Here's a logic puzzle for any aspiring card counters out there from one of the top counters in Australia.

A high stakes blackjack player is playing at a $500 minimum table, just one box, on his own.

His only information is from a simple plus-minus count.

During one hand he places an insurance bet on a negative count. He is not playing for disguise.

PUZZLE:

What circumstances could justify this play?

Insurance is ONLY correct when the probability of a dealer 10,J,Q,K is >= 1/3.

Consider a single deck card.

Without any additional circumstances that probability is 16/51 - which is too low.

But if the player had 2 Aces - then split to receive a 9 apiece:

The probability of a dealer 10,J,Q,K = 16/47 - higher than 1/3.

And the count is -3.

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I haven't played this millenium so my memory of the rules and strategy is rusty.

But assuming the player can insure at any time while completing his hand there would be a number of similar situations MANDATING insurance.

give the man the prize,,,it sounds right ,,,,,cheers..gaz..

The person who set this puzzle didn't think this was the correct solution, but if JFC wants to argue the point I'll pass on his details. Here is the intended solution.

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Nearly at the end of the shoe the player might have, say, a 3% disadvantage. He has 16 vs Ace. He places a $1 insurance bet.

His chance of busting on 16 is over 50%; so is the chance of 'eating' an extra card (to be drawn by the dealer because of the insurance bet.)

Since a 3% disadvantage would mean a negative expectation of $15 per $500 bet (table minimum) each card 'eaten' would save over $2 in negative expectation, hence the $1 insurance bet would be justified.

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I'm posting a new and more difficult blackjack problem now, from the same source.

Why can't you publicly give the reason why I've been jdged incorrect, presumably because (some) rules state "insurance bets must be placed prior to any players receiving a third card".


http://www.starcity.com.au/dir013/i...ackPerfectPairs

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In that case, consider a modest variation to my original.

In a single deck, with a +2 count, the probability of a 10-point card would be presumed to be 16/50.

If then the 3 Aces are dealt, that probability drops to 16/47 i.e. > 1/3.

Hence insurance should be taken.

I'm not sure if this was the reason, but it sounds right. I'll pass on the modified answer and see what he says. It's a bit tricky for me playing Chinese Whispers like this when I don't fully understand it all myself!

I spoke to the person who set this problem and he pointed out that there are no single deck games in Australia, which is why Australia was specified as the location. However, perhaps it would have been helpful to mention this, as not everyone knows that.

He also reiterated that a side count of aces is not allowed; the person was using only the simple plus-minus count specified.

A top counter in Australia composed a puzzle about a blackjack game in an UNSPECIFIED location.

Australia was NOT specified as the location.

A side count of Aces was not necessary. The 3 Aces in question are glaring at the player.

The player's stunt of making a $1 bet at $500 table should get him barred for life. But having lost his livelihood, at least he could then boast about his triumph to his companions at Belmore Park. (If in Sydney).

Unless you are Bob Hawke.

Okay, fair enough :-) We stuffed up the description.