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jfc 24th July 2006 06:37 PM

Quote:
Originally Posted by crash
...
Peter's gamble
Peter asks me for a bid on the following gamble. I get to
flip a coin up to 10
times. If I get heads on the kth
flip, 1 . k . 10, I collect 2k1 and stop. If I
manage to
flip tails 10 times in a row, I collect 1024.
How much should I offer Peter for this gamble? In theory, the value of
this gamble is
· 210
(1=2 · 1 + 1=4 · 2 + 1=8 · 4 + :::+ 1=210 · 29) + 1=210
= 10 · 1=2 + 1
=6.


The Peter's Gamble example should make sense if you try to work through it with a spreadsheet.

The amazing thing to me, is that it illustrates Huygens propensity for inventing incredibly useful things.

After he invented Expectation he then had to prove it. So to do that he invented Hedging! I'd always thought that Hedging was a modern concept.

The example illustrates Hedging as a proof of Expectation.

It shows how you can convert an Expectation of 6, with huge volatility into a guaranteed return of 6. By using appropriate hedge bets.

Right now there appear to be around 10 documented investor/gamblers each heading towards a billion from scratch. In no small part thanks to Huygens.

Chrome Prince 24th July 2006 06:39 PM

Quote:
Originally Posted by jfc
Do you have any idea who you're trying to put down?



Yes, I'm not putting down the mathematicians or scientists, I'm putting down the use of superfluous explanations of simple facts - it's simply to make them feel superior.

The courts and police do exactly the same thing trying to stamp superiority - unfortunately it doesn't work!

I read overcomplicated formulae and soon realize that these people are not as intelligent as they try to make out, if they were, they would make things easier for themselves and stop wasting time trying to fluff their own feathers.

Chuck 24th July 2006 07:09 PM

not sure if this has been done before, but anyway i love this little scenario

you are in a gameshow where you are shown 3 doors. you can't see whats behind them. the host says to you that behind each of the doors there is either a car or a goat. of the 3 doors there are 2 goats and 1 car. Obviously you want to pick a car. So you pick a door (for arguments sake door 1). The host opens a different door (e.g. door 2) and there is a goat behind it. He asks you do you want to change doors.

Do you, or does it not matter in your quest for the car?

mathematically you should change doors, although prima facie it is 50/50

KennyVictor 24th July 2006 07:22 PM

OK I'll ask, why should we change doors?

jfc 24th July 2006 07:31 PM

Quote:
Originally Posted by Chuck
not sure if this has been done before, but anyway i love this little scenario

you are in a gameshow where you are shown 3 doors. you can't see whats behind them. the host says to you that behind each of the doors there is either a car or a goat. of the 3 doors there are 2 goats and 1 car. Obviously you want to pick a car. So you pick a door (for arguments sake door 1). The host opens a different door (e.g. door 2) and there is a goat behind it. He asks you do you want to change doors.

Do you, or does it not matter in your quest for the car?

mathematically you should change doors, although prima facie it is 50/50


This is the notorious Monty Hall Problem.

It is dealt with comprehensively in this 520 page book.

http://www.dartmouth.edu/~chance/te...amsbook.mac.pdf

As are:

Woof's earlier controversy - BG,BB,GB

The St Petersburg Paradox

Markov Chains

Expectation


It is a remarkably readable accumulation of profound wisdom.

For CP's benefit it ain't something to speed read through.

Chuck 24th July 2006 08:13 PM

Quote:
Originally Posted by KennyVictor
OK I'll ask, why should we change doors?


The chance of winning the car is doubled when the player switches to another door rather sticking with the original choice. The reason for this is that to win the car by sticking with the original choice the player must choose the door with the car first, and the probability of initially choosing the car is 1/3. Whereas, to win the car by switching the player must originally choose a door with a goat first, and the probability of choosing a goat door first is two in three.

At the point the player is asked whether to switch there are three possible situations corresponding to the player's initial choice, each with equal probability (1/3):

The player originally picked the door hiding goat number 1. The game host has shown the other goat.
The player originally picked the door hiding goat number 2. The game host has shown the other goat.
The player originally picked the door hiding the car. The game host has shown either of the two goats.
If the player chooses to switch, the car is won in the first two cases. A player choosing to stay with the initial choice wins in only the third case. Since in two out of three equally likely cases switching wins, the odds of winning by switching are 2/3. In other words, a player who has a policy of always switching will win the car on average two times out of the three.


Lovely

crash 25th July 2006 02:42 AM

Chuck,

A smart presenter [Bob Dyer of 'pick-a-box' fame ....yes I'm that old] would of course know that theory and for obviously smart contestants like Barry Jones [was the all time pick-a-box King], Bob would use reverse psychology and offer the door with the car, not the goat. Barry Jones would counter with reverse, reverse psychology of course and take it :-))


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