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Gaming Guru
Hedging Can Be a Costly Way to Protect Your Bets10 October 1994
Someone throws two dice and everyone bets on how they'll land. Sounds simple. And boring. So why is craps a gambling classic? Much of the mystique emanates from the provisions to bet on multiple outcomes simultaneously, and in the varying amounts the same size bets pay on results with differing probabilities of occurrence. Usually, players bet on several complementary outcomes for increased action. For instance, instead of wagering $40 that a shooter will throw nine before "missing out" with a seven, a player might place $10 each on 10, nine, five, and four. All's lost on seven either way. Profits when individual numbers hit will be less than that for the whole $40 on nine, but there'll be more frequent wins. Players also use complementary bets to introduce high-payoff longshots into otherwise conservative strategies involving bets paying close to even money. Say a shooter is trying for a point of eight. A player might augment a $10 line bet on this number with $1 hard eight. This offers a 9-to-1 return for a four-four in addition to the even money on the "flat" portion of the line bet and 6-to-5 payoff on the odds for any eight. Within the margin arising from different house "edge" on various bets, complementary wagers don't affect overall long-term chances of winning or losing. And, getting paid every time the dice land, from money spread all over the layout, sure pumps the adrenalin. Some players bet on opposing rather than complementary outcomes. They think that by cleverly "hedging" in this manner, they get their bets to "protect" one another. The most common hedge is $1 or $2 on any craps to protect $10 or $15 on the pass line during a "come-out" roll. Any craps pays 7-to-1 on the very numbers that make the pass line an instant loser. It may seem like cheap insurance, but protection afforded by this and other hedges turns out to be expensive. Here's what I mean. Say you play through 3960 come-out rolls over a long period. Betting $10 on the pass line and taking no odds, the statistically correct distribution of results predicts you'd lose $560. This means you'd have to win 28 more and lose 28 fewer $10 bets than expected from the laws of probability just to break even. The following table shows how to find this result:
For the same play, but hedging with $1 any craps on the come-out, the statistically correct distribution of results predicts a $1000 loss. The additional $440 comes from collecting $3080 on 440 any craps winners and sacrificing $3520 on 3520 any craps losers. This means you'd have to win 50 more $10 bets than predicted by the laws of probability just to break even. Some players carry notions of opposing bets well past hedging as a form of protection. They try to wed wagers into sure-fire fool-proof "systems," guaranteed to win if only they could tweak the proportions a bit better. Before you try any of these screwball ideas yourself, bear in mind the musings of Sumner A Ingmark, crooner of the craps pit: A player who alleges Recent Articles
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