GTS, honestly, I think anyone who does not yet agree with the 1/3-2/3 probabilities is a total retard. There is no other excuse in the world. I mean, come on, how stupid does one have to be with all the resources given to still believe in any other solution. Its probably the the same people who plunk down thousands of bucks to buy some technical analysis software and training courses.....idiots!!!
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Morgan Stanley Traders Lost $390 Million in One Day in August
- Thread starter ASusilovic
- Start date
GTS, honestly, I think anyone who does not yet agree with the 1/3-2/3 probabilities is a total retard. There is no other excuse in the world. I mean, come on, how stupid does one have to be with all the resources given to still believe in any other solution. Its probably the the same people who plunk down thousands of bucks to buy some technical analysis software and training courses.....idiots!!!
Quote from Equalizer:
Well, I can tell you that many from NuclearPhynance and Wilmott reading this thread are having a giggle or two!
no question. I doubt our "specialists" have ever heard of Wilmott or NP. But hey, this website caters to "Elite" traders. No need for probability knowledge, intuition suffices....lol
Quote from sjfan:
I'm not trying to be rude, but will you write out the four branches in words? I'm having a hard time following what you are doing - with the spacing at all. I do want to try to give you a valid reply once I figure out what you are doing.
Sure.
Tree
____Step1____Step2___Reformulated to match problem
A________* 1/2 = 1/6 -> 1/6 -> (1/6)/(3/6) = 1/3
_____1/3
B________* 1/2 = 1/6 -> N/A
C________* 1/2 = 1/3 -> N/A
_____2/3
D________* 1/2 = 1/3 -> 1/3 -> (1/3)/(3/6) = 2/3
__________________________________________
Total
______1__________1___ 3/6_________________1
Case
A -> odds chosen cup with ball and decided to stay put
B -> odds chosen cup with ball and decided to switch
C -> odds other cup-1 with ball
D -> odds other cup-2 with ball
[A or B interchangeable C or D interchangeable in truth important point is that the odds are given i.e. 50%]
Quote from Capablanca:
...
I see where you are going wrong. You have correctly computed the unconditional probability of these events. What you are missing is that when you are at the switch/no switch point, you need to figure out your conditional probability. By this, I mean you need to figure out the probability of the true cup being the one you have or the other conditioned on the fact that one cup has already been turned over.
Think of it this way (a very extreme case). Let's say that the probability of living to 100 years old is 0.01%. You just found yourself on your 100th birthday. What's your probability of being alive? Unconditionally, it's 0.01%. But conditional on the fact that you are still alive, it's 100%. See the difference between thet two?
Unfortunately I think I get where you are coming from only vaguely. Did you take into account that there is an assumption inherent in the Monte Hall problem that the host/facilitator knows with a certainty where the ball is not and isn't only guessing?
Also doesn't eliminating the probability associated with the revealed cup from consideration and accordingly reducing the universe of probabilities adjust for the said condition?
Perhaps you could illustrate with another decision tree how it should go. Thanks.
Also doesn't eliminating the probability associated with the revealed cup from consideration and accordingly reducing the universe of probabilities adjust for the said condition?
Perhaps you could illustrate with another decision tree how it should go. Thanks.
Quote from Capablanca:
Unfortunately I think I get where you are coming from only vaguely. Did you take into account that there is an assumption inherent in the Monte Hall problem that the host/facilitator knows with a certainty where the ball is not and isn't only guessing?
Also doesn't eliminating the probability associated with the revealed cup from consideration and accordingly reducing the universe of probabilities adjust for the said condition?
Perhaps you could illustrate with another decision tree how it should go. Thanks.
Yes. I explicitly considered the fact that the host knows the truth and made a decision based on it. It's what the "conditional" part.
As for a decision tree, I don't think that's the right way to go. In a sense, I'm not making a decision based on what the decision I made. Instead, I'm making a decision based on what I previously believed (that all cups are equally likely) and my new belief updated by my observation that the host has turned over one of the other cups.
Now, I'm going to act rude and refer you to the wikipedia entry after the previous paragraph. I simply don't want retype everything.
The probability that you chose the cup with the coin is: 1/3. Put that cup to the right side of the table. The chance that the coin is under one of the cups that you did not choose is: 2/3. Put those two cups to the left side of the table.
Now, turn up the one cup (of the two) without the coin on the left side of the table. The probabilities of which side of the table the coin is have not changed. There is still a 1/3 chance that the coin is on the right side of the table and a 2/3 chance that the coin is on the left side of the table (under the cup that was not turned up).
You will be correct twice as many times if you always switch.
Joe.
Now, turn up the one cup (of the two) without the coin on the left side of the table. The probabilities of which side of the table the coin is have not changed. There is still a 1/3 chance that the coin is on the right side of the table and a 2/3 chance that the coin is on the left side of the table (under the cup that was not turned up).
You will be correct twice as many times if you always switch.
Joe.
From a trader's perspective,
Let's say you place a bet of $1 on the first cup...
After removing one of the cups, do you switch your bet?
How about placing $2 on the second cup...while keeping your
$1 bet on the first cup?
(a) In this way, if coin is under first cup, you lose $1... prob=1/2
if coin is under 2nd cup, you win $3... prob=1/2
if coin is under 3rd cup, you lose $3...prob=1/3
Net expectancy=-1*1/2 + 3*1/2 + -3*1/3 = 0
(b) If you had just switch the $1 bet to 2nd cup,
coin under first cup, you lose $1, prob=1/2
coin under 2nd cup, you win $1, prob=1/2
coin under 3rd cup, you lose $1, prob=1/3
Net expectancy = -1/3
so as FED meeting/earnings call draws closer, and the view becomes clearer
let's say the FED will decrease or keep interest rates constant leaving increase option out,
I would place a new bet (double amt) on the only choice left while keeping my first bet there...save myself commissions on removing an order and loss on spread too esp. big orders.
Let's say you place a bet of $1 on the first cup...
After removing one of the cups, do you switch your bet?
How about placing $2 on the second cup...while keeping your
$1 bet on the first cup?
(a) In this way, if coin is under first cup, you lose $1... prob=1/2
if coin is under 2nd cup, you win $3... prob=1/2
if coin is under 3rd cup, you lose $3...prob=1/3
Net expectancy=-1*1/2 + 3*1/2 + -3*1/3 = 0
(b) If you had just switch the $1 bet to 2nd cup,
coin under first cup, you lose $1, prob=1/2
coin under 2nd cup, you win $1, prob=1/2
coin under 3rd cup, you lose $1, prob=1/3
Net expectancy = -1/3
so as FED meeting/earnings call draws closer, and the view becomes clearer
let's say the FED will decrease or keep interest rates constant leaving increase option out,
I would place a new bet (double amt) on the only choice left while keeping my first bet there...save myself commissions on removing an order and loss on spread too esp. big orders.