in the dice problem, solution can be worded even simpler. it is given that prob of rolling 1,2,3 is twice that of 4,5,6. so lets assume prob of 4 = prob of 5= prob 6 = p. then prob of 1= ..of 2 = ..of 3= 2p (given). now, we know that total prob must be 1. ( i mean when you roll the dice, you are bound to get a number from 1 to 6, so). therefore, 2p+2p+2p+p+p+p=1 which gives, p=1/9. now we can find all sorts of occurrence prob. i mean you did the same thing, mine just looks easier to me lol. but your method is ofc more useful when problems get harder , intuition fails many times. anyways, good vid.
You are correct! Essentially, your approach is the same as what I do in the video, just based more on intuition and less on mathematical rigor. During my daily work as an engineer, my solutions look much more like what you're writing in this comment and less like what it looks like in this video ^^ However, as you point out, intuition only takes you so far, and it can be misleading. Many times, I convince myself of a solution at the white board, and then when I go to implement the resulting algorithm in real life, it doesn't perform as expected. Usually, because I made some mistake in the intuitive solution. Then, you have to go through the slog of looking at every detail and figuring out where your intuition failed :P
I found the probability by thinking of a dice probability being 1/6. Since P(omega) needs to add to one we get 6*(1/6) = 1. I then replaced the 6 with x. 6*(1/x)=1 and expanded 3*(1/x) + 3*(1/x) = 1. I then multiplied the first term by two since 1,2,3 are twice as likely. and got 6*(1/x) + 3*(1/x) = 1. solved for x and got x = 9. Even though they are pretty much the same thing, I like your way of thinking about this solution better haha.
If the die was a normal die (all six sides show up with equal probability), then the answer 1/2 would be correct! However, note that the die is not fair. This die is loaded such that the outcomes "one", "two", and "three" turn up twice as often as the remaining three outcomes. I describe the problem at 0:37 Intuitively, since there are more odd numbers than even numbers in the set {1,2,3}, we should expect even numbers to show up less often than 50% of the time.
![[Discrete Mathematics] Axioms of Probability](http://i.ytimg.com/vi/xuv6BCR-iNc/mqdefault.jpg)
in the dice problem, solution can be worded even simpler.
it is given that prob of rolling 1,2,3 is twice that of 4,5,6. so lets assume prob of 4 = prob of 5= prob 6 = p. then prob of 1= ..of 2 = ..of 3= 2p (given).
now, we know that total prob must be 1. ( i mean when you roll the dice, you are bound to get a number from 1 to 6, so).
therefore, 2p+2p+2p+p+p+p=1 which gives, p=1/9.
now we can find all sorts of occurrence prob. i mean you did the same thing, mine just looks easier to me lol. but your method is ofc more useful when problems get harder , intuition fails many times.
anyways, good vid.
You are correct! Essentially, your approach is the same as what I do in the video, just based more on intuition and less on mathematical rigor. During my daily work as an engineer, my solutions look much more like what you're writing in this comment and less like what it looks like in this video ^^
However, as you point out, intuition only takes you so far, and it can be misleading. Many times, I convince myself of a solution at the white board, and then when I go to implement the resulting algorithm in real life, it doesn't perform as expected. Usually, because I made some mistake in the intuitive solution. Then, you have to go through the slog of looking at every detail and figuring out where your intuition failed :P
very cool
I found the probability by thinking of a dice probability being 1/6. Since P(omega) needs to add to one we get 6*(1/6) = 1. I then replaced the 6 with x. 6*(1/x)=1 and expanded 3*(1/x) + 3*(1/x) = 1. I then multiplied the first term by two since 1,2,3 are twice as likely. and got 6*(1/x) + 3*(1/x) = 1. solved for x and got x = 9. Even though they are pretty much the same thing, I like your way of thinking about this solution better haha.
That's awesome, Rashawn! Finding a new way to solve the problem really reinforces that you understood it.
i could not understand that why the Last question's answer is 4/9 instead of 1/2? i think it is supposed to be 1/2. can you explain it to me.
If the die was a normal die (all six sides show up with equal probability), then the answer 1/2 would be correct!
However, note that the die is not fair. This die is loaded such that the outcomes "one", "two", and "three" turn up twice as often as the remaining three outcomes. I describe the problem at 0:37
Intuitively, since there are more odd numbers than even numbers in the set {1,2,3}, we should expect even numbers to show up less often than 50% of the time.
@@henrikhellstrom1241 oh yes now i understand, i missed that point i guess.
thank you for replying to me.