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Thread: It was two years ago today...

  1. #21
    Here's why the answer is 1/6 and it comes from the original problem:

    ...and tells you, truthfully, "At least one of the dice is a 2."


    As soon as you have that information, it becomes a one die question.

    You see, it's a reading comprehension issue. You can't overlook the information that one of the two dice is TRUTHFULLY showing a 2.

  2. #22
    Originally Posted by Alan Mendelson View Post
    Here's why the answer is 1/6 and it comes from the original problem:

    ...and tells you, truthfully, "At least one of the dice is a 2."


    As soon as you have that information, it becomes a one die question. You see, it's a reading comprehension issue. You can't overlook the information that one of the two dice is TRUTHFULLY showing a 2.
    This is a an excellent question to debate here, except Rob, per the usual, jumping in and flaming the thread with his childish outbursts.

    Alan, can you explain the mystery of why it is twice as likely you will roll a 3 than a hard 4? Remember, with both the 3 and the hard 4 "at least one of the dice is a 2." But by your logic the chances of rolling a 3 and a hard 4 should be the same.
    Last edited by mickeycrimm; 05-13-2017 at 01:13 AM.

  3. #23
    There are two ways to make a 3, but there is only one way to make a hard 4.

    3: 1, 2 and 2, 1
    hard 4: 2, 2

    But this has nothing to do with the question.

    There are two dice. We are told at least one of the two dice is a 2. That eliminates one of the dice. The chance of the remaining die also being a 2 is one out of 6. There are only two dice in the problem. It doesn't matter which of the two dice is showing a 2.

    If you recall, the Wizard, in order to prove his point, had one die showing a 2 and then ROTATED that die to show the other five faces. In the real world once a die lands showing a 2 it cannot rotate again. I called him on that, and that prompted his comment: "(Alan) simply can't get past a die having six sides." And that's right. A die has only six sides and once the die lands it cannot keep spinning or resume spinning.

  4. #24
    Originally Posted by Alan Mendelson View Post
    There are two dice. We are told at least one of the two dice is a 2. That eliminates one of the dice. The chance of the remaining die also being a 2 is one out of 6. There are only two dice in the problem. It doesn't matter which of the two dice is showing a 2.
    If you set one die on the table showing a 2 then roll the other die then yes the answer is 1 in 6. But when you roll both the dice at the same time the answer is 1 in 11.

    Here's a simple little game you can play on the kitchen table at home if you have some dice. You would need paper and pencil to record the stats. Roll the dice and record everytime "at least one of the dice is a 2". Look to see what the other die is. Forget the numbers 3,4,5,and 6. If the other die is on those numbers pick the dice back up and roll them again.

    Only record the stats for when the other die is a 1 or 2. When you get enough rolls in you will start to see the pattern. The other die will be a 1 about twice as often as it is a 2.

    And the stats should show that the other die will be a 2 only about 1 in 11 times.
    Last edited by mickeycrimm; 05-13-2017 at 02:42 AM.

  5. #25
    Originally Posted by regnis View Post
    I hate to start this argument over again, but jbjb, that is so stupid I don't even believe you believe it. If a 2 is showing on one die, the other spots on that die are out of the equation, leaving one die with 6 spots, one of which is a 2.

    Unless you have x-ray eyes and can see all the sides of the dice.
    The question is what are the odds of BOTH dies showing a two. The answer is 1/11. If the question was what are the odds of the OTHER die showing a two then the answer would be 1/6. Simple. The amazing thing about this thread to me though, is where the heck did the last two years go?
    Take off that stupid mask you big baby.

  6. #26
    Originally Posted by Alan Mendelson View Post
    There are two dice. We are told at least one of the two dice is a 2. That eliminates one of the dice. The chance of the remaining die also being a 2 is one out of 6. There are only two dice in the problem. It doesn't matter which of the two dice is showing a 2.
    So, in the original question, the guy looks under the cup and could see one of the following:

    Dice One Dice Two
    2...............1
    2...............2
    2...............3
    2...............4
    2...............5
    2...............6

    1...............2
    2...............2
    3...............2
    4...............2
    5...............2
    6...............2

    11 possible unique outcomes satisfy the statement "At least one of the dice is a 2."
    only one of those outcomes is 2,2.

    So... why isn't it 1 in 11?

  7. #27
    Originally Posted by a2a3dseddie View Post
    Originally Posted by Alan Mendelson View Post
    There are two dice. We are told at least one of the two dice is a 2. That eliminates one of the dice. The chance of the remaining die also being a 2 is one out of 6. There are only two dice in the problem. It doesn't matter which of the two dice is showing a 2.
    So, in the original question, the guy looks under the cup and could see one of the following:

    Excellent example, Eddie. The best I've seen.

    Dice One Dice Two
    2...............1
    2...............2
    2...............3
    2...............4
    2...............5
    2...............6

    1...............2
    2...............2
    3...............2
    4...............2
    5...............2
    6...............2

    11 possible unique outcomes satisfy the statement "At least one of the dice is a 2."
    only one of those outcomes is 2,2.

    So... why isn't it 1 in 11?
    Excellent example.

  8. #28
    Originally Posted by a2a3dseddie View Post
    Originally Posted by Alan Mendelson View Post
    There are two dice. We are told at least one of the two dice is a 2. That eliminates one of the dice. The chance of the remaining die also being a 2 is one out of 6. There are only two dice in the problem. It doesn't matter which of the two dice is showing a 2.
    So, in the original question, the guy looks under the cup and could see one of the following:

    Dice One Dice Two
    2...............1
    2...............2
    2...............3
    2...............4
    2...............5
    2...............6

    1...............2
    2...............2
    3...............2
    4...............2
    5...............2
    6...............2

    11 possible unique outcomes satisfy the statement "At least one of the dice is a 2."
    only one of those outcomes is 2,2.

    So... why isn't it 1 in 11?
    Alan thinks 2...1 and 1...2 are the same (for some reason), although he recognizes there are two ways to make a 3, "1...2 and 2...1". But hey, this isn't the first time Alan has played both sides before or had no idea what the subject was about.

    He says he uses conventional strategy on bonus poker, yet he holds 3 to a RF over a high pair.
    He says hole carding is generally worthless, yet doesn't understand what it is.
    Sometimes he recognizes using win goals / loss limits wouldn't yield an advantage, then other times argues it makes perfect sense to do so "No one ever went broke selling their stock at a profit", or something like that.

  9. #29
    Originally Posted by a2a3dseddie View Post
    So, in the original question, the guy looks under the cup and could see one of the following:

    Dice One Dice Two
    2...............1
    2...............2
    2...............3
    2...............4
    2...............5
    2...............6

    1...............2
    2...............2
    3...............2
    4...............2
    5...............2
    6...............2
    Your chart is exactly right and you must consider only ONE of the following. If it is Die 1, the answer would be (as your graph shows) 1/6. If it is Die 2, it is also 1/6. YOU CANNOT SWITCH BACK AND FORTH.

    In order to switch back and forth, you have to do what the Wizard did on his forum in his video: you must CHANGE the result of a die after it comes to rest. In the real world that doesn't happen.

    In fact, the original problem is simply this: a shooter throws two dice on a craps table. One die immediately comes to rest showing a 2. The second die is a spinner and keeps spinning. What are the chances that the spinner will come to rest on a 2 as well? The answer is 1/6.

    Guys, that's the same thing as the stated problem. When the "observer" looks and says one of the dice is a 2, what are the odds that the other die is also a 2, it is the same thing as the "spinner" problem.

    If you don't interpret it as being the same as being the "spinner" problem you are making up your own conditions. The only way to interpret the original question is to consider it to be the same as the spinner problem. Because we know at least one of the two dice is showing a 2, just like with the spinner.

    Sorry, but you are going to have to deal with this reading comprehension issue.

  10. #30
    Originally Posted by RS__ View Post
    He says he uses conventional strategy on bonus poker, yet he holds 3 to a RF over a high pair.
    Another example of your reading comprehension problem. I explained that yes, I play conventional strategy, but when I play video poker my goal is to hit a royal flush -- not to save high pairs.

  11. #31
    Originally Posted by Alan Mendelson View Post
    Originally Posted by a2a3dseddie View Post
    So, in the original question, the guy looks under the cup and could see one of the following:

    Dice One Dice Two
    2...............1
    2...............2
    2...............3
    2...............4
    2...............5
    2...............6

    1...............2
    2...............2
    3...............2
    4...............2
    5...............2
    6...............2
    Your chart is exactly right and you must consider only ONE of the following. If it is Die 1, the answer would be (as your graph shows) 1/6. If it is Die 2, it is also 1/6. YOU CANNOT SWITCH BACK AND FORTH.

    In order to switch back and forth, you have to do what the Wizard did on his forum in his video: you must CHANGE the result of a die after it comes to rest. In the real world that doesn't happen.

    In fact, the original problem is simply this: a shooter throws two dice on a craps table. One die immediately comes to rest showing a 2. The second die is a spinner and keeps spinning. What are the chances that the spinner will come to rest on a 2 as well? The answer is 1/6.

    Guys, that's the same thing as the stated problem. When the "observer" looks and says one of the dice is a 2, what are the odds that the other die is also a 2, it is the same thing as the "spinner" problem.

    If you don't interpret it as being the same as being the "spinner" problem you are making up your own conditions. The only way to interpret the original question is to consider it to be the same as the spinner problem. Because we know at least one of the two dice is showing a 2, just like with the spinner.

    Sorry, but you are going to have to deal with this reading comprehension issue.
    The problem is you quoted the question wrong (again). It's not the "other die" it's "both dies". You have the reading comprehension issue.
    Take off that stupid mask you big baby.

  12. #32
    Sorry quahaug. It's a two dice problem. The observer reports on one of the dice and says truthfully it is a 2. You cannot consider that same die again. You must consider and only consider the second die in the problem.

    You must face reality.

    It is the same as the spinner problem. You cannot shift from one die to another. You cannot change the face of one die which is known to be a 2.

    You have a real world problem.

    If you take two real dice as I did in my video there is only one conclusion: 1/6

    The wizard in order to prove 1/11 had to CHANGE the face of a die. And in the real world you CANNOT change the face of a die after it comes to rest.

    It just doesn't happen.

    Deal with it.

  13. #33
    Originally Posted by Alan Mendelson View Post
    Originally Posted by a2a3dseddie View Post
    So, in the original question, the guy looks under the cup and could see one of the following:

    Dice One Dice Two
    2...............1
    2...............2
    2...............3
    2...............4
    2...............5
    2...............6

    1...............2
    2...............2
    3...............2
    4...............2
    5...............2
    6...............2
    Your chart is exactly right and you must consider only ONE of the following. If it is Die 1, the answer would be (as your graph shows) 1/6. If it is Die 2, it is also 1/6. YOU CANNOT SWITCH BACK AND FORTH.
    Not too sure what you mean about "switching back and forth". You said yourself that the chart is exactly right. These are the 11 unique outcomes that can occur when 2 dice are thrown and "At least one of the dice is a 2."

    Originally Posted by Alan Mendelson View Post
    I even provided the "correct question" that would have given the 1:11 answer.
    Could you re-post your version of the correct wording that would yield a 1:11 answer?

  14. #34
    Originally Posted by Alan Mendelson View Post
    In fact, the original problem is simply this: a shooter throws two dice on a craps table. One die immediately comes to rest showing a 2. The second die is a spinner and keeps spinning. What are the chances that the spinner will come to rest on a 2 as well? The answer is 1/6.

    Guys, that's the same thing as the stated problem. When the "observer" looks and says one of the dice is a 2, what are the odds that the other die is also a 2, it is the same thing as the "spinner" problem.

    If you don't interpret it as being the same as being the "spinner" problem you are making up your own conditions. The only way to interpret the original question is to consider it to be the same as the spinner problem. Because we know at least one of the two dice is showing a 2, just like with the spinner.
    I thought this was the original problem:

    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?


    Last edited by a2a3dseddie; 05-13-2017 at 02:36 PM.

  15. #35
    Originally Posted by a2a3dseddie View Post
    I thought this was the original problem:

    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."
    Yes, it is. And it's the same thing that happens when you throw two dice and one of them turns out to be a spinner while the other comes to rest as a 2.

    Here's a lengthy thread from two years ago which discusses how the Wizard posted a video and had to change a die in order to prove his 1/11 result. And as we all know, dice don't change after they come to rest:

    https://vegascasinotalk.com/forum/sh...s+dice+problem

  16. #36
    Originally Posted by Alan Mendelson View Post
    Originally Posted by jbjb View Post
    Two dice equal 12 sides. If a 2 is showing on one, there are 11 sides left. Only one of which is a 2. The damn answer is 1 in 11.

    Case closed you wimps who work for a living and can't cut it in the gambling world.
    Are you fucking serious? If a 2 is showing on one die, you set that die aside. It's showing a 2. That leaves only the other die with six sides. So the chance of 2-2 can only be 1 out of 6.

    The problem with your interpretation is that we know ONE DIE IS SHOWING A 2. We are told ONE DIE IS SHOWING A 2. It doesn't matter which because there are only two dice. If one of the dice is showing a 2 we are only concerned with the second die -- and the chance of a 2 on that other die is 1/6.

    That leaves only the OTHER die which has SIX sides.

    And don't start flipping over the dice the way the Wizard did in his video. That's the biggest laugh in the world. Dice don't continue flipping after they come to rest. If a die came to rest, it's done.
    You just can't seem to get past the idea that a die has six sides.

    No, but seriously...

    The problem here is that you can't set aside the one with a 2, because you don't know which one.

    That's the trick here.

    If they said "Die #1 shows a 2, what are the odds that Die #2 also shows a 2", then you're correct -- the odds are 1 in 6.

    The problem here is that the 2 can be on either die, so nothing can be "set aside".

    There are 36 outcomes to 2 dice being rolled.

    All the ones involving 2 are as follows (in first/second format):

    1/2
    2/2
    3/2
    4/2
    5/2
    6/2
    2/1
    2/3
    2/4
    2/5
    2/6

    That's 11 combinations where one or more dice show a 2.

    The other 25 combinations don't show a 2, so we can rule them out of this problem.

    So basically the problem is, "Which of the above combinations were rolled?"

    And the answer is, "WE DON'T KNOW!" Because we are only told that one die has a 2, which could be any of the 11 above.

    That's what makes the answer "1 in 11". Because we are given information about the roll that gives us 11 possible outcomes, and the one we're looking for (2/2) is only one of those 11.
    Check out my poker forum, and weekly internet radio show at http://pokerfraudalert.com

  17. #37
    Dan you're making the same mistake. There are only TWO dice. If you tell me that at least one die is showing a 2, I MUST set aside one die. Since it's a two dice problem it makes no difference WHICH of the two dice I set aside and that's because I was told at least ONE of the TWO dice is showing a two.

    So here's what I asked the Wizard to do and he didn't, and I am going to ask you and everyone else to do: act out the question using your own dice and your own cup.

    Do it: take two dice, put them in a cup, shake rattle and roll, and when at least one die is showing a 2, figure the odds that the second die is also showing a two.

    If you do that -- which is what the original wording of the problem said -- you can ONLY come up with the answer 1/6.

    Want a question with an answer of 1/11? Well here it is:

    Using two dice, how many different combinations will show at least one 2? Answer: 11
    Of those 11 combinations, how many will show 2-2? Answer 1/11

    THAT is the question that will give you the answer 1/11.

    But when you shake two dice and you know that at least one of the two dice has landed on a 2, then the odds that the other die is also a 2 can only be 1/6.

    Again, it's a matter of reading comprehension and properly interpreting the question.

    The question does not ask you to consider all of the combinations of two dice -- which is what the Wizard did in his video. The question asks to do what I did in my video which is the same as throwing two dice and having one a spinner while the other lands on 2.

    Deal with it. It's the English language.

  18. #38
    Alan,

    Imagine this was somehow a game you could bet on.

    A dealer has two 6-sided dice in a cup. He shakes the dice, and slams the cup down onto the table, hiding the result. The dealer peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2."

    Your bet is a push if there are no 2s.

    If at least one of the dice is a 2 (11 out of 36 combinations or 2.27-1 against you) you win... let's say 2 to 1 on your bet factoring in a house edge.

    If both dice are 2 what do you think the true payoff should be? 5-1?

    There are 11 combinations of 2 dice where at least 1 of them is a 2. If you only get paid on (2,2) knowing ahead of time that at least 1 of the dice is a 2, you have 1 win and 10 losses don't you? Shouldn't the true payoff be 10-1?
    Last edited by a2a3dseddie; 05-13-2017 at 02:28 PM.

  19. #39
    To everyone who thinks I'm wrong: please get yourselves two dice, and a cup, shake rattle and roll, and then figure what the odds are for the second die to also be a two.

    That's what the question is.

    Don't bother me with side bets, don't bother me with anything else. It's stupid. Read and comprehend English. Act out the question.

  20. #40
    I'd love to see Alan analyze the Monty Hall problem. That would be a real barn burner.
    Take off that stupid mask you big baby.

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