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[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]
