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[PSL 8] Week Seven


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Just now, kevola said:

Yeah, it's really counter intuitive. Idk why I brought it up 

I hate it because when ever i think about it with 3 doors its still just 50/50 in my mind

Just now, gbwead said:

I always pretended I understood the solution, but in reality I think it's bullshit.

the point is if you upscale the problem to say 100 doors, you pick a door, then they remove 98 incorrect doors, obviously you switch doors because the odds go from 1/100 to 99/100 

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Just now, gbwead said:

I always pretended I understood the solution, but in reality I think it's bullshit.

You can go on deal or no deal and beat the banker when you understand it

 

If you run a simulation for it, the probability will most likely get closer and closer to 66.67% if you switch

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Alright here is a little lesson because today is one of my good days : 

 

The intersection of two events is when you can say Event1 AND Event2 (it is used here to represent the turns one after the other)

Let's call P a function which gives the probability (takes values between 0 and 1)

2 events A and B are independant is equivalent to : P(Event1 intersection Event2) = P(Event1) x P(Event2) 

 

The union of two events is when you can say Event1 OR Event2 (Inclusive OR)

It's always true that : P(Event1 union Event2) = P(Event1) + P(Event2) - P(Event 1 intersection Event2)

If the events are disjoint, then you can only keep the addition because P(Event 1 intersection Event2) = 0

 

This explains the calculation of lord gbweak

 

 

Don't thank me tho

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1 minute ago, XPLOZ said:

Alright here is a little lesson because today is one of my good days : 

 

The intersection of two events is when you can say Event1 AND Event2 (it is used here to represent the turns one after the other)

Let's call P a function which gives the probability (takes values between 0 and 1)

2 events A and B are independant is equivalent to : P(Event1 intersection Event2) = P(Event1) x P(Event2) 

 

The union of two events is when you can say Event1 OR Event2 (Inclusive OR)

It's always true that : P(Event1 union Event2) = P(Event1) + P(Event2) - P(Event 1 intersection Event2)

If the events are disjoint, then you can only keep the addition because P(Event 1 intersection Event2) = 0

 

This explains the calculation of lord gbweak

 

 

Don't thank me tho

TY XPLOZ

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Just now, kevola said:

You can go on deal or no deal and beat the banker when you understand it

 

If you run a simulation for it, the probability will most likely get closer and closer to 66.67% if you switch

Ya, I know, but still don't get it. I feel it goes agaisnt the independance law XPLOZ just talked about.

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11 minutes ago, XPLOZ said:

Alright here is a little lesson because today is one of my good days : 

 

The intersection of two events is when you can say Event1 AND Event2 (it is used here to represent the turns one after the other)

Let's call P a function which gives the probability (takes values between 0 and 1)

2 events A and B are independant is equivalent to : P(Event1 intersection Event2) = P(Event1) x P(Event2) 

 

The union of two events is when you can say Event1 OR Event2 (Inclusive OR)

It's always true that : P(Event1 union Event2) = P(Event1) + P(Event2) - P(Event 1 intersection Event2)

If the events are disjoint, then you can only keep the addition because P(Event 1 intersection Event2) = 0

 

This explains the calculation of lord gbweak

 

 

Don't thank me tho

Yeah, I get that, but I still don't understand how Gb did it.

It's coming up to 2am here so maybe I should just look at it in the morning, I'm probably just being silly 

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8 minutes ago, kevola said:

Yeah, I get that, but I still don't understand how Gb did it.

It's coming up to 2am here so maybe I should just look at it in the morning, I'm probably just being silly 

=(1/16)+((1-(1/16))*(1/16))+(((1-(1/16))^2)*(1/16))+(((1-(1/16))^3)*(1/16))+(((1-(1/16))^4)*(1/16))

 

Chance to crit on the first try: (1/16)

Chance to crit on the second try: + ((1-(1/16))*(1/16))

Chance to crit on the third try: + (((1-(1/16))^2)*(1/16))

Chance to crit on the fourth try: + (((1-(1/16))^3)*(1/16))

Chance to crit on the fifth try: + (((1-(1/16))^4)*(1/16))

 

Since the chance to crit on the second try only matters if you don't crit on the first try, you need to take it into account when you calc the chance to crit on the second try. Same logic applies for the third, fourth and fifth tries. I hope that makes sense.

 

4 minutes ago, kevola said:

You can't apply the same law (Independence) I think, as it's conditional probability

XPLOZ just explained that to me. I get it now thanks.

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5 minutes ago, Moetal said:

Did we just went through 7 pages of Nik vs Kimi debates?

no we've moved past that and now we're discussing mathematics. i'm certain that at this rate we'll have founded a new religion by page 20 and then reached the moon at page 23. 

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3 minutes ago, Rigamorty said:

no we've moved past that and now we're discussing mathematics. i'm certain that at this rate we'll have founded a new religion by page 20 and then reached the moon at page 23. 

I ain't dragging Eric to the moon. No one can carry that trashboi.

 

Edit: Inb4 Eric tries to fold a piece of paper on itself 42 times in order to reach the moon...

Edited by gbwead
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