We first pose a quick question: If you drive fifty miles in fifty minutes, must there be some ten minute interval in which you drive exactly ten miles?

Of course there must — mustn’t there? Well prove it!

Our main feature this week is an interview with Michael Breen, from the American Mathematical Society, who came and hosted a game show “Who Wants To Be A Mathematician!” About a hundred high school students from all over northwest Arkansas came to cheer on their classmates; Kyle Strong of Har-Ber High in Springdale came in first, winning $1250, and Karan Batra, of Bentonville placed second with $250.

Our interview includes a few sample problems… I guess we shouldn’t list too many of them, in case Micheal wants to recycle them! Mike’s also responsible for the great series of Mathematical Moments posters— check them out!

PS: We opened with the Up To One Million Dollars In Prize Money May Be Given Away gag… Always fun!

Mark A. recently wrote us:

The contestant started with 25 cases each with a value of dollars inside. They select one case with the hopes that it has the “Grand Prize” (GP) amount inside. Cases are opened and other amounts eliminated. In the end only two cases remain. The contestant has the one they selected and the other is still in the “Game Field” across form them. One case has the 1,000,000 dollars GP and the other has 10 dollars. As I understand it there is a higher probability that the GP is in the case not selected by the contestant in the beginning so that contestant should always swap cases. I do not see how this can be so. I understand that in the beginning the case selected only had a 4% chance of having the grand prize (GP) but so did the other remaining case. As cases were eliminated the odds of these cases having the grand prize raised at the same rate to the point that now they should each have a 50% probability of having the GP. In addition in the beginning each case also had a 4% chance of having the 1 dollar and that chance also rose to 50% for each case. I do not see any preferred state where the decision of the contestant has an effect on the distribution of the odds.

I see this interpretation of Bayesian analysis often and I can not see the validity. Am I wrong?

This is a really fascinating question, and the answer depends on the critical phrase:

…Cases are opened and other amounts eliminated.…

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.

All the pirates are eminently logical; the last pirate knows that if he gets a chance to make a proposal, he will get all of the loot:

If we’re playing where ties are enough, the second to last pirate knows he can get all the loot if he gets a chance to make a proposal [There are many famous woman pirates, but to keep things simple, we will stick to masculine pronouns].

What should the third to last pirate propose, should he get a chance? The last pirate expects nothing if the second pirate gets a chance to propose, and so will gladly support the third pirate’s proposal for a single gold piece:

How does this continue?

etc. It is clear that the 100th to last pirate (i.e. the first pirate in line) should dole out a single gold piece to the 98th, 96th, etc pirates. The game is odds vs. evens!

Failed proposals = death, ties are not enough

etc. Again the pattern is clear; the 100th to last pirate should give 1 gold piece to the 98th to last, two to the 97th to last, and 1 to the 95th, 93rd ,etc.

Failed proposals = expulsion, ties are not enough

First, here are some of the details of the calculation for mismatched pennies.

The expected payouts (from Kyle’s point of view) are
HH -> -3
HT or TH -> 2
TT -> -1

1) Is it really true, if Kyle steadily plays 3/8ths Heads and 5/8ths Tails, that Chaim can’t adjust his own strategy to take advantage of Kyle? Suppose Chaim plays some fraction p Heads and then (1-p) Tails. Then we expect to see

HH exactly 3/8 * p of the time, for an expected payout of 3/8 * p * (-3)
TH exactly 5/8 * p of the time, for exp. payout of 5/8 *p *2
HT exactly 3/8 * (1-p) of the time, for exp. payout of 3/8 * (1-p) * 2
TT exactly 5/8 * (1-p) of the time for an expected payout of 5/8 * (1-p)*(-1)

The total expected payout is thus

3/8 * p * (-3) + 5/8 *p *2+3/8 * (1-p) * 2+5/8 * (1-p)*(-1) =

-9/8 p + 10/8 p – 6/8 p + 5/8 p – 5/8 + 6/8 = 1/8

No matter what p is chosen, Kyle will come out 1/8th of a penny ahead, per game, on average, in the long run.

Similarly, if Chaim chooses to play 3/8ths H and 5/8ths T, Kyle cannot exploit this and do any better.

As discussed in the podcast, there is a quick trick for calculating this, at least for simple games like this one.

2) What’s really going on? Why is von Neumann’s theorem true, for more complicated games?

Really, one can think of the expected payout as a function of the various probabilities assigned to each outcome. If you think about it, this function is quadratic– each term will have degree at most two– and the von Neumann minimax is a search for a saddle point. The existence of such saddle points– unless the game is always tilted to one inevitable outcome– is straightforward enough, but is beyond what we can discuss here.

This week, to celebrate the recent “rock scissor paper” World Championship, Kyle and Chaim play a game of “mismatched pennies”. Each puts down a penny on the table, choosing to lay the penny down showing heads or tails.

If the pennies both show heads, Chaim wins 3¢; if both show tails, Chaim wins 1¢, and if one is heads and the other tails, Kyle wins 2¢.

Thinking this through, second and triple guessing one’s opponent, is ultimately fruitless, as wonderfully illustrated in this scene from The Princess Bride
(Incidentally, game theory shows up in many movies, television shows and works of fiction!)

John von Neumann tells us, in his celebrated minimax theorem, that there is an optimal strategy for both players; each assigns a percentage to each of his options; the choice of which option to use is made randomly, by these percentages. Von Neumann tells us that there is no way to take advantage of knowing what the opponent’s optimal strategy is– that’s what makes it optimal!

But the game still might favor one player or the other, even if both are using their optimal strategy. This week’s puzzle then, is to answer: does this game of mismatched pennies favor Chaim or Kyle?

(So, if everyone could just hold it together and send in the number “1″, we’ll be out a lot of dough! Of course, if everyone else sends in “1″, you might consider being a rascal and sending in “2″… this is a slippery slope though…)