Probability of HCP Distributions

Conditions and Constraints

Imagine you are playing in a 2♠ contract without the opponents’ interference. Dummy come down and you count up 20 HCP for each side. Well bid! The defenders lead clubs and sooner of later you ruff and draw trumps. Now you are faced with playing on diamonds with KJxx in dummy and xxx in hand. So far the RHO has shown up with 8 HCP and the LHO with none. Who is the more likely to hold the ♦A? On the hundreds of hands you have played 2♠ against silent defenders holding 20 HCP your memory tells you that most of the time the HCP are pretty evenly divided between them. Does that mean the LHO is more likely to hold the ♦A, because so far he has shown up with fewer HCP? No, even though the a priori odds tell you the HCP are likely to be split evenly.

This is similar to a classic problem that arose with coin tosses. It wasn’t until the 17th Century that thinkers got it right. Consider a sequence of tosses with all heads, say, HHHH. How often does it arise? Not very often – the a priori odds are definitely against such an occurrence. So, if someone has thrown HHH would you think that tails (T) is now more likely than heads? Of course not. The probability of a tails is 50%, just as it was at the beginning of the sequence and all the way through. At this point the relevant odds are the a posteriori odds, and HHHH and HHHT are equally likely. This is still very difficult for some to grasp. They think the more heads that appear in sequence, the greater the chances of a tails on the next toss. They give odds against the string continuing. Wrong. One has to abandon the a priori odds, and concentrate on the current odds.

Consider 4 players sitting around the table as a dealer gives them cards one-by-one. The expectation is that by the end of the deal each will receive 4 court cards (A,K,Q,J). As the cards are dealt one-by-one you find that your first 4 cards are court cards. Does that mean you are likely subsequently to receive fewer court cards than a player who has received none as yet. No. The chance of receiving a court card in the next 9 rounds is the same for all players regardless of their current holding.

Of the 20 HCP missing initially East has shown up with 8 HCP so far, thus there are 12 HCP remaining. These are likely to be split in accordance with the number of vacant places. If there is a balance in vacant places, the odds favor 6 HCP on each side for an overall split of 6 HCP in the West and 14 in the East. However, suppose that East has passed in first seat. This constrains East to at most 12 HCP. He cannot have 14 HCP and pass. This constraint acts to eliminate the possibility of some card combinations. To calculate the resulting probabilities one merely discards the now impossible combinations, and adds up the numbers of combinations still within the bounds of the constraint. This is the way of conditional probability, and it is the proper guide to the play the hands.

The best process with regard to estimating probabilities is to get (safely) as much information as possible concerning the distribution of cards before making a decision. Ideally to find the ♦A, say, one would like to know how the diamond suit is split. If one can’t accomplish that, then one may have to work with 2 suits, treating them as equals during the deal. This is the situation we shall treat to show how the calculation proceeds in several simple examples. One may be able to use the number of vacant places to calculate the current probabilities, but sometimes constraints apply and the vacant places won’t yield a good approximation.

South	North	Case 1: South opens	Case 2: East doubles
♠ AK765	♠ QJ84	1♠ | Pass | 4♣ | Pass	1♠ | Pass | 4♣ | Dbl.
♥ K106	♥ A752	4♠ | All Pass	4♠ | All Pass
♦ 865	♦ KJ72	4♣ is a splinter raise	Dbl. is lead directing
♣ A3	♣ 8

 

NS Bid Game, EW Have 15 HCP

In Case 1 West leads the ♣ J (Jack Denies, so the top of a sequence). South takes his ace, ruffs a club and draws trumps in 2 rounds. With some luck in diamonds declarer just might get rid of a heart loser. There are 9 HCP remaining in the red suits. West has shown up with 1 HCP (♣J) and East with 5 (♣KQ). Does that mean West is more likely to hold the ♦A than the ♦Q? No.

The key to the locations of the ♦A and ♦Q lies in the number of vacant places remaining before declarer breaches the diamond suit. Let’s assume the following splits to this point based on the evidence of the opening lead and East’s plays in the club suit.

	West	East
Spades	2	2
Clubs	5	5
Diamonds and Hearts	7	7

 

One cannot be certain of the club split as the defenders may be false-carding, but when there exists a high degree of uncertainty, it is best to assume the most even split consistent with the play to this point, there being nothing to suggest otherwise. In the absence of interference this is the most probable condition. Under that assumption, the probability of the ♦A being in the West is 50%. The same applies to the ♦Q, or any other red card. On the basis of a random deal of the missing red cards, we can calculate the probabilities of the number of HCP held in the red suits by West.

0 HCP	3%	5 HCP	13.6%
1 HCP	6%	6 HCP	15%
2 HCP	12%	7 HCP	12%
3 HCP	15%	8 HCP	6%
4 HCP	13.6%	9 HCP	3%

 

The symmetry is evident. The median number of HCP held is 4.5 for each side. The chance of finding the ♦A or the ♦Q in the West is 50%. This is because the vacant places are equal at 7 -7. It doesn’t matter that East has shown up with just 1 HCP to West’s 5. The same figures would apply if West had led the ♣ K from ♣KQJ. A random deal is blind with regard to the ranks of the cards. The evidence of the bidding slightly affects the odds. With 14 HCP and KQJxx, some West players might overcall, but East would be less inclined to take action over 4♣ with a similar holding.

The equality of vacant places is a very significant feature. The a priori odds are also a consequence of symmetry with 13 vacant places to a side. The reduction from 13 a side to 7 a side changes the probabilities, but the conclusions are not unexpected. The 9 HCP that are missing are most likely to be evenly divided 4-5 or 5-4. Before play in the suit starts, there is nothing to choose between ♦A on the right or ♦A on the left. Next we look at a case where the bidding tells us there is an imbalance in vacant places.

Case 2: East has 7 clubs and doubles for the lead

The ♣ J is led, declarer ruffs a club and draws trumps that prove to be split 2-2. The evidence of the bidding and play point to an imbalance of vacant places for the red suits.

	West	East
Spades	2	2
Clubs	3	7
Diamonds and Hearts	8	4

 

Here are the probabilities for red suit HCP held by West:

0 HCP	0%	5 HCP	17%
1 HCP	1.5%	6 HCP	11%
2 HCP	3%	7 HCP	23%
3 HCP	11%	8 HCP	11%
4 HCP	7%	9 HCP	14%

 

Obviously symmetry is destroyed and West is likely to hold more HCP than East. The median value is 6.5 for West, hence 2.5 for East. The chance of finding the ♦A or the ♦Q in the West is 2:1, in accordance with the ratio of the vacant places (8 against 4). East has not limited his HCP by his double, so there is no restriction in that regard. Next we shall look at the situation where East has limited his HCP range.

Case 3: East preempts and a constraint applies

South	North	East opens 3♣ preemptively
♠ AK765	♠ QJ84	West leads the ♣J
♥ K106	♥ A752
♦ 865	♦ KJ72	3♣ | (3♠) | 4♣ | (4♠) | All Pass
♣ A3	♣ 8

 

Now the bidding tells us something about the distribution of the HCP. Let’s bravely assume that East cannot hold the ♦A along with 7 clubs to the KQ. Here are the probabilities for red suit HCP held by West.

0 HCP	0%	5 HCP	8%
1 HCP	0%	6 HCP	17%
2 HCP	0%	7 HCP	34%
3 HCP	0%	8 HCP	17%
4 HCP	2%	9 HCP	21%

 

Based on our working assumption the probability of the ♦A in the West is 100%. The probability of the ♦Q being with West is 63.6% based on a compilation of the possible card combinations without an ace in the East. This result agrees exactly with a vacant place calculation. Here we have placed the ♦A in the West, so the vacant places are now 7-4. The probability of any red card other than the red ace being with West is 7/11.

Case 4: A conflicting constraint

The above cases are simple illustrations of how probabilities are calculated using vacant places when there is no conflicting constraint. The mathematical procedure conforms to what can be classified as a common sense approach by a competent declarer. The distribution of the cards is the foremost consideration. In the case where the bidding has gone (P) 1♠ (P) 2♠ All Pass one may assume initially the hands are balanced in shape and HCP content, but maybe not. The play of the cards may reveal that an abnormal constraint applies. We’ll now examine a case where the distribution of sides is the most common 8-7-6-5 configuration.

South	North
♠ AK765	♠ Q98	Case 4: Pass | 1♠ | Pass | 2♠ | All Pass
♥ K106	♥ 875
♦ 865	♦ KJ72	West leads the ♣J
♣ A3	♣ 875

 

South ducks the opening lead and West, after some thought, helpfully continues with the ♣T leaving declarer in peace to work things out for himself. We don’t mind that. South wins the second club, goes to dummy in trumps, ruffs a club, and draws trumps, West having been dealt ♠Jxx. East has dropped the ♣KQx so we determine the vacant places to be 5 with West and 8 with East. Because he passed initially, we assume East cannot hold 2 red aces to go alone with his club honors. Under these restrictions the distributions for the 13 remaining HCP in the East hand are:

0 HCP	0%	5 HCP	14%
1 HCP	1%	6 HCP	14%
2 HCP	2%	7 HCP	27%
3HCP	7%	8 HCP	14%
4 HCP	5%	9 HCP	17%

 

East has more red cards than West, but cannot hold more than 9 red HCP (AQQJ), so the constraint on HCP in the East acts contrary to the tendency of the vacant place imbalance to favor East as the holder of the ♦A. It is very important that declarer note this conflict which arises when a defender has revealed a long, topless suit. The HCP distribution appears truncated at the top end of the scale due, of course, to the constraint placed on the total number of HCP. East holds 7-9 red HCP in 58% of the cases, but did not open the bidding with12+HCP. This points to a flat shape: ♠xx ♥QJxx ♦Axxx ♣KQx ? It was not very sporting of him to pass, reasonably conservative perhaps, albeit a rare and unexpected occurrence. Immediately a long topless suit is revealed, declarer should become aware that something unusual has occurred.

Double dummy ‘errors’ occur most commonly when a player holds a topless long suit. It is more probable that a long suit contains its fair share of court cards. When a long suit is deficient in HCP the distribution of HCP will not conform to expectations. Bidding systems are designed for normal conditions. If a player preempts in a long, topless suit with honors in his short suits, the other players will reasonably assume otherwise. Swing results are likely to occur. Similar considerations apply during the play of the hand. A declarer who finds evidence of a long, topless suit, must adjust his prior expectations.

The probability of the ♦Q in the East is 65%, of the ♦A in the East, 40%. How to play the diamond suit? If one simply plays to the ♦K and that loses, the ♥A is sure to be offside. But if the ♦K holds, there is a very good chance the ♥A is also onside. So the optimist plays to the ♦K and then to the ♥K and claims 8 tricks! This approach has a 48% chance of success.

Finally we note that East made a discard on the 3rd spade. Was it a low diamond? That points to a hand like this: ♠ xx ♥ AJx ♦ Qxxxx ♣ KQx. Usually one ignores such slim indications, however, an attempt should be made to speculate without being unduly swayed. Here a diamond discard supports the direct approach, so merely reinforces a decision rather than being an essential element in its construction.

Case 5: a normal situation

In the previous example West led from ♣J109xx opposite ♣KQx, an abnormal configuration. This led to constraints on East that resulted in an abnormal distribution of HCP far different from the prior expectations. Now we consider the more normal situation where West leads the ♣10 (coded 9’s and 10’s) from ♣KJ109x opposite ♣Qxx. This is a somewhat venturesome attacking lead, but we assume West didn’t have a better lead. The upshot of this is that the constraints on the total HCP held in the West or East are not onerous. In the previous Case 4, one had to count up the excluded combinations, but in this case West could have opening points and not wish to overcall or balance with a club suit. Consequently, due to the lack of constraint, uncertainty is at a high level and declarer may use the vacant places as the primary tool in estimating probabilities.

The probability that the ♦A sits in the West and the ♥A sits in the East is 26%, so the optimistic approach of drawing trumps and playing of low to the ♦K has lost its glow.

[The odds of success are (8/13) times (5/12), or 26%] On the other hand, the chance that West holds at least one of the diamond honors is 64%. [The odds are 1 minus (8/13) times (7/12).] So rather that adopting a Kamikaze attacking approach as before, it makes better sense for declarer to play a diamond to the jack initially and hope West holds a doubleton honor. Try to get the cooperation of the defenders with regard to the play in the red suits. This is a hand where declarer wishes he had better spot cards.

It pays to envision the distribution of the sides based on the evidence. What is the single most probable distribution?

	I	II	III
	♠ 3 – 2	♠ 3 – 2	♠ 3 – 2
	♣ 5 – 3	♣ 5 – 3	♣ 5 – 3
	♥ 4 – 3	♥ 3 – 4	♥ 2 – 5
	♦ 1 – 5	♦ 2- 4	♦ 3 – 3
Weights	40	100	80

 

Condition I looks to be much less likely on the bidding and play than indicated by the weights which are based solely on the number of card combinations on the deal. Rule it out. It is better to play West for a doubleton diamond rather than a doubleton heart. If the diamonds are split 2-4, the chance of West holding at least one top honor is 60%. So choosing to play for the single most likely distribution (Condition II) makes sense as it is in accordance with the vacant place analysis. Both methods are based on the assumption of an unconstrained random deal in the red suits. So, declarer should finesse the ♦J, expecting to lose, duck the heart return, hoping to hold the losses to 2 hearts and 2 diamonds. The ♥10 may prove crucial.

Concluding Remarks Probabilities should be no more or less than a numerical expression of common sense. If an expert, like Bob Hamman, say, plays the hands well consistently over a decade or more, one may assume that he has followed sound mathematical principles even though he may or may not express his procedures in mathematical terms. Artificial intelligence computer programs that adapt to the expert’s procedures may grow to simulate his performance with greater and greater efficiency without providing less gifted players with insight into how to improve their game. It is the task of the mathematician to provide an explicit model that functions as a general guide for good decision making. Simplifying assumptions are necessary to make the model useful at the table, or, at the very least, to provide insight into an expert’s approach.

The application of vacant place estimations of the probability of the location of a card of interest is a consequence of such a model. Vacant places yield a good approximation under the right conditions, namely, when there is maximum uncertainty with regard to the distributions of suits which are yet to be played. The degree to which a mathematical model fits reality is the true test of its worth. Those who insist in applying the a priori odds inappropriately are doing a disservice, because their simplistic model is inflexible with regard to changing circumstances, especially with regard to imbalances in the vacant places.