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rbx270j
Profile Joined November 2010
Canada540 Posts
Last Edited: 2011-06-10 08:19:32
June 10 2011 08:16 GMT
#221
On June 10 2011 12:29 Count9 wrote:
I don't understand how people can know the paradox, see that it's worded in the exact same way (i.e. purposely ambiguous) and still argue that it's for sure one or the other.



It's because It's worded fine, he just misunderstands what he wrote. If I ask you what are the chances I flip a coin and it'll be heads, you'd say 1/2. if I told you I flipped it twice and at least 1 was heads, the chance that the other was heads is still 1 in 2.

by the OPs logic, it would be

HH
TH
HT

therefore 1 in 3 to get heads from 1 flip of a coin. That's wrong.

I don't know if the OP has changed the wording since I read it hours ago, but if he has, I maintain this about the original wording.
Deleted User 3420
Profile Blog Joined May 2003
24492 Posts
Last Edited: 2011-06-10 08:19:13
June 10 2011 08:16 GMT
#222
I see how it's 1 out of 5, I enjoyed the question.

(don't get what all the hubbub is about)
Hamster1800
Profile Blog Joined August 2008
United States175 Posts
June 10 2011 08:23 GMT
#223
It's things like this
On June 10 2011 16:34 garbobjee wrote:
Show nested quote +
On June 10 2011 10:19 oxidized wrote:

Limiting your options to ZP, ZZ, ZT, PZ, TZ (order preserved).

That means there is only a 1/5 chance you got the ZZ, so that the other match was zerg.


When you said, ZP, ZZ, ZT, PZ, and TZ,
you stated the number of possible combinations incorrectly. The order of the games doesn't actually matter, the only thing that does matter is how many times you played as a certain race. So now, ZP and PZ should be one outcome, and ZT and TZ should also be one outcome.

With those outcomes combined, there should only be three outcomes: ZZ, ZP, and ZT, so the chance of playing as zerg twice given that you played as zerg at least once, is 1/3.

that make me think that it's a bad idea to teach probability as P(A) = (# ways A can happen)/(# ways anything can happen), simply because this assumes that every ``way'' is equally likely, which would lead to the conclusion that if we say the ``way'' A can happen is just that A happens and the ``ways'' anything can happen is that either A happens or A doesn't happen, then we get that there is 1 ``way'' for A to happen and 2 ``ways'' for anything to happen, so the probability of A happening is 1/2.

I'm sure everyone here agrees that is absurd (except for that poster many pages back who made a joke about exactly that).

The problem is that it can be very unclear when two different ``ways'' of something happening are actually different or the same, and I haven't come up with a good way to resolve it. Something about the ``size'' of a ``way'' is sort of necessary, but this seems to bring in probability circularly.

Here's another example of how that way of thinking can screw you up. Let's say I play a game with you. I have three cards, one with two red sides, one with two blue sides, and one with one red side and one blue side. I have a machine that shuffles them and randomly flips them over, then deals one out such that neither of us can see what is on the other side. If it comes up red, I'll bet you (at 1:1 odds) that the other side is red, and if it comes up blue, then I'll bet you that the other side is blue.

Now from your perspective, you see a red card and someone saying ``I bet you that the other side of the card is red''. You might think, well the other side is either red (which can happen in one way, with the red/red card), or the other side is blue (which can happen in one way also, with the red/blue card). So therefore I don't lose anything by taking this bet.

After a while, you lose all your money and wonder why. It's because the ``way'' of a red side coming up with a red/red card is twice as likely as the ``way'' of a red side coming up with a red/blue card.

Does anyone know of a good way to explain probability without using this idea of counting ``ways''?
D is for Diamond, E is for Everything Else
Dimagus
Profile Joined December 2010
United States1004 Posts
Last Edited: 2011-06-10 08:27:08
June 10 2011 08:26 GMT
#224
On June 10 2011 17:23 Hamster1800 wrote:
Does anyone know of a good way to explain probability without using this idea of counting ``ways''?


You mean permutations versus combinations?
StarDrive
Profile Joined September 2010
90 Posts
June 10 2011 08:30 GMT
#225
Isn't it just?

ZZ - 1/9
ZX - 2/9
XZ - 2/9

so 1/(1 + 2 + 2) = 1/5
Hamster1800
Profile Blog Joined August 2008
United States175 Posts
June 10 2011 08:32 GMT
#226
On June 10 2011 17:26 Dimagus wrote:
Show nested quote +
On June 10 2011 17:23 Hamster1800 wrote:
Does anyone know of a good way to explain probability without using this idea of counting ``ways''?


You mean permutations versus combinations?


No, I mean more basic than that. Like someone wants to understand what I mean when I say ``the probability of a die landing on 3 is 1/6''. Then the standard answer is to say ``Well, the die has 6 ways it can come up (1, 2, 3, 4, 5, or 6), and only one of those ways is a 3, so the probability is 1/6.''

But this leads to the misconceptions that I pointed out earlier, since this doesn't give any notion of relative size, which may well be important. So I want a better method of defining probability, which I haven't really seen.
D is for Diamond, E is for Everything Else
mr_tolkien
Profile Blog Joined June 2010
France8631 Posts
June 10 2011 08:37 GMT
#227
I couldn't think of a worst asked problem :/ This is not clear AT ALL.
The legend of Darien lives on
Cambium
Profile Blog Joined June 2004
United States16368 Posts
June 10 2011 08:37 GMT
#228
On June 10 2011 17:07 Ivs wrote:
Show nested quote +
On June 10 2011 17:04 Cambium wrote:
On June 10 2011 16:59 Geo.Rion wrote:
Monthy python paradox reloaded?


Monty Python is very different. Monty Python is a "paradox" because the state space was never fully defined.

You mean the Monty Hall paradox?

Zerg tears (Wiki)Monty Hall

Oops, obviously

He typed monty python, and I followed his suit T_T
When you want something, all the universe conspires in helping you to achieve it.
Piggiez
Profile Joined March 2011
393 Posts
June 10 2011 08:38 GMT
#229
On June 10 2011 17:16 rbx270j wrote:
Show nested quote +
On June 10 2011 12:29 Count9 wrote:
I don't understand how people can know the paradox, see that it's worded in the exact same way (i.e. purposely ambiguous) and still argue that it's for sure one or the other.



It's because It's worded fine, he just misunderstands what he wrote. If I ask you what are the chances I flip a coin and it'll be heads, you'd say 1/2. if I told you I flipped it twice and at least 1 was heads, the chance that the other was heads is still 1 in 2.

by the OPs logic, it would be

HH
TH
HT

therefore 1 in 3 to get heads from 1 flip of a coin. That's wrong.

I don't know if the OP has changed the wording since I read it hours ago, but if he has, I maintain this about the original wording.


No. By the OPs logic, the question would be worded (essentially) "In 2 coin flips what is the probability that one flip yields two heads?" In which case the answer is 1/3
Defrag
Profile Joined February 2010
Poland414 Posts
Last Edited: 2011-06-10 08:43:02
June 10 2011 08:42 GMT
#230
Original post is terribly explained and messy as hell, jesus.
Dimagus
Profile Joined December 2010
United States1004 Posts
Last Edited: 2011-06-10 09:20:33
June 10 2011 08:53 GMT
#231
The more popular form of this question is about a family that has two children, and it asks the probability that both are boys if you know either A) at least one of them is a boy or B) there are not two girls. It's meant to show how people make assumptions about a starting condition, and don't consider order.

You will see answers for 1/2, 1/3, and 1/4. Yes, all three. Their reasoning:

1/2 - You know one child is a boy so there is a 1/2 chance the other child is a boy too
1/3 - You know both children can't be girls, so that narrows the possible scenarios to 3, only one of which is two boys.
1/4 - Knowing information about a particular family's children after both have been born does not actually change the probability that both children would be boys.
Cyba
Profile Joined June 2010
Romania221 Posts
Last Edited: 2011-06-10 09:17:31
June 10 2011 09:14 GMT
#232
On June 10 2011 17:38 Piggiez wrote:
Show nested quote +
On June 10 2011 17:16 rbx270j wrote:
On June 10 2011 12:29 Count9 wrote:
I don't understand how people can know the paradox, see that it's worded in the exact same way (i.e. purposely ambiguous) and still argue that it's for sure one or the other.



It's because It's worded fine, he just misunderstands what he wrote. If I ask you what are the chances I flip a coin and it'll be heads, you'd say 1/2. if I told you I flipped it twice and at least 1 was heads, the chance that the other was heads is still 1 in 2.

by the OPs logic, it would be

HH
TH
HT

therefore 1 in 3 to get heads from 1 flip of a coin. That's wrong.

I don't know if the OP has changed the wording since I read it hours ago, but if he has, I maintain this about the original wording.


No. By the OPs logic, the question would be worded (essentially) "In 2 coin flips what is the probability that one flip yields two heads?" In which case the answer is 1/3


Isn't it 1/4 ?
p(event)=p(flip1)*p(flip2)=1/4

afaik the easiest way to handle chain events and probabilities is, if you know the probability of each smaller event and the relationship is "and" you can just multiply them. As long as order doesn't matter.

Elaborating:
It makes sense to be 1/4 because you have 4 posible versions of the large event:
HT
TH
HH
TT
I'm not evil, I'm just good lookin
neo_sporin
Profile Blog Joined August 2010
United States516 Posts
June 10 2011 09:21 GMT
#233
On June 10 2011 18:14 Cyba wrote:
Show nested quote +
On June 10 2011 17:38 Piggiez wrote:
On June 10 2011 17:16 rbx270j wrote:
On June 10 2011 12:29 Count9 wrote:
I don't understand how people can know the paradox, see that it's worded in the exact same way (i.e. purposely ambiguous) and still argue that it's for sure one or the other.



It's because It's worded fine, he just misunderstands what he wrote. If I ask you what are the chances I flip a coin and it'll be heads, you'd say 1/2. if I told you I flipped it twice and at least 1 was heads, the chance that the other was heads is still 1 in 2.

by the OPs logic, it would be

HH
TH
HT

therefore 1 in 3 to get heads from 1 flip of a coin. That's wrong.

I don't know if the OP has changed the wording since I read it hours ago, but if he has, I maintain this about the original wording.


No. By the OPs logic, the question would be worded (essentially) "In 2 coin flips what is the probability that one flip yields two heads?" In which case the answer is 1/3


Isn't it 1/4 ?
p(event)=p(flip1)*p(flip2)=1/4

afaik the easiest way to handle chain events and probabilities is, if you know the probability of each smaller event and the relationship is "and" you can just multiply them. As long as order doesn't matter.

Elaborating:
It makes sense to be 1/4 because you have 4 posible versions of the large event:
HT
TH
HH
TT

Yes but then he gives you the detail you are looking for at least 1 head, so TT is irrelevent since any time 2 tails pops up you trash that result entirely as for the purpose of the problem, it doesnt exist. In the example you are talking about he left out that detail but you have it right, he was just thinking in terms of the 1 of them is a heads situation which the problem this thread is based on.
Geo.Rion
Profile Blog Joined October 2008
7377 Posts
June 10 2011 09:24 GMT
#234
On June 10 2011 17:37 Cambium wrote:
Show nested quote +
On June 10 2011 17:07 Ivs wrote:
On June 10 2011 17:04 Cambium wrote:
On June 10 2011 16:59 Geo.Rion wrote:
Monthy python paradox reloaded?


Monty Python is very different. Monty Python is a "paradox" because the state space was never fully defined.

You mean the Monty Hall paradox?

Zerg tears (Wiki)Monty Hall

Oops, obviously

He typed monty python, and I followed his suit T_T

sorry, i ve meant Monty Hall indeed, i posted before drinking coffe, my mind is not reliable before coffe
"Protoss is a joke" Liquid`Jinro Okt.1. 2011
Akta
Profile Joined February 2011
447 Posts
Last Edited: 2011-06-10 09:53:15
June 10 2011 09:29 GMT
#235
On June 10 2011 17:23 Hamster1800 wrote:Does anyone know of a good way to explain probability without using this idea of counting ``ways''?
Coin flips is kind of classic but it's still about assigning correct probability to possible outcomes, or "counting ways" as you put it. Example:

Flipping two coins there are 3 possible outcomes:
Head and head.
Tail and tail.
Head and tail.

Then we can illustrate the probabilities:

Have 2 people flip one coin each at the same time. Possible outcomes are:
Person 1 Head, person 2 Tail
Person 1 Tail, person 2 Head
Person 1 Head, person 2 Head
Person 1 Tail, person 2 Tail

Therefore if a person flips 2 coins the probabilities are:
Head and head 25%.
Tail and tail 25%.
Head and tail 50%. This is because coin 1 head/coin 2 tail and coin 1 tail/coin 2 head both produce this result.

Another way to illustrate could be: If you bet on a normal double coin flip there are 3 possible outcomes, but if you flip one coin then another, what is the probability that the second coin will land on the same side as the first?
Answers obviously depend on the questions asked and a lot of funny "problems" are based on trying to make our brains answer the wrong question. So if you for example use the answer to my last question in other 2 coin flip scenarios you need to understand that the question doesn't separate head/head and tail/tail just like a normal double coin flip does not separate head/tail and tail/head.
Liveon
Profile Blog Joined September 2010
Netherlands1083 Posts
Last Edited: 2011-06-10 09:32:15
June 10 2011 09:31 GMT
#236
I'm probably a total noob now, but I still don't understand. Read the boy/girl paradox on wikipedia, still don't get it.

In the boy/girl paradox there's a difference between the two children outside of their gender, namely their age. One is older than the other, which gives us indeed three options.
boy girl, where boy is older
boy girl, where girl is older
boy boy.
then yes, it's 1/3.

but with your race, you're explicitly stating that *you* are the one playing zerg. Which makes the only possible matchups zvz, zvp and zvt. If you cound tvz and pvz as well, that would mean that you're not the one playing zerg.
That is only when I assume you're naming yourself first in the xvx format.

so wouldn't it just be 1/3 then?
Hearthstone manager ECVisualize, Head Admin DSCL
rbx270j
Profile Joined November 2010
Canada540 Posts
Last Edited: 2011-06-10 09:36:03
June 10 2011 09:32 GMT
#237
On June 10 2011 17:38 Piggiez wrote:
Show nested quote +
On June 10 2011 17:16 rbx270j wrote:
On June 10 2011 12:29 Count9 wrote:
I don't understand how people can know the paradox, see that it's worded in the exact same way (i.e. purposely ambiguous) and still argue that it's for sure one or the other.



It's because It's worded fine, he just misunderstands what he wrote. If I ask you what are the chances I flip a coin and it'll be heads, you'd say 1/2. if I told you I flipped it twice and at least 1 was heads, the chance that the other was heads is still 1 in 2.

by the OPs logic, it would be

HH
TH
HT

therefore 1 in 3 to get heads from 1 flip of a coin. That's wrong.

I don't know if the OP has changed the wording since I read it hours ago, but if he has, I maintain this about the original wording.


No. By the OPs logic, the question would be worded (essentially) "In 2 coin flips what is the probability that one flip yields two heads?" In which case the answer is 1/3


Did you read before he edited it (before Nestea, TLO and watching games got involved)? If so and that's what you're talking about, you're wrong. If you're not talking about that, I said I was talking about it's original iteration, so we're not really discussing the same thing.

I'm arguing against the people who say they agree with 1/5 with how it was originally phrased.
SolC361
Profile Joined July 2010
United States184 Posts
June 10 2011 10:35 GMT
#238
On June 10 2011 17:53 Dimagus wrote:
The more popular form of this question is about a family that has two children, and it asks the probability that both are boys if you know either A) at least one of them is a boy or B) there are not two girls. It's meant to show how people make assumptions about a starting condition, and don't consider order.

You will see answers for 1/2, 1/3, and 1/4. Yes, all three. Their reasoning:

1/2 - You know one child is a boy so there is a 1/2 chance the other child is a boy too
1/3 - You know both children can't be girls, so that narrows the possible scenarios to 3, only one of which is two boys.
1/4 - Knowing information about a particular family's children after both have been born does not actually change the probability that both children would be boys.
An interesting variation on this question is: What is the probability of a family having two girls if one of the children is a girl named Kerrigan?
althaz
Profile Joined May 2010
Australia1001 Posts
June 10 2011 10:53 GMT
#239
To those getting confused, think of it like this:

For any two games that a random player plays, given that in one of the games he was Zerg, what is the probability that he was also Zerg in the other? The answer here is clearly not 1/3.
The first rule we don't talk about race conditions. of race conditions is
piegasm
Profile Joined August 2010
United States266 Posts
June 10 2011 11:01 GMT
#240
On June 10 2011 16:11 Abenson wrote:
erp.
1/5 - i don't see how hard this can be lol
All you have to do, if you're confused, is to list out all the possibilities.

As I'm posting right now, 61% have answered 1/3 lol


That's because the OP has been edited. His wording was bad initially and it looked like he was asking the probability of getting zerg in a single game.
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