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Hello I will come again.

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This lesson is going to be an exercise on what I want you to do is this I want you to write the program

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that the rules a dice for one million times and put uses a number between 1 to 6 just like hard Dyce's.

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And that tells us either the numbers of 1 to 6.

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How many times they showed up.

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So once again run a dice or roll the dice for one million times and find out how many times they'd want

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to show up.

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How many times did you show up on the Trishul.

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So on and so forth.

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Now to get us started I'm going to give you one piece of hands or one piece of advice essentially this

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jump into X hold in the expert I'm going to start an application and that's going to be a command line

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tool.

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I'm going to call it dice project and the dice project I'm going to go ahead get my.

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Clean it up better formatting and everything in here I'm going to show you one thing and that is if

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I have ints random for instance random instead and that's ARTC for random this will essentially will

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produce a random number for me.

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That's all I'm going to tell you.

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So if I say slight whatever you find in the value random ints and 5 run bill then run this application

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it will show you that it has produced one a random number.

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Let's see that together the random number is minus 10 billion the world is read only again this time

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is going to give us a different random number.

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That's all I'm going to tell you.

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So you can use the arc for random to produce a random number and based on that.

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Here is your task.

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You have to essentially write a program that runs the dice for a one million times and tells us how

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many times 5:59 showed up.

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So pause the video here.

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Given the choice on your own.

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And then once you're done you can just you know work along with me and you can see how I'm going to

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do it again.

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So I'm going to go back to school and I'm going to show you the solution myself in scores.

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I'm going to do this.

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I'm going to say my random end which is ARTC for Andong.

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And it produces this.

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Let's do this.

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So many times so for eons it is zero or a smaller than let's say 100 times for an hour and not a million

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times and then I'm going to go ahead and say let's reinvent everything clean them up.

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Number one this is showing me so any random numbers 100 random numbers but none of them is essentially

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what I want which is within a range of 1 to 6.

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So what I'm doing is I'm going to go ahead and say this random number that you produce.

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Let's get a marginal loss of faith against six.

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If we do this we get numbers that are between zero to five.

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So we get 0 1 2 3 4 5.

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Any of those numbers will show up.

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But that's not what they want either.

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Simply because I want numbers of 1 to 6 I want the ones I want sixes.

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I also don't want any zeros such as this one that's going up in here.

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So what I'm going to do is I'm going to add the one tweet in here.

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I don't remember because of the presidents of the operations this first happens first then that the

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spot happens.

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If it is difficult Remember you can just pantheists this on here.

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So you remember what's going on or you could actually do it the other way on say one plus the odds.

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So it's going to produce any number between 0 2 5 1 1 2 8 so essentially becomes any number between

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1 to 6.

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That's Rhonda.

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So now I have all of these different results 5 6 2 6 4 6 1 3 3 to you.

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All the numbers are crunched between 1 to 6 which is acceptable numbers for a dice.

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Now the next thing we have to do is we have two accounts that we have to find out how many times have

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one happen when it comes to who's up in threes.

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So on and so forth.

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So for that I'm going to go ahead and say I'm going to call it count of one that is currently zero and

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counter 2 is also currently 0.

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Let me actually call it this.

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It's going to be easier for me.

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So counter 3 is also 0 count four is also zero and then I count five.

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And then I have finally counter 6 so these are the different counters that each of them is going to

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count.

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One of the many numbers of the of the dice.

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So when I go through my for loop I'm going to go ahead.

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I'm safe.

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If the random INS was one if I won it what should I do.

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I should say counter one increment itself and I don't mean to print the values exactly there so I'm

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going to draw that.

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So let me just get myself a little bit further spacing in here because this is going to be a bit of

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a longer call.

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I want to make sure that you know I can somehow zoom into everything for you guys.

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And now I have to write this if statement.

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I know that five times.

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So I have to write else if how random it was to then obviously counter to gets incremented it duplicated

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so many times.

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Coffee coffee coffee coffee coffee.

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So if it was true if it was theory then sorry that I was yes.

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If it was 3 then three if it became four Dan four if it was five then five.

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And then I don't need to really check for six because obviously if it is not one to five it's certainly

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the sixth one.

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So like this I'm going to run for a loop a hundred times and 100 times I'm going to produce a random

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number between 1 to 6 and then I'm going to count them once the for loop is done.

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And I have to print them I'm going to say and that's long and I'm going to say what happened.

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How many times and times and for that I'm going to use counter one.

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So one happened counter one times and of course we have to duplicate this six times as well.

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So I'm going to say two three four five six.

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And of course that is going to be two three four or half and capital F 5 6 happened that too many times.

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So let's run let's see if we get what we want to get.

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So it says 1 happened 11 times to have it 19 times so on and so forth.

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Let's run this for a million times.

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That is one hundred thousand that's 1 million times so almost all of them are happening within Iran.

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One hundred sixty six thousand that's kind of makes sense because they all should have a similar chance

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of showing up again.

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Now that this is working I'm going to keep this up in here for a second so if you want to you know look

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at my quote for any reason you can pause there be doing here because I'm going to change all of these

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and I'm going to rewrite this record so I'm going to pause here for a few seconds and then I'm going

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to rewrite this statement with a switch statement.

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OK.

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Let's get to it.

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So instead of this knowing if else if else if else if a situation delete all of that I'm saying when

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you find a new random number switch the switch not the random number random.

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And in case it was a one obviously counter one odds to itself and breaks.

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And now let's duplicate that case one more time.

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It didn't.

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OK.

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Two three four five and six.

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So if when we switch if you got two then counter two should go up.

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If we got three then counter two should go up.

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If we got four then counter four would go up five out of five.

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I have six counter six.

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And then of course at the end again you have two princedom we have to find out how many times counter

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1 happened how many times counter to this fourth essentially is going to stay intact.

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So let's run this go it is working perfectly fine this time as well.

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Like I said when we were discussing the switch lesson switches not necessarily less code is just more

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organized so you can see it easier in this age if cases when you do this thing.

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And if that's the case the break we don't continue with any of these.

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And because of that its performance is a little bit better than an if else statement.

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So this is writing it with a switch.

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So I'm going to keep it here for a second.

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And you know what just so that you can all you can see all of the content on my screen.

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I'm going to drop that 3 4 5 and 6 so you can see it only for the first three.

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And you know the rest of them are exactly identical.

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So I'm going to pause the video here for a second and then I'm going to rewrite this another way.

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OK.

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So now so far we managed to simplify some parts of our switch and now I want to simplify our counters

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so I count them individually.

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One two three four or five.

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And that doesn't really seem like a very intuitive way of doing it.

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So I'm going to delete all of this for now I'm going to delete all of that.

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As that as I'm going to delete all of these guys for now.

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I'm going to go ahead and say I have only one integer up there called Count headers and that is an array

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field 6 I'll use.

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Which are all of the current Sile's zero.

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So these are my counters so my counter's is an array.

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This first one is for one.

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This one is for two.

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This one is for three.

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This one is full for this one for a while.

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This one is for six.

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Now here is what they wanted to do I want to say if I got one then this first one should go up.

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Should become one.

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If I got one again then the first one should become two three so on and so forth not elements in the

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array should increase themselves as they should increments themselves as we find new dice numbers that

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they represent them.

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However I have a problem and that problem is this the end I don't see my array are numbered 0 1 2 3

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4 5 1.

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I actually look for 1 to 6.

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So because of that I'm going to change my counters to 7 I'm going to imagine that is one at the very

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beginning that I don't use for anything.

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And then there's another one at the end as well.

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So now I've got six elements in my array.

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The first one I'm going to completely ignore.

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Let's play that minus one.

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So we know that that one is always going to be ignored.

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I don't use it for anything.

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So now if in my switch I get chased one I can say counters one increment yourself if case is on by two

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to what am I typing.

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And then counters 2 will increment itself.

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And of course we break after we do the same thing for all six of them.

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So for now I'm going to say two three four five six.

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So in case of three then that becomes three in case of four.

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Then Number four increases itself in case of five then five increases it's not a case of six then six

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increases it's.

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So now that they can count up correctly how am I going to show them I'm going to say let's go outside

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the for loop.

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So that's here.

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I'm going to say I'm not there for loop this time for it and let's call it C equals 1 because we start

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from 1 and then see the smaller require six.

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That's the last object in here.

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I'm going to say for that C++ to do this and this log

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something happens something times.

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So what is the first something the first thing you see for instance or what happens how many times it

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happens count thare's one time which is see again.

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So let's see what I have written.

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It says go from 1 to 6.

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Say for instance if it is 1 1 happens counter of 1 times 2 happens counter of 2 times so on and so forth.

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Let's on this it says one happens one hundred sixty six thousand or so on and so forth all the way to

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solve it quickly.

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What we have done in my array I have 7 elements ignore the first one from 1 to 6.

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Each of them are zeroes.

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Whenever we find one count there one this guy counts up.

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So for instance this one becomes one and we find that two.

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This one becomes one when we find another one.

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This one becomes two when we find another one.

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This one becomes three so on and so forth they count themselves until the for loop is finished.

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So let me reset these viruses and let's try to simplify it to one last time.

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So when I look at my statement it feels like one case is one counter one counts up in cases to counter

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two counts up all the way to the end.

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So how about I do this.

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How about I say Well when you produce a random number and remember that number is either 1 2 3 4 5 or

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6.

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I'm going to say when you produce a random number say counters elements at that location or random in

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location increments yourself all right.

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Let me drop one of these and I have to re-explain it really quickly for you.

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So I don't hear any of that.

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I don't need any of the comments I'd rather opt out.

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And this is my code in its entirety.

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This is all you have to do.

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So let's see what's happening if you're doing something a million times.

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Every time we put using a random number between 1 to 6 and we have an array of counters that has elements

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from 1 to 6 they are all zeroes.

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So when I get the random numbers such as 5 it says counter's elements.

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Number five this line this one increased increment yourself.

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Increase yourself when I produce another random number.

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And that is don't know.

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To me it says this small increments yourself go up.

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So on and so forth until my entire for loop is finished.

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And then at the end I'm going to say print everybody who is there let's run this.

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So one happens one hundred sixty six thousand two happens so on and so forth.

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So as you can see we started from a more complicated approach with so many if else statements then we

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simplify it a little bit by adding a switch then we simplify it a little bit by removing all the separate

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counters and putting them all in array.

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And then we finally simplify it by not counting them based on if else's statements are rather based

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on their index in the array.

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So I'm going to keep this here for a second if you want to have another final look you know fix your

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code by any chance.

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And then I will see you in the next lesson.