Cut-Rate Insurance

A reader says: A card-counting friend of mine playing at a six-deck \$3 table at Atlantic City takes insurance every time the dealer shows an ace as it only costs \$1. I wait for a count per deck of +3 before insuring, and believe I am correct but cannot justify my belief mathematically. Could you help?

The answer depends on whether \$1 of insurance protects the whole \$3 bet or just \$2 of it. When your friend has a bet of \$3 and a winning insurance bet of \$1, does the dealer take a dollar or does your friend keep his \$3 bet as well as his \$1 insurance?

If \$1 of insurance protects only \$2 of the bet, then you are correct in waiting for a count per deck of +3 before buying insurance. The mathematics, to a non counter, shows an average of 96 wins of \$2 each for every 215 losses of \$1 each, for a net loss of \$23 in 311 insurance bets.

However, if the \$1 of insurance saves all \$3 of the bet, then the correct decision is to insure virtually all of your \$3 bets. Those 96 wins become \$3 each, yielding a profit of \$73 per 311 insurance bets. Insurance would be worthwhile at a count per deck as low as -7.

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The Inside Straight

It might even be that Joe would have remained as a long-term losing player if no one had allowed him to get heavily enough in debt to make it worth his emotional while to skip out on the debt burden. The Joes of the world are often willing to lose \$50 a week forever (as long as they win occasionally), but when faced with a single large payoff, the temptation to avoid responsibility becomes too much to handle.