Calculate the odds

[Christopher]

Calculate the odds

You have three coins you flip the first one it comes up heads, you flip the second it comes up heads.

Question #1

what are the odds that coin 3 will come up tails?

a. 1 in 2

b. 1 in 3

Lets say it does come up tails and all three coins are covered by nut shells and moved around like in a shell game. The object is to chose the shell containing the face up tail if you pick correctly you will win a billion dollars!

You have shell a, b, and c. To chose from.

You chose ©, however before you can look under shell © you are shown what’s under shell (a) and its heads.

 

 

You are told you can change your mind and pick either b or keep your original chose of c.

 

Question # 2

 

What are the odds that coin c. will be tails?

a. 1 in 2

b. 1 in 3

 

Question #3

 

What are the odds that coin b. will be tails?

a. 1 in 2

b. 1 in 3

[Turtle]

What are the odds the coin will land(come to rest) on its edge? :eek: (Devils advocate) :evil:

[Jay-qu]

1. a - its still going to be 1 in 2 even if you toss 100 heads, for question 2 and 3 it will be 1 in 2 because you know that one heads is out of the picture leaving only 1 head and 1 tail remaining

[Edge]

For A is 1 in 2

 

For B and C, it depends.

 

Did the guy know that under A was heads?

 

If he did, then the answer to 2 is 1 in 3, and the answer to 3 is 2 in 3

 

Kinda like the Monty Hall problem.

 

http://www.comedia.com/hot/monty.html

http://www.comedia.com/hot/monty-answer.html

[Turtle]

Jay & Edge, et al, what about the odds of landing on edge? Come on now, you can't discount it. :evil:

[Erasmus00]

Jay & Edge, et al, what about the odds of landing on edge? Come on now, you can't discount it. :evil:

 

The odds are so small as to be discounted. Consider that as the coin spins, it encompasses a sphere of surface area 4*pi*r^2. Where r is the radius of the coin. Of that only 2*pi*r*t (t is the thickness of the coin) is encompassed by the edge of the coin.

 

So, to a very gross first approximation, probablitiy is proportional to t/r, hence if the thickness is considerably smaller then the radius, this probability heads to 0.

-Will

[Turtle]

The odds are so small as to be discounted. Consider that as the coin spins, it encompasses a sphere of surface area 4*pi*r^2. Where r is the radius of the coin. Of that only 2*pi*r*t (t is the thickness of the coin) is encompassed by the edge of the coin.

 

So, to a very gross first approximation, probablitiy is proportional to t/r, hence if the thickness is considerably smaller then the radius, this probability heads to 0.

-Will

:eek: Discount it at your peril. :evil: I saw it happen with a nickel in a bar years ago & it was one hell of a fight after that.

[Christopher]

For A is 1 in 2

 

For B and C, it depends.

 

Did the guy know that under A was heads?

 

If he did, then the answer to 2 is 1 in 3, and the answer to 3 is 2 in 3

 

Kinda like the Monty Hall problem.

 

http://www.comedia.com/hot/monty.html

http://www.comedia.com/hot/monty-answer.html

 

Yes the guy can see coin a is heads. So you would change the pick to b., correct

?

if so why ?

[Erasmus00]

Yes the guy can see coin a is heads. So you would change the pick to b., correct

?

if so why ?

 

So we have three shells, one of them is a winner. I pick a shell, so there is a 1/3 chance its a winner, and a 2/3 chance its wrong. Then the game master (monty hall or what have you) flips over a shell containing a loser. Now, odds are still 2/3 that I'm wrong. That means its in my best interest to switch.

 

So, I would change the pick, because 2/3 of the time that'll be in my best interest. I think the most straightforward way to arrive at this conclusion is the method above, though you can also work it out with Bayes' theorem.

-Will

[Christopher]

So we have three shells, one of them is a winner. I pick a shell, so there is a 1/3 chance its a winner, and a 2/3 chance its wrong. Then the game master (monty hall or what have you) flips over a shell containing a loser. Now, odds are still 2/3 that I'm wrong. That means its in my best interest to switch.

 

So, I would change the pick, because 2/3 of the time that'll be in my best interest. I think the most straightforward way to arrive at this conclusion is the method above, though you can also work it out with Bayes' theorem.

-Will

 

Correct :evil:

 

 

Its about how we perceive information. I set this up to show that there are levels to which we associate linear thinking with non-linear thinking.

How information paradigms are constructed by information as opposed to patterns that are random.

 

Question #1

Some assume that two consecutive out comes would increase the odds of a opposite outcome. (believe me I know people that think this way) Some would assume this is a pattern or a trend will increase the odds of repeating itself. Of coarse most of us realize that each flip of the coin has no connection to one another. So the answer is 1in 2

But in real life people make this mistake all the time not only in games of chance but in how we perceive the world.

 

Question #2 # 3

What are the odds that coin b. will come up tails?

1 in 2

What are the odds that coin c. will come up tails?

 

1 in 3

We should change from c to b increasing our odds of becoming a billionaire from 1 in three to 1 in two.

 

Our first choice of c is from the paradigm of 1 to 3 chance model. When we are then shown a is heads it changes the model for c but not b.

the new information can only be applied by giving up the first choice that was not relying on new information but only chance.

The reason it is counter intuitive is that we do not recognize a. being heads as new information as it relates to our current position. The one being nothing more than a choice based on an assumption.

 

 

WikipediaBayesian statisticians claim that methods of Bayesian inference are a formalisation of the scientific method involving collecting evidence that points towards or away from a given hypothesis. There can never be certainty, but as evidence accumulates, the degree of belief in a hypothesis changes; with enough evidence it will often become very high (almost 1) or very low (near 0).

 

As an example, this reasoning might be

The sun has risen and set for billions of years. The sun has set tonight. With very high probability, the sun will rise tomorrow.

Bayesian statisticians believe that Bayesian inference is the most suitable logical basis for discriminating between conflicting hypotheses. It uses an estimate of the degree of belief in a hypothesis before the advent of some evidence to give a numerical value to the degree of belief in the hypothesis after the advent of the evidence. Because it relies on subjective degrees of belief, however, it is not able to provide a completely objective account of induction. See scientific method.

 

Bayes' theorem also provides a method for adjusting degrees of belief in the light of new information

[cwes99_03]

I've heard this before. I knew the answer to #1 because you can forget about the previous two flips of the coin. Each new flip is a new 1 in 2 chance.

 

However, what makes the shell game different.

Your initial choice has a 1 in 3 chance of being correct.

 

One shell is removed as it is a wrong answer. Now you have a 1 in 2 chance of picking the right one. Should you chose the same shell as before, you have a 1 in 2 chance. Should you chose the other shell that is left, you still have a 1 in 2 chance. The fact that one shell is removed means nothing except that you didn't chose a shell that contained a wrong answer. Of course this is not new because there were two wrong answers out there. Regardless of which shell you chose, there would be at least one shell to remove. However if you chose the correct shell the first time, that increases the ability of the master of the game to chose either wrong shell remaining. This again changes nothing. The master is only removing one wrong answer. By removing a wrong answer (not changing any statistics in making the new decision of which of 2 possible shells is the correct one) the game has not changed, just elongated the eventual outcome. There is no benefit to changing shells mid way through the game.

[Erasmus00]

The fact that one shell is removed means nothing except that you didn't chose a shell that contained a wrong answer. Of course this is not new because there were two wrong answers out there. Regardless of which shell you chose, there would be at least one shell to remove. However if you chose the correct shell the first time, that increases the ability of the master of the game to chose either wrong shell remaining. This again changes nothing. The master is only removing one wrong answer. By removing a wrong answer (not changing any statistics in making the new decision of which of 2 possible shells is the correct one) the game has not changed, just elongated the eventual outcome. There is no benefit to changing shells mid way through the game.

 

There is a definate benefit to switching, and it is established mathematical fact. Google Bayes theorem or Monty Hall Problem. By always eliminating a wrong answer, the master of the game adds information to the system. If it helps you see it, enumerate all the possible situations.

 

Consider this, if you pick the wrong shell at the start, and then switch, you'll win. You have a 2/3 chance of picking a wrong door, so if you switch, you'll win 2/3 of the time.

-Will

[cwes99_03]

You are not taking both sides of the coin (to use a figure of speech) into consideration. By removing the second shell you have no new information. You knew picking the shell at the beginning that among the other two shells that weren't picked there would be at least one wrong answer. So by removing that shell, you have your initial information confirmed.

What new information then do you have? That you picked a shell with either a heads or a tails in it. Nope knew that before one shell was removed. So yes your odds of having picked either shell that remains is still 1 in 3 because whichever shell you picked, the master would have removed one wrong shell.

I've never read the Bayesean proof, so I guess I shall have to. BRB.

[cwes99_03]

Nope no problem with Bayes theorum. All it says is that if you add information, that you increase the accuracy of your prediction by recalculating with the new information. I still fail to see what new information you get by having one shell removed.

[Enescu]

This is just a simple probality question.

 

our law's of physics is based on these probabilities...you cant be exact in it ..that's why the laws of physics are currently beeing changed (in vancouvre i belive)

 

your question cannot have an answear..

[Edge]

Jay & Edge, et al, what about the odds of landing on edge? Come on now, you can't discount it. :evil:

Well, I guess it's one in 10000. LOL!!

[Enescu]

well if you people wana but it this way...

 

Well what are the odds for the coin to fall throw the floor?

1/100000000000000000000000000000000000000...

but its possible ..

so.. like i said before ther is no answear to your question.