Red's Deal

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Math Exercise

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TV gives a heavy emphasis to mathematics in poker. Hold'em games entrance viewers with a chance to ooh and ahh as the two decimal odds flicker by on every card dealt.

Unfortunately this over-emphasizes the importance of math in the game. It counts, sure, but it's not nearly as big a deal as knowing what your opponents hold and what they play.

Of course this doesn't keep punks at the table from brow-beating you over the odds. What's funny, though, is I find this particular playing subset actually don't know math that well. They're memorizers, not theoreticians.

So if you're a brow-beater, just wanna be, are interested in seeing the kinds of math the more hardcore can do in their head at the table, or just want an interesting math exercise, this one's for you...

This is the scenario: Imagine sitting at a round table with six seats. There are three men and three women who are randomly going to be set around the table. The table being talked about here in my problem is not a poker table. It's a round table with six chairs (think cylinders in a 6-shooter if it makes it easier) -- meaning that if you numbered the places 1-6, seat 1 and seat 6 are adjacent. (In other words, if women were sitting in positions 5, 6, and 1, they would be sitting next to each other.)

This is the question: What are the odds that the women will be seated next to each other?

I'll post the answer in a couple of days -- once I'm out of the computing hell I currently find myself residing in.

I'm a bigtime cheapskate. I hate paying full price on anything that can be had for less -- especially in something as fast-and-loose as an airline ticket, where the guy sitting next to me could be paying half the price I am.

The process I outline here is what I use myself. It may not give you the absolute cheapest ticket, but it'll get you to within probably 5%, which is good enough for me. The steps here take about 20 minutes from start-to-final-purchase.