1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Interesting question about throwing of dice

  1. Aug 28, 2005 #1
    Can someone help me? Why the argument is false ?

    In this problem you must explain the results in intuitiva way. One formal prove or verbal description of a prove is not satisfactory answers.

    1) It considers the following events:

    A: To get one 6 in four throwing of a die;
    B: To get one double 6 in twenty and four throwing of a pair of dice.

    Argument: We have P(A) = P(B), because the probability of if getting one 6 in the throwing of a die is six times bigger than the probability of if getting double 6 in the throwing of a pair of dice, but the number of throwings of event B is six times bigger than the number of throwings of event A.

    It is true or false. Explain in intuitive way.
     
  2. jcsd
  3. Dec 23, 2005 #2
    It turns out this is a famous problem called, deMere's Problem,http://mathworld.wolfram.com/deMeresProblem.html.
    deMere was a careful gambler who kept notes and had come to the correct conclusion that it is easier to throw a 6 in 4 throws than a 12 with two dice in 24 throws. But he was not certain, and certainly could not prove it.

    Of course, in those days nobody could prove it and Pascal was just developing the theory of probability. To explain it intutively without the math is hard, as it is clear that deMere did not know.
    With the math it is fairly easy to explain it, nothing more than showing and solving some simple equations. This is done in the article. That is about as "intutive" as it gets, I guess.

    BUT now to get involved in your reasoning: Is it just as easy to get a 6 in four throws of the dice as double six in 6 times as many throws?
    The difficulity here is very simple, the problem is not exactly one throw, but one or more throws. That is why as deMoive did over and over in "The Doctrine of Chance," you have to work from the other side or the possibility of no sixes in four throws and then subtracting that from the total possibility giving 1-(5/6)^4.

    Look at it this way, instead of throwing a six four times, consider throwing it 24 times. Would we expect to get one or more sixes six times as frequently, or would that be impossible since the total probablity is just 1?
     
    Last edited: Dec 24, 2005
  4. Dec 24, 2005 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    "In this problem you must explain the results in intuitiva way. One formal prove or verbal description of a prove is not satisfactory answers."

    In other words we are required to explain it badly? Sounds to me like a teacher asking you to state it in your own words so he/she will know you are not just copying a proof. That means it is homework and giving our words rather than your own would be cheating.
     
  5. Dec 24, 2005 #4

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper
    2015 Award

    i do not understand the problem. do you mean to get one or mmore 6's in 4 throwings? or exactly one?



    the traditional way to calculate the chance of geeting one or more, is to calculate instead the chance of not getting any, and then subtract that answer from 1.


    so the chance of getting one or mor is 1 minus the product of 4 fractions each equal to (5/6).

    see if you can figure out the chance of getting one or more double sixes in 24 throws.

    it seems rather obvious the two probabilities aren different without calculating anything, using the uniqueness thorem on factoring integers.

    I learned this theory as an 11 year old from the book Scarne on Dice, and found it rather fascinating.

    the basic principle of comuting probabilities of repeated trials yielding a certain result, is that the chance of getting the same result twice in a row is the square of the chance of getting it in one trial. assuming success on one trial has no effect on the likelihood of success in another trial.

    in real life this is not true as repeated success can either discourage or motivate ones opponent and possibly affect your odds. or repeated success can lull you into security, or motivate you to keep up your record.
     
  6. Dec 25, 2005 #5
    mathwonk: I learned this theory as an 11 year old from the book Scarne on Dice, and found it rather fascinating.
    Scarne's work is today probably still the classic work, though it is rather long and involved since it covers so many variations and possibilities.
    Scarne is probably the most mathematical account of this available, all of which is very elementary math.

    I notice that middle class players here in Vegas tend to play somewhat differently than locals who make more proposition bets, which books say are bad. (I assume that visitors are more likely to be well-educated and read books about dice.) I have noticed bettors who take notes on what happens, I attribute this to writers who advises this strategy, which seems to me ridiculous since every throw is random and so little can be learned from tracking previous play. It does seem that some players follow fads and are influenced by whatever book recently written advises. Few, I think, actually study the odds or much of an idea how to calculate them, though most are familiar with the payouts.

    I remember when computers first came out, people were told that everyone would have to learn simple programing. Later there was a program to milk cows and I heard, "The farmer does not want to learn even one single line of programing." For dice I think it is pretty much the same, the average player does not have the self-assurance to trust his own attempts at calculation.

    In fact, quoting "Scarne on Dice," p99....(As proof of the fact that) most gamblers and some dealers know almost nothing about figuring percentages (let me tell.......)
     
    Last edited: Dec 25, 2005
  7. Dec 28, 2005 #6
    It makes sense though, to find the probability of getting 1 six in a roll of 4 die would be exactly as robert Ihnot explained. Find the total number of possibilities that can be rolled subtracted from the possibilities that cannot roll a six at all. That will give you AT LEAST one six.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Interesting question about throwing of dice
  1. Throw three dice (Replies: 2)

Loading...