Prove it - Ep3: Terminating Dice Sums

00:10:04
https://www.youtube.com/watch?v=YKE5SRXK_nI

Summary

TLDRIn this engaging walkthrough, the narrator explores the expected number of times one has to roll a standard six-sided dice to achieve a sum of six. The estimation shows an expected value slightly above two rolls, specifically around 2.16, when applying probability theory and combinatorics. The video employs techniques such as the stars and bars method and the Christmas stocking theorem to determine the probabilities of different roll outcomes. It concludes with a mathematical thought experiment regarding whether it's better to roll first or second in a dice game aiming to reach a sum of six. The viewer is encouraged to apply these principles independently.

Takeaways

  • 🎲 Expected rolls to sum six: slightly above 2.
  • 🔢 Uses stars and bars method for probability calculation.
  • 📚 Introduces Christmas stocking theorem for combinatorial sums.
  • 🤔 Thought experiment: Roll first or second in a dice game?
  • 🧮 Generalizes the method using binomial theorem.
  • 🎯 Highlights an intuitive understanding of expectation.
  • 🧊 Demonstrates probability calculations using dice.
  • 👨‍🏫 Encourages deeper exploration of geometric series.
  • 🔍 Connects to exponential number e in an extended example.

Timeline

  • 00:00:00 - 00:10:04

    The video is about understanding the probability and expected values related to rolling a six-sided die until the sum reaches at least six. It begins by breaking down the probability of stopping at each roll and determining that a significant number of scenarios terminate by the second roll. The expected number of rolls to reach a sum of six using dice is calculated through a mathematical approach involving cumulative probabilities and binomial coefficients, simplified with mathematical theorems like the Christmas stocking theorem. It concludes by deriving the expected number to be around 2.16 rolls, presenting a challenge question about strategy in probability games.

Mind Map

Video Q&A

  • How many times do you expect to roll a dice to reach a sum of six?

    The expected number of times is slightly above two rolls, around 2.16.

  • What mathematical concepts are employed in this video?

    The video uses probability theory, combinatorics, and concepts like the stars and bars method and the Christmas stocking theorem.

  • How is the probability that multiple dice rolls sum to five calculated?

    The stars and bars method is used to find ways to split a number as a sum of positive integers.

  • What is the 'Christmas stocking theorem'?

    It is a method for finding the sum of a diagonal in Pascal's triangle, useful for probabilities in combinatorics.

  • Does the video include practical exercises?

    Yes, there is a challenge to determine whether you prefer going first or second in a dice game.

  • Can this solution approach be generalized?

    Yes, the video explains a method to generalize finding expected rolls for any number n using binomial theorem.

  • What was the philosophical question asked at the end of the video?

    It asked whether you'd prefer to go first or second in a dice game where the sum reaching six wins.

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  • 00:00:01
    [Music]
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    hi good to have you back at this point
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    your legs are probably a toucha from all
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    the random walking we've been doing so
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    let's take a break to roll some dice
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    [Music]
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    [Music]
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    I'm going to keep rolling a standard
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    sixer dice tracking the sum of my dice
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    rolls as I go and I'm going to stop
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    rolling as soon as the sum is at least
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    [Music]
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    six how many times do I expect to roll
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    this
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    dice probably not many times what do you
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    think maybe like two or three rolls
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    let's do our usual thing of seeing what
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    we can get without trying too hard for
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    our first rle there is a one and six
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    chance that we get a six so it's pretty
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    unlikely we hit the stopping threshold
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    of six on roll one however with two
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    roles the sample space of the sum is
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    this familiar from high school
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    grid we can see that it is more likely
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    than not that we have terminated by the
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    second
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    roll actually there's less than a one in
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    three chance we make it past two rolls
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    what is the chance we make it past roll
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    three very small to keep going Beyond
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    three rolls the first three rolls are
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    required to sum to at most five this
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    forces all of the dice rolls to be quite
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    small in value in fact none of them are
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    allowed to be four or greater and that
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    knocks off half the outcomes for each
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    dice we can use this to quickly NAB an
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    upper Bound for the probability here
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    each dice has a 1 half chance of rolling
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    a number less than four so there is a
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    one on eight chance that we roll three
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    Dice and they all show something less
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    than four we can already see that the
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    bulk of this roll distribution is close
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    to two rolls and so that's probably
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    going to be around where the expected
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    value is and will the actual answer be
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    slightly above or slightly below two see
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    if you can reason your way to a decision
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    on this
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    it's time for us to prove it the
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    solution is kind of neat and I'm going
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    to keep things fairly General so that
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    you can extend the solution later on to
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    do things right we will let X denote the
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    thing of interest that is X is the
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    number of times the dice has to be
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    rolled so that the sum of the rolls is
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    at least six the formula for the
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    expected value of x looks like this we
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    take each outcome K that's the number of
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    RS it takes for the sum to be six or
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    greater and we multiply by its
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    Associated probability and then we take
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    the sum of all of these now in our case
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    we know that it will take at least one
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    roll and at most six rolls so we can
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    attach a range of 1 through six to our
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    [Music]
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    sum let me show you a little trick that
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    comes in handy with such calculations
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    it's easier to see if we first write the
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    sum in full and then stack it all up
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    like a triangle now what are the sums of
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    each of these columns each column sum is
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    in itself a sum of probabilities and
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    therefore a probability for example the
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    First Column is the probability that X
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    is greater than zero the second column
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    is the probability that X is greater
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    than 1 and so on therefore we have
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    established that the expected value of x
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    is equal to the sum of these cumulative
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    probabilities as k goes from 0 to 5 why
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    did we do this well it turns out in this
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    case that these cumulative probabilities
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    are easier to calculate let's take a
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    look with an example say we want to
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    calculate the probability that X is
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    greater than three then we just need to
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    count all of the ways we can roll three
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    Dice and still not have gotten to six
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    yet for then we know that X is greater
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    than 3 basically I need to count
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    solutions to this inequality
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    where dice one dice two and dice three
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    are the values of the three dice I've
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    rolled of course I will eventually need
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    to divide out by all of the possible
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    dice rolls to transfer over to
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    probability land to reduce further we
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    can just count the solutions to each of
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    the following equations now we could
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    Brute Force this but I want to do this
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    in such a way that you can see how you
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    might structure the general solution
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    first three dice must sum to at least
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    three so we only care about the bottom
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    three equations and now we have a more
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    General mathematical thing to work out
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    given some positive integer M how many
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    ways can I express it as the sum of
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    three positive
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    [Music]
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    integers the so-called stars and bars
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    method Works a treat I'll illustrate it
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    with the bottom equation how many ways
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    are there to write five as the sum of
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    three positive integers
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    well simply take five stars we want to
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    split this up into three smaller pieces
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    and we can do this by inserting two bars
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    into the four possible spaces between
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    the
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    stars for example something like this
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    illustrates that 1 + 2 + 2 is a way of
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    summing to five therefore to count all
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    possible ways that three dice can sum to
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    five we just count all possible ways we
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    can insert two bars into the above row
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    of stars well there are four spaces so
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    the answer is four choose
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    [Music]
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    two now in general this approach lifts
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    and the number of solutions to each
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    equation can be found reminder our
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    current goal is to find the probability
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    that X is greater than 3 as such we
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    basically need to add up the above
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    solution counts and divide by how many
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    possible ways there are to roll three
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    dice it won't come to you as news that
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    there are 6 to the power of three
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    possible outcomes from three dice rolls
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    so we can see that the probability that
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    X is greater than 3 is equal to 1 6
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    cubed multiplied by 2 choose 2 plus 3
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    choose 2 plus 4 choose 2 summing
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    binomial coefficients it feels like
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    we're back in video one we are not not
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    these are a little bit different and we
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    can actually use something called the
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    Christmas stocking theorem to smush this
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    sum into a single
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    coefficient the Christmas stocking
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    theorem tells you that whenever you take
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    a diagonal sum of numbers in Pascal's
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    triangle starting from a top edge of the
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    triangle you can find the answer to the
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    sum by completing the Christmas stocking
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    it's also sometimes called the hockey
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    stick
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    theorem thus for our sum we can simply
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    rewrite it as five choose
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    [Music]
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    three I'll leave the details to you but
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    you can generalize this method to find
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    the other probabilities for our dice in
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    a similar way you can show that the
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    probability that X is greater than K is
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    equal to 1 6 K * 5 choose k for K
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    ranging from 0 to
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    5 we done the analysis on the
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    probability and we now apply the direct
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    substitution oh is there anything we can
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    do with this indeed there is and I'll
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    give you a two-word hint binomial
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    theorem we can recognize that our
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    expression is what you get if you expand
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    1 + 1 16 all to the power of
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    5 ah thus we have the answer and this
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    comes out around 2.16 or something like
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    that which is pretty close to where our
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    intuitive estimate placed
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    us okay if you followed all of that well
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    done and if your Curiosity has flared up
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    looking at this expression then well
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    done again because it looks very close
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    to the form 1 + 1 on N all to the power
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    of n which many of you know tends to e
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    the exponential number as n tends to
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    Infinity okay our has a slightly
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    different form but you can prove in a
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    couple of lines that this still tends to
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    e and actually you can use what we've
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    learned today to show that the expected
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    number of times an insid a dice must be
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    rolled so that the sum of the rols is at
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    least n is equal to 1 + 1 on N all to
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    the power of
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    [Music]
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    nus1 finally here's something a bit more
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    challenging for you to work on and send
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    in you and your friend take turns
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    rolling a six-sided Dice and you keep a
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    joint running total the player who first
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    gets the sum to six or greater gets paid
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    out the sum my question is do you want
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    to go first or second when you get the
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    answer make sure you prove it as usual
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    send it in for me to look at until next
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    time enjoy the maths
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    [Music]
Tags
  • probability
  • dice
  • expected value
  • combinatorics
  • Christmas stocking theorem
  • stars and bars method
  • binomial theorem