Anyone good at probability?

[mudman]

Re: Anyone good at probability?

This is a problem where you have three doors which behind them sits hidden objects. One of the doors contains a valuable prize and the other do not (normally I hear this problem involve a car and two goats). The game show host will ask you, the contestant, to choose one of the three doors without zero penalty. The door options are #1, #2, and #3. Let's say that you choose door #2. The game show host, who know what is behind each of the doors, will then show you one door which he knows has a goat behind it, so let's say that he shows you door #1. Here is the kicker: the game show host will now ask you whether you want to switch your choice.

A lot of people will be instinctive, paranoid, or superstitious, and say that because it's a 50/50 chance, it might as well be in your best interest to stick with your gut choice. But of course, since this is simply a probability problem, this is where you are tricked into thinking that you are really looking at a 50/50 probability. What we all have to remember is that the game show host isn't just showing us where one goat is, he is showing us where the car isn't, and is asking us whether we want to switch our choice to one other door or not. When we first chose a door, our probability was 33.3% of being correct. When he showed us that a car was not in the other door, that now increases our chances of the car being in the other door by another 33.3%, as our choice, no matter how you look at it, was always and will always be 33.3%. That's basically how you have to look at it.

To better get a grasp of this example, let's make it 100 doors with 99 goats and 1 car. If we choose a door, the probability we landed on the car is 1%. Now let's say that the game show host shows us 98 other doors revealing goats, and we're left with the door that we chose and one other door. Is it in our best interest to switch? Hell yeah it is! Now see, does that make sense that our probability is now 50/50 since we're reduced to two doors? No way! He simply showed us where the car isn't, and he's asking us whether we want to switch from our 1% probability to our 99% probability.

Always account for variable change. But the question is... does this problem want us to do so as well? If so, mudman provided the correct answer, and my answer would be for a problem given that the probability is fixed.

This is referred to as the Monty Hall problem and you explained perfectly why there is zero ambiguity. The only people who debate it are those who don't know what they're talking about. I also much prefer using 100 instead of 3 as it better illustrates just how absurd it is to think that there's a 50/50 chance when there's 2 doors left.

[Speedy]

Re: Anyone good at probability?

I'm subscribed to this thread and keep reading new posts even though my intelligence has been surpased several posts back. This is where intelligence and common sense separate.

Adieu.

[Tomahawk]

Re: Anyone good at probability?

This logic is flawed. What you just calculated was the probability that you will draw at most three Republicans, given the events are not mutually exclusive (you did 1 minus the probability that the first, second, third, and fourth selections are all Republicans... again, without any sort of mutual exclusion, which is still this thread's hot topic debate).

I'm not sure I agree with the "the probability that I would draw three Republicans". I stated on the first draw you have a 11/20 chance of drawing a Republican. Then on the 2nd draw if you didn't draw a Republican on the first one then your odds lower to 11/19 that you draw a Republican then to 11/18 and 11/17 for the next two draws.

You only need one Republican to make the statement true so don't you need to basically calculate the odds of drawing 4 times and not getting a Republican? The first time you draw the odds would be 11/20 then the odds of not drawing a Republican go to 11/19 for the 2nd draw then 11/18 then 11/17, right? Multiply the four odds together and subtract from 1 which gives you the odds of drawing four times without getting one Republican right?

[Tomahawk]

Re: Anyone good at probability?

#2 No, as Blzr said, this is wrong.

The best way to think about this is to realize that the only way to NOT have 1 republican is to have all 4 democrats. P(all Dem) = (9/20)*(8/19)*(7/18)*(6/17) = .026006.

So P(not all Dem) = P(at least 1 repub) = 1 - .026006 = .973994 or 97.3994%

I see and understand the math behind this, but I'm not exactly sure why you couldn't calculate the odds of making four straight draws without pulling a Republican by saying 11/20 of Republicans then 11/19 then 11/18 then and 11/17? Multiply your odds and subtract from one to finish the problem. Isn't saying you made four draws and got all Democrats the same as saying you pulled four straight cards without pulling a Republican?

[Blzer]

Re: Anyone good at probability?

I'm not sure I agree with the "the probability that I would draw three Republicans". I stated on the first draw you have a 11/20 chance of drawing a Republican. Then on the 2nd draw if you didn't draw a Republican on the first one then your odds lower to 11/19 that you draw a Republican then to 11/18 and 11/17 for the next two draws.

You only need one Republican to make the statement true so don't you need to basically calculate the odds of drawing 4 times and not getting a Republican? The first time you draw the odds would be 11/20 then the odds of not drawing a Republican go to 11/19 for the 2nd draw then 11/18 then 11/17, right? Multiply the four odds together and subtract from 1 which gives you the odds of drawing four times without getting one Republican right?

No, I think that the question is kind of asking something differently. It's asking this in a proverbial sense: "What is the probability that you will select exactly one Republican + What is the probability that you will select exactly two Republicans + What is the probability that you will select exactly three Republicans + What is the probability that you will select exactly four Republicans?"

This is where my binomial distribution formula actually makes sense. It considers all of the varying possibilities that you select a certain number of Republicans provided that you select a certain number of Democrats, and the probabilities are added up. But I did it the quicker way... I simply said "What is the probability that you will select exactly zero Republicans?" and I did 1 minus that answer.

I may do another video that kind of better explains it, but I do believe that my 95.9% answer is correct. I don't want to sound arrogant when I'm saying that, I just do believe that is how the problem is done.

[Blzer]

Re: Anyone good at probability?

Okay, I said I wouldn't waste my time doing it, but I did anyway. I'll just type out my solution since filming and uploading takes too long. Refer to my video of showing the "opposite" way to do this problem (which was 1 minus the probability of selecting no Republicans):


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Okay, here is the other way of doing it, as I kind of demonstrated in my previous post.

Note: if I show the syntax n C x where n and x are values, that is the same thing as saying [n!]/[(x!)(n-x)!]


P(4,0.55,1) = (4 C 1) * (0.55)^1 * (0.45)^3 = (4)*(0.55)*(0.091125) = 0.200475
P(4,0.55,2) = (4 C 2) * (0.55)^2 * (0.45)^2 = (6)*(0.3025)*(0.2025) = 0.3675375
P(4,0.55,3) = (4 C 3) * (0.55)^3 * (0.45)^1 = (4)*(0.166375)*(0.45) = 0.299475
P(4,0.55,4) = (4 C 4) * (0.55)^4 * (0.45)^0 = (1)*(0.09150625)*(1) = 0.09150625

0.200475 + 0.3675375 + 0.299475 + 0.09150625 = 0.95899375

= 95.9%

Which is the same as the answer in the video I put up.




I am convinced that this is the correct answer, but I suppose I may have to re-explain it again if it's still not making sense.

[mudman]

Re: Anyone good at probability?

I see and understand the math behind this, but I'm not exactly sure why you couldn't calculate the odds of making four straight draws without pulling a Republican by saying 11/20 of Republicans then 11/19 then 11/18 then and 11/17? Multiply your odds and subtract from one to finish the problem. Isn't saying you made four draws and got all Democrats the same as saying you pulled four straight cards without pulling a Republican?

because 11/20, 11/19, 11/18, and 11/17 are probabilities of drawing republicans. You're talking about drawing democrats and yet you're writing probabilities for drawing republicans.

[mudman]

Re: Anyone good at probability?

Okay, I said I wouldn't waste my time doing it, but I did anyway. I'll just type out my solution since filming and uploading takes too long. Refer to my video of showing the "opposite" way to do this problem (which was 1 minus the probability of selecting no Republicans):


<object width="640" height="385"><param name="movie" value="http://www.youtube.com/v/sCFngTnNHPg&hl=en_US&fs=1&"></param><param name="allowFullScreen" value="true"></param><param name="allowscriptaccess" value="always"></param><embed src="http://www.youtube.com/v/sCFngTnNHPg&hl=en_US&fs=1&" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="640" height="385"></embed></object>




Okay, here is the other way of doing it, as I kind of demonstrated in my previous post.

Note: if I show the syntax n C x where n and x are values, that is the same thing as saying [n!]/[(x!)(n-x)!]


P(4,0.55,1) = (4 C 1) * (0.55)^1 * (0.45)^3 = (4)*(0.55)*(0.091125) = 0.200475
P(4,0.55,2) = (4 C 2) * (0.55)^2 * (0.45)^2 = (6)*(0.3025)*(0.2025) = 0.3675375
P(4,0.55,3) = (4 C 3) * (0.55)^3 * (0.45)^1 = (4)*(0.166375)*(0.45) = 0.299475
P(4,0.55,4) = (4 C 4) * (0.55)^4 * (0.45)^0 = (1)*(0.09150625)*(1) = 0.09150625

0.200475 + 0.3675375 + 0.299475 + 0.09150625 = 0.95899375

= 95.9%

Which is the same as the answer in the video I put up.




I am convinced that this is the correct answer, but I suppose I may have to re-explain it again if it's still not making sense.

It's not a binomial distribution as the events are not independent http://en.wikipedia.org/wiki/Binomial_distribution . There are 4 events (the picking of each person). Event 1 is picking the first person, event 2 is picking the 2nd person, etc. The probability of event 2 depends on what you picked in event 1, thus not independent and not a binomial distribution.

What I posted is correct.

What I did is usually taught in the first week or two of a probability class and is very basic.

P(at least 1 rep) is what we're looking for.

The basic principal to be used is P(A) = 1 - P(A'), where A' is the complement of A (opposite for those who don't know the term complement).

Let A = at least 1 rep. Therefore, A' = 0 rep = all dem.

P(A') = P(0 rep) = P(4 dem). P(4 dem) = P(1st one is a dem)*P(2nd one is a dem)*P(3rd = dem)*P(4th = dem) = (8/20)(7/19)(6/17)(5/16).

So P(A) = 1 - (8/20)(7/19)(6/17)(5/16) = what i posted earlier.

[TheLetterZ]

Re: Anyone good at probability?

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[Blzer]

Re: Anyone good at probability?

It's not a binomial distribution as the events are not independent http://en.wikipedia.org/wiki/Binomial_distribution . There are 4 events (the picking of each person). Event 1 is picking the first person, event 2 is picking the 2nd person, etc. The probability of event 2 depends on what you picked in event 1, thus not independent and not a binomial distribution.

What I posted is correct.

You're probably right, what I keep forgetting is that these events aren't independent. I don't know why that slipped by me again, but I just felt like perhaps I stumbled across the correct answer somehow because it seemed right. Oh well, mine would be right if it was coin flips lol.

Then it makes me wonder how you would answer this question if they asked for "at least two" instead of "at least one", because now getting the complement wouldn't be quite as easy as it would be in this case.

[Blaxican8504]

Re: Anyone good at probability?

This thread makes my head hurt. I will take my B- in Stats that I had no business getting and call it a day. Thank you very much.

[deaduck]

Re: Anyone good at probability?

There is absolutely zero ambiguity in this question as it's crystal clear. People's inability in this thread to understand the question doesn't mean it isn't clear.

The question isn't crystal clear...it's dependent on the unexplained manner in which the 4 are picked from the group of Dem/Rep. Replacement issues aside...if you know the result of each draw as a single action...the remaining probability changes.

Mathematics is based on things that are proven to be true, it doesn't get any more solid than that.

I use to think that until I watched a top notch mathematician destroy the assumptions of math. She shreaded all the axioms of Geometry/Trig...evicerated Calculus and about gave me a concussion verbally by ruining the concept of zero for me forever. Plus she was kind of cute.

Also, your blue ball red ball problem is not related in any way to this dem/repub question. It may sound similar but it's not.

I wasn't relating it too the problem as much as the nature of assumed probability built into the question

[Kevin26385]

Re: Anyone good at probability?

The best way to think about this is to realize that the only way to NOT have 1 republican is to have all 4 democrats. P(all Dem) = (9/20)*(8/19)*(7/18)*(6/17) = .026006.

So P(not all Dem) = P(at least 1 repub) = 1 - .026006 = .973994 or 97.3994%

A tree diagram will also get you this result, although it's a bit tedious.

This is correct

[Blzer]

Re: Anyone good at probability?

This is correct

Cool, glad it's been resolved then.

Don't hesitate to come back to this thread for more questions.

[Kevin26385]

Re: Anyone good at probability?

Cool, glad it's been resolved then.

Don't hesitate to come back to this thread for more questions.

haha, I started laughing when I came back the next day and saw 4 pages on posts...I would surprised to say the least.

Thanks for taking the time to try and figure these out guys, much appreciated