Probability & Statistics

About group

Probability is one of the most interesting topics in math as it is closely related to our life and decisions we make on a daily basis. It is also a lot of fun as a number of games are based on probability & statistics theory. We will post some of the most interesting problems as well as answer the questions you may have on the subject.

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Anyone planning to major in Statistics here? I plan to major or minor in it but not sure what else to take, just doesn't seem like a great major
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2 months ago
I did MA in Stats - it is a great major irrespective of what you want to do later in life as it is relevant for many disciplines including finance, politics, biology. As an example, you can take it together with Economics for a double major
2 months ago
I might do a minor
4 weeks ago
Might do a minor
Has anyone taken AP Stats here?
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4 months ago
No but planning to next year
3 months ago
I will do next year
Anybody interested in Game Theory and Financial Engineering in this group?
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9 months ago
Am interested in game theory and was actually looking for a class i can take (nothing around me)...dont know much about financial engineering.
9 months ago
@James feel free to post some game theory problems here. Re financial engineering, I did take a couple of fin engineering/math finance classes but ultimately believe that market behavior cannot be explained using just stochastic models...
8 months ago
John coursera has game theory course. apparently very good
This is an old problem and we should have probably started our group with it but here we go...

Monty Hall Problem
You are on a TV game show and given the choice of three doors:

Behind one door is a car which you want to win and behind the others, goats.

You pick a door, say Door 1, but the host (who knows what's behind the doors) instead opens another a door which has a goat.

He then asks you, "You have one more option: do you want to pick another door or stay with Door 1 that you originally picked?" Is it to your advantage to switch your choice?
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3 months ago
It is in your advantage to pick another door. When the game started you had a 33.3% chance to pick correctly. If you pick door 1 then the host shows a door which contains a goat that means the other door has a 66.6% chance of containing the car. Therefore it is in you best interest to switch doors.
Not sure if this the right place to ask this but what are my career options with Statistics degree? I like Stats and thinking to major in it in college. Any alternative majors that are similar?
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11 months ago
Lots of options like actuary, economist, statistician (across various industries) but also many finance related jobs require good understanding of statistics. If economics or finance is of interest to you going for econometrics major is an option too. And of course you can choose to do applied math.
If you are planning to attend university in the US you should probably know the answer!

Assuming you are one of approximately 30,000 of equally qualified students applying to top 10 US universities with an average acceptance rate of around 7%, what is your chance of being admitted to at least one of these schools?
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Now that we have solved birthday paradox problem an easy one for you. For those of you who understand the game of tennis it will be more fun...this problem was posted by Peter Winkler (math professor at Dartmouth University) in his math puzzles collection

Winning Wimbledon

As a result of temporary magical powers, you have made it to the Wimbledon finals and are playing Roger Federer for all the marbles. However, your powers cannot last the whole match. What score do you want it to be when they disappear, to maximize your chances of hanging on for a win?
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1 year ago
Yes the correct answer is to be two sets up and tiebreak at 6-6 in the third one with 6-0 score then you have more chances (6) to win the match than if you are just up 40-0 in the final game
1 year ago
Wouldn’t it be better to tiebreak the 1st set then get as close to winning the 2nd set as possible so that even if you lose the 2nd set you still have a chance to win the 3rd?
1 year ago
No because then you would have to win many more points if you are playing for the whole set and if you are in the last set (by the way Wimbledon is 3 out 5 sets but it doesn't matter) and up 6-0 in the tiebreak you just need to win 1 out of 6 points to win the match. The probability of winning 1 out of 6 is higher (even assuming you always only have 0.1% chance of winning any single point from Federer).
Did you know that in the room of 20 to 30 people you are very likely to find two people who share the same birthday? This is called birthday paradox. Try to calculate the probability of two students having the same birthday in the class of 30. This is counter-intuitive and you will be surprized by the result.
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1 year ago
To follow...anyone has an idea how to solve it?
1 year ago
I've always found it easiest to calculate the complement of this problem rather then solve it directly, it makes it easier to understand to me. So in a room of 30 people find the probability that none of them have the same birthday.

First we need to find the entire sample space Ω or all the choices in which no condition occurs in the event. If we have 30 people, they can choose any birthday in ( 365 ) ^ 30 ways.

Now we can choose the number of people without the same birthday in 365 * 364 * 363 * ..... * 336 ways. ( First person can choose from 365 days and each subsequent has 1 less choice in order for none to have the same birthday )

Hence the desired probability is
( 365 * 364 * 363 * ..... * 336 ) / ( 365 ^ 30 ) = 29.4%

Thus the Probability 30 people have the same birthday in one room is

1 - .294 = 70.6%

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Jay your right when you start counting each possible pair of people, since each person compares their birthday to each other person. However in doing so each of these pairs has a probability of 1 / 365 of being a match meaning it must be taken to the power of ( 29 + 28 + ... ).

The probability 2 people have 2 different birthdays would then be

1 - ( 1 / 365 ) = 364 / 365

If we take this probability to the nth power for its n amount of pairs
( 420 in this case) we get :

( 364 / 365 ) ^ 420 = 31.6 %

// With the 2% error margin from my calculator not being able to accurately compute such large #s

This probability being the probability that 2 people had different birthdays we subtract it from one to get its complement.

1 - .316 = 68.4 %
1 year ago
Harris - excellent! And even better explanation!