This is one of the most simple, yet mind-boggling areas of card play. Rather than reading words, look at this typical example:
K 10 5 4
A 7 6 3 2
You lay down your A and LHO (West) follows low and the queen drops from RHO (East). You play a small spade from hand and West follows low. Should you put up the K and hope East started with QJ doubleton? Or, should you finesse dummy's 10, playing for West to have started with J98?
The theory of restricted choice says you should finesse—don't play for the drop. Two times out of three, the honor that dropped from East will be a singleton. Only one time in three will it be from queen-jack doubleton.
Why? Many of you will be happier if you stop reading now. Just be aware that, when missing queen-jack-low-low, and an honor drops on the first round, play it to be a singleton. End of story.
For anyone still reading, here is why it works. When East (in the example above) plays the queen, you should assume he played it because he had to play it. (That is where the term "restricted" comes into this discussion.) If he had been dealt queen-jack doubleton, he might have played the jack (half the time). When the queen plops onto the table, it is more likely that it appeared because East's choice was restricted. He had no other card he could have played. If he had been dealt queen-jack doubleton, he would have had the freedom to play the queen 50% of the time and the jack 50% of the time.
Do you remember the Monty Hall problem from Let's Make a Deal? This link might still work. In the finale, the contestant always picks one of three doors. Behind one door is a car (let's say). Behind the other two doors are goats.
Let's say the contestant chose Door #1. Monty would always say, I am first going to reveal what is behind Door #3 (let's say). Door #3 would always have a goat behind it (NEVER a car). Now, Monty would say, "Do you want to stick with #1, or change to #2?" This is the exact same thing as the Queen/Jack decision. Monty's choice was restricted! You should assume he displayed door #3 because he had to (singleton queen).
If your door (#1) contained a goat (as it will 1 time in 3), Monty has to reveal the other goat door. He knows the car is behind #2. He has to show you #3 (he has to play the queen).
So, when he offers you a switch, you should switch. You will WIN THE CAR 2/3 of the time! Yes—you will win 2/3 of the time by always switching.
Here is why. When you picked Door #1, you had a 1/3 chance to win the car. If you have actually picked the car, that is bad news for you—because your strategy is always going to be to switch, so you aren't keeping the door with the car. You will be a loser the 1 time in 3 you start out by (unluckily) picking the car. But, if you've picked a goat (a 2/3 chance), Monty will now always have to reveal the other goat door and leave the car hidden behind the other door. You now abandon #1 (if it was right, tough luck) and go for the 2/3 chance. Whenever the car is behind #2 or #3, you win. If it is behind your original pick you lose. By starting out with a goat door you have forced Monty into a restricted choice of opening the other goat door. He can't ever reveal the car door.
When you have indeed picked the car to begin with, Monty does have a choice. He can show you either of the other 2 doors. That's when you will be losing. But, of course, 2 times in 3 (the 2 you start with a goat), he has restricted choice–and will end up revealing the car for you (the non-goat).
Back to bridge. East is dealt 1 of these 3 holdings:
The QJ doubleton, is in effect the equivalent of your picking the Car Door to start with. You are going to be a loser. But, when the, say, Queen appears, you should assume that East HAD TO SHOW YOU THE QUEEN. It is as if you picked a goat (West has it), and East has to show you the other goat. He had no choice. If he happens to have both goats (Q-J doubleton), you are dead. You will lose that 1 time in 3.
By now, you may wish you had stopped after the first paragraph. If so, just take my word for it (or try the link). Furthermore, many theorists and bridge mathematicians say all this "restricted" talk is hogwash. They say it is simple math: J98 or Q98 is twice as likely to have been originally dealt than 98 doubleton.
By the way, there are many other restricted-choice combinations. For example:
A K 8 6
Q 5 4
You lay down the ace then queen and RHO drops the nine and then the jack. There are only 2 cards left. Should you play for the 10 to drop (J109 tripleton on your right) or should you finesse and assume LHO started with 10732 and East with J9? This is actually "double-restricted" choice. Assume East played the jack and nine because he was dealt exactly (only) those 2 cards. If he started with J109, he could have played any two of them (J9, J10, or 109). When you see specifically the jack and nine, presume he had to play those two. He probably does not have the 10. If he had the 10, you might have seen it. This is even more than a 2 in 3 chance when you finesse the 8 on the third round of the suit.
Try taking a look at a few more layouts:
A Q 4
K 9 3 2
A 9 3
K Q 8 4
K Q 2
A 9 5 4
In each layout above, after you lay down two top honors, and the jack (or ten) falls on the second round, you should finesse on the third round. So, in 1), cash the ace-queen, then play low to the nine–losing only to J10x on your left. In 2), cash the ace-king (unblocking the 9) and then lead low to the 8—losing only to J10x on your left. In 3) cash the king-queen, then lead low to the 9—losing only to J10x on your left. In all cases, you lose to the 3-3 break, but gain any time RHO started with J(10)xxx. The theory of restricted choice will be with you.