Author Topic: TT's Maths Thread (Read 149536 times) Tweet Share

the.watchman · « **Reply #975 on:** May 04, 2010, 09:07:03 pm »

Ah but you know that one of the other doors is a goat, right?
So in each case it is a question of 50:50, but in the overall scheme, the three cases yield a 2/3 chance IF you switch

I love this question, don't you TT and brightsky?

TrueTears · « **Reply #976 on:** May 04, 2010, 09:09:59 pm »

no i dont like probability.

anyways i dont get it...

"If you pick door B and you switch, you'll get the car."

u dont tho, u can pick the goat still

So if you pick door C first, you have a probability of 1 of picking the goat if u switch, so u lose, so why would u wanna switch here?

But the answer is switching is always better, but clearly here, if u switch ur gonna lose instead of win

kamil9876 · « **Reply #977 on:** May 04, 2010, 09:22:21 pm »

What if the host himself does not know where the car is, and may accidentally remove it. what should you do?

the.watchman · « **Reply #978 on:** May 04, 2010, 09:24:07 pm »

After an overall view of all cases, it is obvious that switching is better in general
Oh, and with the pick door B case, you are shown that one of the doors (A in this case) has a goat behind it, so if you switch you will pick C which has the car etc.

Quote from: kamil9876 on May 04, 2010, 09:22:21 pm

What if the host himself does not know where the car is, and may accidentally remove it. what should you do?

Lol, not gonna happen...

TrueTears · « **Reply #979 on:** May 04, 2010, 09:25:44 pm »

ahhh i don't care anymore, dont get this Q, dont care either now

brightsky · « **Reply #980 on:** May 04, 2010, 09:34:49 pm »

Quote

"If you pick door B and you switch, you'll get the car."

u dont tho, u can pick the goat still

So if you pick door C first, you have a probability of 1 of picking the goat if u switch, so u lose, so why would u wanna switch here?

The other door is already open. So you know what's in one door already, the only options are to stick with door B or switch to whatever door isn't open.

Again, picking door C is the only case where when you switch you don't get the car. In the remaining two cases, you do get the car if you switch.

And yes I do hate probability...and statistics too. xD

TrueTears · « **Reply #981 on:** May 04, 2010, 09:36:30 pm »

OMFG

I READ THE FUKN QUESTION WRONG, LOL I THOUGHT THE HOST DOESNT OPEN ANY DOORS AFTER U PICK UR DOOR, I THOUGHT HE'D JUST ASK U IF U WANNA SWITCH TO EITHER OF THE 2 OTHER DOORS

OMFG

FKN.........

shit shit no wonder, yeah i get it now totally lolololol epic fail ><

the.watchman · « **Reply #982 on:** May 04, 2010, 09:41:17 pm »

Quote from: TrueTears on May 04, 2010, 09:36:30 pm

OMFG

I READ THE FUKN QUESTION WRONG, LOL I THOUGHT THE HOST DOESNT OPEN ANY DOORS AFTER U PICK UR DOOR, I THOUGHT HE'D JUST ASK U IF U WANNA SWITCH TO EITHER OF THE 2 OTHER DOORS

OMFG

FKN.........

shit shit no wonder, yeah i get it now totally lolololol epic fail ><

Lol, it is trippy though...

Yitzi_K · « **Reply #983 on:** May 04, 2010, 10:20:34 pm »

It's counter intuitive, but it is better to switch. There's an iPhone app which runs simulations of this scenario, and if you switch, the likelihood of getting the car is 66.67%

/0 · « **Reply #984 on:** May 04, 2010, 11:50:03 pm »

I think this is called the Monty Hall problem

TrueTears · « **Reply #985 on:** May 04, 2010, 11:53:21 pm »

yeah it is, kamil just told me on msn, also showed me a neat tree diagram which explains it instantly

Ahmad · « **Reply #986 on:** May 10, 2010, 11:11:26 pm »

Probability is awesome.

TrueTears · « **Reply #987 on:** June 27, 2010, 04:32:49 am »

Find the Taylor series for $\sqrt{x}$ centered around $x=1$ .

I've gotten pretty close but I don't know how to put it back into summation format. This is my working:

$f(x) = \sum_{n=0}^{\infty} \frac{f^n(1)}{n!}(x-1)^n$

$f^1(1) = \frac{1}{2}$

$f^2(1) = \left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)$

$f^3(1) = \left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)$

$f^4(1) = \left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)$

So we have:

$\sqrt{x} = 1 + \frac{\frac{1}{2}}{1!}(x-1) + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}(x-1)^2+\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{3!}(x-1)^3 + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{4!}(x-1)^4 + \cdots$

So I know the summation form must have $1+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}\frac{1}{2^n}\text{something}}{n!}(x-1)^n$ but I can't find the pattern for $(1), (1)(-1), (1)(-1)(-3), (1)(1)(-3)(-5)$

Help please, thanks!

googoo · « **Reply #988 on:** June 27, 2010, 08:46:13 am »

To me it is so simple. There is a greater chance (2/3) of picking the goat at the first pick. If you are given a second chance to switch, why not.

Damo17 · « **Reply #989 on:** June 27, 2010, 09:53:55 am »

Quote from: TrueTears on June 27, 2010, 04:32:49 am

Find the Taylor series for $\sqrt{x}$ centered around $x=1$ .

I've gotten pretty close but I don't know how to put it back into summation format. This is my working:

$f(x) = \sum_{n=0}^{\infty} \frac{f^n(1)}{n!}(x-1)^n$

$f^1(1) = \frac{1}{2}$

$f^2(1) = \left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)$

$f^3(1) = \left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)$

$f^4(1) = \left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)$

So we have:

$\sqrt{x} = 1 + \frac{\frac{1}{2}}{1!}(x-1) + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}(x-1)^2+\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{3!}(x-1)^3 + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{4!}(x-1)^4 + \cdots$

So I know the summation form must have $1+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}\frac{1}{2^n}\text{something}}{n!}(x-1)^n$ but I can't find the pattern for $(1), (1)(-1), (1)(-1)(-3), (1)(1)(-3)(-5)$

Help please, thanks!

Notice that the numerators starting from the term $-\frac{1}{4}$ are like the factorials we normally see with an odd positive integer multiplied by all the odd positive integers down to 1.

From Wikipedia:

Quote

A function related to the factorial is the product of all odd values up to some odd positive integer n. It is often called double factorial (even though it only involves about half the factors of the ordinary factorial, and its value is therefore closer to the square root of the factorial), and denoted by n!!.

So we get as part of the summation:

$\frac{(-1)^{n-1}(2n-3)!!}{2^n \cdot n!}(x-1)^n$

But this is only for $n \ge 2$ . So the full summation form is:

$\sqrt{x}= 1 + \frac{1}{2}(x-1) + \sum_{n=2}^{\infty}\frac{(-1)^{n-1} \cdot (2n-3)!!}{2^n \cdot n!}(x-1)^n$

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