The Math Factor Podcast

We present a recording of Raymond Smullyan’s lecture at the Gathering for Gardner, March 30, 2008; Newcomb’s paradox really is a stumper.

 

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We also asked, on this week’s segment how to label the faces of some ordinary dice, with twelve different numbers (we did say different didn’t we?) so that every roll produces a prime number. This puzzle is from the fascinating site www.primepuzzles.net. Don’t peek!

We can say a bit more about the Princess’s escape.

Amazingly, an optimal path for the Princess is to swim in a half circle of radius 1/8 that of the lake, then dash out to the edge.
We’ll give an analytic proof, but we could give a totally synthetic (geometric) proof as well.

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What numbers can 1,2,4,8,16,… etc “form”? Well, every number can be “formed” by summing various powers of 2. For example, 13 = 1 + 4 + 8.

In this way, we could say that a power of 2, say 64, is “full of divisors” since it has enough divisors to form any number up to 64. Its divisors are of course 1, 2, 4, 8, 16 and 32, and we can form any number from 1 to 63 by summing up these divisors as needed.

But what other numbers of “full of divisors”?

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A correspondent writes:

Greetings,

I think that in the long run both strategies are equivalent. This game doesn’t favor any player.

Demonstration

Chaim Expected Gains = 3 * 1/4 + 1 * 1/4 = 1

Kyle Expected Gains = 1/4 * 2 + 1/4 * 2 = 1

This is so if both of us pick H half of the time, and pick T half of the time.

But!

If I know Kyle is going to pick H half of the time and T half of the time, I should adjust my strategy. I can do better by always picking H; the payout would then be

C: 3*1/2 = 3/2
K: 2*1/2 = 1
Net 1/2 in my favor!!

Conversely, if I am picking H half of the time and T half of the time, Kyle should adjust his strategy and choose T all of the time; this comes out to

C: 1*1/2 = 1/2
K: 2*1/2 = 1
Net 1/2 in Kyle’s favor– rats!

John von Neumann’s celebrated result is that both players have an optimal strategy, one that cannot be exploited by the other player. If we both play optimally, is the game balanced?

Amusingly, this problem has exactly the same solution as the proof that there are as many rational numbers as there are counting numbers. And the proof generalizes: one stork can catch three frogs, or ten or fifty.

Here are some bonus problems:

1. The stork can catch the frog even if it can start at any rational number and hop any fixed rational distance each step.
2. However, if the frog can start at any real number or hop any real distance, the stork has no strategy that guarantees a catch. This is, in effect, the same as proving that the real numbers are not countable.
   

Graham’s number, as huge as it is, can be “described” or “named” in a very few symbols. Several people sent us programs that (in principle!) calculate Graham’s number— you can think of any of these programs as notation for Graham’s number.

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