Are there Mathematics models to determine probabilty of determinism vs randomness?

Discussion in 'Secondary Education' started by oldstudent, May 1, 2009.

  1. oldstudent

    oldstudent Comrade

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    May 1, 2009

    As an individual who has experienced countless events that have been sequenced, synchronized, and interwoven to create realities that I feel could never happen by mere chance, I have come to wonder whether mathematical models can determine the probabilty of events being random as opposed to determined by outside forces.

    I therefore pose the following question:

    If a coin were flipped three consecutive times, and each time it came up heads, one would assume that this was only coincidence and totally random; yet if the coin were flipped 50 consecutive times and came up heads each time, one would conclude that there is a causative factor to explain this phenomenom, since the chances of this happening by coincidence would be practically zero.
    I would therefore like to know if it can be mathematically determined how many consecutive coin flips of heads must take place before the chances of there being a causative factor is more likely than chance?
    If such a model or equation exists, does anyone know how to determine the answer?
     
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  3. MathManTim

    MathManTim Companion

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    May 1, 2009

    The outcome of flipping a fair coin is what we in the probability racket call an independent discrete random variable. For each and every flip of the coin, the probability of the coin showing heads or tails is always and forever 1/2 (or .5) regardless of what the coin showed in the previous flips. If you are going to flip a coin 10 times, then there are exactly 2^10 = 1024 outcomes, each equally possible.

    This may sound counter-intuitive, because right now you're thinking, "There is no way that 10 consecutive heads is as equally probable as 5 heads and 5 tails."

    And you would be correct. But hold on a second...

    There is only 1 way to flip 10 heads in a row: HHHHHHHHHH

    But there are 10 ways to flip 9 heads and 1 tail:
    THHHHHHHHH HTHHHHHHHH HHTHHHHHHH etc...

    Assuming that what matters to you is simply how many heads or tails you get--as opposed to the order in which they arrive--there is, in fact, a fairly straightforward formula for figuring it out. It's called the Binomial Theorem.

    Since the probability of heads = the probability of tails in this case, the Binomial Theorem simplifies considerably. What we need to know is how many ways we can flip the desired number of heads and tails. To figure that out in this case, use the following formula:

    Let n = the number of times you will flip the coin.
    Let r = the number of heads you want to show up.

    The number of ways in which heads can show up r times in n flips is given by:

    nCr = n!/[r!(n-r)!]

    ! is the factorial sign. 4! = 4 * 3 * 2 * 1, 6! = 6 * 5 * 4 * 3 * 2 * 1, etc.

    So, the number of ways in which 10 flips of the coin will give you exactly 5 heads is 10!/(5!5!) = 252.

    We mentioned before that for 10 flips there are 1024 different outcomes. We've just shown that exactly 5 heads will show up on exactly 252 of those 1024 outcomes. Therefore, the probability of getting exactly 5 heads on 10 flips of the coin is 252/1024 or only about 24.6% of the time. It's the most likely outcome, but 75% of the time you will not get exactly 5 heads.

    How about 4 heads? nCr = 10!/(4!6!) = 210
    How about 6 heads? nCr = 10!/(6!4!) = 210

    When we add these together, we see that 672 times out of 1024 we will get either 4, 5, or 6 heads. That works out to 65.6% of the time. Think about how this would look in reality. If you flipped a coin 10 times and it came up as 6 heads and 4 tails, or vice versa, would you really be all that surprised?

    What you first need to ask yourself is at what threshold would you be willing to cry "the fix is in!" If an event with a probability of 1 chance in 1000 occurred, would you conclude a fix, or would you accept the fact that people do get lucky and win the lottery on occasion?

    Here are the probabilities of getting all heads on specific numbers of flips of a fair coin:

    5 heads in a row - 1 chance in 32
    7 heads " " " - 1 in 128
    10 heads - 1 in 1024
    20 heads - 1 in 1,048,576
    30 heads - 1 in 1,073,741,824

    To generalize, 10 heads in a row is about a 1000-to-1 chance. For each additional 10 heads in a row, those odds go up about another 1000 times (1024 to be exact).

    30 heads in a row is 1 chance in a billion. 40 heads is 1 in a trillion, and 50 is 1 in a quadrillion.

    This entire argument, though, is predicated on the fact that you can actually measure the probability of the event in question. Would you be willing to be more specific about what event concerns you?

    MathManTim
     
    Last edited: May 1, 2009
  4. TeacherGroupie

    TeacherGroupie Moderator

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    May 1, 2009

    6! = 6 * 5 * 4 * 3 * 2 * 1, yes, Tim?

    I proofread because I can't not, but this is a fascinating discussion on which to eavesdrop.
     
  5. MathManTim

    MathManTim Companion

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    May 1, 2009

    Thanks for catching that! I went back and fixed it. Looks like my brain was ahead of my fingers right there.

    If you're trying to test the fairness of a coin, you would probably have to perform hundreds of trials to see if you get a different distribution than the expected one.

    MathManTim
     
  6. oldstudent

    oldstudent Comrade

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    May 2, 2009

    I am not a math person, but I do find it fascinating. I will have to examine the responses more thoroughly. I appreciate the thought and effort put forth.
    However, I am not sure my question has been answered, but this could be because there is no true answer, and thus my quest could be more philosophical than mathematical.
    I will therefore now present my question a little differently.

    How many consecutive flips of a coin that come up heads would be necessary before one could conclude that the chances are greater than 50% that the coin is rigged in some way (e.g heads on both sides, magnetized coin etc) as opposed to it coming up heads all the time by mere chance? Can the probabilty of a "normal vs. rigged" coin be mathematically determined, and if so, how many consecutive flips of heads would it take to conclude the chances are greater than 50% that the coin is rigged?
     
  7. HMM

    HMM Cohort

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    May 2, 2009

    You could always do a Hypothesis Test

    Hypothesis Test For a Proportion

    You could also look at the Standard Error
     
    Last edited: May 2, 2009
  8. American

    American Rookie

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    May 2, 2009

    I think that I got your question. Theoretical probability is also developed based the practical probability. To conclude that causative factor is more likey to occur you have to do a large number of trials and it should show that 50% probability of getting a H or T is false. i.e., As you increase the number trials the probability should move away from 50% rather than coming closer to 50% to get a H or T and in my experience it never happened when I used TI calculator to do a large number of trials of flipping a coin.
     

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