Gamblers Fallacy

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Have I been doing this all wrong???

[h=2]Definition of 'Gambler's Fallacy'[/h]
When an individual erroneously believes that the onset of a certain random event is less likely to happen following an event or a series of events. This line of thinking is incorrect because past events do not change the probability that certain events will occur in the future.

For example, consider a series of 20 coin flips that have all landed with the "heads" side up. Under the gambler's fallacy, a person might predict that the next coin flip is more likely to land with the "tails" side up.

This line of thinking represents an inaccurate understanding of probability because the likelihood of a fair coin turning up heads is always 50%. Each coin flip is an independent event, which means that any and all previous flips have no bearing on future flips.

This can be extended to investing as some investors believe that they should liquidate a position after it has gone up in a series of subsequent trading session because they don't believe that the position is likely to continue going up.
 

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makes perfect sense to me

I agree.

Probability for the results of any given event are not dependent on previous results, unless there is a direct relationship between previous results and current event.

Basically, the probability of a coin falling on one side will always be 50% regardless of previous results. This is because a previous result will not have any influence on the next coin toss, unless the coin is altered in some way due to the last toss; for example, if for some inexplicable reason the coin lost a small piece near the edge and that allows for one side to become more likely to land as the result.

However, if you're playing blackjack, things will change on every hand. The more you play, the less cards there are in the shoe; and depending on which cards have been given out you will be more or less likely to get a low or high card next time. So, if a lot of low cards have been dealt, the chances of seeing a high card soon increases. That's the basic idea behind counting cards.

So basically, if the previous event doesn't affect the next one, probability ignores the previous result.
 
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Another fallacy is that by using some bet-sizing "system" that you can change the (EV) expected value of the outcome.

I.e. Martingale Progression, doubling down after successive losses in order to lock in a win.

If you're playing blackjack, and the house edge is 1%, it doesn't matter if you bet $10,000 a hand or $1 a hand,
(or if you Martingale) your expected return is 99% of your total wagers (unless you are counting cards of course).
 

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Personally, I prefer the "Gambler's Phallus, see"

Which states that anyone that slams their dick on the table gets a free drink

cheersgif
 

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Somebody said that gamblers only see the wins as real and forget about the losses but isn't that almost everything in life?
 

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good topic Fresh..............<><>
 

"Who's winning?"
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How about the lets make a deal question. You choose door #1. Monty Hall says we are going to eliminate door #2. Do you change your pick? I say yes, your original pick had odds of 33%. Now your odds are 50%
 
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How about the lets make a deal question. You choose door #1. Monty Hall says we are going to eliminate door #2. Do you change your pick? I say yes, your original pick had odds of 33%. Now your odds are 50%

correct
 

A Separate Reality
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How about the lets make a deal question. You choose door #1. Monty Hall says we are going to eliminate door #2. Do you change your pick? I say yes, your original pick had odds of 33%. Now your odds are 50%[/QUOTE


Not as simple. What if door number 1 was a winner? (3 doors, behind 1 door is a Maserati, goats behind the other 2) Changing your pick would be wrong. The issue is that this not a random process for Monty knows what's behind the doors, See full explanation and an introduction to the 3 main rules of probability in Mlodinow's book The Drunkards Walk, a must for any gambler who has ever flipped a coin.
 
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How about the lets make a deal question. You choose door #1. Monty Hall says we are going to eliminate door #2. Do you change your pick? I say yes, your original pick had odds of 33%. Now your odds are 50%[/QUOTE


Not as simple. What if door number 1 was a winner? (3 doors, behind 1 door is a Maserati, goats behind the other 2) Changing your pick would be wrong. The issue is that this not a random process for Monty knows what's behind the doors, See full explanation and an introduction to the 3 main rules of probability in Mlodinow's book The Drunkards Walk, a must for any gambler who has ever flipped a coin.

Um, yes it is that simple. You should switch, although the previous answers show the wrong probabilities.

If you switch, you have a 2/3 chance of "winning." If you don't switch, you have a 1/3 chance of "winning." Fact. Your statement about the possibility that the door you chose first could have been the winner shows you don't understand the basics of probability theory.
 

Retired; APRIL 2014 Thank You Gambling
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Um, yes it is that simple. You should switch, although the previous answers show the wrong probabilities.

If you switch, you have a 2/3 chance of "winning." If you don't switch, you have a 1/3 chance of "winning." Fact. Your statement about the possibility that the door you chose first could have been the winner shows you don't understand the basics of probability theory.

this is correct,, saw the full breakdown on MYTHBUSTERS,,, frigging AWSOME explanation,,,
 

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Um, yes it is that simple. You should switch, although the previous answers show the wrong probabilities.

If you switch, you have a 2/3 chance of "winning." If you don't switch, you have a 1/3 chance of "winning." Fact. Your statement about the possibility that the door you chose first could have been the winner shows you don't understand the basics of probability theory.

You're testy know it all pimple on my ass aren't you, Festy. Careful or I'll sit down.

Read the book I mentioned for full explanation.
 
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You're testy know it all pimple on my ass aren't you, Festy. Careful or I'll sit down.

Read the book I mentioned for full explanation.

http://en.wikipedia.org/wiki/Monty_Hall_problem

"Contestants who switch have a 2/3 chance of winning the car, while contestants who stick to their choice have only a 1/3 chance. One way to see this is to notice that, 2/3 of the time, the initial choice of the player is a door hiding a goat. When that is the case, the host is forced to open the other goat door, and the remaining closed door hides the car. "Switching" only fails to give the car when the player picks the "right" door (the door hiding the car) to begin with. But, of course, that will only happen 1/3 of the time."
 

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I don't need to read a book, it's basic probability theory 101.

hahaha. ur a fuken genius, a legend in ur on mind. Festy where is the Malaysian jet? who took it down? Does God exist? If you choose the winning door and Monty tells you so, should you still switch to take advantage of 2/3? hahaha Oh wait, the process is not random if Monty is giving u choices right? Does a coin choose if its going to land heads or tails?
 

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Disregard my last post Festy. ur wiki reference makes the point I was trying to make: "Switching" only fails to give the car when the player picks the "right" door (the door hiding the car) to begin with. But, of course, that will only happen 1/3 of the time."
 

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