Friday, April 20, 2007

Social Workers Attempting Math

One of the social workers in my department sent out one of those "amazing" math puzzles, that tells you to multiply by 5, add 20, etc, and the result is a number that reveals your age and some other original number you started with. They always say breathlessly "This is the only year it will work!"

She sent it to the whole department - 30 people - asking if anyone could explain how this amazing math stunt worked. I have never found these particularly interesting. The principle is obvious: you you add and multiply stuff up, and you subtract and divide the same numbers back down, just disguised. To take an over-simple example, you add 9 and 2 at the beginning, then subtract 6 and 5 later on. 11 up, 11 down.

I don't think I had ever bothered to take one apart completely, though I may have years ago. But I thought it my duty to fight innumeracy in the social sciences, so I made an effort to explain the solution in as simple a way as possible. I pointed to the unusual number 1757 that was stuck into this whole mess and noted it was a nice round 250 from the current year. I reminded folks that the year of their birth, which was asked for in the problem, plus their age always equals the current year. It's a trick, folks. They got you to reveal you age via your year of birth. Duh.

It "only works this year" because last year 1756 worked, and next year 1758 will work. Shazaam.

I was mildly distressed at the people who were so grateful and impressed that I had explained it to them. Sheesh. Then later I heard about the people who still couldn't follow it. CP Snow's Three Cultures, extended out 40 years later. Ouch.

8 comments:

Doug W. said...

It drives me crazy. It has become socially acceptable to be bad at math. How many times have you heard someone say, "I hate math," or "I stink at math." I have even heard of business majors switching to liberal arts majors so they did not have to take a statistics class.
Well, it isn't ok not to know how to read or write is it?
I just don't get it.

Woody said...

I have a statistics problem.

On the TV game show "Deal or No Deal," the board showed something like these dollar amounts left, from which to select and eliminate the lower dollar amounts to win the largest amount:

$1
$10
$50
$100
$1,000,00

The host said that the contestant had a four out of five chance of selecting something other than $1,000,000. However, the contestant was selecting from only four closed briefcases, because he had taken one out of play at the beginning of the game--the closed one sitting next to him.

Were his chances really one out of five of selecting something other than $1,000,000 (the numbers remaining) or one out of four (the number of choice that he had) or something in between--and, how would that be computed?

I have always tried to figure the odds and realize that the chances of getting the "big dollars" are pretty small, because they show only two sides of the board, and the "big dollar side" is also loaded with small dollar amounts. If you add to it the statistical question above, that makes the contestant's chances even less. So, what are the real odds?

This is extremely important, because I have a one in 200,000,000 chance of going on the show.

Thanks.

Assistant Village Idiot said...

I haven't watched the show, so I don't get what is meant by the big dollar amounts on the other side and all that. Also, who took one briefcase out, and is it coming back into play at any time?

What I think your question is, however, gives an answer of 1 in 5 - 20% - for each of the remaining briefcases. You can either figure that by taking 100% distributed over the five original briefcases or 80% distributed over the four remaining briefcases (because when you took one out, there was then only an 80% chance that the four remaining have the big dollar amount). It works out the same either way.

You could also look at it as two separate probability events: 4 out of 5 that you chose the $1M in your first choosing, then 1 out of 4 to get it on your second choosing. 1/4 x 4/5 = 1/5, or 20%.

Woody said...

If there are thrity-two briefcases at the beginning, then there is only a 1:32 chance of selecting the $1,000,000 at the beginning, and those odds never change. See the Monty Hall Dielmma.

http://www.cut-the-knot.org/hall.shtml

http://mathforum.org/dr.math/faq/faq.monty.hall.html

Neither the 1:4 and 1:5 seem to be correct. Let's see, you had a 1:32 chance to pick the $1,000,000 at the beginning, so I'm going to say that your chance of picking it out of the remaining four is something like 1:8.

I can't believe that you've never watched the show.

bs king said...

I agree with Doug W. As an engineer, we were constantly being monitored for any slips in our writing/communicating skills. However, a journalism major can announce proudly that they were terrible at math. Gah.

Interestingly enough, when engineers at my school would go to our career counselor for advice about the GREs, they would tell us that any time we spent studying for the math section would be time wasted. The math section of the GREs apparently assumes one semester of college math. I don't know how I feel about that being the standard for the "educated" class.

bs king said...

Oh, and Woody....the AVI has no TV. Never has in the 20 or so years I've known him.

Assistant Village Idiot said...

Where did the 32 briefcases come from? There were four, then there was another one taken out of play. There are some missing steps here.

Woody said...

AVI has no TV?! Then you're missing FOX News, too! And Ernest Angely! And the Braves!

Here's the game's link. It may explain it better than I can. I assume that you have internet.

http://www.nbc.com/Deal_or_No_Deal/game/

They have a sample game, and I randomly picked the $0.01 brifcase.

This really isn't important, but I'm intrigued by how the "banker" figures the amounts that he will pay and what a person's odds are with each pick.