RS___ made the same mistake on the WOV forum that Arc made here. The phrase "at least" does not mean the 2 can change which cube it is on. "At least" means that in a two dice problem one or both shows a 2. It doesn't mean the 2 on the first die can move the second die for convenience.
Really guys. Speak English and improve your reading comprehension and stop "over thinking" the problem.
I can't decide if it would be good or bad to play craps with those dice that Arci uses. What do you think Alan?
I'm afraid that with his dice the faces can change to increase 7-outs.Originally Posted by regnis
At this point, it might be helpful to re-work the same exercise but with different parameters.Originally Posted by regnis
If at least two of three dies show a 2, then what is the probability all three dies show a 2? If at least one of three dies shows a 2, then what is the probability all three dies show a 2? (Does the latter question revert to the original question?)
Would rolling one die consecutively three times change the results?
Last edited by OneHitWonder; 05-10-2015 at 01:15 PM.
One hit now you are sounding like the hacks on the WOV forum. There is no need to change the question. They either misinterpreted the question or can't read or understand English or -- as they have shown -- they can't use two real, physical dice to understand that the "2" can't flip from one cube to another to fulfill their 1/11 answer.
Still on the WOV forum they won't look at two real, physical dice. The Wizard has revealed that his video was not an examination of the problem using real dice but was showing him roll dice hundreds of times tracking results.
They are dodging and weaving and just won't come to grips with the condition of the question. And OneHit I'm not going to let you dodge and weave around the question either.
I'll say it again: the answer 1/11 does not fit this question as asked. 1/11 is valid math but for a different question.
In the closed thread, http://vegascasinotalk.com/forum/showth...-vs-1-11/page7 , I asked only about extending yours and the Wizard's thinking to three dies, to gain a better appreciation for both sides of the argument. I was and still am in the camp of the one-sixth chance.
However, I have lost faith in the forum process here. I doubt that the Wizard as a forum host would meet such an honestly innocuous attempt at further enlightenment with, "I'm not going to let you dodge and weave around the question either," much less therefore promptly close the thread.
As far as I am aware, the Wizard's thread is still very much open to you after your numerous replies there.
Sorry that the thread was accidentally closed. It's reopened. Both Dan and I have closed threads accidentally when we've posted. We'll have to find out why. OneHitWonder's post from another thread has been moved here -- just above -- which is where it belongs.
OneHit I am sorry the thread was closed.
It appeared to me that when you suggested considering a three dice problem it was a way to dodge the question, because WOV forum members have suggested changes to the original question both here on this forum and on the WOV forum.
If you do in fact support the 1/6 it is better if you clearly state why you do. Thanks.
I received an email that on WOV they are saying I closed this thread. My explanation about what happened is above. Still they have not created a video explaining their 1/11 answer but continue to create side issues. Of course they won't create a video with two dice explaining their 1/11 answer because with two real dice the answer 1/11 is impossible.
Not a vid, because I'm not daft enough to waste the time and energy, Besides, I can't find an actor capable of portraying Alan in all his belligerent glory. http://wizardofvegas.com/forum/quest...46/#post456034
Kudus for the followup table layout from Indi in a later post there.
Last edited by OnceDear; 05-10-2015 at 11:18 PM. Reason: fixing a typo
I read your dramatic script. That's all it was.
Instead describe your proof using descriptions of two dice, OnceDear.
And you can tell "indignant" that his layout and bet do not justify the 1/11 answer either.
When will one of you apply the question and answer to two real, physical dice that cannot magically shift the 2 from one cube to another and where one die can only be included in the answer one time? Ya know... like in the real world?
The reason some have presented a three-dice question or a 2-coin problem or the 2-child problem is not to change the question -- but to show you how your logic/reasoning isn't correct.
The 2-coin and 2-child problem are EXACTLY the same (except one uses heads/tails and the other uses boy/girl). These two problems are almost the exact same thing as the 2-dice problem, except instead of having something with a probability of 50% (heads/tails or boy/girl), the dice problem has a probability where the outcomes are 16.66%. But just because the chance of any individual coin being a head or tail (at 50%) or a child being a boy or girl (also 50%)....that doesn't mean the other coin or the other child has a 50% chance [of being a heads/tails or boy/girl]. And, in the same way, the other die does not have a 16.66% chance of being a 2.
Now, instead of ignoring the 2-coin puzzle, you can try to learn from it. You are told a mother has at least one boy, and are asked what is the chance the second is a boy. When you realize the probability/chance of the other child being a boy is not 50% but is 33.33% -- at this point, you can (hopefully) relate the two problems and learn that 1/6 is not the correct answer to the question......just like 50% isn't the answer to the 2-coin or 2-child question. Baby steps, Alan. Baby steps.
The two coin or child question is not the same as the dice question. What makes you think it is? In the coin question the coins can be flipped this way:
HH, TT, HT, TH
but in the dice question you use each die only once. Once a die is identified as a 2 that die does not get used again in figuring the solution. That 2 face is frozen.
The dice can be rolled three ways with X = a number other than 2 and they are:
22, 2X, X2. But once "at least one of the dice" is known to be a 2 that die is removed from further use. And that leaves ONE die with SIX faces.
If you would simply roll two dice you would see that once at least one die shows a 2 it cannot be used again to come up with the 1/11 answer.
Roll the dice. Show me the video.
Edited to add: with the dice the rolls of XX aren't used as defined by the problem. When there is no 2 there is nothing to consider.
I am going to break down the dice rolls in detail to show you again why the answer is 1/6.
Roll 2,2. Use one of the dice and the remaining die has six faces to be 2,2.
Roll 2, X and you have a 2 that is frozen leaving six faces on one die.
Roll X,2 and the 2 is frozen leaving six faces on one die.
In all cases the answer can only be 1/6 in the real, physical world.
RS___ I have a question for you. Are these two questions the same?Originally Posted by RS__
1. Maria has two kids what are the chances both are boys?
2. Maria has two kids. One is a boy. What are the chances the other is a boy?
Please sir, no sir... Not a chance. you are not even quoting a puzzle correctly.
The answer to q1 is simple and determined. 1/4 (in a gender neutral world)
The answer to q2 depended on what population sample Maria was chosen from and so is indeterminate.
One perfectly valid answer to q2 is 1/2
One other perfectly valid answer to q2 is 1/3
No perfectly valid answer is 1/4
http://www.curiouser.co.uk/puzzles/kids3.htm
The last paragraph gives the answer to the second question.
Did you get it right Alan, did you? All on your own? Do you now understand probability a bit better?
Or is it back to the drawing board...
...Or is everyone else in the world just plain wrong?
Now.... In my video script..... what proportion of bets on 'pair of deuces' would you expect to win? This isn't maths or charts or excel. A simple question. Do you expect to win 1 in 6 or 1 in 11.
Really no need for your explanation at this point.
Maybe also, what proportion of your friends' bets would you expect to win... the bets on other outcomes?
Last edited by OnceDear; 05-11-2015 at 08:59 AM.
OnceDear I am glad you read the Maria questions. Yes the answer to #2 is 1/2. There is no reason to consider the first child so it's not 1/3. This is the same problem you are having with the dice question. You aren't reading properly.
Once again Alan, you show your ignorance.Originally Posted by Alan Mendelson
Read the very, very clear explanation on that web site. Until you can grasp where you are wrong on this you will never qualify to answer questions about dice.
By the way, speaking of ignorance. Why are you ignoring my question
I expect to win a pair of deuces 1/36 times. When one deuce is showing on one die the chance of a deuce showing on the second die is 1/6.
OnceDear regarding the children question until you understand what you are reading you will continue to make errors as you have with the dice question. You offered two answers to #2. Make up your mind BASED ON THE WORDING because there can't be two answers.
I didn't want to, but I returned to the WOV forum to ask the Wizard a question since it seems I am now being painted as ignorant. Since they are now talking about boy children and girl children, and heads or tails on coins, I asked this:
Wizard: If I am wrong, the same way you can visualize the coin problem by flipping heads and tails, and the children problem with stick figures of boys and girls, please use two dice to help me visualize the dice problem.
Thanks.
I am ready to be shown how when at least one die is a 2 how it is a 1/11 chance that both will show 2? As I've said and others have said, with one die a 2 (and in the case of both dice showing a 2 you just choose one) the "other die" only has 6 faces to choose among -- and not 11.
Again I am being criticized for referring to "the other die."
So, a visual presentation will help me understand what they're talking about.
Just look at the bloody table layout that indi kindly drew for you. place your money on pair of deuces or any other bet you like and then figure out how you can make a profit. It's as real world as any scenario that you can conceive of.Originally Posted by Alan Mendelson
http://wizardofvegas.com/forum/quest...46/#post456141
Stop thinking about bloody dice, and start thinking of odds.
Last edited by OnceDear; 05-11-2015 at 03:07 PM. Reason: fixing up link
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