I wonder how many people spotted the faulty maths in yesterday's suit combination post?
Although the result was correct, the likelihood of success was miscalculated. With such a simple suit combination it is fairly easy to compute the odds as we are just comparing the 4-2 breaks of the suit - so we just count all the combinations as each specific combination has the same probability of occuring.
How many ways can we arrange the missing clubs (QJ8754) so that West has four and East has two, with no honour appearing when we play the king and lead towards the ace. One option is QJxx and there are six combinations (QJ87, QJ85, QJ84, QJ75, QJ74, QJ54); the other option is Hxxx and the easiest way to count these is to consider the two-card holding, Hx - there are clearly eight combinations, the queen with the four small cards and the jack with the four small cards.
So this means that the odds on finding Hxxx/Hx split is 8:6, or 4:3. Slightly worse that the 2:1 quoted yesterday, but still good enough to make playing the ace the right play when you need four tricks.
James has already covered this well in the comments to yesterday's post but many might not read the comments. He was not the only one to notice.