Need help conceptualizing the constant "e"

[Fly]

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I'm quite happy that boris was nice enough to post a question that perfectly shows why clever amateur will sometimes beat not so clever pros


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If I call every bet, I will never be bluffed!!!!!!!!!!!!

[Fly]

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Here is a cooler problem, imo: show that the expected value of the # of people who get their hat back is 1, independent of n.

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Hmm, how about n(1/n)=1!
uA

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Ok, mr genius, show that the outcomes are independent. I know this is "intuitively obvious" but actually providing a proof is an altogether different matter.


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Why do we care about whether or not the outcomes are independent? The question is about expected values, and the expected value of the sum is the sum of the expected values regardless of whether or not the variables are independent.

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n(1/n) = 1/n + 1/n + .... + 1/n (n terms in the sum) where the ith term represents the probability that the ith person receives his hat. If the events were not independent, the 2nd through nth terms could be different, yielding a different sum.

[PairTheBoard]

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Here is a cooler problem, imo: show that the expected value of the # of people who get their hat back is 1, independent of n.

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Hmm, how about n(1/n)=1!
uA

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Ok, mr genius, show that the outcomes are independent. I know this is "intuitively obvious" but actually providing a proof is an altogether different matter.


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Why do we care about whether or not the outcomes are independent? The question is about expected values, and the expected value of the sum is the sum of the expected values regardless of whether or not the variables are independent.

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n(1/n) = 1/n + 1/n + .... + 1/n (n terms in the sum) where the ith term represents the probability that the ith person receives his hat. If the events were not independent, the 2nd through nth terms could be different, yielding a different sum.

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Actually, the events are not "independent" as the term is normally used. To be "independent" would mean that the probality of me getting my hat stays the same regardless of whether Joe gets his hat or not. But it does change. If Joe gets his hat my chance improves to 1/(n-1) instead of 1/n.

So the concept being talked about is not "independence" as the term is used in probability. I'm not sure it has a name. But it's the same concept as the one we invoke when we say that the unseen cards in other players' hands do not affect my probability of hitting my flush card on the river. With the hats, each person can say ahead of time that he will have 1/n chance of getting his own hat regardless of what place in line he is in when they are handed out. That defines n dependent indicator functions each of whose EV can be added despite their not being independent.

What is the correct term for the concept that unseen cards don't affect my probabilities regardless of whether they are in other players' hands or at the bottom of the deck? I'm not sure. I don't think it's something we ask to be proved all the time though.



PairTheBoard

[David Sklansky]

Two in a row. Unbelievable.

[thylacine]

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Here is a cooler problem, imo: show that the expected value of the # of people who get their hat back is 1, independent of n.

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Hmm, how about n(1/n)=1!
uA

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Ok, mr genius, show that the outcomes are independent. I know this is "intuitively obvious" but actually providing a proof is an altogether different matter.


[/ QUOTE ]

Why do we care about whether or not the outcomes are independent? The question is about expected values, and the expected value of the sum is the sum of the expected values regardless of whether or not the variables are independent.

[/ QUOTE ]

n(1/n) = 1/n + 1/n + .... + 1/n (n terms in the sum) where the ith term represents the probability that the ith person receives his hat. If the events were not independent, the 2nd through nth terms could be different, yielding a different sum.

[/ QUOTE ]

Actually, the events are not "independent" as the term is normally used. To be "independent" would mean that the probality of me getting my hat stays the same regardless of whether Joe gets his hat or not. But it does change. If Joe gets his hat my chance improves to 1/(n-1) instead of 1/n.

So the concept being talked about is not "independence" as the term is used in probability. I'm not sure it has a name. But it's the same concept as the one we invoke when we say that the unseen cards in other players' hands do not affect my probability of hitting my flush card on the river. With the hats, each person can say ahead of time that he will have 1/n chance of getting his own hat regardless of what place in line he is in when they are handed out. That defines n dependent indicator functions each of whose EV can be added despite their not being independent.

What is the correct term for the concept that unseen cards don't affect my probabilities regardless of whether they are in other players' hands or at the bottom of the deck? I'm not sure. I don't think it's something we ask to be proved all the time though.



PairTheBoard

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FWIW when I wrote "Hmm, how about n(1/n)=1!" I was fully aware that the events were not independent (for n>1) and I was fully aware that it didn't matter.

[Fly]

PTB,

concept that is not independence = symmetry?

[borisp]

ugh, yea I misused the term independence. I know nothing about formal statistics and hence I misused the vocabulary. What I really meant was the concept PTB alluded to, which is to observe that there was a symmetry among the hat recipients.

But my real point was that a solution is incomplete when the author has not accounted for all relevant considerations, even when some of them are obvious. Of course, this is a subjective judgment in any case.

[borisp]

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Now please make yourself useful and go over to my thread "Improving On Buffet And Desert Cat" on the Business Forum and explain to DeserstCat that I'm right. (Even though my argument doesn't meet your brand of rigor.)

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It now occurs to me that I have no idea what you are talking about in that thread.

[David Sklansky]

If your estimate of the future price of all stocks is on average more accurate than the present market price, but at the same time you are not sure you have an edge investing, unless your estimate is at least 30% higher or lower than the market price, (because you admit you are not perfect) then barring weird discontinuous functions, you can conclude that the actual EV of a stock is on average somewhere in between your estimate and the market's estimate. (As long as the market price is not totally random.)

The subject arose because Buffett and DesertCat say the right way to invest is to know stock analysis, look at the information about a stock before knowing its price, estimate its true value, and then invest if there is a large discrepancy. I pointed out that this is fine but that the original estimate should be moved to varying degrees toward the market price. DesertCat disagreed. I said if you disagree you can't claim you need a large cushion to invest.

[jay_shark]

Here is a rigorous solution to the expected number of hat matches .

Let X denote the number of matches we can compute E(x) by writing X= X1 + X2 + X3+ ...+ Xn

where Xi = { 1 if the ith person selects his own hat , or 0 otherwise }

For each i , the ith person is equally likely to select any of the N hats .

E(Xi]= P(Xi=1)= 1/N

E(X) = E(X1) + E(X2) + ... + E(Xn) = 1/N*N = 1

Hence on average , exactly one person selects his own hat .