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[quote=Anonymous][quote=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.[/quote]

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 [u]today[/u] 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 [u]seven days in a row[/u].

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 [u]why[/u] 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 [u]two[/u]  heads flips is 50% [u]of[/u] 50% ... or half of the half time when I flipped heads on my first try.  So mathematically expressed, that's [u]50% x 50%[/u].  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.

[img]http://i.imgur.com/s0JoT.png[/img][/quote]

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