LOTT as a decision assessment tool

I found this question of interest, because it spurred a LOTT analysis of a decision made at the table.

http://bridgewinners.com/article/view/would-you-bid-5d/?cj=296538#c296538

So we were essentially asked to rate bidding 5 after a 4 bid, compared to a pass, or I would have added a penalty double.

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My thought is to apply Law of Total Tricks to the analysis. Assume that LOTT is accurate 50% of the time, in that if there are N total trumps, then there are N total tricks to be had on proper defense and play. Also assume that if LOTT is inaccurate, it will only be off by one trick in either direction with a 25% probability in either direction.

So to consider the problem posed by PANJA here, I'll presume there are 17 total trumps in the deal (the the cases where there were 16, 18, or 19 total tricks all would be computed the same way.) So perhaps they have a 4-4 trump fit, and that partner has a 7 card diamond suit. Or maybe they have 9 trumps, and partner a 6 card diamond suit for his preempt.

Now, postulate that the 4 contract makes 10 tricks with probability P10, and 11 tricks with probability P11. So 4 goes down with probability 1-P10-P11. Assume that 4 never goes down more than 1 trick.

We now have sufficient information to assess questions like bidding 5, or even making a penalty double. I'll assume IMP scoring here. What is the expected score differential for bidding 5, assuming they double that contract, and that the other table plays the contract in a calm 4?

If LOTT is accurate, then the score differential is:

(620 - 800)*P10 + (650 - 1100)*P11 + (-100 - 500)*(1-P10-P11)

= 420*P10 + 150*P11 - 600

If LOTT under-predicts the number of tricks available by 1 trick, then the score differential is:

(620 - 500)*P10 + (650 - 800)*P11 + (-100 - 300)*(1-P10-P11)

= 520*P10 + 250*P11 - 400

If LOTT over-predicts the number of tricks available by 1 trick, then the score differential is:

(620 - 1100)*P10 + (650 - 1400)*P11 + (-100 - 100)*(1-P10-P11)

= -280*P10 - 550*P11 - 200

If LOTT is accurate 50% of the time, and it over/under predicts the result by 1 trick with 25% chance each way, then the weighted gain is simply:

270*P10 - 450

Interestingly, it did not matter if 4 makes an overtrick. All that matters is the probability that 4 can make exactly 10 tricks. As it turns out, we see that conditioned on the event that there are 17 total tricks available, we ALWAYS have a negative expected score for bidding 5 if they double. While my gut says that 4 will make roughly about 50% of the time, you can choose your own value for P10. If P10 is roughly 50%, then it costs us roughly 315 points (7 or 8 IMPs) to bid 5 if they double (I think they might.)

All of the above was conditioned on exactly 17 total tricks. We can do a similar expected score differential computation for 16, 18, or 19 total tricks, getting an expected gain or loss for each of those cases. In the end, we would have a set of expected gains for each case. If your partner is an upstanding citizen who could ALWAYS be expected to have a 7 card suit there, then you can narrow down the set of results based on that information, since then the 16 total trick case will never happen. If I am your partner, then a 6 card suit is at least conceivable.

Similar analysis could also be applied to a direct penalty double, 4X, although LOTT does not factor in here. Thus if P10 and P11 are the probabilities of making 10 and 11 tricks respectively...

(620 - 790)*P10 + (650 - 990)*P11 + (-100 + 200)*(1 - P10 - P11)

= 100 - 440*P11 - 270*P10

So P10 and P11 need to be fairly small for a direct penalty double to be successful over the long term, unless you might be able to set the contract by more than one trick. Again, if 10 tricks happens 50% of the time, 11 tricks 5% of the time, then double has an expected gain of -57, or about 2 IMPs. It is something you might try if you are looking for a swing, but this won't be much of a swing unless you can take 4X down 2 tricks.

Likewise, we can see what happens to if you did bid 5, and they chose to compete to 5. (Would you double that bid? Lets say that you will.) Thus 5X would have an expected differential (compared to 4 at the other table):

(620 + 200)*P10 + (650 - 790)*P11 + (-100 + 500)*(1 - P10 - P11)

= 420*P10 - 540*P11 + 400

So if you think they can make 11 tricks 5% of the time, and 10 tricks 50% of the time, then the expected score differential if they do bid over 5 is +583. Again that assumes they can make at least 9 tricks even when they go down. This is one good consequence of bidding 5, if you think you might stampede them into 5X. Remember, we said we were risking -315 against a plus score of 583 (based on 17 total tricks) if we bid 5 and they can be stampeded into 5X. A similar analysis would need to be done for the other total tricks cases to understand the overall expected gain or loss. But take a good look at your opponents. What do you know about them? Are they readily stampeded? I do know some opponents who will always bid one more.

I'll readily admit that none of this analysis is something one could do at the table, unless you are a bit quicker with these computations than me. At the same time, this idea of conditioning the return on a LOTT analysis seems to be a valuable approach to analyze a question after the fact. A rational analysis is always better than "Partner, were/are you insane?"

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