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Posts Tagged ‘probability’

For a 3 sets tennis game, would you bet on it finishing in 2 sets or 3 sets?

Posted by guptaradhesh on September 19, 2012

This question was asked to a friend of mine as a part of interview process for DE Shaw & Co. I liked the question. Pretty basic probability question to test ones thinking process.

My two cents:

If the two players have equal probability of winning, the chance that the game will finish in 2 sets or 3 sets is same.

If one of the players is better then other, the chances of game finishing in 2 sets is more then chances of game finishing in 3 sets.

Say, p is probability of Player A winning a set ; q is the probability of Player B winning a set

and lets assume p > q

If game ends in 2 sets => Player A wins first two games or player B wins first two games;  P1 = p*p + q*q

If game ends in 3 sets => Player A and B win first or the second game; P2 = p*q + q*p

And we know that,   p^2 + q^2 >= 2 p*q  [with minima when p=q]

 

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Some very simple “Probability” problems

Posted by guptaradhesh on November 26, 2010

1. In a lotto game, one has to choose 6 numbers from a total of 40.  Julio has chosen 1, 2, 3, 4, 5, 6.  Alison has chosen 39, 1, 17, 33, 8, 27.  Who has a greater chance of winning?

A. Julio has a greater chance of winning.
B. Alison has a greater chance of winning.
C. Julio and Alison have the same chance to win.

2. If you roll two dice simultaneously, which of the following has a greater chance of happening?
A. Getting the pair 5-6.
B. Getting the pair 6-6.
C. Both have the same chance.

3. The likelihood of getting heads at least twice when tossing three coins is…

A. Smaller than
B. Equal to
C. Greater than

…the likelihood of getting heads at least 200 times out of 300 times.

Answers:
1. C. Julio and Alison have the same chance to win.  (B option looks tempting though)
2. A: Getting the pair 5-6
3. C: The likelihood of getting heads at least twice when tossing three coins is greater than the likelihood of getting heads at least 200 times out of 300 times

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Birthday Paradox

Posted by guptaradhesh on November 26, 2010

Something amazing paradoxical given by probability which is hard to accept in real life.

I had to sit and analyze the proof to accept this:

“In a room of just 23 people there’s a 50-50 chance of two people having the same birthday. In a room of 57 there’s a 99% chance of two people matching.”

Here is the explaination:  (reference: http://mathworld.wolfram.com/)

Consider the probability Q_1(n,d) that no two people out of a group of n will have matching birthdays out of d equally possible birthdays. Start with an arbitrary person’s birthday, then note that the probability that the second person’s birthday is different is (d-1)/d, that the third person’s birthday is different from the first two is [(d-1)/d][(d-2)/d], and so on, up through the nth person. Explicitly,

Q_1(n,d) = (d-1)/d(d-2)/d...(d-(n-1))/d
=  ((d-1)(d-2)...[d-(n-1)])/(d^(n-1)).
 Q_1(n,d)=(d!)/((d-n)!d^n),  

so the probability P_2(n,d) that two or more people out of a group of n do have the same birthday is therefore

P_2(n,d) = 1-Q_1(n,d)
= 1-(d!)/((d-n)!d^n).

If 365-day years have been assumed, the number of people needed for there to be at least a 50% chance that at least two share birthdays is the smallest n such that P_2(n,365)>=1/2. This is given by n=23, since

P_2(23,365) = (38093904702297390785243708291056390518886454060947061)/(75091883268515350125426207425223147563269805908203125)
 approx 0.507297.

Hence the astounding Birthday Paradox..

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Pigeonhole Principle

Posted by guptaradhesh on November 25, 2010

I wondered why I didn’t study this in my academics. Although it seems I have found this mentioned somwhere in the discussions. I liked the description given on wiki and some other urls.

I am imitating some part text for the places which I believe should be useful for a ‘peek look and move ahead’ crawler.

<<<

The pigeonhole principle states that if n items are put into m pegeonholes with n > m, then at least one pigeonhole must contain more than one item.

That is, if there is a mapping between two finite sets of unequal size, then at least one element in the smaller set must be paired with more than one element in the larger set.

The extended pigeonhole principle says that if p > nh for some integer n, then at least one hole will contain n + 1 pigeons.

(despite seeming intuitive it can be used to demonstrate possibly unexpected results)

Some common fun applications of Pegionhole Principle are as:

– At any given time in NY there lives at least two people with the same numbe of hairs

– Some prople reading this blog will have the same birthday (birthday paradox)

– If you pick five cards from a standard deck of 52 cards, then atleast two will be of the same suit

. . . . .

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