From Drew Mathieson comes an exploration of basketball’s historied hot hand:

This season, on the way to winning her fourth WNBA championship in her 17-year career, Sue Bird made approximately 50 percent of her field goal attempts. Suppose she and Seattle Storm teammate Breanna Stewart are interested in testing whether Bird has a “hot hand” — that is, if her chances of making a basket depend on whether or not her previous shot went in. Bird happens to know that her chances of making any given shot is always 50 percent, independent of her shooting history, but she agrees to perform an experiment.

In each trial of the experiment, Bird will take three shots, while Stewart will record which shots Bird made or missed. Stewart will then look at all the trials that had at least one shot that was preceded by a made shot. She will randomly pick one of these trials and then randomly pick a shot that was preceded by a made shot. (If there was only one such shot to pick from, she will choose that shot.)

What is the probability that Bird made the shot that Stewart picked?

Inspired, Vikrant Kulkarni has an electoral enigma for you:

On Nov. 3, the residents of Riddler City will elect a mayor from among three candidates. The winner will be the candidate who receives an outright majority (i.e., more than 50 percent of the vote). But if no one achieves this outright majority, there will be a runoff election among the top two candidates.

If the voting shares of each candidate are uniformly distributed between 0 percent and 100 percent (subject to the constraint that they add up to 100 percent, of course), then what is the probability of a runoff?

Extra credit: Suppose there are N candidates instead of three. What is the probability of a runoff?

 

Can You Parallel Park Your Car? Riddler Express from 538

Every weekend, I drive into town for contactless curbside pickup at a local restaurant. Across the street from the restaurant are six parking spots, lined up in a row. While I can parallel park, it’s definitely not my preference. No parallel parking is required when the rearmost of the six spots is available or when there are two consecutive open spots. If there is a random arrangement of cars currently occupying four of the six spots, what’s the probability that I will have to parallel park?

Riddler Express from 538 FiveThirtyEight

From Zack Beamer comes a baffling brain teaser of basketball, just in time for the NBA playoffs:

Once a week, folks from Blacksburg, Greensboro, and Silver Spring get together for a game of pickup basketball. Every week, anywhere from one to five individuals will show up from each town, with each outcome equally likely.

Using all the players that show up, they want to create exactly two teams of equal size. Being a prideful bunch, everyone wears a jersey that matches the color mentioned in the name of their city. However, since it might create confusion to have one jersey playing for both sides, they agree that the residents of two towns will combine forces to play against the third town’s residents.

What is the probability that, on any given week, it’s possible to form two equal teams with everyone playing, where two towns are pitted against the third?

Extra credit: Suppose that, instead of anywhere from one to five individuals per town, anywhere from one to N individuals show up per town. Now what’s the probability that there will be two equal teams?

Solution from 538

Credits

Puzzle:
Three envelopes are presented in front of you by an interviewer. One contains a job offer, the other two contain rejection letters. You pick one of the envelopes. The interviewer then shows you the contents of one of the other envelopes, which is a rejection letter. The interviewer now gives you the opportunity to switch envelope choices. Should you switch?

Common solution:
The answer is yes. Say your original pick was envelope A. Originally, you had a 1/3 chance that envelope A contained the offer letter. There was a 2/3 chance that the offer letter was either in envelope B or C. If you stick with envelope A, you still have the same 1/3 chance. Now, the interviewer eliminated one of the envelopes (say, envelope B), which contained a rejection letter. So, by switching to envelope C, you now have a 2/3 chance of getting the offer and you’ve doubled your chances.

Discussion:
One critical piece of information is missing, and the answer actually depends on it. Does the interviewer know the content of her envelopes, or opens one at random?

• If the the interviewer knows the content of her envelopes, then the common solution is correct. This is related to the so called Standard assumptions (wikipedia article on the Monty Hall problem)
• If the the interviewer does not know the content of her envelopes, then the common solution is not correct. In this case we are dealing with conditional probabilities, given that the interviewer at random eliminated one of the rejection envelopes. This gives you a 50% conditional probability of success, and switching envelopes will not increase your chances.To make it more intuitive, consider the “law of large numbers” arguments. Suppose you play this random game 9 million times. Approximately 3 million times the interviewer at random will eliminate the offer letter. Out of the remaining approximately 6 million times, you will choose an offer letter about 3 million times. Your chance of success will not depend on switching envelopes (which is a great example on conditional probabilities).

This discussion is related to the Monty Hall problem: The probability puzzle that makes your head melt (by the BBC)

 

Links and References:

http://betterexplained.com/articles/u… (Regarding the birthday paradox)

http://www.bbc.com/news/magazine-2783… (more on the birthday paradox)

https://en.wikipedia.org/wiki/Birthda… (More on the birthday paradox)

http://dictionary.cambridge.org/dicti… (Paradox definition)

https://en.wikipedia.org/wiki/Pigeonh… (Pigeonhole principle)

Credits

You have 100 balls (50 black balls and 50 white balls) and 2 buckets. How do you divide the balls into the two buckets so as to maximize the probability of selecting a black ball, if 1 ball is chosen from 1 of the buckets at random?

$$\begin{pmatrix}n\\k\end{pmatrix}=\frac{n!}{k!(n-k)!}=\frac{n\cdot(n-1)\cdot\ldots\cdot(n-k+1)}{k!}$$

\( E=mc^2 \)

\( E=mc^2 \)

$$ i\hbar\frac{\partial}{\partial t}\left|\Psi(t)\right>=H\left|\Psi(t)\right> $$

\[ i\hbar\frac{\partial}{\partial t}\left|\Psi(t)\right>=H\left|\Psi(t)\right> \]

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Using LaTex Formats in Aurora