Hey this is Presh Talwalkar
A standard dice has the numbers 1 to 6 each appearing with the probability of 1/6
If the dice is modified, so the number 6 appears with half the frequency of the other numbers
What is the new probability of rolling a 6?
I want to mention, I came across this problem in a new book "Power-Up" by Matthew Lane
The book uses video games to discuss really interesting mathematical concepts
and I learned quite a few new things
I did receive a free advanced copy of the book for review
and it's really a great book! You should check it out. There's a link in the video description
So returning to the problem
Can you figure it out?
Give this problem a try and when you're ready keep watching the video for the solution
In a standard dice, the probability of rolling the 6 Is equal to 1/6, which is the same as rolling any other number
In this loaded dice, we want the probability of 6 to equal half the probability of rolling any other number
So one way you might think about solving the problem is that you might simply decrease the probability in half
That is, you would say the probability of rolling a 6 will be 1/2 times 1/6, which equals 1/12
But there's a problem when you do this
If you add up all the other sides, which still have a probability of 1/6,
then the total probability will be too small
It will not add up to 1
You will actually get that the probability of these numbers will be equal to 11/12
So this is the wrong answer because we need the sum of all the probabilities to be equal to 1
But no worries there's a way to fix this
Since we need the probability to be 1, we can scale each term by 12/11
This is so that we have 11/12
being equal to 1 once we scale it
and if we scale all the other terms by the same factor, we'll get the correct probability
so 12/11 times 1/12 is equal to 1/11
and 12/11 times 1/6 is equal to 2/11
and when we sum these new probabilities,
we do in fact get that the total probability is equal to 1
so this is the correct answer
The probability of rolling a 6 will be equal to 1/11
but the lesson here is that you can't simply adjust one probability
if you decrease the probability of rolling a 6 you have to compensate
by adjusting all the other probabilities
There's another way you could've come up with this answer
Let's suppose P is equal to the probability of rolling a 6
We then have the probability of rolling any other number will be equal to 2P
Now we need the total probability of all the rolls to be equal to 1
so, we can algebraically add up these variables
We have one term of P and five terms of 2P to be equal to 1
This will be a total of 11P that's equal to 1, which means P is equal to 1/11
So we get to the same answer but it was done in an algebraic way
rather than scaling up the probabilities and adjusting
There's another way you could have solved this problem
You could have also thought about the odds of the rolls or the relative probabilities
So the odds of rolling a 6
will be 2 to 1 against rolling a 6. It'll be twice as likely to roll any other number than rolling the 6
Now in order to convert these odds into probabilities, what we do is we add up all of these numbers
We have one for the six and we have five terms of two
This will be equal to 11
In order to convert these odds into probabilities, we divide each of the relative odds by 11
so we again get to the same answer that the probability of rolling a 6 is equal to 1 divided by 11
Did you figure out this problem?
and which method did you use?
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You can catch me on social media at Presh Talwalkar and if you like this video please check out my books.
There are links in the video description