High School Math Brain Teaser...

[Anonymous]

0.4?

[Anonymous]

Anonymous wrote:I'm OP, I'm a mom, my kid is at Deal and this a math problem the kids are doing in their little impromptu math club--it was presented to them as being high school level.

OK, OP, I'll try. First, notice that the probabilities all add up to 1.0 or 100% -- that makes things easier right off the bat. So for shuffling, the probability each and every day is 0.4, or 40% chance I'm shuffling. If the first day of the week is Monday, the probability I'm shuffling today is 0.4 or 40%. But what the problem is really looking for is not the probability on any one given day, but rather the probability of shuffling seven days in a row.

Mathematically, when you're stringing together independent probabilities like that, you learn to multiply each independent probability together to get the final solution. So here, where there's a 40% chance of shuffling on each day, and we're trying to string together seven days, the formula will look like this ... 40% x 40% x 40% x 40% x 40% x 40% x 40% = 0.0016384 = 0.16%

It's a little hard to explain in writing why you multiply all the probabilities together. Here is a diagram that shows why, based on a 50% coin-flip chance -- http://www.mathsteacher.com.au/year10/ch05_probability/06_further_representation/Image3890.gif . The odds of one head is 50% or one-half. Half the time I'll get heads. But the odds of two heads flips is 50% of 50% ... or half of the half time when I flipped heads on my first try. So mathematically expressed, that's 50% x 50%. When you do the same thing with shuffling over seven days, you just have to string the multiplication out longer.

Does that make sense? There's probably an easy and elegant way to explain it, but I can think of one right now.

[Anonymous]

It's easier to explain if the probability is 1/3 and we work with a two-day weekend instead of a seven-day week.

Think of each day as a separate pile of beans, three beans in each day, two piles of beans. In each pile there is one red bean, one blue bean, and one green been. I close my eyes and pick a bean. The probability that I pick a red bean is 1/3.

From the first pile I pick a bean. Here are the beans I may have picked:
red green blue

Now I pick a bean from the second pile. Here are all the possible combinations of beans that I may have picked:

red, red
red, green
red, blue
green, red
green, green
green, blue
blue, red
blue, green
blue, blue

nine possibilities, and only one of them has two red beans. So the probability of picking a red bean on both days is 1/9, or 1/3 times 1/3.

(This is an example, not a proof, but I think you can see that it would just go on and on....)

[Anonymous]

I'm OP, I'm a mom, my kid is at Deal and this a math problem the kids are doing in their little impromptu math club--it was presented to them as being high school level.

Maybe you should let them do their own work or activities or whathaveyou.

[Anonymous]

Didn't someone ask recently whether DCPS teaches probability? I guess they do -- in their own distorted way.