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
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.