Statistics riddle problem continued.

[Sir Kenyth]

quote:

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Originally posted by 250:
it is 100% right as long as there is no false testify

also, you BROKE down the problem in half, and the two halves cannot be resembled back together. it became a new problem, which should be rephrased.

I see no point of arguing with you. if you want a serious answer, argue without changing problems.

also, a quick kill:

to the first person, he ALWAYS get 2/n chance
why? because he got to pick TWICE out of n doors

2/n

simple, huh?

---

You just luv the numbers bud. What about the reasoning and perception that came up with them? The conditions of a problem are perceived by someone. Another person can perceive things differently. Your solution is correct, from that point of veiw. Explain in words to me, why two people with an exact same physical task should have different probabilites for it. What is the physical difference between the two at that point?

[Waluin]

I'm getting the impression you guys are arguing apples and oranges. 250, I dont think he's disputing the fact that you're right, just that there are always different ways to approach a problem. That's true, last I checked. Statistics wise, you're totally on the ball. I just think he was trying to expand the horizon to other possibilities.

[250]

quote:

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Originally posted by 250:
their possibility differs because

the first person will be able to pick twice

the second person can only pick once

therefore, their possibility differs

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this is the difference, need more prove?

[250]

quote:

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Originally posted by Waluin:
I'm getting the impression you guys are arguing apples and oranges. 250, I dont think he's disputing the fact that you're right, just that there are always different ways to approach a problem. That's true, last I checked. Statistics wise, you're totally on the ball. I just think he was trying to expand the horizon to other possibilities.

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no no no, I dont think so. I dont care if he is disputing I am right or wrong, but the starting point of his arguement is not exactly correct.

[Ronn_Bman]

I've already explained your both wrong using the rule of "Barker's Beauties". Can't you see it?

[Sir Kenyth]

Yes, we did get a bit off topic. The initial argued point was that the formula may not apply to this problem. Which is exactly where we needed to go. The point has been proven that subsequent choices have a better probability than first ones. Even if the number of doors didn't narrow down with each one. This was the point of 250s formula. The person in the problem is still on their first choice though, so why would this apply?

[Sir Kenyth]

Anyone? This thread was jumping a little while ago. I guess I must have bored everyone to death!

[250]

the first person choose 2 out of 10

the second person choose 1 out of 9

that is the difference

what do you mean by "it is not ended"? it is ended THE MOMENT you add one more person in. then the problem changed into something completely different: "whats the probability that one can pick a prize out of nine door?" it is no longer concerned with previous numbers

for the first person:
yes, the first time, he picks 1/10
the secon time he picks 1/9
IN TOTAL: he picks 2/10

[Sir Kenyth]

quote:

---

Originally posted by 250:
the first person choose 2 out of 10

the second person choose 1 out of 9

that is the difference

what do you mean by "it is not ended"? it is ended THE MOMENT you add one more person in. then the problem changed into something completely different: "whats the probability that one can pick a prize out of nine door?" it is no longer concerned with previous numbers

for the first person:
yes, the first time, he picks 1/10
the secon time he picks 1/9
IN TOTAL: he picks 2/10

---

Ummm... I don't see where I posted anything about anything ending?

Anyway, you're avoiding the harder questions in my posts. Here they are, for your perusal.

These are two related to our off-topic conversation:

Just to tickle your brain. Ask yourself why? Physically there's little difference at that point in time of choice. The first and second person each have 9 doors to choose from and one choice to make. From that perspective, their chances are equal. From the perspective that the first person has done this before. His total chances because of his past are higher than the second. How is it possible to have equal chance and unequal chance at the same time? It's not. It's simply your perspective on the problem. Here's another tickler. What if the first person had done this before and got the prize. Does this mean his chances are reduced in comparison to someone who has never gotten a prize? The chances of winning two times are less than winning once.

You just luv the numbers bud. What about the reasoning and perception that came up with them? The conditions of a problem are perceived by someone. Another person can perceive things differently. Your solution is correct, from that point of veiw. Explain in words to me, why two people with an exact same physical task should have different probabilites for it. What is the physical difference between the two at that point?


This one is related to our primary conversation:

Yes, we did get a bit off topic. The initial argued point was that the formula may not apply to this problem. Which is exactly where we needed to go. The point has been proven that subsequent choices have a better probability than first ones. Even if the number of doors didn't narrow down with each one. This was the point of 250s formula. The person in the problem is still on their first choice though, so why would this apply?

[250]

grrrr

physically, the first person picked twice

the whole crap about "switching" is just to SIMPLIFIED the idea of picking the door twice. (in fact, they made it more complex)

so the simple question is "whats the probability for a person to pikc a prize out of 10 doors if he has two chances?"

what he would do is, pick one door, then switch. correct?

so 2/10

I dont see whats so difficulty to see this

about in the second time, comes another person

yeah, he does have 1/9 chance, but that is for THE SECOND TIME ONLY. chances are, he might get NOTHING because the first person already picked the prize. the two people combine together has a 2/10 chance

and is that what the problem is about??? NOPE IT IS NOT! so dont change the problem

answer me the hard question:

is the problem about "whats the probability for a person to pikc a prize out of 10 doors if he has two chances?"

(see the procedure yourself)