I had a dream last night involving — (?) well I am not really sure, except that it left me wondering if there is a simple proof (if indeed it is true) that there must be a common factor of

m choose i = m!/(i! (m-i)!)
m choose j = m!/(j! (m-j)!)

for all counting numbers i,j,m with 1 < i,j < m Another way to state this same thing is: any pair of entries, on any row of Pascal's triangle (except for the 1's on the edges) will have a common factor. With facts of this sort, often there is a clever way to cast things in terms of counting something a couple of different ways which makes things clear.

Peter Winkler tell us which full house to choose, and asks: How long must we wait until all the ants fall off the rod?

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Peter Winkler brings us a short poker puzzle, from his new collection Mathematical Mind Benders: What is the best full house?

(The answer is not three aces and two kings…)

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Robert Schneider, of The Apples In Stereo discusses his logarithmic tonal system and why he loves mathematics.

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Strangely, it depends on WHO is doing the eliminating and with what knowledge!

In other words, are they eliminated in such a way that the GP must be in play at the end, or in such a way the game might have been aborted prematurely?

A) If the contestant chooses, or the cases are chosen randomly (i.e. the GP was at risk at every stage), then the probability is the same for each case, at each stage, right to the end. It doesn’t matter either way if the contestant switches. This is the way Deal or No Deal is played.

B) If the game-show host, or some knowledgeable party removes cases from play (knowing they do not contain the GP), then it is better to switch. Incidentally, this version is known as “The Monty Hall Problem”, after the host of the 70’s game show Let’s Make a Deal (In which a contestant would be offered three doors, one of which conceals a fantastic prize; the contestant chooses one door, and then Monty Hall would eliminate one of the remaining doors that doesn’t have a prize; the contestant is then given a chance to switch to the last, unopened door— an opportunity which should always be taken!)

This seems paradoxical, doesn’t it? The knowledge and intention of the person removing cases from play seems to change the probabilities.

But this really does make sense.

—-

In (A) the probabilities remain equal, in effect, because no action has been taken that changes the relative likelihood of any outcome. Suppose we have, at a given stage, N equally likely possibilities, and one is removed at random, if the game does continue (which it might not) then there now (N-1) possibilities— all of which are still equally likely, etc.

In (B) the actions change the relative probabilities. This is a little harder to explain, but in a nutshell, the host sweetens the deal: your original choice is just as likely to hold the prize, but the other choices have become more likely to be winners, since a losing choice has been removed. Let’s count out the possibilities:

Suppose we have three briefcases are a, b, c, and the prize is in case a. We will list them in the order of

“case chosen by the contestant, case eliminated, case remaining”

The game would have ended if case a had been eliminated, so this leaves only

a b c (contestant should keep)
a c b (contestant should keep)
b c a (contestant should switch)
c b a (contestant should switch)

In (A) each of these is equally likely, since each of the choices was made completely at random. Any of the six sequences
a b c
a c b
b a (stop)
b c a
c a (stop)
c b a
was equally likely (Since there is 1/3rd chance the contestant will pick a,b or c; then there are two equally likely possible ways for one of the remaining case to be eliminated; the final case, if there is one, is determined)

Now 1/3rd of the time the game ends prematurely, but if the game finishes, there is 1/2 probability that the contestant should switch– it’s 50-50 either way.

In (B) though, the choices are not equally likely.
There is still a 1/3rd chance that the contestant chooses a, 1/3rd b, 1/3rd c.
If the contestant chooses a, it is equally likely that the host opens case b or c;
On the other hand, if the contestant chooses b, the host will certainly open c; if the contestant chooses c, the host will certainly open b.

so this gives

a b c 1/6th of the time
a c b 1/6th of the time
b c a 1/3rd
c b a 1/3rd

2/3rds of the time, the contestant is better off switching.

Prof. Bernie Madison discusses his innovative course at the University of Arkansas; thinking straight about mathematics is fundamental for an informed and capable citizenry– but why doesn’t mathematics education seem to foster this?

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The ‘expected’ answer is not always the one people choose: Dennis Shasha explains that psychology plays a role in the answer to last week’s puzzle.

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Dennis Shasha answers his cake conundrum and poses a new puzzle: should you switch envelopes given the chance?

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Dennis Shasha, author of Puzzles for Programmers and Pros
joins us once again, posing a cake conundrum!

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How does this simple trick work?

Ask a friend to pick, silently, a three-digit number, then “double” it to make a six-digit number. For example, if she picks 412, the new number would be 412412. Then dividing by 7, then by 11, then by 13, presto! The original number!

Interestingly, there is no decent trick for two-digit numbers; and for four-digit numbers the trick is not so great. But for nine and fifteen digits (for the right kind of people only!!) there is a relatively simple variation.

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