The procedure is as follows:

  1. Enumerate all possible combinations of answers
  2. Combine outcomes whose distinction is irrelevant
  3. Score each outcome by plausibility
  4. Combine the two least plausible outcomes, marking one as the left sub-option and the other as the right sub-option
  5. Repeat (4) until all options are combined.

For example, let us take the questions above.

  1. If we call for aid, will our ally arrive in time to help?
  2. If we send our army to defend O—, will they meaningfully impede the invading army?
  3. Is there any chance of our victory?

The possible combinations of answers are as follows:

  A. Yes, Yes, Yes
  B. Yes, Yes, No
  C. Yes, No, Yes
  D. No, Yes, Yes
  E. Yes, No, No
  F. No, Yes, No
  G. No, No, Yes
  H. No, No, No

As we identified above, however, if the last answer is No, we don't care about the answers to the first two questions. Let us combine those options.

  A. Yes, Yes, Yes
  B. Yes, No, Yes
  C. No, Yes, Yes
  D. No, No, Yes
  E. —, —, No

Now, we can assign plausibilities. Think of this in the same way an Abadarian might bet "one against three" or "two against five", but in this case you assign a plausibility to each option against all of the other options combined. One possible assignment of plausibilities is as follows:

  A. YYY = 5
  B. YNY = 3
  C. NYY = 3
  D. NNY = 1
  E. ––N = 5

In the above plausibility scoring, we find it most plausible that holding the river O— and receiving reinforcements from the ally leads to our victory, or there is no way to win and we lose anyway. Not being able to hold the river and not getting any reinforcements, but somehow winning anyways, is scored the least plausible.

Now we can move on to step 4. First, we merge the items D and C (or D and B, it does not matter which one we choose) and merge them into F, identifying D as the left sub-option and C and the right sub-option. The score of F is the sum of the scores of D and C.

  A. YYY=5
  B. YNY=3
  C. ———
  D. ———
  E. ––N = 5
  F. = 4
      Left: NNY = 1
      Right: NYY = 3

Now we repeat, merging F and B into G

  A. YYY=5
  B. ———
  C. ———
  D. ———
  E. ––N = 5
  F. ———
  G. =7
      Left: = 4
          Left: NNY = 1
          Right: NYY = 3
      Right: YNY=3

And again, merging A and E into H

  A. ———
  B. ———
  C. ———
  D. ———
  E. ———
  F. ———
  G. =7
      Left: = 4
          Left: NNY = 1
          Right: NYY = 3
      Right: YNY=3
  H. =10
      Left: ––N = 5
      Right: YYY=5

And now we finally merge G and H into our final node.

  I. =17
      Left: =7
          Left: = 4
              Left: NNY = 1
              Right: NYY = 3
          Right: YNY=3
      Right: =10
          Left: ––N = 5
          Right: YYY=5

How does transforming our original set of three questions into a tree of this shape assist us? Now, we can ask the question "At the root node of my question tree, is the answer along the left branch?". If we continue asking questions about which direction to take along the branches of the tree, we can see that three of the endpoints of the tree only require two questions, instead of the original three. In addition, we can see that the two cases with the lowest plausibility score are the ones assigned to three-question routes through the tree. Assuming our plausibility scores are accurate, we spend on average 2 and 4/17ths = 2.235 questions using this tree instead of the three questions using a naive strategy.