Then accept the 8:1 bet. I'm 100% certain you'll go broke.
Then accept the 8:1 bet. I'm 100% certain you'll go broke.
Hard to understand. First the WoV forum badmouths Frank for being a member of this forum, he's rudely mocked along with me for having enough money to be able to give away our books while AP wizard had to beg for money when his skills just weren't there, then Frank's criticized for not opining on the issue.Originally Posted by jbjb
You guys need to get not only the original problem straight--you need to get your stories straight.
Alan, at this point I am trying to understand how you read the question. I understand fully well, that based on your understanding of the question, the answer is 1/6. Based on my understanding of the question, the answer is 1/11. Neither of these two facts are arguable.
You contend that the wording is unclear, and I contend that it is not. The wording as presented in the problem is exactly the wording that I would expect to see, and is the wording that I would use in a simple conditional probability exercise. If this wording is unclear, I would like to understand why so that I can avoid this issue.
I would like to understand how you would word the problem with the answer 1/11 in a clear and concise manner.
I'll accept that bet. And I'll bet you 10X what I win or lose that I won't go broke. And no....this time I will not pay for anyone's airfare to LV, hotel, or food.Originally Posted by jbjb
I've already answered this: the 8:1 bet is a lousy bet for the player. There is no way in hell I would make that bet. The reality of the situation is that throwing 2-2 is a 1/36 event.Originally Posted by jbjb
And once again: the damn bet has nothing to do with the original question/problem.
Wake up -- we're not stupid. And we might be almost as smart as you are.
Thank you. This is probably the first sensible post from those who are in the 1/11 "camp."Originally Posted by 1in11
Honestly, I don't know how to phrase the question so that the appropriate answer would be 1/11, but I am sure there is a properly phrased question that would elicit that answer as the correct answer.
I can only say this:
When someone tells me that there are two dice, and at least one die shows a 2, then common sense and my knowledge of what is on a six sided die tells me that the chance of a 2 showing on the "other die" has to be 1/6. To put it another way (which I have said before) the information provided to me only allows me to answer 1/6. The answer is defined by the question.
That doesn't exclude 1/11 as being an answer -- it's just that it can't be the answer given the information that was provided in the stated original question.
Thank you again for asking this.
I take a bit of offense here, I think my asking you if you would tell me which of my steps don't match up your original question was plenty sensible. I'd still appreciate your doing so, actually.Originally Posted by Alan Mendelson
Please don't be offended. But the original question made it clear that the only thing to be considered was the "other die" and with the other six-sided die it was a 1/6 answer.Originally Posted by synergistic
Over on the Wizard's forum they are continuing their mantra about 1/11 with all of the combinations of two dice. So one of you guys who is here reading this please send them this message:
I understand that there are 11 combinations of two dice that include 1 die with a 2. But that wasn't the question that was asked and so, those 11 combinations of two dice do not fit what is needed to answer the question. The question doesn't ask for the number of dice combinations that contain a two. The question asks: "What is the probability that both dice are showing a 2?"
Now you can take the "long way" around and explain that there are 11 different combinations of dice that include a two, but I am answering the specific question: What is the probability that both dice are showing a 2? And the answer to that specific question is 1/6.
I am sorry that such a simple answer is all that is needed. Now go figure out when that big meteor is going to crash into earth.
edited to add:
Over on the other forum there are several who posted the various combinations of two dice that include a two. And then there is this fellow Ibeatyouraces who wrote:
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!!!
Clearly this is overthinking the problem. And by overthinking they make a mistake because it's not a question of any two dice combination -- it's what are the odds of a 2 showing on the other die.
Well. Alan, I never thought that I would hear you say that. Thank you.Originally Posted by Alan Mendelson
Do you actually know what probability is and how it is worked out? You do know that probability is a numberr don't you, generally expressed as one number divided by another number?Originally Posted by Alan Mendelson
Do you know where the top number comes from and the bottom number comes from?
I'll give you a clue.
The probability of an event=Count of the ways that event can happen/Count of all possible events that can happen
We agreed that there is one way that a 2-2 can happen so that makes the top bit, the numerator, a 1.
The count of all possible events that can happen.... well, you just admitted it. there are 11 possible ways that we could have ended up in this stupid mess where at least one of the dice is a two. We might reasonably conclude that probability in this instance, and not in any other instance, not with 5 dice or one dice or one spinning dice, but with two resting dice under a cup...
that probability = count of number of ways 2-2 can happen/ count of number of ways two dice can be at rest under a cup with at least one dice showing
= 1/11
Now.... explain it to me. what part of my logic is flawed? It should be possible without any bluster about knowing anything else.
OnceDear your logic is fine, but you are not answering the original question. The original question makes it clear that at least one of the two dice is a two. And once again (how many times, Lord, how many times?) given that information you only have to consider the six faces on the other die.
Tell your buddies over on the Wizard's forum who enjoy insulting me and the others here who also say it's 1/6 that we appreciate your logic but your logic does not answer the question and the conditions posed in the question.
And frankly, I wish the original poster would come out of the bushes, desert, closet, Siberia, or wherever else he might be to clarify why he asked the question that way? I don't know if it was in error, an accident, or intentional to mislead? But whatever he did and whatever his motive, he sure created a mess out of a simple exercise.
Alan, a WoV poster called dalex (who may also be posting under a different anonymous handle here) has put up a very good, very comprehensive post about how he gets to the 1 in 11 conclusion. I guess he did it with the assumption that people who know the answer to the original question to be 1 in 6, just can't figure out how the geniuses arrived at 1 in 11.
Well, just as we've been saying from moment one, we all have always known where 1 in 11 came from, except that it is not answering the OP's question. The way they interpret it is as if there always needs to be two dice in the formula even though the OP clearly eliminated one die from the problem. I can only summize that the reason these people need to keep two dice in play is due to their incessant requirement to make complicated math problems out of the simplest of events.
It is all so very similar to the Monty Hall problem. If that question clearly identified that all 3 doors were in play when the question about odds was asked, of course the answer would be 2 in 3. But the way it was asked appeared to be "what are the odds NOW" which naturally would be 1 in 2. As my articles said, it is all in how one interprets the wordsmithing. Simple, more sensible people would prefer to take the shortest, surest route. Those forever in conflict-mode who enjoy appearing like the smartest in the room, will always take the lengthier, more complicated route.
Here's another example of their stubborn thinking that cannot consider ONE die but must consider only two dice answers. This comes form "Dalex64":
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:
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.
Again, their only answer seems to come from the idea that the answer must come from a two dice combination of numbers. They don't even consider the idea that the answer can come from a single die with six faces.
Is this something taught in school? That the easy, simple answer can't be right?
I think this is the most important part of that post on WOV:
I'll listen to explainations as to why all 36 possibilities couldn't be under the cup
Well how about this one: there is one other die with only six faces on it, 1 thru 6?
first you tell me there are only 6 numbers on one die. Now they say there's no santa claus. I am a broken man.
Wizard is now inviting anyone not in the 1 in 11 camp to his MonekyFest, and to bring "plenty of money" while he brings the dice.
AGAIN, WoVers....I will GLADLY show up. PLEASE tell him! All he need do is follow the parameters of the original question: set one of the dice--either one, doesn't matter--to a 2; then roll the other die. 1 in 6 will yield a 2nd 2. And I will happily send him back to begging the pseudo-mensas for "mo money".
Count me in too Rob. Can't let you have all the fun,
If the Wizard conducts his bet the way Rob describes and is offering 8 for 1 OR 8 to 1 pays for every hard-4 (2-2) I am also in. But I am afraid that is not what's going to happen. They will want two dice rolled simultaneously like in an everyday game of craps -- and in that case 2-2 occurs 1/36 times. And in that case the Wizard is only paying 8 to 1 or 8 for 1 for a 1/36 event.
I vowed to stay out of this any further, but I don't find the language or question ambiguous at all. One die is a 2--that is clearly stated. Therefore, the question is what are the odds that the other is a 2. Answer-1 of 6. As a licensed arbiter for many years, I am required to look at the simplest and plain meaning of a contract or other document in resolving a dispute. In this case, it is so clear as to not even constitute a dispute. I must take the stated fact--that one die is a 2--in its simplest form. that leaves only the other die in question.
Math, excel spreadsheets, all the other hypothecations are meaningless to the simple question that was presented. If you want to answer the question of what are the odds of rolling two dice and one or more of them being a 2, then go ahead and do the math and create your fancy spreadsheets. But that is totally unnecessary and unreasonable in this context.
Actually--just roll one die as the other is a 2. Then pay 8-1 on every 2 on the die that is rolled. I'm in all day.
Which one is a 2 though?
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