Confidence

Discussion in 'Probability and Statistics' started by Jonathan, Oct 3, 2021.

  1. Jonathan

    Jonathan

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    Example problem
    A casino offers you a gamble with a 1% chance of winning a try. How many tries will it take to win at least once?

    Solution
    For this example, I chose 95% confidence, a willingness to be wrong once in twenty:
    tries = log(chanceToBeWrong) / log(chanceOfFailureEachTry) = log(1/20) / log(99%) ≅ 300 tries

    But what about that decision to choose 95% confidence?
    It was an arbitrary choice, governed by nothing more than a general rule of thumb. Can we do better?

    Here's what I've figured out so far
    Confidence is a linear variable, a Real in the range 0-1.
    It can be expressed in this example problem as confidence = -0.0099 × tries + 0.9999

    If I choose confidence = 0.5, then I'm making a prediction of the future designed to be wrong half the time
    If I choose confidence = 1, then it would take infinite tries
    So the range of admissible values for confidence appears to be (0.5, ..., 1) excluding end points
     
    Last edited: Oct 3, 2021
    Jonathan, Oct 3, 2021
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  2. Jonathan

    Jonathan

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    Another scenario, to provide a different perspective

    A jar contains five marbles, one white and four black.
    I perform an experiment of three draws from the jar, putting the marble back after each draw, aka sampling with replacement.
    • Let a try be one draw of a marble from the jar of five marbles
    • Let success be the probability the drawn marble was white, and failure be 1 - success
    • Let risk be the probability of running the entire experiment of three tries and not getting a white marble outcome, and confidence be 1 - risk
    This is a Bernoulli trial where there are two possible outcomes each try, and three tries, so there are a total of 2^3 = 8 possible outcomes to the experiment.

    [​IMG]

    Of these eight possible outcomes, only the last one (Outcome #8 in the table) results in failure on all three tries. And the risk of this experiment resulting in outcome #8 can be calculated just like any of the outcomes as the product of it's component parts:

    risk = (4/5) × (4/5) × (4/5) = (4/5)^3 = 0.512

    The neat part of this is that we can represent the probability risk in the compact form (4/5)^3. The two component parts of this expression are (4/5) → failure and 3 → tries. So we can restate the above equation as:

    risk = failure ^ tries

    And from there we can re-arrange the formula so that given any two variables we can calculate the third

    risk = failure ^ tries
    tries = log(risk) / log(failure)
    failure = risk^ (1/tries)

    So in this scenario we were able to calculate confidence (1-risk) as a function of the probability of success (1-failure) each try and the number of tries.
     
    Jonathan, Oct 21, 2021
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  3. Jonathan

    nycmathguy

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    Is this a question or a lesson?
     
    nycmathguy, Oct 21, 2021
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  4. Jonathan

    Jonathan

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    It's a study of Confidence as clearly as I can express it, because I find the arbitrary selection of confidence level to be somewhat less than a measurement.

    If you can add to it then please do.
     
    Jonathan, Oct 23, 2021
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  5. Jonathan

    nycmathguy

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    Allow me to be honest with you. Probability is not my strongest math skill. I took probability back in high school. I am 56 years old.
     
    nycmathguy, Oct 24, 2021
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  6. Jonathan

    HallsofIvy

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    ??? This problem, as stated, has nothing to do with "confidence intervals"!

     
    HallsofIvy, Nov 8, 2021
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  7. Jonathan

    HallsofIvy

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    I hope you realize that as long as there is any chance at all, on each trial, of NOT winning, there is NO number of trials where you will "win at least once".
     
    HallsofIvy, Nov 10, 2021
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  8. Jonathan

    nycmathguy

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    Can you provide an example?
     
    nycmathguy, Nov 10, 2021
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  9. Jonathan

    HallsofIvy

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    Seriously? You flip a fair coin, you win if you flip a head. The probability you will flip a tail is 1/2. The probability you will flip two tails in a row is (1/2)(1/2)= 1/4. For any n the probability you will flip n tails in a row is 1/2^n. For large n that will be very small but there is NO n that makes it 0.
     
    HallsofIvy, Nov 17, 2021
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  10. Jonathan

    nycmathguy

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    The probability of flipping 4 tails in a row is 1/(2)^4.
    The same can be said for heads. Probability is one of my biggest weaknesses in math. Why? Most probability questions are what I call FUZZY applications. I plan to do a thorough study of probability after calculus 3. Currently, as you know, reviewing precalculus. One course at a time.
     
    nycmathguy, Nov 17, 2021
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  11. Jonathan

    HallsofIvy

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    Yes, the probability of flipping 4 tails in a row is 1/2^4= 1/16. The probability of flipping 4 heads in a row is also 1/16. The remaining 1- 1/16- 1/16= 1- 1/8= 7/8 is the probability of the other possible combinations heads and tails. There are 16 such combinations. They are
    HHHH
    HHHT
    HHTH
    HHTT
    THHH
    THHT
    THTH
    THTT
    TTHH
    TTHT
    TTTH
    TTTT

    Each of those 16 possible outcomes is "equally likely" so each has probability 1/16
    HHHH and TTTT both have probability 1/16
    HHHT, HHTH, HTHH, and THHH
    THH
    HTHT
    HTTH
    HTTT
     
    HallsofIvy, Nov 17, 2021
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  12. Jonathan

    nycmathguy

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    I look forward to someday learning probability to a comfortable level. What does "equally likely " mean?
     
    nycmathguy, Nov 18, 2021
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  13. Jonathan

    HallsofIvy

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    "Equally likely" means pretty much what it sounds like- two events are "equally likely" if the probabilities each is to happen are the same. A coin is called "fair" if flipping it, heads and tails have the same probability, 1/2. Rolling a single die, each of the six sides is "equally likely" to come up so each has probability 1/6.
     
    HallsofIvy, Nov 18, 2021
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  14. Jonathan

    nycmathguy

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    Thank you for making this clear to me.
     
    nycmathguy, Nov 18, 2021
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