## EZ. Google, Flutes and Monopoly

We ask: What do Google, flutes and monopoly have in common? In fact, important principles behind this question apply to an astounding array of phenomena!

(Since we’ve been offline for a week or so, due to a tremendous ice storm that has paralyzed the town, we add a special bonus: the very first Math Factor episode ever aired, from January 25, 2004.)

1. ### jecampbell319 said,

February 3, 2009 at 2:16 am

Just like to say my name is Jonathan, and I am a big fan of what you guys are doing with Math Factor. I am new to listening to Math Factor, though I have been listening to the old podcasts, and trying to figure them out. Here is my idea for the King’s Decree:  [spoiler]
– I know you said that it could be solved using common sense, but I prefered to take an approach using algebra since you did not specify the population or number of couples.
1. His decree would not work because there would be slightly more boys than girls.
2. There would be an average of 2 children per couple.  [/spoiler]

Here is how I did it:
[spoiler] I realised that since we are using an unknown population, though number of couples is more accurate for the method I believe I solved it. I figured that with a given population, say 4096 couples you could figure out the number of boys to girls that would be born. You said there was a 50-50 shot of whether there was a boy or girl. So if 4096 couples were to have children, then 2048 would have a boy and stop and the other 2048 would have a girl, and keep trying so I came up with a list (that unfortunally due to formatting I can not post) granted the total number of couples adds up to 4095 out of 4096, in the theory of X number of couples, I theorise all the numbers would add up:   [/spoiler]
*A chart would be here if formating allowed*  Spoiler Continued [spoiler]
Total Number of Children = 8178
Boys = 4095
Girls = 4083
Average Number of Children per Couple = 1.997–> 2 children per couple [/spoiler]

My Question: [spoiler]  So I do have a question for you guys if you could help me out with it,   I decide to use an exponetial of 2 since there was a 1/2 chance (since I was going opposite was 1/2->2/1=2)

So I figure I could figure out how many boys and girls given any population using this system of equations:
# of Couples             n  = 2^X
# of Boys                    b  = n  or   b = 2^x
# of Girls                    g = b – x or g = (2^x) – x

X being exponetial growth from 2 to the # of couples

I was wondering if this system would work; if given any number of couples would it produce the correct number of boys and girls?  And if so then the boys to girls ratio would be:
n : n – x  [/spoiler]

Anyways that is how I figured it out, and that is what I believe my answer to be, and thank you for taking a look. Let me know,
Jonathan

2. ### avgbody said,

February 5, 2009 at 1:59 pm

Just using common sense
[spoiler]
it ends up being a 1:1 ratio
50% boys | 50% girls (50% boys | 50 % girls(etc…))
The count will always be equal.
[/spoiler]

3. ### Bob Nachbar said,

September 10, 2009 at 10:40 am

Regarding resonance, after listening carefully to the podcast a second time, I was struck by the similarity of the description of the Stop & Go game to Boolean network models (also called graphical or algebraic models). Also, as the 2 cups of water problem was being described, I immediately started thinking about casting the problem as a dynamical system that can be solved with a differential equation (DE) model. Here, the resonances are just the steady states of the system of DEs. I use DE models in much of my work modeling biological systems and diseases. Just yesterday I read 2 very nice papers in Science (2009, vol. 325, pp. 541-542 and 542-543), and the latter described both kinds of models.

I find making these kinds of connections among seemingly disparate topics or concepts fascinating. I wonder what kind of model could be used do this in an automated way?

4. ### strauss said,

September 10, 2009 at 10:52 am

That’s all correct: there is an underlying unity and each of these is a different take on the same kind of phenomena; this is one of the amazing and beautiful — and powerful — aspects of mathematical thought! I hope there’s no way to automate this: we’d be out of jobs!

5. ### Shawn said,

March 7, 2012 at 1:26 am

Did anyone figure out how much water will be in each cup at the end? [spoiler] It seems to me that if there were 8/3 ounces in Cup One and 16/3 ounces in Cup Two, then all 8/3 ounces will go from One to Two and 1/2 of the 16/3 ounces will go from Two to One. Since 8/3 = 1/2 * 16/3, then it seems that it would be stable. [/spoiler] And about Monopoly… I sort of figured that Illinois Avenue is landed on with such frequency because it is fourteen spaces after Jail. With all the ways players can wind up at Jail, it doesn’t seem too surprising that Jail would be landed on quite a bit. Also, fourteen is twice seven, which I guess is the most common sum for two rolled dice. No matter what number the first die is, there is always a chance that the sum will be seven.