Math question - probability

[pipigrande]

No, you got it wrong. I agree that the answer is 1/6. I just don't agree with other things.

other things being what?

Both case has the same probability of getting a number, but in reality, one method has more chance on getting that number than other method.

That's nonsense.

That's like saying "The expected probability of getting a head in a fair coin is 50%, but in reality it isn't because I just did the experiment with 100 coins and I actually got 40 heads and 60 tails, so it's actually 40% probability of getting heads"

[Plutonium]

other things being what?

Both case has the same probability of getting a number, but in reality, one method has more chance on getting that number than other method.

That's nonsense.

That's like saying "The expected probability of getting a head in a fair coin is 50%, but in reality it isn't because I just did the experiment with 100 coins and I actually got 40 heads and 60 tails, so it's actually 40% probability of getting heads"

dont be so mean to NuclearSilo..... many people in this thread can already infer that he's not very bright at math(probability).

[takolin]

other things being what?

Both case has the same probability of getting a number, but in reality, one method has more chance on getting that number than other method.

That's nonsense.

That's like saying "The expected probability of getting a head in a fair coin is 50%, but in reality it isn't because I just did the experiment with 100 coins and I actually got 40 heads and 60 tails, so it's actually 40% probability of getting heads"

That's why people always add confidence intervals to show how probable a statistic test is.
You can toss a fair coin 100 times and it can be 90 heads and 10 tails. It's possible, but not really probable.

[Goobronicus]

Both case has the same probability of getting a number, but in reality, one method has more chance on getting that number than other method.

The two cases have the same probability provide these conditions.

Either you are choosing the numbers blind and don't get to see the results of 1a before drawing 1b OR you aren't told which number you are looking for prior to making your selections. If you know you are searching for the number 1, and you can visibly see the results (whether before drawing or after you first draw) then the probability is skewed.

What's the probability of drawing the queen of spades out of a completely shuffled full deck (52 cards) face down? 1/52
What's the probability of drawing the queen of spades out of a completely shuffled deck laid out card by card face up? I would hope 1/1.

[SM-Count]

I guess someone should explain why algebraically:

Spoiler!

I use the term obviously because my hint is that you should do the problem with 't' times. What if t=1? Then you have (1/7) chance to get it the first time and (100%) chance every time after that, the pattern is based off that simple fact and that's also why it works for any t.

[NuclearSilo]

Spoiler!

I use the term obviously because my hint is that you should do the problem with 't' times. What if t=1? Then you have (1/7) chance to get it the first time and (100%) chance every time after that, the pattern is based off that simple fact and that's also why it works for any t.

what??

[SM-Count]

Just choosing the number is also choosing one number at a time (t=1), then choosing a number from that group of one. Conceptually choosing two numbers at a time from the whole, then choosing one number from that group is the same thing. Basically I'm saying that the two "cases" aren't distinct because algebraically it's just writing the same equation (1/n) differently.

[klokus]

And now calculation as I should do in class:

total amount of possibilities:
p = 2
n = 6
repitition allowed
sequence not important
=> combination of 2/6 = (6!)/(2!*4!) = 1440/48 = 30

amount of right cards:
p = 1
n = 6
repetition (not important)
sequence (not important)
=> 6 are right.

draw 1 of 2:
(1/2)*6 = 3

=> Chance is 3/30 = 1/10 = 10%

[SM-Count]

@above
Except (6!)/(2!*4!)=(6*5)/2=15

[woutR]

we just use the nCr and nPr button. Ezmode ftw.

[pipigrande]

And now calculation as I should do in class:

total amount of possibilities:
p = 2
n = 6
repitition allowed
sequence not important
=> combination of 2/6 = (6!)/(2!*4!) = 1440/48 = 30

amount of right cards:
p = 1
n = 6
repetition (not important)
sequence (not important)
=> 6 are right.

draw 1 of 2:
(1/2)*6 = 3

=> Chance is 3/30 = 1/10 = 10%

This isn't a combination or permutation question. It's rather a mxn question. Meaning that you get the percentage of one event happening (m) times the percentage of another event happening (n). Basically, you multiply the base rate probability times the probability rate.

When you say 'repitition allowed', it doesn't make sense according to what NuclearSilo provided. Since question said you have n=1,2,3,4,5,6 and since most of probability questions assume a fair random test unless stated otherwise then you would have to assume that you select 2 numbers randomly out of those 6. Say you pick 1 first, then the 2nd possible option is only 2,3,4,5 and 6. This is why you can't solve it with permutation.

You can't solve by combination because order is not important. Picking 1 in your first pick and 2 in your second pick is the same thing as if you picked 2 first and then 1 second.

[SM-Count]

You're picking 2 numbers simultaneously, that's why it is a combination question.

[Plutonium]

You're picking 2 numbers simultaneously, that's why it is a combination question.

no, the person above is correct. combination is a counting problem but this is a probability question.

[pipigrande]

You're picking 2 numbers simultaneously, that's why it is a combination question.

"Combinations" in Math means, that order matters. Now, referring to the question proposed by NuclearSilo... Do you think that picking 1 in the first attempt or picking 1 in the second attempt matters? Actually, it doesn't. Since order does not matter in this problem, then it isn't a combination question.

For example, picking 1 first attempt and then picking 2 second attempt would give the same result as picking 2 first attempt and then 1 second attempt.

[Plutonium]

You're picking 2 numbers simultaneously, that's why it is a combination question.

"Combinations" in Math means, that order matters. Now, referring to the question proposed by NuclearSilo... Do you think that picking 1 in the first attempt or picking 1 in the second attempt matters? Actually, it doesn't. Since order does not matter in this problem, then it isn't a combination question.

For example, picking 1 first attempt and then picking 2 second attempt would give the same result as picking 2 first attempt and then 1 second attempt.

actually combination means order don't matter. but u dont apply combination becuz for the 1st pick order dont matter but it does for the 2nd round. anyways u just do (2/6)*(1/2) = 1/6

why the fck people people have to make shit so complication jebus christ. most stuff are really that simple dont have to make it any harder people

[NuclearSilo]

What i'm trying to say is that, 2 methods have the same probability to obtain a number, but one method has more chance (you can call it 'luck' in real life). Specifically, the method "pick only one" is more lucky than the other method.

[pipigrande]

You're picking 2 numbers simultaneously, that's why it is a combination question.

"Combinations" in Math means, that order matters. Now, referring to the question proposed by NuclearSilo... Do you think that picking 1 in the first attempt or picking 1 in the second attempt matters? Actually, it doesn't. Since order does not matter in this problem, then it isn't a combination question.

For example, picking 1 first attempt and then picking 2 second attempt would give the same result as picking 2 first attempt and then 1 second attempt.

actually combination means order don't matter. but u dont apply combination becuz for the 1st pick order dont matter but it does for the 2nd round. anyways u just do (2/6)*(1/2) = 1/6

why the fck people people have to make shit so complication jebus christ. most stuff are really that simple dont have to make it any harder people

Yes, you are correct. Combination means order do not matter. My bad there. I guess my example would refer to permutation.

[klokus]

You're picking 2 numbers simultaneously, that's why it is a combination question.

"Combinations" in Math means, that order matters. Now, referring to the question proposed by NuclearSilo... Do you think that picking 1 in the first attempt or picking 1 in the second attempt matters? Actually, it doesn't. Since order does not matter in this problem, then it isn't a combination question.

For example, picking 1 first attempt and then picking 2 second attempt would give the same result as picking 2 first attempt and then 1 second attempt.

actually combination means order don't matter. but u dont apply combination becuz for the 1st pick order dont matter but it does for the 2nd round. anyways u just do (2/6)*(1/2) = 1/6

why the fck people people have to make shit so complication jebus christ. most stuff are really that simple dont have to make it any harder people

It's mathematics, not logics.

[woutR]

What i'm trying to say is that, 2 methods have the same probability to obtain a number, but one method has more chance (you can call it 'luck' in real life). Specifically, the method "pick only one" is more lucky than the other method.

No.

[Jantje]

What I'm trying to say is that, 2 methods have the same probability to obtain a number, but one method has more chance (you can call it 'luck' in real life). Specifically, the method "pick only one" is more lucky than the other method.

I think you are assuming you will always pick number 1 in case 1. But you can also pick 3 and 4, for example. Still, picking 1 out of 6 numbers the chance you pick number 1 is 1/6 no matter what way.

[wicked]

But anyways, I think this is easy math leading up to real probability maths with things like there are 3 red and 7 blue balls, you can pick 5, how big is the chance you get 3 red and 2 blue?
But that's also not very hard.

Total of 10 and u get 5, so (3C3)*(7C2)/(10C5)=0.083=8,33%

we just use the nCr and nPr button. Ezmode ftw.

Exactly

[NuclearSilo]

What i'm trying to say is that, 2 methods have the same probability to obtain a number, but one method has more chance (you can call it 'luck' in real life). Specifically, the method "pick only one" is more lucky than the other method.

No.

reason?

[klokus]

That's just because you think wrong (and what your feeling says)

Maths is always right...

[NuclearSilo]

Tested using computer simulation. Test with picking one number a time gives an accurate result, while other doesn't. Math is just theory, while computing is real.

[SM-Count]

You're picking 2 numbers simultaneously, that's why it is a combination question.

no, the person above is correct. combination is a counting problem but this is a probability question.

Did you even look at my explanation? Wait, I can answer that, no. Maybe if you'd bother read my explanation and how combinations were used you wouldn't be so eager to jump in with an obviously wrong opinion. Order doesn't matter for the second round because there is no second round! Do you not understand the word simultaneously?

@above
You're computer simulation is wrong then. Math is fact, computing is avoiding luck and looking for statistical significance while always yielding a little to chance.

[raphaell666]

For me it's 1/6 for both. It doesn't matter if you divide (or take) two from the total, the chance of you getting number one among numbers one, two, three, four, five and six is one in six, 1/6.
You could divide it into three groups of two numbers, the chance of getting number one from any of these is still 1/6 imo. I fail to understand why would it vary.

@NuclearSilo
Math =/= reality (in some cases). Math = fact/theory, and in this case, as it's percentage, reality may never be equal to the theory in my opinion, even if tested by anything, as it varies according basically to luck, chance.

What i'm trying to say is that, 2 methods have the same probability to obtain a number, but one method has more chance (you can call it 'luck' in real life). Specifically, the method "pick only one" is more lucky than the other method.

No.

+1
I fail to understand that, why would one method be more lucky than another if both have 1/6 chances of happening?

[Plutonium]

Tested using computer simulation. Test with picking one number a time gives an accurate result, while other doesn't. Math is just theory, while computing is real.

what computer simulation software or source code did u use?

[cuchulainn]

Ok, this thread is full of fail.

Case 2 is the easy one: 1/6.

Case 1 is more interesting.

Here are the possibilities for attempt 1:

1-2, 1-3, 1-4, 1-5, 1-6, 2-3, 2-4, 2-5, 2-6, 3-4, 3-5, 3-6, 4-5, 4-6, 5-6

So you have 5/15=1/3 chance of picking a 1.

Now, for attempt 2, you have.

1/2 chance of picking a 1 if there's a 1 present. 0 chance of picking a one if there's not a 1 present. So, if there's a 1 present 1/3 of the time, you have 1/2 a chance of picking it.

So, 1/3*1/2=1/6. Probability is the same.

A side note to the people saying probability is "blah blah" but the chance irl is "something else". You are retarded. You do not understand statistics, so go away.

So yes, the answer is both methods have the same probability. In probability, when you're in doubt about whether you did it right, list every single possibility. From that you can easily find out if you're right.

[SM-Count]

Here are the possibilities for attempt 1:

1-2, 1-3, 1-4, 1-5, 1-6, 2-3, 2-4, 2-5, 2-6, 3-4, 3-5, 3-6, 4-5, 4-6, 5-6

So you have 5/15=1/3 chance of picking a 1.

6 choose 2 = all possible pairs of numbers
5/6 choose 2 = 1/3, exactly what I did except you wrote it out >.>

[cuchulainn]

Here are the possibilities for attempt 1:

1-2, 1-3, 1-4, 1-5, 1-6, 2-3, 2-4, 2-5, 2-6, 3-4, 3-5, 3-6, 4-5, 4-6, 5-6

So you have 5/15=1/3 chance of picking a 1.

6 choose 2 = all possible pairs of numbers
5/6 choose 2 = 1/3, exactly what I did except you wrote it out >.>

Wrote it out for the people that have no clue what combination/permutation stuff is, but yeah, you were 100% right!