Robert Heinlein And Evolving Probabilities

Writer Stephen Dawson, a long-time participant at this blog, wrote and asked the following:

In his The Moon is a Harsh Mistress, Robert Heinlein had his intelligent computer calculate the odds of a successful revolution for independence of the moon, which turned out to be a low number. The conspirators embarked on their quest and after a while, although successful to that point, were disheartened to find that the odds of ultimate success had fallen, and continued to fall. From memory, Mike, the computer, explained that there were various paths to success, but that selection of any one thereby eliminated alternatives, reducing the odds. As they proceeded, the odds continued to reduce.

Eventually, of course, they neared their goal sufficiently for the odds to begin increasing.

Completely putting aside the question of whether such a calculation could reasonably be made in the first place, this idea that making a choice from one of several mutually exclusive paths will lower the odds of success has always troubled me. Do you think that, wonderful as he was, Heinlein was wrong on this?

Heinlein was right. Here’s why. Moon is a Harsh Mistress

All statements of logic, including probability statements about the success of revolution, are conditional on certain evidence. Thus we cannot say, “The probability of showing a 5 spot on a die roll” is 1 in 6 without adding the evidence, “Given that we have a six-sided die, only one spot of which will show when tossed, and only one side of which is labeled 5.”

Keep that evidence in mind and consider this scenario, which we can call the revolution game: you will roll a die twice; if you roll at least a 5 on the first, you are allowed a second roll (else not), and if this second roll is a 6 you win.

What are the chances your revolution succeeds? Well, rolling at least a 5, given our evidence and game rules, is 2/6, and then rolling a 6 the second time is 1/6, so together the chance is 2/36 = 1/18. Not so high.

But once you “go down the path” of rolling a 5 or 6 on the first roll, then your chance of winning soars by three times to 6/36, or 1/6. Because, of course, your evidence that you use to compute the probability has changed based on what has happened.

This is no different than considering a situation where a mythical traveler is presented with several doorways, all but one of which open to his destination. If the traveler has more information than just the number of doorways, then he can increase his chance of arriving by applying that information. Perhaps, for instance, he has learned that the real doorway will be “hand crafted” and that the false will be machine made. That additional information changes the probability of succeeding, though perhaps not in a perfectly quantifiable way.

(One delusion to which we Moderns succumb is that quantification is always possible, usually to arbitrary precision.)

Or think of it this way. Suppose you want to operate on a fellow with an operable disease. Given our knowledge of surgery, what are the chances you save him? First consider that you haven’t gone to medical school. The odds of success are low. But after slogging through for four years of school, seven or eight years of residency and a fellowship—all choices you made—then the odds of a successful outcome increase dramatically. Because the evidence used to the compute that probability has changed.

Update I read the problem backwards; it should be a decreasing, not increasing, probability along a path, but one which turns into increasing probability. This is still possible.

I don’t want to include any complex drawing. But imagine in front of you are several paths. You take one and then learn, via Mike and given the information available, your probability of success has decreased; it would have increased or not decreased as much had you taken another path. Perhaps on this path, no matter what, you to win the lottery at least once to continue.

You take another path (already having progressed down the first) and learn that you now have to win the lottery twice. And so on.

If I have time later, I’ll include a numerical example.


  1. Rich

    But the question was, is it possible that the initially low odds continue to fall as you move along the path, only rising as you near the end. It’s not obvious to me that that’s possible but I don’t know.

  2. Ken

    For a statistical analysis I believe the last paragraph got overly specific. Instead of presenting the case as “Suppose YOU [emphasis added] want to operate on a fellow with what are the chances YOU [emphasis added] save him?”

    Wouldn’t a better a ‘better’ thought model be something like: “Suppose someone were to operate on a person in need, what are the chances of saving him/her?” With the same uneducated vs. medical school education, and residency, assessment.

    Of course the odds would go up, in general (& for the focus of this blog essay that’s probably good enough). However, in this real world there’s a number of other very significant factors at play — and an even better model ‘ought’ [my opinion] consider explicitly the maximum possible probability for achieving the result of interest.

    Why? This helps flush out other significant factors that otherwise lie lurking in the shadows as implicit assumptions–and such implicit assumptions are commonly both wrong & influential.

    One such factor, that the best students [and presumably those that will perform best once into the labor force] actually get into medical school. Such a fundamental assumption, and [in the U.S., once upon a time, a core value largely taken for granted] is no longer:

  3. Briggs


    Well, it’s just as easy to construct an example where the probabilities decrease as increase. I’ll leave it as a reader exercise. Hint: you start with several paths which might lead to success, but after choosing the number of future paths decrease, leading to a decreased probability of success.

  4. Matt


    All probabilities, like politics, is local.

  5. William Sears

    Your example doesn’t prove the point given. A better one is to ask what is the probability of throwing three dice, one at a time, such that the total is six or less. At the start the probability is 20/216 or about 9.3%. However, if the first die shows a four then the probability becomes 1/36 or about 2.8%. If the first die had been a one then the probability shifts to 27.8%. For an initial two or three the result is 16.7% or 8.3% respectively. This demonstrates Heinlein’s point. They have chosen a poor path to success. The question now becomes: is it possible to produce an example where the odds always reduce, regardless of path chosen?

  6. William Sears

    An addendum to my previous comment.
    If the first two dice show a four and then a one the probability has risen again to 16.7%. Thus for the path illustrated for throwing three dice to get a six or less the probabilities shift from 9.3% to 2.8% to 16.7%. This is a fuller illustration of Heinlein’s thesis.

  7. JH

    Keep options open for alternative solutions, and persistence can lead to success! My simplistic view.

  8. SteveBrooklineMA

    I can see that going down a path might decrease your chances, and that as you continue you might string together enough right choices to get your chances rising again. But Mike The Computer’s explanation seems odd. It sounds a bit like he is suggesting that with two paths your odds are P1+P2, but that after you choose path 1 your odds are lowered to P1.

  9. Doug M

    Can the probability of success fall and then rise again? Think of a candidate for president. Over the next 12 months, he will participate in debates, slough off rumors, propose policy ideas, fight in primaries, attempt to secure the party nomination and face the voters on election day. He will stumble and rise, as will his opponents. It is impossible to precisely say what his probability of success actually is. However, there will be a sense of how taht is changing over time.

    There is a bookmaking site that will pay 7/3 if Mitt Romney is elected president in November of 2012. Are these his actuall odds of success? Not really, this is a reflection of what the bookie and the bookies most recent clients think is his probability of success.

    I think the real question is if a computer even as sophisticated as Mike could make an assement of the probability that was any more accurate than an intelligent man’s intuition.

  10. Ken

    Isn’t this just a variation of the Monty Python game show problem?

  11. max

    The door example fails because, unlike Mycroft, it supposes a single possible correct solution.

    Let me propose a dice based example. There are 4 paths to successful revolution each with a 1 in 6 chance of success. Of however you select 1 path, you double the chance of success of that path to 1 in 3 while making the probability of all other paths being successful 0. In dice terms, rolling a 6 on any of 4 dice equals success OR by selecting 1 die success can be obtained by rolling a 5 or 6 on that die. The chance of rolling a 6 on at least 1 of 4 dice is 671/1,296 or about 52%, while the chance of rolling 5 or 6 on a single die is 1/3 or about 33%.

    This seems to be the reasoning Mycroft was using. Mycroft’s model must include parameters that include a low increase in probability for early steps along the path, when the total increase in probability exceeds Ppath1/Pallpaths x Ppath1 then the result will be an increase in probability by advancing along path1. I doubt you can model this with dice (usually dice models are obvious to me but not here) and not easily even without dice, a Babbage machine would help.

  12. Will

    I am confused.

    Say the rebels have 3 paths, each with a .05 chance of success. Regardless which of the three paths they chose, they will still have a .05 chance of success– the paths are mutually exclusive (according to Mike the computer).

    Think of it in terms on certainty. If I make a move in chess that will result in mate in 6, my odds of winning don’t decrease because that move means sacrificing a queen. The path itself has likelihood of success; it is not dependent on the other possible paths.

    Am I totally in left field??

  13. Thanks Matt for looking into this. My mind has kept returning to this from time to time for thirty five years because it seems so counter-intuitive. But one thing my rudimentary acquaintance with statistics has taught me is that relying on intuition in statistical matters is dangerous.

    Anyway, to clarify the problem. Heinlein’s suggestion, as I read is, wasn’t that wrong choices would be made on a given path, thereby lowering the odds. Nor that right choices would be made, but that these failed to yield the desired effect, also therefore lowering the odds.

    My understanding was that Heinlein was saying that if there are multiple (presumably, in the real world, not totally independent) paths to a successful outcome, embarking of one of those paths necessarily reduced the probability of success.

  14. Will

    Stephen: That would be impossible. Doing nothing, presumably, gives you a zero percent chance of success. Embarking on any of the correct paths by necessity would increase your chances.

  15. Greg Cavanagh

    I havn’t read the book, though I have read much of Robert Heinlien.

    If during the course of the rebelion; some rebels were cought, some are killed, some are routed, some are infiltrated ect. And at the same time security was strengthened and wanted posters were having success, intelegence gathering was paying off ect. Then it makes sence that odds would decrease over time.

    Witness Islamic attacks in the western world, their successes have decreased over time. Plane high-jacking is almost unheard of in the modern world, whereas in the 70’s it was perfectly common.

  16. Tom S

    Have you noticed that you make mistakes in almost every post on the topic of probability?

  17. Alex

    It’s very simple. Take path A and the revolution succeeds with probability 1. Take path B and the revolution fails. You have an equal chance of choosing A or B, so before you choose the probabilty the revolution succeeds is 1/2. You choose A the prob. increases to 1, you choose B the prob. decreases to 0.

  18. DAV


    no need for complicated examples. Suppose you want to go to the store. There are two paths: 1) over a lake covered by thin ice and 2) the normal road. Taking path (1) is attractive because it cuts across a horseshoe in the road. If you take path (1) your chances of success are lower but once you reach the road again on the other side they will improve.

    The situation is similar for a revolution. When you start out, here are many paths to take each with its own dangers contributing to failure but each having an attractiveness making them likely paths. As each is successfully navigated The number of bad paths decrease while the number of paths leading to success increase.

    Heinlein’s Mike must have been giving generic answers. If instead he could follow all of the paths to arrive at an answer, Da Hero could just have asked Mike to pick one for him. In that case, the chances of the success of the story would have been dismal: Humans want to revolt; push button to get how; succeed — Boooooring.

  19. DAV

    Greg Cavanagh ,

    Plane high-jacking is almost unheard of in the modern world, whereas in the 70′s it was perfectly common.

    Not the best example. The only benefits from plane-jacking are: advertising and increasing inconvenience to the point of caving. The primary dangers are backlash and ennui. The number of unexpected trips to Cuba were such that on one trip a passenger was heard to mutter, “If it isn’t one damn thing it’s another.”

    I’ve never heard of anyone getting their wishes granted through plane-jacking so the primary benefit has to be advertising. The need for that will obviously decrease over time. Outside of advertising, plane-jacking is pointless.

    Heinlein’ revolutionaries launching rocks at the Earth was far more poignant. “I can smash you to smithereens anytime I choose” is by far a more powerful statement than “I’m going to make you wish you took the train.”

  20. Briggs

    Tom S,

    Have you noticed that you’re prone to making unsubstantiated generalizations? But if you give me your address, I’ll see you get a refund.

  21. JH

    Interesting comments!

    Is there any reason that Heinlein has to be right?

    Maybe he was thinking about the parallel system that will fail only if all its components (alternatives) fail. And each component has a decreasing probability of surviving until a certain time. Sooner or later, the system will fail.

    Or maybe he was thinking about his dating experience when writing this book.

  22. Wayne

    It seems to me that there are two cases:

    1. The computer is omniscient. It knows all possible actions and options and it knows the exact odds of each of them. (In the real world, this seems impossible.) In this case, we can try to come up with dice roll or other scenarios and I’m not sure if decreasing-then-increasing is possible.

    2. The computer is not omniscient. It does not know all possible actions and options. It does not know precisely the odds of the actions and options it knows about.

    In this case, it seems rather Bayesian: its initial priors and early evidence push posteriors in one direction, but at some point the evidence starts pushing the posteriors in the opposite direction.

  23. Matt

    In the case of the story, the characters working with Mike weren’t the only ones choosing paths. Also, the outcome of a particular choice isn’t always known, especially when human reactions are involved.

  24. Hasdrubal

    Suppose you roll two dice in succession.

    If you roll a 1 first, you will be successful if your second roll is 2 or higher.
    If you roll a 2 first, then you need a 3 or higher next.
    If you roll a 3 first, then you need a 4 or higher.
    With a 4, you need a 5 or 6 on your second roll.
    On a 5 or 6, you need to roll a 6 on the second roll.

    So, before you roll your chance of success is 16/36. But after your first roll your chances change, they might increase to 5/6 or fall to 1/6.

    I don’t know, but I would imagine that if you chain a lot of rolls together you could have paths where your chance of success increases after some rolls and decreases after others.

  25. Greg Cavanagh

    yea I agree. I kinda regretted putting any examples in at all. It wasn’t needed after my description of increasing difficulty (reducing odds of success).

    Which was my point; the chance of a successful high-jack these days is considerably smaller than it was in the 70’s. i.e; the odds of success are far smaller as time has progressed. The purpose of the high-jack is arbitrary, though similar, since the original story is about trying to make a successful revolution.

    The PLO did it purely for advertising and they were the most prolific high-jackers. They were quite successful in their own revolution because of it though. But there were some poeple who genuinely thought they’d jump off the plane and catch a taxi home.

  26. Alan D McIntire

    As I recall from the book, the original chance of success was about 1 in 6 or 1 in 7.

    After various diplomatic efforts, the odds of success dropped to 1 in 20 or 1 in 30. I think Stephen Davis was correct that Heinelin erred. By picking a specific course of action one doesn’t NECESSARILY reduce one’s chances of winning. Another scenario could have
    given odds of 1 in 7, 1 in 5, 1 in 3, with constantly increasing probability of success.

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