You have to be careful with probability scenarios. It is easy to find paradoxes and make mistakes. In this video, the material as presented for question b) is incorrect. Lizzie is correct - the answer is 1/6 not 1/11. Scenario b) as presented: 1) Two fair dice are rolled. Neither Eddie or students know the outcome. The sample space of outcomes to both Eddie & students are 36 possibilities. 2) Eddie randomly shows one dice. This turns out to be a 6 (it could equally have been any number 1 through 6). The sample space of possible outcomes for both Eddie & students now reduced to 6. The probability of the second dice being a 6 is now 1/6 (Lizzie's answer). The alternative scenario, which I think Eddie was intending to show is: 1) Eddie continually rolls both dice without showing the results to the students until Eddie sees that at least one of the dice rolls a 6. 2) Eddie tells the students what he has done. The sample space of outcomes for the students now has 11 possibilities. The sample space of outcomes for Eddie has been reduced to 1. 3) Eddie (knowing what the dice are) shows one that is a 6. 4) To the students, the probability that the second dice is a 6 is now 1/11. There is only 1 of the 11 possible outcomes with a double 6. It is easy to come across paradoxes in conditional probability questions - have a look at the famous Monty Hall problem which is well documented.
Why does this not work for scenario b) as presented? If Eddie randomly shows one dice which happens to be a six, it could still be either die, couldn't it?
This is very interesting point that has been brought up and is worthy of further discussion. Here is my guess. First of all, the way it is stated in the whiteboard is correct and the answer is 1/11. Based on Eddie's description to the class, I would say that we have insufficent information. Specifically, the information we require is what would've Eddie done had he not rolled a 6? Lets suppose he instead rolled a 1 and a 2 for example. If Eddie had pre determined that he would not show any dice to the class if neither of them were a 6, then the probability would be 1/11 as stated in the video. If instead, Eddie rolls a 1 and a 2 and then decides to reveal the die with the value 1 to the class, then the probability of the 2nd die being a 6 is not 1/11.
If you as the teacher label the dice A and B without the students watching, then roll the die until at least a six appears, show the die with 6, but dont tell them whether its A or B, its 1/11 (Which is what he shows) (Can either be dice A or B with 6, students dont know, so the array thingy shows that) If he tells them each dice has the six, then it becomes 1/6... crystal math
sample space = {1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6} independent events: reduced space = {1, 2, 3, 4, 5, 6} Why? Since we have one more dice left. P(two 6 | one 6) = 1/6 dependent events: reduced space = {1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, -} Why? Since we do not know if it's the first or second dice. P(two 6 | one 6) = 1/11
As a fellow Maths teacher; I love this channel (keep it up!!) for inspiration on how to explain/introduce topics. I'm not sure the example here as highlighted below is the clearest example as the action of rolling a die is an independent event which means you have to be careful with your words. Maybe it would be better to say; I have rolled a die and it's an even number. What is the probability of rolling a 6; given the information that an even number has been rolled?
I had to have a long hard think about this. At first I didn't think you could use Bayes' theorem in this case because rolling a dice is always an independent event. So I went and and did what every sane person would do. Make a spreadsheet with 10'000 random numbers to let the Law Of Large Numbers decide. I made a table consisting of 10 columns with 1000 rows in each with a RNG between 1-6. Paired them up in pairs of 2 to simulate a 2 dice roll. Next I went and made a similar table showing what what number number dice B had if dice A=6, and vice versa. Then all that was left was to do a count and add up how many you had of each* This is where the tricky part come in. When I did a normal count, it looked as if the answer was 1/6. Only after I looked at the data in my 2nd table I realize why this was. My count counted the numeric value of numbers in my dataset. If I had manually gone in and counted how many "pairs" there was, I would have gotten a slightly smaller number. The reason for this is that every time I got two 6's, my pair in the data set would say (6,6). While every other time it was a 6 and another number it said ("blank",1-5) or (1-5,"blank"). The double count had to be made into a single count. Dividing the number of 6's I got by two fixed this problem, and at that point the Law of large numbers showed that the probability was converging towards 1/11. It's a subtle difference, as it is with the Monty hall problem. Even though a dice is independent, you don't get to know what dice was a 6, just that a 6 is there. Hope my explanation made sense to someone.
hey for question (a) the probability should be 13/36 as the probability of one dice being 6 + probability of second dice being 6 + probability of both being 6. 1/6+1/6+1/36 = 13/36. the probability of both dice being 6 is calculated by 1/6 * 1/6 as they are events are independent events.
its 1/6+1/6-1/36, since the two 1/6 includes the 1/36 and is being counted twice if you dont subtract it. also P of zero 6 is 5/6*5/6=25/36 thus P of st least one is 1-25/36=11/36
"There is only a single way to throw 2 sixes" in that case there is only a single way to throw 1,6. 2,6, 3,6 etc as that is also the same as 6,1,6,2,6,3 etc ...
You have to be careful with probability scenarios. It is easy to find paradoxes and make mistakes.
In this video, the material as presented for question b) is incorrect. Lizzie is correct - the answer is 1/6 not 1/11.
Scenario b) as presented:
1) Two fair dice are rolled. Neither Eddie or students know the outcome. The sample space of outcomes to both Eddie & students are 36 possibilities.
2) Eddie randomly shows one dice. This turns out to be a 6 (it could equally have been any number 1 through 6). The sample space of possible outcomes for both Eddie & students now reduced to 6. The probability of the second dice being a 6 is now 1/6 (Lizzie's answer).
The alternative scenario, which I think Eddie was intending to show is:
1) Eddie continually rolls both dice without showing the results to the students until Eddie sees that at least one of the dice rolls a 6.
2) Eddie tells the students what he has done. The sample space of outcomes for the students now has 11 possibilities. The sample space of outcomes for Eddie has been reduced to 1.
3) Eddie (knowing what the dice are) shows one that is a 6.
4) To the students, the probability that the second dice is a 6 is now 1/11. There is only 1 of the 11 possible outcomes with a double 6.
It is easy to come across paradoxes in conditional probability questions - have a look at the famous Monty Hall problem which is well documented.
Why does this not work for scenario b) as presented?
If Eddie randomly shows one dice which happens to be a six, it could still be either die, couldn't it?
This is very interesting point that has been brought up and is worthy of further discussion. Here is my guess.
First of all, the way it is stated in the whiteboard is correct and the answer is 1/11.
Based on Eddie's description to the class, I would say that we have insufficent information. Specifically, the information we require is what would've Eddie done had he not rolled a 6? Lets suppose he instead rolled a 1 and a 2 for example.
If Eddie had pre determined that he would not show any dice to the class if neither of them were a 6, then the probability would be 1/11 as stated in the video.
If instead, Eddie rolls a 1 and a 2 and then decides to reveal the die with the value 1 to the class, then the probability of the 2nd die being a 6 is not 1/11.
That was my guess, part B is the Monty Hall problem
@@mmhhmm8374 that's in-line with my point.
If you as the teacher label the dice A and B without the students watching, then roll the die until at least a six appears, show the die with 6, but dont tell them whether its A or B, its 1/11 (Which is what he shows) (Can either be dice A or B with 6, students dont know, so the array thingy shows that) If he tells them each dice has the six, then it becomes 1/6... crystal math
sample space = {1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6}
independent events:
reduced space = {1, 2, 3, 4, 5, 6} Why? Since we have one more dice left.
P(two 6 | one 6) = 1/6
dependent events:
reduced space = {1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, -} Why? Since we do not know if it's the first or second dice.
P(two 6 | one 6) = 1/11
I now love this guy. Making a mistake shows he really is a human being.
As a fellow Maths teacher; I love this channel (keep it up!!) for inspiration on how to explain/introduce topics. I'm not sure the example here as highlighted below is the clearest example as the action of rolling a die is an independent event which means you have to be careful with your words.
Maybe it would be better to say; I have rolled a die and it's an even number.
What is the probability of rolling a 6; given the information that an even number has been rolled?
I had to have a long hard think about this.
At first I didn't think you could use Bayes' theorem in this case because rolling a dice is always an independent event.
So I went and and did what every sane person would do. Make a spreadsheet with 10'000 random numbers to let the Law Of Large Numbers decide.
I made a table consisting of 10 columns with 1000 rows in each with a RNG between 1-6. Paired them up in pairs of 2 to simulate a 2 dice roll.
Next I went and made a similar table showing what what number number dice B had if dice A=6, and vice versa.
Then all that was left was to do a count and add up how many you had of each*
This is where the tricky part come in. When I did a normal count, it looked as if the answer was 1/6.
Only after I looked at the data in my 2nd table I realize why this was.
My count counted the numeric value of numbers in my dataset. If I had manually gone in and counted how many "pairs" there was, I would have gotten a slightly smaller number.
The reason for this is that every time I got two 6's, my pair in the data set would say (6,6). While every other time it was a 6 and another number it said ("blank",1-5) or (1-5,"blank").
The double count had to be made into a single count. Dividing the number of 6's I got by two fixed this problem, and at that point the Law of large numbers showed that the probability was converging towards 1/11.
It's a subtle difference, as it is with the Monty hall problem. Even though a dice is independent, you don't get to know what dice was a 6, just that a 6 is there.
Hope my explanation made sense to someone.
wtf is happening to me... I am watching this at 4 am randomly without any reason
Hahaha....Me Tooo🤣
It is may be you got interest in learning math. which Indeed is a positive sign.
Incredibly cool video, love it !
Rapid Science thats true.
Maths with Muneer like your channel, just subscribed!
Rapid Science thank you so much for the amazing feedback. It means a lot to me :)
hey for question (a) the probability should be 13/36 as the probability of one dice being 6 + probability of second dice being 6 + probability of both being 6. 1/6+1/6+1/36 = 13/36. the probability of both dice being 6 is calculated by 1/6 * 1/6 as they are events are independent events.
its 1/6+1/6-1/36, since the two 1/6 includes the 1/36 and is being counted twice if you dont subtract it. also P of zero 6 is 5/6*5/6=25/36 thus P of st least one is 1-25/36=11/36
What time does your school start? It’s just gone 7.30 and already in class. Damn
Brian Holder lol good question
@Shreya T are you studying in his class ?
"There is only a single way to throw 2 sixes" in that case there is only a single way to throw 1,6. 2,6, 3,6 etc as that is also the same as 6,1,6,2,6,3 etc ...
You came to my school
I hope someday I will too.
Hey I’m from gpps
Vortex same
:D
You came to the right place.
@@primary_magic1227 I give the help to others :D
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