Monday, July 17, 2006
P for Probability
The mathematical foundations of the ashram have been shaken. As one can imagine, this has caused a bit of a stir. It's all because of this post.
Now everybody's know that the name's Baba. Mental Baba. And that he likes it shaken, but not stirred.
However, since it IS stirred, here goes nothing.
WHAT THE HELL?
Let me set the record straight. I am an EXpert in probability and statistics. Because, as I mentioned in my previous post, I was awarded a P in this subject. P for PROFICIENT. By the most venerable (bows his head in silent reverence as a lump forms in his throat) Indian Institute of Technology, Kharagpur. Phds? Huh? What's that? We're talking P here. P. To put things in perspective, the probability that a person on the third rock from the sun knows probability at least as well as I do, is precisely 0.418 x 10^-8.
Having said that, I refuse to have my ass kicked by anybody. That includes chicks. Even one with an IQ of 198 (is that in the decimal number system?). I might, rather magnanimously, consider having it licked though. Preferably by ones with real IQ.
If this explanation were to be believed, it would be nothing short of sacrilege. It would be defiance of logic and mathematics and the scientifically-inclined Baba's way of life.
The solution appears to be one contrived for the sake of convenience.
I. Baba's first observation
P(O1) = P(C1) x P(O1|C1) + P(C2) x P(O1|C2) + P(C3) x P(O1|C3)
What does the problem state? It states that the third door is opened by the host and , GIVEN THAT SCENARIO, the probability of the car being behind door #2 needs to be determined.
So why consider the piece P(C3) x P(O1|C3) in the first place? There is no freaking P(C3) in question or in existence any more.
Therefore, P(O1) = P(C1) x P(O1|C1) + P(C2) x P(O1|C2)
And. Why would P(C1) or P(C2) be 1/3 now?
Two doors. One car. One goat. P(C1) = P(C2) = 1/2.
II. Baba's second observation
a) Now let's take a look at the definitions of P(O1|C1)and P(O1|C2). As depicted in the solution, I would relate them to the mislead that the host puts on the contestant. Going by that logic, P(O1|C1) = 0 and P(O1|C2) = 1.
Thus, P(O1) = P(C1) x P(O1|C1) + P(C2) x P(O1|C2)
= 1/2 x 0 + 1/2 X 1
= 1/2
b) Again, there is enough reason to dispute this. The host does not decide which door to open. He merely tries to mislead the contestant. Eventually, it is the contestant who tells the host which door needs to be opened. If the contestant is not a retard or does not think too much (because he/she do not have an IQ that measures up to Marilyn vos Savant's), they will remain uninfluenced by the host's gimmicks(which could go either way in order to influence them). This would set P(O1|C1) = P(O1|C2) = 1/2.
Thus, P(O1) = P(C1) x P(O1|C1) + P(C2) x P(O1|C2)
= 1/2 x 1/2 + 1/2 X 1/2
= 1/2
Ah, both scenarios throw up the same number.
III. Baba's third observation
a) If, and only if, we're talking about somebody which an extremely high IQ,
P(C2|O1) = {P(O1|C2) x P(C2)} / P(O1)
= {1 x 1/2} / {1/2}
= 1
What it means is that somebody with high IQ will always be conned by Monty Hall.
b) If, and only if, we're talking about an average Joe on the street,
P(C2|O1) = {P(O1|C2) x P(C2)} / P(O1)
= {1/2 x 1/2} / {1/2}
= 1/2
So that's my take on this 'problem'. But really, the problem would be if the contestant actually gets the car. With rising petrol prices, one may perhaps need to swap the car for a hefty goat in order to ride to work everyday.
Nightwatchmen and The Alternate Moebyus - I know that you were not Privileged enough to be awarded a P in this exalted branch of mathematics. However, this is what I ask of you - what did that idiot Bayes actually say?
And don't tell me it was "Baba, you just got your ass kicked."
On a different note though, Lisa, you may want to change the name of the Kick Ass Chicks Society to the Head Butt Chicks Society. The times, they are a-changing.
Management Class : News
Now everybody's know that the name's Baba. Mental Baba. And that he likes it shaken, but not stirred.
However, since it IS stirred, here goes nothing.
WHAT THE HELL?
Let me set the record straight. I am an EXpert in probability and statistics. Because, as I mentioned in my previous post, I was awarded a P in this subject. P for PROFICIENT. By the most venerable (bows his head in silent reverence as a lump forms in his throat) Indian Institute of Technology, Kharagpur. Phds? Huh? What's that? We're talking P here. P. To put things in perspective, the probability that a person on the third rock from the sun knows probability at least as well as I do, is precisely 0.418 x 10^-8.
Having said that, I refuse to have my ass kicked by anybody. That includes chicks. Even one with an IQ of 198 (is that in the decimal number system?). I might, rather magnanimously, consider having it licked though. Preferably by ones with real IQ.
If this explanation were to be believed, it would be nothing short of sacrilege. It would be defiance of logic and mathematics and the scientifically-inclined Baba's way of life.
The solution appears to be one contrived for the sake of convenience.
I. Baba's first observation
P(O1) = P(C1) x P(O1|C1) + P(C2) x P(O1|C2) + P(C3) x P(O1|C3)
What does the problem state? It states that the third door is opened by the host and , GIVEN THAT SCENARIO, the probability of the car being behind door #2 needs to be determined.
So why consider the piece P(C3) x P(O1|C3) in the first place? There is no freaking P(C3) in question or in existence any more.
Therefore, P(O1) = P(C1) x P(O1|C1) + P(C2) x P(O1|C2)
And. Why would P(C1) or P(C2) be 1/3 now?
Two doors. One car. One goat. P(C1) = P(C2) = 1/2.
II. Baba's second observation
a) Now let's take a look at the definitions of P(O1|C1)and P(O1|C2). As depicted in the solution, I would relate them to the mislead that the host puts on the contestant. Going by that logic, P(O1|C1) = 0 and P(O1|C2) = 1.
Thus, P(O1) = P(C1) x P(O1|C1) + P(C2) x P(O1|C2)
= 1/2 x 0 + 1/2 X 1
= 1/2
b) Again, there is enough reason to dispute this. The host does not decide which door to open. He merely tries to mislead the contestant. Eventually, it is the contestant who tells the host which door needs to be opened. If the contestant is not a retard or does not think too much (because he/she do not have an IQ that measures up to Marilyn vos Savant's), they will remain uninfluenced by the host's gimmicks(which could go either way in order to influence them). This would set P(O1|C1) = P(O1|C2) = 1/2.
Thus, P(O1) = P(C1) x P(O1|C1) + P(C2) x P(O1|C2)
= 1/2 x 1/2 + 1/2 X 1/2
= 1/2
Ah, both scenarios throw up the same number.
III. Baba's third observation
a) If, and only if, we're talking about somebody which an extremely high IQ,
P(C2|O1) = {P(O1|C2) x P(C2)} / P(O1)
= {1 x 1/2} / {1/2}
= 1
What it means is that somebody with high IQ will always be conned by Monty Hall.
b) If, and only if, we're talking about an average Joe on the street,
P(C2|O1) = {P(O1|C2) x P(C2)} / P(O1)
= {1/2 x 1/2} / {1/2}
= 1/2
So that's my take on this 'problem'. But really, the problem would be if the contestant actually gets the car. With rising petrol prices, one may perhaps need to swap the car for a hefty goat in order to ride to work everyday.
Nightwatchmen and The Alternate Moebyus - I know that you were not Privileged enough to be awarded a P in this exalted branch of mathematics. However, this is what I ask of you - what did that idiot Bayes actually say?
And don't tell me it was "Baba, you just got your ass kicked."
On a different note though, Lisa, you may want to change the name of the Kick Ass Chicks Society to the Head Butt Chicks Society. The times, they are a-changing.
Management Class : News
mental baba 12:40 AM
