Projectile motion puzzle

This problem is inspired by Jackie Bradley, Jr., outfielder for the Boston Red Sox, who last week in warm-up threw a baseball from near home plate over the 17-feet-high wall in deep center field, 420 feet away.  (Here is a video clip of the throw.)

It’s a pretty amazing throw… but just how amazing is it?  That is, how hard would you have to throw a baseball to clear a 17-foot wall 420 feet away?

This is an interesting question in its own right, with the usual appeal of encouraging both pen-and-paper as well as computer simulation for a solution.  I’ll get to the answer shortly– but while working on it, I encountered an interesting relationship between some of the variables that has a nice geometric interpretation, which I think is best illustrated with the following slightly different version of the problem:

Problem: Suppose that you are standing on the outer bank of a moat surrounding a castle, and you wish to secretly deliver a message, attached to a large rock, to your spy inside the castle.  A wall h=11 meters high surrounds the castle, which is in turn surrounded by the moat which is d=19 meters wide.  At what angle should you throw the rock in order to have the best chance of clearing the wall?

At what angle should you throw an object to clear a wall 19 meters away and 11 meters high?

At what angle should you throw an object to clear a wall 19 meters away and 11 meters high?

The intent of the large rock is to allow us to ignore the relatively negligible effects of air resistance, thus preventing the calculus problem from becoming a differential equations problem.

We can’t afford to do that with a baseball, though.  Coming back to the original problem at Fenway Park, there are two important atmospheric effects to consider.  First, air resistance significantly increases the speed at which Bradley must have thrown the ball to clear the outfield wall.  But second, the Magnus force resulting from backspin on the ball (also responsible for curve balls and surprisingly hard-to-catch pop-ups) actually makes the ball travel farther, thus decreasing the required speed compared with a ball thrown with no backspin.

Accounting for both of these effects, by my calculations (which I can share if there is interest), Bradley would have had to throw the ball at over 99 miles per hour, at an angle of a little over 31 degrees.

 

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Using a watch– or a stick– as a compass

A couple of years ago, I wrote about a commonly cited method of direction-finding using an analog watch and the sun.  Briefly, if you hold your watch face horizontally with the hour hand pointing toward the sun, then the ray halfway between the hour hand and 12 noon points approximately true south.  (This is for locations in the northern hemisphere; there is a slightly different version that works in the southern hemisphere.)

The punch line was that the method can be extremely inaccurate, with errors potentially exceeding 80 degrees depending on the location, month, and time of day.  I provided a couple of figures, each for a different “extreme” location in the United States, showing the range of error in estimated direction over the course of an entire year.

Unfortunately, I ended on that essentially negative note, without considering any potentially more accurate methods as an alternative.  This post is an attempt to remedy that.  In recent discussion in the comments, Steve H. suggested analysis of the use of the “shadow-stick” method: place a stick vertically in the ground, and mark the location on the ground of the tip of the stick’s shadow at two (or more) different times.  The line containing these points will be roughly west-to-east.

Illustration of the shadow-stick method of direction-finding.  With a stick placed vertically in the ground, the tip of the stick's shadow moves roughly from west to east.

Illustration of the shadow-stick method of direction-finding. With a stick placed vertically in the ground, the tip of the stick’s shadow moves roughly from west to east.

As the following analysis shows, this shadow-stick method of direction-finding is indeed generally more accurate than the watch method… most of the time, anyway.  But even when it is better, it can still be bad.  It turns out that both methods are plagued with some problems, with the not-so-surprising conclusion that if you need to find your way home, there is a tradeoff to be made between accuracy and convenience.

One of the problems with my original presentation was condensing the behavior of the watch method over an entire year into a single plot (in this case, at Lake of the Woods in Minnesota, at a northern latitude where the watch method’s accuracy is best).  This clearly shows the performance envelope, i.e. the maximum possible error over the whole year, but it hides the important trending behavior within each day, and how that daily trend changes very gradually over the year.  We can see this more clearly with an animation: the following shows the same daily behavior of error in estimated direction using the watch method (in blue), but also the shadow-stick method (in red), over the course of this year.

Accuracy of the watch method (blue) and shadow-stick method (red) of direction-finding, over the course of the year 2014 in Lake of the Woods, Minnesota. The shadow-stick method is more accurate 40.6% of the time.

Accuracy of the watch method (blue) and shadow-stick method (red) of direction-finding, over the course of the year 2014 in Lake of the Woods, Minnesota. The shadow-stick method is more accurate 40.6% of the time.

For reference, following are links to a couple of other animations showing the same comparison at other locations.

  • Florida Keys (a southern extreme, where the watch method performs poorly, included in the original earlier post)
  • Durango, Colorado (discussed in the comments on the earlier post)

There are several things to note here.  First, this is an example where the shadow-stick method can actually perform significantly worse than the watch method.  Its worst-case behavior is near the solstices in June and December, with errors exceeding 30 degrees near sunrise and sunset.  This worst-case error increases with latitude, which is the opposite of how the watch method behaves, as shown by the Florida Keys example above.

However, note the symmetry in the error curve for the shadow-stick method.  It always passes from an extreme in the morning, to zero around noon, to the other extreme in the evening.  We can exploit this symmetry… if we are willing to wait around a while.  That is, we could improve our accuracy by making a direction measurement some time in the morning before noon, then making another measurement at the same time after noon, and using the average of the two as our final estimate.  (A slightly easier common refinement of the shadow-stick method is to (1) mark the tip of the shadow sometime in the morning, then (2) mark the shadow again later in the afternoon when the shadow is the same length.  The basic idea is the same in either case.)

Finally, this issue of the length of time between measurements is likely an important consideration in the field.  A benefit of the watch method is that you get a result immediately; look at the sun, look at your watch, and you’re off.  The shadow-stick method, on the other hand, requires a pair of measurements, with some waiting time in between.  How long are you willing to wait for more accuracy?

Interestingly, the benefit of that additional waiting time isn’t linear– that is, all of the data shown here assumes just 15 minutes between marking the stick’s shadow.  Waiting longer can certainly reduce the effect of measurement error (i.e., the problem of using cylindrical sticks and spherical pebbles, etc., instead of mathematical line segments and points) by providing a longer baseline… but the inherent accuracy of the method only improves significantly when the two measurement times span apparent noon, as in the refinement above, which could take hours.

To wrap up, I still do not see a way to condense this information into a reasonably simple, easy-to-remember, expedient method for finding direction in the field without a compass.  The regular, symmetric behavior of the error in the shadow-stick method suggests that we could possibly devise an “immediate” method of eliminating most of that error, by considering the extent and sense of the error as a function of the season, and a “scale factor” as a function of the time until/since noon… but that starts to sound like anything but “simple and easy-to-remember.”

 

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The Price Is Right puzzle solution

This is just a quick follow-up to capture my notes on an approach to solving the following problem from the last post:

Problem: Suppose you and n other players are playing the following game.  Each player in turn spins a wheel as many times as desired, with each spin adding to the player’s score a random value uniformly distributed between 0 and 1.  However, if a player’s cumulative score ever exceeds 1, she immediately loses her turn (and the game).

After all players have taken a turn, the player with the highest score (not exceeding 1) wins the game.  You get to go first– what should your strategy be, and what are your chances of winning?

Solution: Suppose that our strategy is to continue to spin until our score exceeds some target value t.  The first key observation is that the probability that we do so “safely,” i.e. reach a total score greater than t without exceeding 1, is

q(t) = e^t(1-t)

To see this, partition the successful outcomes by the number k of initial spins with total score at most t, with the final k+1-st spin landing the total score in the interval (t, 1].  The probability of such an outcome is

\frac{t^k}{k!}(1-t)

(which may be shown by induction, or see also here and here), and the result follows from the Taylor series expansion for e^t.

At this point, we can express the overall probability of winning using strategy t as

p_n(t) = q(t) \int_t^1 \frac{1}{1-t} (1 - q(s))^n ds

Intuitively, we must:

  1. Safely reach a score in the interval (t, 1], with probability q(t); and
  2. For each such score s, reached with uniform density 1/(1-t), each of the remaining n players must fail to beat our score.

The optimal strategy consists of choosing a target score t maximizing p_n(t).  Unfortunately, this does not have a closed form; however, after differentiating and some manipulation, the desired target score t can be shown to be the root in [0, 1] of the following equation, which has a nice interpretation:

\int_t^1 (1 - q(s))^n ds = (1 - q(t))^n

The idea is that we are choosing a target score t where the (left-hand side) probability of winning by spinning one more time exactly equals the (right-hand side) probability of winning by stopping with the current score t.

Handing the problem over to the computer, the following table shows the optimal target score and corresponding probability of winning for the first few values of n.

Optimal target score (red) and corresponding probability of winning (blue), vs. number of additional players.

Optimal target score (red) and corresponding probability of winning (blue), vs. number of additional players.

One final note: how does this translate into an optimal strategy for the players after the first?  At any point in the game, the current player has some number n of players following him.  His optimal strategy is to target the maximum of the best score so far from the previous players, and the critical score computed above.

[Edit: Following is Python code using mpmath that implements the equations above.]

import mpmath

def q(t):
    return mpmath.exp(t) * (1 - t)

def p_stop(t, n):
    return (1 - q(t)) ** n

def p_spin(t, n):
    return mpmath.quad(lambda s: p_stop(s, n), [t, 1])

def t_opt(n):
    return mpmath.findroot(
        lambda t: p_spin(t, n) - p_stop(t, n), [0, 1], solver='ridder')

def p_win(t, n):
    return q(t) / (1 - t) * p_spin(t, n)

if __name__ == '__main__':
    for n in range(11):
        t = float(t_opt(n))
        p = float(p_win(t, n))
        print('{:4} {:.9f} {:.9f}'.format(n, t, p))

 

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The Price Is Right… sort of

After a couple of recent conversations about the dice game puzzle proposed a few months ago, I spent some time experimenting with the following game that has a vaguely similar “feel,” and that also has the usual desirable feature of being approachable via either computer simulation or pencil and paper.

Suppose you and n other players are playing the following game.  Each player in turn spins a wheel as many times as desired, with each spin adding to the player’s score a random value uniformly distributed between 0 and 1.  However, if a player’s cumulative score ever exceeds 1, she immediately loses her turn (and the game).

After all players have taken a turn, the player with the highest score (not exceeding 1) wins the game.  You get to go first– what should your strategy be, and what are your chances of winning?

The obvious similarity to the dice game Pig is in the “jeopardy”-type challenge of balancing the risk of losing everything– in this case, by “busting,” or exceeding a total score of 1– with the benefit of further increasing your score, and thus decreasing the other players’ chances of beating that score.

I like this “continuous” version of the problem, for a couple of reasons.  First, it’s trickier to attack with a computer, resisting a straightforward dynamic programming approach.  But at the same time, I think we still need the computer, despite some nice pencil-and-paper mathematics involved in the solution.

We can construct an equally interesting discrete version of the game, though, as well: instead of each spin of the wheel yielding a random real value between 0 and 1, suppose that each spin yields a random integer between 1 and m (say, 20), inclusive, where each player’s total score must not exceed m.  The first player who reaches the maximum score not exceeding m wins the game.

This version of the game with n=3 and m=20 is very similar to the “Showcase Showdown” on the television game show The Price Is Right, where three players each get up to two spins of a wheel partitioned into dollar amounts from $.05 to $1.00, in steps of $.05.  The television game has been analyzed before (see here, for example), but as a computational problem I like this version better, since it eliminates both the limit on the number of spins, as well as the potential for ties.

 

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Reddit’s comment ranking algorithm revisited

Introduction

The “Bayesian/frequentist” coin puzzle discussed in the last couple of posts was really just an offshoot of some thoughts I have been mulling over about Reddit’s current default approach to ranking user comments on a post, based on the number of upvotes and downvotes each comment receives.  (Or more generally, the problem of ranking a collection of any items, whether comments, or consumer products, etc., based on varying numbers of “like/dislike” votes.)  Instead of trying to estimate the bias of a coin based on the observed number of heads and tails flipped, here each comment is like a coin, and each upvote (or downvote) is like an observation of a coin flip coming up heads (or tails).

If we assume that each comment has some fixed– but unknown– probability \theta that a random user will upvote the comment, then it would be convenient to simply sort all of the comments on a particular post by decreasing \theta, so that the “best” comments would appear near the top.  Unfortunately, we don’t actually know \theta, we can only estimate it somehow by using the observed pair (u,d) of upvotes and downvotes, respectively.

A natural first idea might be to “score” each comment using the maximum likelihood estimate

\hat{\theta} = \frac{u}{u+d}

and sort the comments by this score.  But this tends to unfairly compare comments with very different numbers of total votes; e.g., should a comment with votes (3,0) really be ranked higher than (99,1)?

Wilson Score Interval

Evan Miller’s “How Not To Sort By Average Rating” does a good job of presenting this and other approaches, eventually arguing for sorting by the lower bound of the Wilson score interval, which is what Reddit currently does.  Briefly, the Wilson score interval is a confidence interval intended to “cover” (i.e., contain) the true– but unknown– value \theta with at least some guaranteed probability, described as the “confidence level.”  In general, the higher the confidence level, or the fewer the number of observations, the wider the corresponding confidence interval.  By scoring each comment with the lower bound of this confidence interval, we are effectively starting with a point estimate based on the fraction of upvotes, but then penalizing this score according to the total number of votes, with fewer votes receiving a greater penalty.

Reddit’s use of this scheme has evolved slightly over time, initially computing a 70% confidence interval, but then changing to the current wider 80% confidence interval, having the effect of imposing a slightly greater penalty on comments with fewer total votes.  This “fine-tuning” of the scoring algorithm raises the question whether there might not be a more natural method for ranking user comments, that does not require this sort of knob-turning.

A Bayesian Alternative

Last year, James Neufeld proposed the interesting idea of sampling a random score for each comment by drawing from a corresponding beta distribution with parameters

(\alpha, \beta) = (u+1, d+1)

The idea is that this beta distribution is a natural way to express our uncertainty about the “true” value \theta of a comment, starting with an assumed prior uniform distribution on \theta (i.e., a comment is initially equally likely to be great, terrible, or anything in between), and updating based on the observation of (u,d) upvotes and downvotes, respectively.  For example, a comment with 30 upvotes and 10 downvotes yields a beta distribution with the following density:

Probability density of beta distribution with parameters (30+1,10+1).

Probability density of beta distribution with parameters (30+1,10+1).

A key point is that every user does not necessarily see the comments for a post in the same order.  Each time the post is viewed, the comments are re-scored by new random draws from the corresponding beta distributions, and sorted accordingly.  As a comment receives more and more upvotes and/or downvotes, it will “settle in” to a particular position among other comments… but comments with few votes, or even strongly downvoted comments, will still have some chance of appearing near the top of any particular user’s view of the page.

I really like this idea, but the non-deterministic ordering of comments presented to different users may be seen as a drawback.  Can we fix this?

Sorting by Expected Rank

I can think of two natural deterministic modifications of this approach.  The first is to sort comments by their expected ranking using the random scoring described above.  In other words, for each comment, compute the expected number of other comments that would appear higher than it on one of Neufeld’s randomly generated pages, and sort the comments by this expected value.

Although this method “fixes” the non-determinism of the original, unfortunately it suffers from a different undesirable property: the relative ranking of two comments may be affected by the presence or absence of other comments on the same post.  For example, consider the two comments identified by their upvote/downvote counts (0,1) and (1,3).  If these are the only two comments on a post, then (0,1) < (1,3).  However, if we introduce a third comment (7,3), then the resulting overall ranking is (1,3) < (0,1) < (7,3), reversing the ranking of the original two comments!

Pairwise comparisons

Which brings me, finally, to my initial idea for the following second alternative: sort the comments on a post according to the order relation

(u_1,d_1) < (u_2,d_2) \iff P(X_1 > X_2) < \frac{1}{2}

where

X_k \sim Beta(u_k+1,d_k+1)

More intuitively, we are simply ranking one comment higher than another if it is more likely than not to appear higher using Neufeld’s randomized ranking.

Note one interesting property of this approach that distinguishes it from all of the other methods mentioned so far: it does not involve assigning a real-valued “score” to each individual comment (and subsequently sorting by that score).  This is certainly possible in principle (see below), but as currently specified we can only compare two comments by performing a calculation involving parameters of both in a complex way.

Open Questions

Unfortunately, there are quite a few holes to be patched up with this method, and I am hoping that someone can shed some light on how to address these.  First, the strict order defined above is not quite a total order, since there are some pairs of distinct comments where one comment’s randomized score is equally likely to be higher or lower than the other.  For example, all of the comments of the form (u,u), with an equal number of upvotes and downvotes, have this problem.  This is probably not a big deal, though, since I think it is possible to arbitrarily order these comments, for example by increasing total number of votes.

But there are other more interesting pairs of incomparable comments.  For example, consider (5,0) and (13,1).  The definition above is insufficient to rank these two… but it turns out that it had better be the case that (13,1) < (5,0), since we can find a third comment that lies between them:

(13,1) < (70,8) < (5,0)

This brings us to the next open question: is this order relation transitive (in other words, is it even a partial order)?  I have been unable to prove this, only verify it computationally among comments with bounded numbers of votes.

The final problem is a more practical one: how efficiently can this order relation be computed?  Evaluating the probability that one beta-distributed random variable exceeds another involves a double integral that “simplifies” to an expression involving factorials and a hypergeometric function of the numbers of upvotes and downvotes.  If you want to experiment, following is Python code using the mpmath library to compute the probability P(X_1 > X_2):

from mpmath import fac, hyp3f2

def prob_greater(u1, d1, u2, d2):
    return (hyp3f2(-d2, u2 + 1, u1 + u2 + 2, u2 + 2, u1 + u2 + d1 + 3, 1) *
            fac(u1 + u2 + 1) / (fac(u1) * fac(u2)) *
            fac(u1 + d1 + 1) * fac(u2 + d2 + 1) /
            ((u2 + 1) * fac(d2) * fac(u1 + u2 + d1 + 2)))

print(prob_greater(5, 0, 13, 1))

John Cook has written a couple of interesting papers on this, in the medical context of evaluating clinical trials.  This one discusses various approximations, and this one presents exact formulas and recurrences for some special cases.  The problem of computing the actual probability seems daunting… but perhaps it is a simpler problem in this case to not actually compute the value, but just determine whether it is greater than 1/2 or not?

In summary, I think these difficulties can be rolled up into the following more abstract statement of the problem: can we impose a “natural,” efficiently computable total order on the set of all beta distributions with positive integer parameters, that looks something like the order relation described above?

 

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A coin puzzle revisited

This is a follow-up to some interesting discussion in the comments on my previous post, involving a coin-flipping probability puzzle, and a comparison of Bayesian and frequentist approaches to “solving” it.  For completeness, here is the original problem:

You have once again been captured by pirates, who threaten to make you walk the plank unless you can correctly predict the outcome of an experiment.  The pirates show you a single gold doubloon, that when flipped has some fixed but unknown probability of coming up heads.  The coin is then flipped 7 times, of which you observe 5 to be heads and 2 to be tails.  At this point, you must now bet your life on whether or not, in two subsequent flips of the coin, both will come up heads.  If you predict correctly, you go free; if not, you walk the plank.  Which outcome would you choose?

A typical puzzle-solver would (rightly) point out that necessary information is missing; we cannot determine the optimal action without knowing how the coin (and thus its bias) was selected.  Instead of providing that information, I stirred the Bayesian vs. frequentist debate by showing how each might reason without that information, and come up with differing conclusions.

One of the reasons that I like this problem is that the “Bayesian vs. frequentist” perspective is a bit of a ruse.  The frequentist in the original post computes the maximum likelihood estimate of the probability of the coin coming up heads… and makes a betting decision based on that estimate.  The Bayesian performs a slightly more complex calculation, involving updating a prior beta distribution using the observed flips, doing some calculus… but then makes a similar “threshold” betting decision based on that calculation.

The key observation is that any deterministic betting strategy whatsoever, whether wearing a frequentist hat, a Bayesian hat, or a clown hat, may be specified as a function

f:\{0, 1, 2, ..., n\} \rightarrow \{0, 1\}

mapping the number of heads observed in n=7 total flips to 1 indicating a bet for two subsequent heads, and 0 indicating a bet against.  Neither the underlying statistical philosophy nor the complexity of implementation of this function matters; all that matters is the output.

Actually, we can simplify things even further if we only consider “monotonic” strategies of the form “bet for two heads if k or more heads are observed, otherwise bet against.”  That is,

f_k(h) = H[h-k]

where H[] is the unit step function.

As mendel points out in the comments on the previous post, the frequentist MLE strategy is equivalent to f_5 (i.e., bet on two heads with “5 or more” observed heads), and the Bayesian strategy is equivalent to f_6 (“6 or more”).  We can compare these strategies– along with the seven other monotonic strategies– by computing the probability of their success, as a function of the unknown probability p of heads for each single coin flip.  That is, the probability of surviving the game with strategy f_k is

\sum_{h=0}^n {n \choose h} p^h (1-p)^{n-h}(f_k(h)(2p^2-1) + 1-p^2)

The following figure shows the results for all nine strategies:

Comparison of monotonic strategies as a function of probability of heads in a single coin flip.  The frequentist MLE strategy is "5 or more," and the Bayesian strategy is "6 or more."

Comparison of monotonic strategies as a function of probability of heads in a single coin flip. The frequentist MLE strategy is “5 or more,” and the Bayesian strategy is “6 or more.”

The MLE strategy (green) and Bayesian strategy (blue) are certainly contenders for the best reasonable approach.  However, neither of these, nor any other single strategy, dominates all others for all possible values of the unknown probability of heads in a single coin flip.  In other words, whether the Bayesian or frequentist has a better chance of survival truly does depend on the information that we are explicitly not given.

 

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A coin puzzle

Are you a Bayesian or a frequentist?  What do these terms mean, and what are the differences between the two?  For me, these questions have never been terribly interesting, despite many attempts at answers given in the literature (see the references below for useful and entertaining examples).

My problem has been that explanations typically focus on the different approaches to expressing uncertainty, as opposed to different approaches to actually making decisions.  That is, in my opinion, Bayesians and frequentists can argue all they want about what “the probability of an event” really means, and how much prior information the other camp has or hasn’t unjustifiably assumed… but when pressed to actually take an action, when money is on the table, everyone becomes a Bayesian.

Or do they?  Following is an interesting puzzle that seems to more clearly distinguish the Bayesian from the frequentist, by forcing them both to put money on the table, so to speak:

Problem: You have once again been captured by bloodthirsty logical pirates, who threaten to make you walk the plank unless you can correctly predict the outcome of an experiment.  The pirates show you a single irregularly-shaped gold doubloon selected from their booty, and tell you that when the coin is flipped, it has some fixed but unknown probability of coming up heads.  The coin is then flipped 7 times, of which you observe 5 to be heads and 2 to be tails.

At this point, you must now bet your life on whether or not, in two subsequent flips of the coin, both will come up heads.  If you predict correctly, you go free; if not, you walk the plank.  Which outcome would you choose?  (The pirates helpfully remind you that, if your choice is not to play, then you will walk the plank anyway.)

I think this is an interesting problem because two different but reasonable approaches yield two different answers.  For example, the maximum likelihood estimate of the unknown probability that a single flip of the coin will come up heads is 5/7 (i.e., the observed fraction of flips that came up heads), and thus the probability that the next two consecutive flips will both come up heads is (5/7)*(5/7)=25/49, or slightly better than 1/2.  So perhaps a frequentist would bet on two heads.

On the other hand, a Bayesian might begin with an assumed prior distribution on the unknown probability for a single coin flip, and update that distribution based on the observation of h=5 heads and t=2 tails.  For example, using a “maximum entropy” uniform prior, the posterior probability for a single flip has a beta distribution with parameters (h+1, t+1), and so the probability of two consecutive heads is

\int_0^1 \frac{x^h (1-x)^t}{B(h+1, t+1)} x^2 dx = \frac{7}{15} < \frac{1}{2}

where B(h+1, t+1) is the beta function.  So perhaps a Bayesian would bet against two heads.

What would you do?

(A couple of comments: first, one might reasonably complain that observing just 7 coin flips is simply too small a sample to make a reasonably informed decision.  However, the dilemma does not go away with a larger sample: suppose instead that you initially observe 17 heads and 7 tails, and are again asked to bet on whether the next two flips will come up heads.  Still larger samples exist that present the same problem.

Second, a Bayesian might question the choice of a uniform prior, suggesting as another reasonable starting point the “non-informative” Jeffreys prior, which in this case is the beta distribution with parameters (1/2, 1/2).  This has a certain cynical appeal to it, since it effectively assumes that the pirates have selected a coin which is likely to be biased toward either heads or tails.  Unfortunately, this also does not resolve the issue.)

References:

1. Jaynes, E. T., Probability Theory: The Logic of Science. Cambridge: Cambridge University Press, 2003 [PDF]

2. Lindley, D. V. and Phillips, L. D., Inference for a Bernoulli Process (A Bayesian View), The American Statistician, 30:3 (August 1976), 112-119 [PDF]

 

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