Dubious Logic

Multiple Choice

Long ago when I was school, multiple choice examinations were considered something of a soft alternative to the hard slog of essay writing. I've noticed that with tricky phrasings and "none of the above" type answers, the format had got steadily more difficult over the years, but that's probably to make up for the fact that no-one writes essays any more. Try your hand at this problem:

To show that we are in the modern idiom you open those West cards after South deals and passes. North makes an ostensibly weak jump overcall in spades and you declare 4♥ on the king of spades lead. After winning the ace do you:

1. Play a spade back and attempt to ruff two more in dummy?
2. Play two top trumps then concede a spade?
3. Play two top trumps then play clubs?
4. Do something else? (like take one top trump then (B) or (C) or play clubs immediately)?

Your preliminary analysis recognises that with a club and a spade to lose, you will have find trumps 3-2. Even then your trick count is short: with three trumps and a ruff, you only have a four certain tricks in the minors and the spade ace for nine. Drawing trumps, though necessary on some counts – to stop South over-ruffing spades for example – will leave you a trick short. But South can have one spade over-ruff if he has three trumps, only then do you can cash the ace and king - that is the plan behind (A).

By playing trumps first you hope that the player who wins the black card next will not have the third heart. True, if you play a spade then North may have the wit to lead a club himself to South's ace for that ruinous trump continuation. If you don't adopt the more immediate plans (A), (B) and (C) what then of (D)? It's tempting to take one round of trumps but there's no real gain: if you guess incorrectly when losing the lead South might still arrange two over-ruffs or North can put you to a guess about how to get back to hand by playing a trump himself. If you play clubs immediately, the defence holds up and you'll have to revert to one of (A) – (C).

Let's create a list of choices ourselves. Let '3♥' represent North, who has the spades, having three trumps and '♣A' denote the same player holding the club ace. Now:

1. 3♥ and ♣A – nothing works [though exactly ♥QJ10 is a win for (A)]
2. 3♥ – (C) is successful
3. ♣A– both plans (B) and (C) work
4. Neither '♣A' or '3♥' – (A) is successful

Now not all these outcomes are equally likely. With six spades, North has only seven slots to hold hearts whereas South has twelve – South is more likely to hold the three card suit, twice as likely in fact. Without knowing that precisely, it's sufficient to note that because of (iii) and (iv), plans (B) and (C) are as pretty much as good as (A). To decide between them all we have to do is note that with (C) we get (ii) for free.

So it's answer (C) that gets the mark. Another wrinkle they introduced was to 'award' a score, minus a quarter in this case, to deter guessing – I hope you didn't go negative.

Published Saturday 14.Jan.2006