Principle of Restricted Choice

There follows a simplified explanation of the principle of restricted choice

Sometimes it is referred to (perhaps more accurately) as the principle of freedom of choice

It is not important, but for convenience say you are South, faced with this combination

(+ sufficient non-Spades to fill out the hands)

You have no clues from the bidding or play to date, and it is time to tackle the Spade suit. Cashing a top honour cannot lose, and might find a singleton Q, so you try that. It doesn't matter which, so say you cash the Ace. Unhelpfully, both opponents follow low. You continue with a low Spade from South and West again follows low. Crunch time. Finesse or drop, to find the only remaining missing card, the Queen?

Frankly there is not a lot in it. East is slightly favourite to hold it because he has one more card remaining in his hand. Had the hands been of infinite size it would have been a straight 50-50 guess (if we assume that no-one would voluntarily jettison the Queen prematurely, for as long as he can put declarer to the test). The actual odds make the drop about 52%, but who cares. This is not the thread to prove it. 50-50 is close enough for what follows, as in the next example that will be blown into touch (or is that kicked out of the water?) making the odd 2% discrepancy irrelevant. Hell, call it 50-50 if you wish.

Now let's change the hand slightly.

You cash the Ace, LHO (West) follows low, and RHO (East) drops an honour, the Q or J, it doesn't matter which. You continue with a low Spade, and West again follows low. Crunch time. Finesse or drop? The only remaining unseen card in the suit is the other honour (Q or J) that was not played on the first round.

You are asked to accept that, when originally dealt QJ doubleton, East will choose to play them in random order. Contrast this with Qx, when he would always start with the "x". Intuitively you would expect random play from QJ to be the best defence, and so it turns out to be. Again, this thread is not the place to prove it.

Now, rather than delve into mathematics, which may not be every novice bridge player's forte, let us consider this from a "brute force" perspective, and OK, simplify it a bit by approximating each possible combination at the outset as equally likely.

Consider 24000 deals in which you meet this particular dilemma. On average you expect, before cashing the initial Ace:

Case 1: East will be dealt a Spade void on about 2000 hands (rather less in reality but who cares)
Case 2: East will be dealt a Spade singleton J on about 2000 hands
Case 3: East will be dealt a Spade singleton Q on about 2000 hands
Case 4: East will be dealt a Spade singleton x on about 4000 hands (there are 2 "x"s remember)
Case 5: East will be dealt Spade doubleton QJ on about 2000 hands. Of these, he will volunteer the J on the first round on 1000 hands (call it Case 5A), and the Q on the first round on the other 1000 hands (Case 5B).
There are 5 more cases (say 6 to 10) that mirror the above but with West having those holdings.

Moving on, you have now cashed the Ace and led another toward the King, West following low on both occasions (crunch time), East playing an honour on the first round.

From the original population of possible cases you can now eliminate all bar cases 2, 3 and 5.
Furthermore, if you allow yourself the "luxury" of noticing precisely which honour East played on the first round, you can also eliminate precisely one of cases 2 and 3, and precisely one of cases 5A and 5B. It does not matter which set is eliminated, the remaining case 2 or 3 outnumbers the remaining case 5A or 5B by a factor of about 2:1.

And it would indeed be a luxury to notice which honour East played. Because if it affected your line, and East were aware of that, he would not play randomly from QJ but would select whichever card influenced your play in favour of the defence. So if you choose not to notice which honour was played, but in your myopic dotage simply noticed that an honour had been played but you do not know which one, then you are left with all of cases 2, 3 and 5, but again the combined frequency of cases 2 and 3 outnumber case 5 by 2:1.

Thus, when East plays an honour on the first round, the chances of his having the other honour at crunch time are halved compared with the corresponding odds when the only missing honour at the outset was the Queen. And that, in a nutshell, is what the principle predicts.