Odd

[caravelair]

So then, what's the total number of natural numbers? 2x infinity, or 3x infinity?

actually, the cardinality of the natural numbers is N (the countably infinite cardinality), and the cardinality of the even numbers is also N, and so is the cardinality of the set of odd numbers. there are the same number of all of them. other sets with cardinality N: the rationals, the integers, the set of prime numbers, multiples of 5, powers of 5, powers of 2, etc, etc. all those sets are the same size.

larger sets include the real numbers, which has an uncountably infinite cardinality.

 

[gluadys]

This reminds me of the Marilyn VonSavaunt (sp) controversy regarding the "Let's make a deal!" problem. As stated by Ms VS, Monty gives you a choice of three doors, one of which contains a fabulous new car, and the other two contain donkeys and rocks and such. You make a choice, and Monty has one of the other doors opened to reveal a pile of coal. Now he gives you a choice, of whether to stay with your original pick, or switch. Ms. VS (the world's most intelligent woman, according to her) says you stand a better chance of winning the car if you change your decision. I (having barely passed differential equations) have to disagree. I say your chances are even. It is obvious to me that Monty is not going to reveal the car, so you know it is behind one of the two doors. It makes no difference what kind of shell gave has transpired in the past, you are now faced with the choice of two doors, one of which contains the corvette. 50/50 says I. What sez youze?

Hey, if you barely passed differential equations you did better than me. I also assumed it had to be an equal chance once one of the doors was opened. But then I figured out Ms.VS was right. Only now I've forgotten how I figured that out and will have to figure it out all over again

So I hope a resident math whiz will explain it properly for both of us.

 

[einstein314emc2]

we never went to the moon, couldn't you see that there were no stars and the sky was black? geez, what some people will believe.

 

[ThePhoenix]

Mrs. Von Savant is right. There are a grand total of four cases you can be in if you pick a random door. Lets arbitrarily call door a right and doors b and c wrong

Possibilities

a -> b opened
a -> c opened
b -> c opened
c -> b opened

These do not have equal chances. Cases 1 and 2 have a 1/6 chance of happening each, and cases 3 and 4 each have a 1/3 chance of occuring. In cases one and two switching is disadvantagous, and in cases three and four switching is advantagous, which is what gives the false illusion of 50/50 split. However cases 1 and 2 are less probable, meaning that the chances of getting it right if you switch are 66% and the chances of getting it right if you don't are 33%

 

[O'Factry]

Phoenix! I'm suprised at you! After you gave such a good response to the OP. You left out two possibilities.

choose A open B stay
A C stay
B A Monty won't do this, but the possibility exists
B C change
C A Monty won't do this, but the possibility exists
C B change

each possiblility has a one in six chance from the beginning. Two of them will result in a win if you stay, and two if you change. You must acknowledge this, or else you would not have arrived at your 1 in 6 chance. The other two possibilities result in an automatic loss, which is no fun, so the game is altered to keep it interesting.

If you toss a coin and it comes up heads 8 out of ten times, what are the chances it will be tails on the next toss? It's still 50/50. What transpired before has no effect. It is a simple choice between two possibilities.

 

[ThePhoenix]

B->A and C-> A are both possibilities, but since Monty won't do this, they have a 0% chance of occuring. Therefore I felt that it was worth neglecting. And your coin toss comparison was wrong, because coin tosses are independent. Doors are not. Lets recap:

Possibilities
You choose A (1/3 of the time) -
_____Door B is opened (1/2 the time)
_____Door C is opened (1/2 the time)

You choose B (1/3 the time)
_____Door A is opened (never)
_____Door C is opened (always)

You choose C (1/3 the time)
_____Door A is opened (never)
_____Door C is opened (always)


As you can see because door A is never opened, it is advantagous to switch if you chose either B or C. But you choose B or C 2/3rds of the time. That means 2/3rds of the time you're right when you switch, and 1/3rd of the time you're wrong.

 

[Jet Black]

Every natural number greater than 1 is a sum of two lesser natural numbers. There are three possiblilities in addition; an even plus an even equals an even, an even plus an odd equals an odd, and an odd plus an odd equals an even. In two of three cases the result is an even number. Therefore there are twice as many even numbers as odd numbers.

this may have been said but you can create a bijective surjective map between the natural numbers and the even numbers only, so the correspondence is 1:1

1 2
2 4
3 6
4 8
5 10
6 12

and so on, where every even number in the list on the right can be represented by n*2

 

[Jet Black]

So then, what's the total number of natural numbers? 2x infinity, or 3x infinity?

this particular infinity is aleph-null, and it is the same for all three. infinity plus one is still infinity too, no bigger or smaller.

 

[StrugglingSceptic]

If you pick the wrong door to begin with (2/3), you will win by switching. If you pick the right door to begin with (1/3), you will lose by switching. Hence, by switching generally, you have a 2/3 chance of winning. It is therefore always better to switch.

 

[O'Factry]

So Monty says, "Ok, then! It's a good thing you didn't choose the donkey! So what will you do, stay with your choice or choose the other door?" The contestant, dressed in a hot and heavy oversized bee costume becomes very excited and falls over having a heart attack. The paramedics rush in, and just as they are getting ready to wheel the giant bee out on a gurney, Monty grabs one of the paramedics and says, "In the interest of show business, I'm going to let you choose a door for the dying drone. Which will it be, door number one or door number two?" The paramedic has no idea what has happened, or if he should sedate Monty. So he just chooses door number 2 and rushes to catch up with his coworkers. So then, what are the chances that he has chosen correctly?

The paramedic has no knowledge of what has happened on the show. At this point it is a new game, a simple choice between two possibilities. Does he have a one in three or a two in three chance of being correct? Or does he have a 50% chance of guessing correctly? Suppose the bee had not eaten so many burgers and fries, and the question posed by Monty was not "stay or change" but rather, "one or two"? Does that change anything? Suppose the paramedic and the bee were both given the same choice, "one or two?" Are their odds different?

 

[ThePhoenix]

You're ignoring the fact that the contest is not blind. Your hypothetical paramedic has a 50% chance of choosing the right one BECAUSE HE DOES NOT KNOW WHAT DOOR WAS PICKED. Monty knows which door is the right one. THEREFORE HIS CHOICE OF WHAT DOOR TO OPEN IS NOT RANDOM! You're applying the logic of random events to a weighted coin. Because Monty's choice of what door to open is predictable (he will never open the right one) you can take certain actions following the predictable event to increase your chances of winning. By opening a door Monty gives you certain information.

So how could it be made random? If Monty could open the door you picked. Because he is doubly limited, to not open the right door and not open your door his choice isn't random at all.

An example of this would be a dice game. Lets say you bet on whether or not a roll of two dice would net a number greater or less then seven (with rerolls for sevens). This is a 50/50 chance. Now lets say after the first die is rolled you are told two numbers that the dice is not, and allowed to alter your choice accordingly. Could you improve your odds of winning over 50%? YEAH!

 

[StrugglingSceptic]

Now it's 50/50.

 

[ThePhoenix]

Now it's 50/50.

?

 

[Loudmouth]

For those that have not worked with infinite number sets,

2 x (infinity) = infinity
AND
(any number)/(infinity) = 0.

 

[StrugglingSceptic]

?

If the paramedic is simply presented with two doors, one of which the contestant originally chose and one of which is left after the host has removed one, but the paramedic does not know which is which and is asked to choose, his chances of choosing the right door are 50/50.

 

[ThePhoenix]

If the paramedic is simply presented with two doors, one of which the contestant originally chose and one of which is left after the host has removed one, but the paramedic does not know which is which and is asked to choose, his chances of choosing the right door are 50/50.

of course.