## Probabilities in Bridge

Often, we have discussed in the old blog topics in probability theory. The main message turned out to be:

It does not make sense to talk about probabilities without a hypothetical random experiment.

Imagine repeating that experiment over and over again. Probabilities should then reflect the predictions on how often an event happens.

In most cases, we can simply study all possible outcomes, which we assume to happen equally often in the long run, because we have no reason to prefer one outcome over the other. The probability of an event is then the proportion of outcomes where the event happens. The simple example is one dice throw with six possible outcomes. The event of throwing a 6 happens only in one of these outcomes. Thus, we give it the probability 1/6, and assume that it happens 1/6 of the time in the long run.

Mathematicians want to have a stricter approach. They define the probability as a number assigned to a set of events. These probabilities follow simple rules, mainly that they add up for disjoint sets of events. It turns out that this definition yields the heuristic approach with the repeated experiment by the theorem of the law of big numbers. We don’t have to dive into this here.

After this boring excursion to the basics, let us talk about probabilities in the card game Bridge. In case you do not know: In Bridge, 4 times 13 hands are dealt to players called N, E, S, W. the pairs N/S and E/W form a partnership. After the bidding, one player becomes declarer. Let’s say it is S. The cards of his partner N are then displayed on the table. Here is a typical distribution as seen by S.

The so-called „contract“ is 6 spades which means you have to make all tricks but one. You will lose the ace of diamonds in any case. Your job is not to lose a trump trick in spades. In that suit, we are missing

Q, J, 6, 3

In Bridge, the player in front of the table, in this case W, starts with the first card. Let us say, he started with a small club. You take the ace and a small one from the hand. Thus, the N hand on the table wins this trick and has to play the next card to start the next trick.

A typical question now is: How do you play the trumps? Here are your choices:

• You can simply cash AK in spades, hoping that E and W both have two of the remaining cards.
• You can play the A in spades, then cross to the hand in S with the king of hearts, and play a small spade from the hand to see what happens.
• It may happen that E and W added a small spade to the first trick. Then, W will produce the Q or J and you will have to take the K of spades to have a chance to get all spade if they break 2-2.
• Or it may happen that W adds a small card and E adds the Q or the J in spades to the first trick. Then you have to decide if you give the remaining J or Q to W or to E. If you give it to W, you finesse with the 10.

It is interesting to compute the probabilities for these outcomes. As I said, we need an experiment which defines the probabilities. In our case, it is a random deal at the start of the board.

An important trick is to discard all deals from the experiment which do not fit to any additional information we have. For an example, assume that E opened with 2 in hearts in the bidding and thus shows 6 cards in hearts already. We would ignore all deals where E does not have 6 hearts. That increases the probability for W to have three spades, simply because W has now 13 cards which could be spades, and E only 7. For now, we assume that we have no further information at the start of the play.

The first strategy, cashing AK, is simple. It wins if the spades are distributed 2-2. The number of cases in a random deal where this happens is easy to compute with the so-called hypergeometrical distribution. To count all cases of a 2-2 distribution, we select 2 of the 4 spades and 11 of the 22 non-spades for W. We count these and divide by the total count of 13 selected out of 26. If we do this, our success rate will be 40% of all random deals.

\binom{4}{2} \binom{22}{11} / \binom{26}{13} \approx 40\%

Just to compare, how much an information about 6 hearts at E distorts this, we recompute the probability under that assumption. In this case, we select our 11 cards only from the remaining 22-6=16 cards. Note, that the number of possible deals is also reduced because we have fixed the 6 hearts at E. Simple cashing AK succeeds now only in 34% of the cases.

\binom{4}{2} \binom{16}{11} / \binom{20}{13} \approx 34\%

Nobody will blindly cash AK if he sees a Q or J at E and if he can cross to the hand to make a finesse. At least, some thought should be spent if that is not the better option.

So, we assume now that E shows the J to our A of spades, we go to the K of hearts in the hand, and try a small spade where W adds a small one too.

Now, we have a new information. There is only one card remaining, the Q of spades. In our thought experiment, we discard all cases where W does not have two small spades or E not at least one spade. With these cards fixed, we see that it is a little bit more likely that E has the Q, because he has now one card more which could be the Q. Thus, cashing the K has a bit more probability. We could add the information that E and W did not have a void in hearts or clubs, but this does not change our conclusion.

Thus, we should avoid the risk of being ruffed in hearts and cash AK right away. Don’t be misled to think that our reasoning proves that the chances are over 50% to succeed. They are still at 40% only. They just increased if we know that E has a J. If E shows a small card only, it does not matter if we try to finesse or cash the K, because W will show Q or J when we try to finesse. Moreover, we did not consider the 4-0 or 0-4 distributions with probability 9.5%.

2 \binom{22}{13} / \binom{26}{13} \approx 9.5%

We can also count all cases where the finesse works and compare with all cases where it does not, just for an exercise.

For the finesse to work, W must have started with Q and two small cards, written as Qxx, and E with J. There is only one 3-1 distribution of that kind. Again, we take the full deal without additional information.

\binom{22}{10} / \binom{26}{13} = \approx 0.0622

For the finesse not to work, E must have started with QJ. There is only one 2-2 distribution again.

\binom{22}{11} / \binom{26}{13} \approx 0.0678

That is exactly 12/11 higher.

Assume, E shows the Q at the first trump trick. Some may argue that E will never do this with QJ. But that is purely psychological. A clever E will try to fool the declarer.

Another argument often includes the cards that we have already seen during previous rounds before drawing trumps. We need to be aware that those cards are mainly forced. So, the individual cards do not count. But the distribution may count. If you discover a long suit on one side, the other player is more likely to have a remaining card.