Also: "If you’d like to teach me to fish, how could I figure it out for any example of odds? (Assuming the math is a few steps below your level

. ) Thanks"

Well, it's not exactly 50% just because you have five of the 10 starters, because you may have the best five, the worst five, or the middle five. You can't just say you have half the players, so it's 50%. With the stated odds, you have to calculate that none of the five will win, and then the probability that one of them will win is 100% minus that calculated value. Actually, you have to invoke that concept twice.

So, first you change each of the odds to win to fractions: 2/11, 2/13, 1/7, 1/7, and 2/15. Those are the chances of each winning. So the chances that each of them loses are: 9/11, 11/13, 6/7, 6/7, and 13/15. The chance of all five losing is the product of those fractions: 9/11 x 11/13 x 6/7 x 6/7 x 13/15 = 0.441. So, the probability that one of them wins is 100% - 44.1% = 55.9%.

If you bet $25 five times (total of $125), your e.v. is 0.559 x $125 = $69.88. In addition, you get four free $25 bets, or $100 in free bets. It matters, of course, how you bet that, but if you were to make four straight point line bets at -110 each, that e.v. would be an additional 100/210 x $100 = $47.62. Together, your total e.v. is $117.50. Sadly, you're $7.50 short of your $125 outlay, so you lose 6%.

Did I teach you how to fish?

Don

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