Current Events > Statistics are REALLY confusing.

(+) Yeah! (+)

Explain this to me.

Okay. Let's say someone keeps attempting some activity, and each time, there's a 50% chance of success. So, after 1 attempt, there's a 50% chance of succeeding, after the 2 attempts, overall, there's a 75% chance the person has succeeded, and after 3 attempts, there's a 87.5% chance of success. Is this right?

Basically, when you add chances of success together, in any scenario, does the "value" keep halving? I don't know how to explain this any better. Seems kinda strange.

Statistics in general are very confusing, and also counter-intuitive.

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Sunhawk posted...

Statistics in general are very confusing, and also counter-intuitive.

yes, i had to buy an album (with money!) just to pass a probability class

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Does my OP make sense?

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Yes, you're right, but the value keeps halving because you picked 0.5 to begin with. It's the probability that at least one out of 3 will be heads.

1-(0.5)*(0.5)*(0.5) = 0.875

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If you're talking about flipping a coin, no. If you're talking about selecting the correct door on Let's Make a Deal, yes.

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Also, explain to me the Monty Hall Problem. It's so confusing that...wtf.

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Ranting Nord posted...

Yes, you're right, but the value keeps halving because you picked 0.5 to begin with. It's the probability that at least one out of 3 will be heads.

1-(0.5)*(0.5)*(0.5) = 0.875

Let's say the chances of success, in the problem described in the OP was 30%. Would it keep diminishing by a third, or two thirds, or what?

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@Ranting_Nord


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Yes

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Would it diminish by a third or two thirds? You answer was, uh, confusing.

You know, I used to be amazing at maths, growing up. Guess I've forgotten quite a few things.

@Ranting_Nord


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I'm not amazing at math by any means.

1-(0.3)*(0.3)*(0.3) = 0.973

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Very interesting.

Here's something that confuses me, however. If you do a task twice, and there's a 1 in 3 chance of succeeding each day, according to the calculator I just fed numbers into, the chances if succeeding at all is 0.91. Is this right? Did I do something wrong? Also, if it's right, how can it be as high as 0.91? Wouldn't it be roughly 0.6?

@Ranting_Nord
@MedeaLysistrata
@vigorm0rtis

Remember when CE talked for years about whether or not 0.9 recurring equaled 1?


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*each time, not each day

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Not exactly. It's just that when your probability is 50% on a single go, it just so happens that it halves every time.

If you go twice, here are your possible outcomes:

WW
WL
LW
LL

Each has a 25% chance of success, since there's no bias toward one or the other. Now, as you can see, first go winning has a 50% chance, just like we expected, and so does second go winning.

The trick is that one of those four cases is an overlap. The WW case is a part of both the first go wins and the second go wins sets, so if you just add the 50%'s, which would be the intuitive thing, you've counted that case twice. To remedy this, just subtract one out. In this case, you add the 50%'s and subtract out a 25%.

This is an idea taken from set theory, on calculating the number of elements in the union of two sets. Which makes sense because what probability really boils down to is a number of desirable possibilities compared to the total number of possibilities.

Count(A union B) = Count(A) + Count(B) - Count(A intersect B)

Again, you subtract out the intersect once because you already added it twice.

A = (A intersect not B) union (A intersect B)
B = (B intersect not A) union (B intersect A)

Since A and not A are complementary, and B and not B as well, you don't need to worry about their intersections. The term is still there, just always an empty set.

But if we then union these together we, we can see why we need to subtract out one instance of the intersect.

A union B = ((A intersect not B) union (A intersect B)) union ((B intersect not A) union (B intersect A))

A union B = (A intersect not B) union (A intersect B) union (B intersect not A) union (B intersect A)

A union B = (A intersect not B) union (B intersect not A) union (B intersect A) union (A intersect B)

Since B intersect A is equal to A intersect B, and any set unioned with itself is itself, this simplifies to:

A union B = (A intersect not B) union (B intersect not A) union (A intersect B)

Thing is you're usually given A and B so it's easier to work with those than to figure out what the sets look like without each other and with each other and stuff, so the easy thing to do is just manually subtract out the count of the intersection.

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If you flip a coin 10 times and get heads every time, there's like a 99% chance your next flip will be tails

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EndOfDiscOne posted...

If you flip a coin 10 times and get heads every time, there's like a 99% chance your next flip will be tails

Stop messing with us.

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Sunhawk posted...

Also, explain to me the Monty Hall Problem. It's so confusing that...wtf.

Yeah, that one's really counter-intuitive. The trick is that Monty always removes a goat and never the car. So if you swap, you always get the opposite of what you have. Then unless you chose the car on your first try, which you probably didn't, you get it if you swap.

Intuitively, you might think that you have a 50-50 either way, but the important thing here is that Monty knows where the car is. Imagine there are a million doors and he opens all but one after you choose. Either you chose right, at 1 in 1000000 probability, and Monty just picked another door to not open at random, or the one he didn't open is the car.

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If you do something twice, and the chances each time of succeeding are 1 in 3, what are the chances you'll succeed at least once?

@DevsBro
@Ranting_Nord

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I'm tapping out. I'm not in the mood to teach a class on something I barely understand as well.

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1/3 + 1/3 - 1/9 = 5/9

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I see. But how did you arrive at that sum? Where did the 1/9 come from?

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@DevsBro

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Sunhawk posted...

I see. But how did you arrive at that sum? Where did the 1/9 come from?

Because it's the probability of the "both win" case.

Remember, you have to subtract out the intersection.

WW
WL
LW
LL

Unlike before, these don't all have the same probability because a L is twice as likely as a W. After you've got your first flip, that puts you in the base state for the next flip, so you multiply the probabilities together:

WW: 1/3*1/3 = 1/9
WL: 1/3*2/3 = 2/9
LW: 2/3*1/3 = 2/9
LL: 2/3*2/3 = 4/9

Notice it still adds up to 9/9. Notice also that WW union WL and WW union LW both still add up to 1/3. So since adding the 1/3's together accounts for WW twice, you just subtract out the 1/9 once.

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Hey @DevsBro , are you a mathematician, or in a career that requires an education in higher mathematics?

Let's say someone did a task 3 times, and the chances of succeeding each time are 25%. What are the chances of succeeding at least once?

You don't (neccesarily) need to show how you got to the answer, I'm just...curious.


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@vigorm0rtis

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Sunhawk posted...

Also, explain to me the Monty Hall Problem. It's so confusing that...wtf.

This can be really confusing at first but here is an easier way to understand it.

Imagine the monty hall problem but instead of 3 doors we have 100 doors.

You pick door A and I tell you ok sure... let me eliminate 98 doors and the only remaining doors are A and B.

One of these doors is the correct door. Which door do you think is more likely to be correct out of A and B?

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No but I always liked math. I had most of the required credit hours for a minor in mathematics, since I took Vector Calculus and Probability and Statistics as electives. But I didn't end up finishing it out since I had to take a fine arts elective my senior year and I didn't want to do extra coursework on top of that.

I also had a class, Random Signals and Systems, that taught me way more probability than Stats did. Go figure.

Anyway, if you have three iterations, you could try to tackle it the way I did above but it would be easier to look at the set you don't want and subtract it from 1. For 1/4, you just do 1-(3/4)^3, which I believe should be 37/64.

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