gambling sources, at least in my mind, are Rob Singer and Frank Scoblete. Case closed.
Yeah, I've had enough.
There have been mathematical explainations, simulations, and descriptions of how to do several sorts of demonstrations.
Quote:Inconcievable!
Quote:You keep using that word. I do not think it means what you think it means.
I'm glad you finally agree.
Quote: Dalex64...What do you do when no two is rolled? ...
I will accept either of these rules:
- If you insist there be a deuce, then any "no-deuce" roll is a PUSH.
- Otherwise, just announce ONE of the pip-counts, and my bet stays live for a DOUBLE of it. I lose my "unit" wager upon non-double. You pay 8-to-1 upon double.
RS wanted to pay 7-to-1. I wanted 9-to-1. But I will compromise at 8.
Quote: indignant99I will accept either of these rules:
- If you insist there be a deuce, then any "no-deuce" roll is a PUSH.
- Otherwise, just announce ONE of the pip-counts, and my bet stays live for a DOUBLE of it. I lose my "unit" wager upon non-double. You pay 8-to-1 upon double.
RS wanted to pay 7-to-1. I wanted 9-to-1. But I will compromise at 8.
Dude. Alan. Just roll 2 dice a few times, and run this gamble with yourself. It's 1/11.
If I lived in Vegas I would come take your money, but I don't. That is all.
Quote: dwheatley... I would come take your money, but I don't. That is all.
Dude, I'm not Alan. What you don't realize, is that I'd be taking your money! The bet I have proposed is essentially a Field-of-Doubles. 5-to-1 against me, but you'd be paying me 8-to-1 upon my wins. Come on and focus!
You and Shackleford can act as "house," and I will bring however much money you think you can rake off of me. I'll have a purple-glowing Barney out there, every roll.
Quote: indignant99Dude, I'm not Alan. What you don't realize, is that I'd be taking your money! The bet I have proposed is essentially a Field-of-Doubles. 5-to-1 against me, but you'd be paying me 8-to-1 upon my wins. Come on and focus!
You and Shackleford can act as "house," and I will bring however much money you think you can rake off of me. I'll have a purple-glowing Barney out there, every roll.
Rules can be really simple don't even need to hide the dice just roll them.
No 2's is a push.
Any 2 you pay me $1.
Pair of 2's I pay you the $8 you asked for.
Quote: indignant99I will accept either of these rules:
- If you insist there be a deuce, then any "no-deuce" roll is a PUSH.
- Otherwise, just announce ONE of the pip-counts, and my bet stays live for a DOUBLE of it. I lose my "unit" wager upon non-double. You pay 8-to-1 upon double.
RS wanted to pay 7-to-1. I wanted 9-to-1. But I will compromise at 8.
(Up to) How much would you like to wager?
I like the above stipulations, but I'd like to add one: We have to play at least some certain amount of rolls (50, 100?). I don't want to get unlucky on the first few rolls, you quitting at that point when ahead, then me getting slaughtered because of it.
OK. You're on. How much and how shall we make it happen?Quote: indignant99You missed my message... I AM WILLING, EVEN EAGER.
Quote: AyecarumbaIs there any merit to this approach:
There is no ambiguity about the value of one of the dice. It is a two.
No No No.
On the contrary. THERE IS ambiguity about the value of ONE of the dice. Great ambiguity indeed.
There is even ambiguity about the values of BOTH of the dice! If we refer to them as 'The die on the Left' and 'The die on the right' and we know that 'At least one of them is a deuce' then 'The die on the Left' may be any of the values of one to six (while 'The die on the Right' is a Deuce) OR 'The die on the Right' may also be any value from one to six ( If it happens that the Die on the Left' is the deuce). We only know that 'At least one of the dice is a deuce.' We don't know which one. Try it. Get a friend to place two dice under a towel, with one to the left and one to the right. Tell that friend to place at least one of them with a Deuce showing but to swap them at random. How often can you reliably reach under the towel and withdraw the deuce, which by your assertion is the one you know to be a deuce? You'll get it right roughly half of the time and wrong roughly half of the time. This is why you and Alan cannot just set the first die aside and rely on the 'other die'.You don't know which die is which. It DOES matter.
That's before we even consider how often two deuces are placed showing.
Quote:
Under these unchanging conditions, the question is then posed, "One die is a two, what is the probability that the other is also a two?" There is no "If".
If we played the wagering game according the original post, we would do it as follows:
1) Take one die out and place it aside with a two showing. You set it this way because this was the final condition in the original post, just prior to posing the pair probability problem. The question was not asked before the value of one die was already established; the die with the two is not a random variable, it is fixed.
This breaks the rules because you have somehow determined which of the dice is a two before setting it aside. How did you do that without cheating.
Quote:
2) Shake the other die in the cup.
3) Reveal. Theoretically, one time in six it will be a two.
The key to me is what facts are available when the question is posed.
No question if you set aside the deuce and then throw the other die then 1/6 it is. But how did you get there? You pulled out the deuce from under the cup/towel in one action without knowing which of the dice to extract. You'd need to be psychic.
Quote: Dalex64
On the craps table, if you throw two dice, one of them lands on a two where you can see it, and one of them lands out of sight - the odds of the one out of sight being a two is 1/11
Because guess what - there are 11 combinations of dice that have one or two twos in them, and you just rolled one of those 11. Only one of the 11 is 2-2, so the odds of fhe hidden die being a 2 is 1/11.
Sadly, you let yourself down there Dalex. If the one that you saw land was a Deuce then the chances of the one out of sight being a deuce was 1/6. You've fallen into Alan's and Ayacarumba's mistake of changing the rules so that you know which of the dice is a deuce.
Quote: OnceDearSadly, you let yourself down there Dalex. If the one that you saw land was a Deuce then the chances of the one out of sight being a deuce was 1/6. You've fallen into Alan's and Ayacarumba's mistake of changing the rules so that you know which of the dice is a deuce.
You are so ridiculous. It doesn't matter which die is the two. There are only TWO dice in the problem. If it's DIE A showing a 2 then it's 1/6 for DIE B. If it's DIE B showing a 2, then it's 1/6 for DIE A. That's how it is when you have two six-sided dice.
Even if BOTH dice were showing 2s when the friend peeked at them under the cup, the correct answer would still be 1/6 because that's how it is with two six sided dice.
And what are the "rules" you speak of? There is no rule about which die the friend saw when he peeked under the cup. If there is a rule you created it AFTER the question was posted.
Once again, here is the original question:
Quote: DweenYou 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?
Tell me-- where is the rule that you speak of?
Stop inventing things.
And whatever math you guys are using to come up with 1/11 does not apply to two six-sided dice when one number is known on either die. And that is the bottom line.
I wasn't replying to you. Dalex made a mistake. He will surely acknowledge that later, unlike you.
I'm not going to engage you further on this. Unless and until you are prepared to work through the logic that has been presented to you dozens of times.
You have dug your heels in and you are so abjectly wrong that you are making yourself look ridiculous. Put up or shut up. Take someone up on a bet. Take me up on a bet if we can work out the logistics.
Quote: indignant99I will accept either of these rules:
- If you insist there be a deuce, then any "no-deuce" roll is a PUSH.
- Otherwise, just announce ONE of the pip-counts, and my bet stays live for a DOUBLE of it. I lose my "unit" wager upon non-double. You pay 8-to-1 upon double.
RS wanted to pay 7-to-1. I wanted 9-to-1. But I will compromise at 8.
'Just announce one of the pip counts' Needs to be more explicit, otherwise when a 1,2 comes up, your rule might allow the announcer to say "One of the dice is a one". That rule must be "'Otherwise just announce whether any twos are showing"
When it comes to a real 'money on the table' bet, we will need even stricter rules such as the results shall be taken from the top horizontal face of the two dice, ignoring what shows on the vertical faces.
Now Indignant. You can do maths.
If a pair of dice are thrown 1000000 times, how many times do you calculate that the result would be a pair of deuces. Calculate it out by any means at your disposal. Award yourself $8 dollars for each of those times.
Can you do that? Come back here with your calculations and result.
Now, slightly harder
If a pair of dice are thrown 1000000 times, how many times do you calculate that the result would have "At least one Deuce" showing face up. (Clue. It's not 1/6 and it's not 2/6) That's pretty explicit, but for the elimination of doubt, you would look at both dice and if EITHER or BOTH was showing a two, that would count as 'Deuce showing'. Calculate it out by any means at your disposal. Award me $1 dollars for each of those times. Magic the money out of thin air, as it's hypothetical at the moment. Do not take money out of either pile.
At the end of your calculation, how much money might I expect to have in my pile and how much money would you expect in your pile.
Show workings out.
Now, do the same where you award $6 for the 'Pair' results.
Now, do the same where you award $11 for the 'Pair' results.
Show all calculations please. It should be easy enough.
UNDER NO CIRCUMSTANCES just take one die and set it up as a deuce and throw ( or simulate) the other die.
This spreadsheet might help you. It's the most insanely simple simulation in Excel Simulation Spreadsheet for Potential Wager. AKA AllansFolly.xlsx
"Whenever I see any roll on the craps table that has a two in it, is it a pair of twos 1/6 of the time?"
And I just don't feel like I see a pair of twos every sixth time (on average) that I see a roll that has a two somewhere in it.
I'm sure I'll probably change my mind again before it's over with. ;)
Quote: blount2000I keep going back and forth between the 1/6 versus 1/11 in my mind, but one thing that makes me lean more towards the 1/11 is if I ask myself this question:
"Whenever I see any roll on the craps table that has a two in it, is it a pair of twos 1/6 of the time?"
And I just don't feel like I see a pair of twos every sixth time (on average) that I see a roll that has a two somewhere in it.
I'm sure I'll probably change my mind again before it's over with. ;)
Hi Blount. At least you are thinking. If you have Excel, download my spreadsheet linked above and let the proof be before your eyes 50000 simulated dice thrown. Results recorded and counted. Probabilities and 'paytable' calculated.
I will grant you that 1/11 is very VERY counterintuitive, but it is correct. I promise it's not magic or trickery or rocket science.
The 1/6 brigade (falling like flies) have it in their head that the outcome is decided by 'the other die' or 'the second die' and that they can extract and disregard the die that they know to be a two. They cannot. That is not the model in the OP. Not even close.
Quote: OnceDearDalex made a mistake. He will surely acknowledge that later, unlike you.
I'll go back on what I said about being done in this thread and respond, since you put this so politely and I made a mistake.
Yes, sorry. Since I have tried to explain this so many different ways from different scenarios, it was bound to happen.
I will explain my reasoning for my answer. It involves differentiating the dice in some way.
I will call the two dice "R" and "G"
I will label the dice in combinations always in RG order, so 13 represents R is a 1 and G is a 3.
So first we count the combinations where R is the die we see, and it is a two. G can be 1-6:
21 22 23 24 25 26
But G may have been the die that landed in view, and R the one that landed behind the chips, so we also have to count
12 22 32 42 52 62
We counted one of the combinations, 22, twice because of our counting method. We have to deduplicate the results.
That leaves 11 possibilities when we see a 2, only one of which is 22.
I can understand how it can be interpreted as R is the die you see, therefore G has a 1 in 6 chance of being a two. I just thought that was undercounting, since G also might be the die that you see, too.
So, sorry if I was wrong, I'd hate to add confusion and misinterpretation to the conversation.
Quote: Dalex64I'll go back on what I said about being done in this thread and respond, since you put this so politely and I made a mistake.
Yes, sorry.
Cheers. I knew you would do the right thing :o)
Indeed. No problem.Quote:it was bound to happen.
Superb. If we always see Die R and it's always a two, as per the Set aside the Die that is a Two because we know one die is a two then we do indeed see a 1/6 probability of the spinning 'other die' being a 2. No argument with that.Quote:I will call the two dice "R" and "G"
I will label the dice in combinations always in RG order, so 13 represents R is a 1 and G is a 3.
So first we count the combinations where R is the die we see, and it is a two. G can be 1-6:
21 22 23 24 25 26
But G may have been the die that landed in view, and R the one that landed behind the chips, so we also have to count
12 22 32 42 52 62
Quote:So, sorry if I was wrong.
That's how it's done Alan. Not so hard.
And just in case any idiot thinks that anyone is fessing up to being wrong about the correct answer to the OP. Thay are not. The correct answer remains 1/11 as proved many times to those with eyes to see..
Quote: AlanMendelson... ridiculous response.
If it's any consolation, that and other subjective responses are common on the internet forums. And, neither would coming up with the wrong mathematical answer entail any apology or admission. It's a discussion first and last. As agreed to in the "read the rules first"?
Secondly, I doubt that any of us hasn't wrestled with something else in the same way, and to the same extent, which seemed as straightforward to someone else with yet more mathematical experience. In fact, it's best to try to keep an open mind to the extent that such can still "haunt" us, instead of blinding seeking and following the math (or on the other hand, an atom smasher).
Quote: IbeatyouracesI'd be willing to bet all of the Wizard $2mm+ that anyone who thinks that it's 1/6 and bets will lose!
I don't have a million, but I do have a decent pension fund and I'm willing to take some chances when I can get such stupidly good odds.
All we need now is an unbiased source of dice results. Since we are a global community, I suggest random.org, or if you need an actual set of dice, then a venue which can be cheaply tied into a live internet feed. Then we need a mutually trusted escrow ( I trust the wizard to an extent). An adjudicator or team of adjudicators and an absolute and explicit definition of the wager. I'd also like to stay well away from any tax liability ( I'm not aware of stupid US tax laws, only that they are complex and stupid) Not beyond the wit of man. Bring it on you sixthers.
Mr. stubtuse will never accept the bet parameters.Quote: OnceDearI don't have a million, but I do have a decent pension fund and I'm willing to take some chances when I can get such stupidly good odds.
All we need now is an unbiased source of dice results. Since we are a global community, I suggest random.org, or if you need an actual set of dice, then a venue which can be cheaply tied into a live internet feed. Then we need a mutually trusted escrow ( I trust the wizard to an extent). An adjudicator or team of adjudicators and an absolute and explicit definition of the wager. I'd also like to stay well away from any tax liability ( I'm not aware of stupid US tax laws, only that they are complex and stupid) Not beyond the wit of man. Bring it on you sixthers.
http://forum.alanbestbuys.com/showthread.php?3441-Question-for-Math-Gambling-Craps-Experts
BTW. This is NOT a craps question. It's a probability question!
Quote: IbeatyouracesContinuation on his forum....
http://forum.alanbestbuys.com/showthread.php?3441-Question-for-Math-Gambling-Craps-Experts
BTW. This is NOT a craps question. It's a probability question!
LOL. Alan has taken his argument elsewhere, possibly hoping the home crowd will back him up. Unfortunately, and it is very unfortunately, he has quoted the bit of a post where a member here got it wrong. That member subsequently came back here and said he'g got it wrong. Great Kudos to Dalex!
And No. It's not a craps question for one moment. I have no knowledge of, or interest in craps.
I REALLY hope that his friends on that other forum continue to encourage him to take a wager.
Quote: IbeatyouracesContinuation on his forum....
http://forum.alanbestbuys.com/showthread.php?3441-Question-for-Math-Gambling-Craps-Experts
Pay particular attention to posts 12 and 13 there.
Post 12 is describing the problem as it was originally stated.
Post 13 is Alan's claim that this isn't what the question is.
As long as he is convinced that his interpretation of the question is correct, there's no changing his mind as to the answer.
Quote: ThatDonGuyPay particular attention to posts 12 and 13 there.
Post 12 is describing the problem as it was originally stated.
Post 13 is Alan's claim that this isn't what the question is.
As long as he is convinced that his interpretation of the question is correct, there's no changing his mind as to the answer.
It gets worse !!!!!
From that post 13 on that other forum he links off to another page ON HIS OWN SITE where the question is BADLY MISQUOTED AS
Quote: A Quote from a page on Alan's web site...a similar "problem" was brought up for discussion on the Wizard's forum. This time it was about two dice, where after the shake one of them was "peeked at" and known to be a 2, and the question was "what is the probability that both die are showing a 2"?...
Someone who doesn't know how to quote has changed the question to include "ONE OF THEM WAS PEEKED AT and known to be a two".
That is not the scenario in the OP on this forum. Even Alan confirmed to me that both dice were peeked at.
Quote: AlanYES. We can agree that the peeker saw both dice. But that doesn't change anything. It doesn't change the fact that if he says at least one die is a 2, then it is still a 1/6 chance that the other die is a 2.
Different question begets different answer.
It's not that Alan's 1/6 is the wrong answer: It's that Alan's question is the wrong and totally different question. I'm damned why he cannot WILL not see that. He CANNOT conceive of the difference between 'One die as come to rest as a two and the other is spinning (Correct answer 1/6) and 'two dice have come to rest and one of them, we don't know which of them, is a two (Correct answer 1/11)' It matters, it seriously matters.
Is this a ploy for Alan to drive traffic to his own monetized website. Does Zuga permit the copying of material from this forum to other forums and websites?
Before banning him, one last chance. Alan. Put your money where your mouth is. Big Time.
Does anyone want to open a book on my chances?
Will he rise to the challenge like a man?
Will he censor or delete my posts there?
Will he ignore my challenge?
Will he try to sidetrack me?
Will he try to gag me?
Will he bury my post in bullsh1t? or let my post be buried in the bullsh1t of his other forum members?
Or will he prove to be open minded and big enough to say he was wrong.
... Or am I wrong?
PM's of support and sponsorship of my entertainment expenses appreciated $:o)
When the player makes the bet, the player chooses either 1, 2, 3, 4, 5, or 6. If the next roll is a pair of the player's choice, the player wins 9:1. If on the next roll neither die is the player's choice then that roll's bet is a push. But if the next roll has only one die of the player's choice, then the player loses.
The house edge for this bet would be 2.78%, expressed on a "per roll" basis, and it would be 9.09%, expressed on a "per bet resolved" basis. (I am sure we can all agree on these numbers, by the way.)
The player may make as many choices as he wants, and he may pick up any of the bets at any time. I am sure it could become a very popular bet!
Quote: ChesterDogThis thread could be the inspiration for a fun new craps bet!
When the player makes the bet, the player chooses either 1, 2, 3, 4, 5, or 6. If the next roll is a pair of the player's choice, the player wins 9:1. If on the next roll neither die is the player's choice then that roll's bet is a push. But if the next roll has only one die of the player's choice, then the player loses.
The house edge for this bet would be 2.78%, expressed on a "per roll" basis, and it would be 9.09%, expressed on a "per bet resolved" basis. (I am sure we can all agree on these numbers, by the way.)
The player may make as many choices as he wants, and he may pick up any of the bets at any time. I am sure it could become a very popular bet!
I agree.
Although I'm sure some players (like certain people here) would be absolutely stumped as to why they're losing $$ on that bet.
