Two Dice Puzzle
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.
No need to copy the entire post. They already have hop bets. If you roll a pair it pays 30 to 1. If you roll a non pair but a specific number such as 3-1 it pays 15 to 1.
It's a one roll bet. There are no pushes.
Why don't you guys take this "2 question" to a craps table.
Quote: 1in11;27120The craps table example - where one die lands on 2 and the other die is hidden - is not equivalent to the question posed. Once you know for certainty the value of a specific die is 2, then the probability of 2-2 is 1 in 6.
The question posed is "at least one of the dice is a 2." For this question, we do not know if it is the first die that is a 2, or if it is the second die that is a 2 (for how ever you want to define "first die" and "second die"). The person who viewed the dice knows this information, but we do not. This is where the 1 in 11 answer comes from - there are 11 ways to roll at least one 2 (2-1, 2-2, 2-3, 2-4, 2-5, 2-6, 1-2, 3-2, 4-2, 5-2, 6-2), and each of these has the same probability. Of these 11 ways, only 1 way satisfies the condition that "both dice are showing a 2." Hence, 1 in 11.
No, I am going to stick with the craps table example and once again I am going to post the original question on the Wizard's Forum:
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?
It doesn't matter which of the dice is a 2 -- the answer is still 1/6.
Even if both dice showed a 2 -- the answer is still 1/6.
This exercise of yours (and everyone else including the Wizard) where you list the 11 ways to roll at least one 2 (2-1, 2-2, 2-3, 2-4, 2-5, 2-6, 1-2, 3-2, 4-2, 5-2, 6-2) is meaningless.
When you have a 2 showing on one die -- there is only a 1/6 chance of a 2 showing on the second die. And again, it doesn't matter which die the 2 is showing on. And it doesn't matter if a 2 is showing on both dice because the answer is still 1/6.
I am afraid that all of you "overthought" this question.
I am going to copy this and post it on the Wizard's forum also.
Thanks for joining my forum.
Quote: AlanMendelsonOne thing I've learned here, especially from quotes such as this, is that you guys don't play craps. Look:
No need to copy the entire post. They already have hop bets. If you roll a pair it pays 30 to 1. If you roll a non pair but a specific number such as 3-1 it pays 15 to 1.
It's a one roll bet. There are no pushes.
Why don't you guys take this "2 question" to a craps table.
So, if you hop the hard ways (aces + twelve included), $1 each ($6 in action), you will win $25 1/6 of the time. You lose $6 5/6 of the time.
($25 * 1/6) - ($6 * 5/6) = -$0.833 : You expect to lose $0.833 per roll.
This new bet that is presented, would be a little different (paying 9:1). If you bet on all 6 numbers, you lose $2 some of the times (since if you roll a 2-4, the '2' bet and the '4' bet will both lose). There are 6 ways to win ($9) and 30 ways to lose $2.
($9 * 1/6) - ($2 * 5/6) = -$0.166 : You expect to lose $0.166 per roll.
Quote: RSSo, if you hop the hard ways (aces + twelve included), $1 each ($6 in action), you will win $25 1/6 of the time. You lose $6 5/6 of the time.
($25 * 1/6) - ($6 * 5/6) = -$0.833 : You expect to lose $0.833 per roll.
This new bet that is presented, would be a little different (paying 9:1). If you bet on all 6 numbers, you lose $2 some of the times (since if you roll a 2-4, the '2' bet and the '4' bet will both lose). There are 6 ways to win ($9) and 30 ways to lose $2.
($9 * 1/6) - ($2 * 5/6) = -$0.166 : You expect to lose $0.166 per roll.I'm just going to assume a 7 does not cause these bets to lose [well, you'd lose the regular $2, as if it was a 2-4.
Great. Sell it to a casino.
I avoid the "inside" bets. Most knowledgeable craps players do.
None of the above.Quote: OnceDearRight, The gloves are off. Sorry to reference another forum. I won't link to it. This can be a simulcast or the debate can continue here or 'Alans place'. I have categorically challenged Alan to let me explain it to him. PROVE IT TO HIM. On his forum, on his website !!!
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)
What I mean is you will continue to misinterpret the original question.Quote: AlanMendelsonSorry Pew, I already invited OnceDear to state his case on my forum. It's an open forum.
Quote: pewWhat I mean is you will continue to misinterpret the original question.
I am proceeding, not to make my case but to take microsteps where Alan will either make the case for himself or he will seal his mind forever. It's not easy, but I think I can do it if Alan will just stay with me.
His mind is already sealed, glued, nailed down,shut tight, closed forever. He will never concede the question.Quote: OnceDearI am proceeding, not to make my case but to take microsteps where Alan will either make the case for himself or he will seal his mind forever. It's not easy, but I think I can do it if Alan will just stay with me.
Quote: AlanMendelson
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?
It doesn't matter which of the dice is a 2 -- the answer is still 1/6.
Even if both dice showed a 2 -- the answer is still 1/6.
That's only if you count "the roll is 2-2, and I am looking at the first 2 when I make my statement" and "the roll is 2-2, and I am looking at the second 2 when I make my statement" as two separate rolls.
Here's a thought:
Suppose the statement,
"At least one of the dice is a 2"
is changed to the following:
"Exactly one of these three statements is true:
One, the red die is a 2 but the blue die is not;
Two, the blue die is a 2 but the red die is not;
Three, both the red and blue dice are 2."
Do you consider these two to be the same, and if not, then why not?
Quote: pewHis mind is already sealed, glued, nailed down,shut tight, closed forever. He will never concede the question.
Maybe, maybe not. I rather suspect that he has locked himself into a certain way of thinking and that the more the thread goes on and the more things get repeated, the more locks he applies. It's possibly become a deep matter of principle to him that his answer is so obviously correct.
I'm not really worried about that. I'm trying the softest and gentlest way to help him see MY proof, and then to make it his proof too. He still has the right to veto my attempts.
He MUST surely wonder why we all disagree with him. I see many times that he tries to show us we are wrong. Maybe when it's proved to him that 1/11 is correct, then he will look at his correct answer of 1/6 and conclude ' I must be doing something different'. Then it's just a short walk to see what he is doing is different. I see it. Many see it, but we first need to have him look for the difference, and he won't do that until he sees two perfectly valid proofs both leading to different answers.
Quote: ThatDonGuyQuote: AlanMendelson
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?
It doesn't matter which of the dice is a 2 -- the answer is still 1/6.
Even if both dice showed a 2 -- the answer is still 1/6.
That's only if you count "the roll is 2-2, and I am looking at the first 2 when I make my statement" and "the roll is 2-2, and I am looking at the second 2 when I make my statement" as two separate rolls.
Here's a thought:
Suppose the statement,
"At least one of the dice is a 2"
is changed to the following:
"Exactly one of these three statements is true:
One, the red die is a 2 but the blue die is not;
Two, the blue die is a 2 but the red die is not;
Three, both the red and blue dice are 2."
Do you consider these two to be the same, and if not, then why not?
Here's my position. If you have two dice, and one die is identified as showing a 2 (it doesnt matter if the die is blue or green or on the left or on the right) then the other die has a 1/6 chance of also showing a two. Now I am willing and open to be shown why my thinking is wrong. But right now I have two dice in front of me and I have set one showing a 2, and I am looking at the other die. That "other die" has 6 faces, and one of those six is a 2. I cannot imagine how with two dice, and one already showing a 2, how you can say there is a 1/11 chance about the other six-sided die showing a 2.
Quote: AlanMendelsonBut right now I have two dice in front of me and I have set one showing a 2, and I am looking at the other die. That "other die" has 6 faces, and one of those six is a 2. I cannot imagine how with two dice, and one already showing a 2, how you can say there is a 1/11 chance about the other six-sided die showing a 2.
Alan,
Woah........ This is where you come quite correctly to the assertion that The other die has a 1/6 chance of being a deuce and making that magical pair
But........ this is not going to help. you are starting from your own question again. Stick with me on my step by step and you will never reach this point. I promise you.
Quote: OnceDearAlan,
Woah........ This is where you come quite correctly to the assertion that The other die has a 1/6 chance of being a deuce and making that magical pair
But........ this is not going to help. you are starting from your own question again. Stick with me on my step by step and you will never reach this point. I promise you.
This is the problem. The ORIGINAL QUESTION deals with TWO DICE under a cup with at least one of them showing a 2. YOU MUST address the problem using a pair of dice with at least ONE DIE showing a 2. If you don't do this you are altering the question.
Quote: AlanMendelsonQuote: OnceDearAlan,
Woah........ This is where you come quite correctly to the assertion that The other die has a 1/6 chance of being a deuce and making that magical pair
But........ this is not going to help. you are starting from your own question again. Stick with me on my step by step and you will never reach this point. I promise you.
This is the problem. The ORIGINAL QUESTION deals with TWO DICE under a cup with at least one of them showing a 2. YOU MUST address the problem using a pair of dice with at least ONE DIE showing a 2. If you don't do this you are altering the question.
So back to my question:
Suppose the statement,
"At least one of the dice is a 2"
is changed to the following:
"Exactly one of these three statements is true:
One, the red die is a 2 but the blue die is not;
Two, the blue die is a 2 but the red die is not;
Three, both the red and blue dice are 2."
Do you consider these two to be the same, and if not, then why not?
Also, if you are serious about this statement:
"The ORIGINAL QUESTION deals with TWO DICE under a cup with at least one of them showing a 2. YOU MUST address the problem using a pair of dice with at least ONE DIE showing a 2."
(which, by the way, is what pretty much every one of us who have figured out that the answer is 1/11 has already done)
then show us the possible rolls of the two dice, each with equal probability, with at least one die showing a 2.
Aren't the possible rolls 1-2, 2-1, 2-2, 2-3, 2-4, 2-5, 2-6, 3-2, 4-2, 5-2, and 6-2?
Am I missing any?
Quote: AlanMendelsonThis is the problem. The ORIGINAL QUESTION deals with TWO DICE under a cup with at least one of them showing a 2. YOU MUST address the problem using a pair of dice with at least ONE DIE showing a 2. If you don't do this you are altering the question.
Alan, We WILL. I promise you we will. I haven't got to it yet, but I assure you that both you and I are interpreting the original question exactly the same way.
Would you bet on a single roll of the dice where:
a) If no 2 shows up, your bet is a push
b) If a pair or 2's are rolled, you get paid 6 to 1
c) if a 2 and any other number EXCEPT another 2 is rolled, your bet loses
If you say yes to taking that bet, you truly are a sucker.
Quote: IbeatyouracesAsk yourself this question:
Would you bet on a single roll of the dice where:
a) If no 2 shows up, your bet is a push
b) If a pair or 2's are rolled, you get paid 6 to 1
c) if a 2 and any other number EXCEPT another 2 is rolled, your bet loses
If you say yes to taking that bet, you truly are a sucker.
Better is on offer
a) If no 2 shows up, your bet is a push
b) If a pair or 2's are rolled, you get paid 7 to 1 or even 8 to 1
c) if a 2 and any other number EXCEPT another 2 is rolled, your bet loses
Roll the dice. Any time you roll 2-2 add 6. Any time you roll 2-1, 2-3, 2-4, 2-5, or 2-6 subtract 1. All of you that think it's 1 in 6 will quickly see how you are wrong when you go negative fast.
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 like this new side bet: Field of Doubles. It's a self-service, one-roll bet. Bettor loses on any roll that's not doubles (i.e. not 1-1, 2-2, 3-3, 4-4, 5-5, 6-6).
Bettor wins 4-to-1 when the roll is, in fact, a "doubles." Just as good/bad as Big Red.
Quote: Dalex64Nevermind, I just said something really wrong again.
Never mind Dalex, You will never be as wrong as Alan.
By the way folks, Is Indignant still looking in on this thread? Is Ayacarumba?
What's your current thinking guys? Care to jump to Alan's defence? He now has
Rob Singer backing him up, so you'll be in good company, I'm sure.
Quote: IbeatyouracesRob Singer has absolutely ZERO credibility in the gambling world. Case closed on that.
Yeah, I know. We could try an alternative tack. I get Rob to condemn my logic as flawed, then we set about convincing Alan that Rob has no credibility. Alan will then not wish to be seen to agree with Rob, Therefore Alan will be forced to agree with me.
Hmmm Not going to happen. Rob is a contributer to Alan's site. Peas in a pod.
The funniest so far, has to be Alan announcing on his forum (about a fellow member's independent proof)...
"Synergistic your application of Excel was just wrong."
Closely followed by
"Sorry, I am not using a regular computer and I am unable to see the spreadsheets. "
And this guy claims to be a business man. What did he do? Fire the cretin that suggested using a spreadsheet application?
"I learned long ago, never to wrestle with a pig. You get dirty, and besides, the pig likes it."
George Bernard Shaw
I think it is hard to get people to agree on an answer when they don't agree on the question.
Quote: OnceDear...Indignant...
Yes.
Quote: Dalex64I've seen that quote on someone's signature line here.
I think it is hard to get people to agree on an answer when they don't agree on the question.
I've actually got agreement with Alan as to the interpretation of the question. I've even agreed 1/6 was the answer to his different question where he sets aside a deuce and rolls the other dice.
There are still a few poor souls that get the two questions mixed up in the belief that they mean the same, but I'm steadfastly refusing to argue with the 'One die set aside, one die thrown' crowd.
Quote:
- These are all eleven rolls that contain a deuce.
- The viewer/announcer says "I see a 2" for any/all of them.
- The normal question would be "What's the probability that the roll is actually a pair of two's?"
- Instead, it is asked deceptively "What's the probability that the other die shows a deuce?"
Alan has not responded to this explanation...
Quote:Let me try this another way. I did not see that the problem lies in the fact that the observer/announcer gets to see both dice. This fact gives him the opportunity to make ELEVEN announcements, rather than just SIX announcements.
If he only got to see one die, he'd only get to make SIX announcements - and you and I would have been correct about the one-in-six probability. Please spend some time digesting this picture...
Quote:Once Dear you owe me an apology. I am traveling and responding to the forum using my smartphone. Excel is not available. Be sure you mention this on the Wizard's site as well.
So there, I have mentioned it on the Wizards's site as well, as requested.
Quote: OnceDearIs Ayacarumba?
I haven't read all of it since the 11's seem stuck on the probability prior to the establishment of the value of one of the dice.
The original post asks the question after the value of one die is fixed. The 11's argue that the fixed value die could be the first or second die in the 12 possible ways to combine two dice when one of their values is going to be known. My position is that since the value of one die is fixed, it is no longer a variable. It is a fixed condition. This is why playing the wager game until "either" die comes up a two is not a true analogy of the sequence of events posed in the original question.
This is what I consider analogous:
I walk up to a table in your casino. On the table is an overturned dice cup. The dealer says (truthfully) the two dice have been shaken, and at least one of the two dice is a two (At least one die a two is always the starting position). Now, please place your wagers:
Your table offers:
Total 3: Lose 1
Total 4: Pay 8
Total 5: Lose 1
Total 6: Lose 1
Total 7: Lose 1
Total 8: Lose 1
Will you stay in business if you give these odds?
I would not bank this game because the payout on the total four is way out of line (but the 11's would call this a good deal). The established value of one die removes the other five possibilities it could have been from consideration. In the original question, this is the condition when the question of probability is posed. Only six totals are possible, each equally likely.
Quote: AyecarumbaWill you stay in business if you give these odds?
I'll happily bank that all day long.
Quote: OnceDearMe too. So long as the only tampering that the banker had done was to peek under the cup
I don't see why there needs to be a peeker. If there is no two then no action.
Quote: WizardI don't see why there needs to be a peeker. If there is no two then no action.
Actually, in a gaming scenario. Yeah I suppose that's true. We just call the banker a liar and it's a push :o)
Hate to say it..... (not)but I told ya so.Quote: pewHis mind is already sealed, glued, nailed down,shut tight, closed forever. He will never concede the question.
Agree. Peek not necessary.Quote: kenarmanI agree the peeking is totally unnecessary. Just throw the dice.
Quote: Ayecarumba
This is what I consider analogous:
I walk up to a table in your casino. On the table is an overturned dice cup. The dealer says (truthfully) the two dice have been shaken, and at least one of the two dice is a two (At least one die a two is always the starting position). Now, please place your wagers:
Your table offers:
Total 3: Lose 1
Total 4: Pay 8
Total 5: Lose 1
Total 6: Lose 1
Total 7: Lose 1
Total 8: Lose 1
Will you stay in business if you give these odds?
I would not bank this game because the payout on the total four is way out of line (but the 11's would call this a good deal).
Here is the latest version of the spreadsheet that Wiz can use to predict his profit. AlansFollyUpdated
Before we go there: Alan has checked the maths of the spreadsheet and asserts
Quote: Alan on his own forumThe problem with the spreadsheet is that it does not apply to the original problem. ... snipped....
So it all comes down to language because all of the math is correct.
I'm through with trying to show him that the spreadsheet does indeed apply to the original problem.
Quote: OnceDearYeah, I know. We could try an alternative tack. I get Rob to condemn my logic as flawed, then we set about convincing Alan that Rob has no credibility. Alan will then not wish to be seen to agree with Rob, Therefore Alan will be forced to agree with me.
Hmmm Not going to happen. Rob is a contributer to Alan's site. Peas in a pod.
The funniest so far, has to be Alan announcing on his forum (about a fellow member's independent proof)...
"Synergistic your application of Excel was just wrong."
Closely followed by
"Sorry, I am not using a regular computer and I am unable to see the spreadsheets. "
And this guy claims to be a business man. What did he do? Fire the cretin that suggested using a spreadsheet application?
There's a reason why guys like Singer and Scoblete give there stuff away for free.... it's WORTHLESS!!
Quote: IbeatyouracesThere's a reason why guys like Singer and Scoblete give there stuff away for free.... it's WORTHLESS!!
I don't know Scoblete although his reputation precedes him. I give kudos to Scoblete for doing nothing to support Alan's standpoint. Scoblete has said nothing yet (that I've noticed) on this topic that would make him look a dick. Time will tell.....
And a shout out to Dan Druff for being a brave admin and to a few senior and junior members there on that other forum for challenging Alan. Whether right or wrong, they showed backbone.
Quote: AlanNow 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.
What's the opposite to Kudos? I would like to award some to a few unnamed idiots who sucked up to Alan with separate and abject stupidity, misquotes and confusion. You know who you are guys. Lap it up.
Call it the "Doubles before Singles" bet and pay it 8:1.
As for Dan Druff. He is Todd Witteles. He's posted here a few times. You can see him here getting busted by Phil Hellmuth. https://youtu.be/y4WVLD4DW6c
Quote: OnceDearQuote: AlanNow 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.
I've worked for several dumb-stains with attitudes like this. The worst was CEO (but a failed attorney) who thought he could take the simple average of many contributor averages. The dummy was clueless that the contributor averages had been calculated against wildly varying populations.
Set on die on a 2 and roll the second one, 1/6 will be a 2
Roll BOTH DICE and only count the times a 2 shows up, 1/11 will be 2-2
