Posted on October 13, 2019
A recent torts cases from class (Byrne v. Boadle) served as the introduction to a class of problems where negligence could be considered probabilistically. This set of problems mirrors the more wellknown "false positive" problem. Here are two ways to state what is mathematically the same exercise.
"False Positive" problem: A disease afflicts $0.1\%$ of a population. A test exists for the disease, and will show positive $90 \% $ of the time if you have the disease and $1\%$ of the time if you don't (false positive). If you test positive, what is the probability you have the disease?
Negligence problem: Employees of a company act negligently $0.1\%$ of the time they load barrels next to a window. If they act negligently, an accident occurs $90\%$ of the time. When they are not negligent, an accident occurs $1\%$ of the time. If an accident occurs, what is the probability it was due to negligence?
Both have the same solution (we'll use the negligence nomenclature). Define $P_1$, $P_2$, and $P_3$ per the table below.
Probability  Disease Problem  Negligence Problem 

$P_1=1\%$  False positive  Accident, not acting negligently 
$P_2=90\%$  True positive  Accident, acting negligently 
$P_3=0.1\%$  Prevalence of disease  Rate of negligence 
Let's say an accident occurred. The probability an accident occurs through negligence is $P_{negligence}=P_2\cdot P_3=0.00999$. The probability an accident occurs without negligence is $P_{nonegligence}=P_1\cdot(1P_3) = 0.0009$. Therefore the probability it was due to negligence is: $$ \frac{P_{negligence}}{P_{negligence}+P_{nonegligence}} = 8.3\%.$$ So $8.3\%$ of the time it is due to negligence (or isomorphically, there is an $8.3 \% $ chance you have the disease despite a positive test).
So should this example drive our intuition? Unless this pattern of low rates of negligence holds for a majority of inputs, the answer is a resounding no.
Let's change the problem slightly by making the probability of accident from nonnegligence $P_1=0.1\%$ and the rate of negligence $P_3=1\%$. Intuition would tell you negligence is more likely to have been the cause, but by how much? If you work through the same math, the probability the accident was due to negligence now becomes $90.1\%$. Clearly, this problem is susceptible to manipulation through small changes in the inputs.
We were able to "flip" the outcome by flipping $P_1$ and $P_3$, so let's look at the probability of negligence as we vary those two variables. Plots can be generated with the following code.
import matplotlib.pyplot as plt steps = 201 array2d = [range(steps) for _ in range(steps)] interval = .0005 for i in range(steps): for j in range(steps): p1 = interval*(i+1) # probability of drop if properly handled p2 = .20 # probability of drop if improperly handled p3 = 1interval*(j+1) # probability of handled properly pa = p1*p3 # handled properly but dropped pb = (1p3)*p2 # handled improperly and dropped p_proper = pa/(pa+pb) array2d[j][i] = p_proper plt.contourf(array2d, 9, vmin=0.0, vmax=1.0, origin='lower', extent=[interval * 100, interval * steps * 100, interval * 100, interval * steps * 100], cmap='seismic') cbar=plt.colorbar() plt.xlabel('$P_1$, probability of accident, not acting negligently (%)') plt.ylabel('$P_3$, rate of negligence (%)') plt.title('Probability that Accident Happened through Negligence, $P_2 = 20\%$') cbar.set_ticks([0, .2, .4, .6, .8, 1]) cbar.set_ticklabels([0, .2, .4, .6, .8, 1]) plt.show()
This trend makes sense in hindsight, but one should hardly be expected to carry this intuition around with them, naturally or otherwise.
Now let's work with $P_2$. Compare the above plots where $P_2=0.9$ to the plots below where $P_2=0.2, 0.5$. We would expect decreasing $P_2$, probability of an accident given negligence, to decrease the probability of negligence given an accident, which the plots below confirm.
With these results, we can develop a general intuition:
That said, a more important takeaway here is simpler: don't let a limited data set drive your intuition. Look at the sensitivity of the variables involved, and look at a wide swath of results before jumping to conclusions. And unless you have a clear reason to, don't arbitrarily assume probabilities. You probably don't know the difference between $0.1 \%$ and $1 \%$ without a good data set to aid you.
Posted on July 14, 2019
Below is a solution to the Classic problem posed on July 12's edition of the Riddler on fivethirtyeight.com, "What’s The Optimal Way To Play HORSE?".
We will first introduce the game, then walk through a simulation, then the analytic solutions for the simplified game of $\mathrm{H}$ then the full game of $\mathrm{HORSE}$. We'll show that the optimal strategy regardless of the number of letters in the game is to always shoot the highest percentage shot (short of a 100% shot).
Two strategies immediately come mind. First, take a shot that maximizes the probability of you making a basket and him missing.
The second, and provably better strategy is to take advantage of the one thing you have (you are equally good shooters). You get to go first, so take full advantage of that turn, and maximize the probability of scoring a point before you miss.
Intuitive strategies are meaningless unless testing, so let's see what works best.
The code below is a bare bones implementation of a simulation. Player A and Player B can be assigned two different strategies, a_prob and b_prob, and the length of the game can be changed.
import random def make_basket(prob): if prob > random.random(): return True else: return False def run_simulation(a_prob, b_prob, horse): a_score = 0 b_score = 0 a_turn = True while True: if a_turn: if make_basket(a_prob): if make_basket(a_prob): pass # both make else: a_score += 1 # a make, b miss else: a_turn = False # a misses, it's b's turn else: if make_basket(b_prob): if make_basket(b_prob): pass # both make else: b_score += 1 # b make, a miss else: a_turn = True # b misses, it's a's turn if a_score >= horse or b_score >= horse: break return a_score, b_score
The plots clearly show the optimal strategy regardless of the number of letters in the game is to always shoot the highest percentage shot (short of a 100% shot).
Let's consider the probability of Player A winning a point before losing their turn. Let $a$ be the probability of Player A making a basket. Then they could make a point in two shots with $p_2(a)=a(1a)$, in four shots with $p_4(a)=a^3(1a)$, or in general, in $2n$ shots with $p_{2n}(a)=a^{2n1}(1a)$. Therefore, the probability $p_A$ of winning a letter before losing their turn is $$p_A=\sum{p_i(a)}=a(1a)+a^3(1a)+... = \frac{a(1a)}{1a^2}=\frac{a}{a+1}.$$
The maximum value of this function on $[0,1]$ occurs at $a=1$, with $p_A(a)=0.5$. We wouldn't want to shoot a basket neither player ever misses, a trivial solution and never ending game, but we want to pick the shot closest to that.
We'll expand from winning a point without losing a turn, to winning the game of $H$, another geometric sequence. Let the player's probabilities of winning a point without losing a turn be $p_A$ and $p_B$ respectively. Player A could win on the first turn with probability $p_{AH_{1}}=p_A$, on the third turn with probability $p_{AH_{3}}=(1p_A)(1p_B)p_A$, and in $2n+1$ turns with probability $$p_{AH_{ 2n+1}}=(1p_A)^{n}(1p_B)^np_A$$
Similarly, the probability that Player A wins $H$, $p_{AH}$, is this sum $$p_{AH}=\sum{p_{AH_{2n+1}}}= \frac{p_A}{1r}.$$ Where $p_A=\displaystyle\frac{a}{a+1}$, $p_B=\displaystyle\frac{b}{b+1}$, and $r=(1p_A)(1p_B)$.
As a sanity check, one could sum the terms of the geometric series for the probability of Player B winning, $\displaystyle\frac{(1p_A)p_B}{1r}$, add it to the probability of A winning, and find their sum is one, as it should be.
Expanding to multiletter games is no longer a game of compounded geometric series, but instead of counting combinations. You can start by thinking of winning $\mathrm{HORSE}$ the same way as winning a fivegame series. Unfortunately, it gets a little more complicated; your probability of winning a letter is based off whether or not you won the previous letter.
Let $A$ denote a Player A point win, and $B$ denote a Player B point win. Consider the game $AAAAA$. On each round, Player A starts, so they have the advantage; their probability of winning each point is $P_{AH}$.
Now consider, $ABAAAA$. The Player A win on point 3 following the Player B win, has the lower likelihood $(1P_{BH})$ of occurring since Player B had the advantage of starting with the ball after winning point 2.
What makes the counting challenging is that the sequences $AABBBAAA$ and $ABABABAA$, for example, do not have the same probabilities for Player A winning. The first only has two "switches" from who is in control (shooting first), where the second has six switches, changing the probabilities in different ways. There is no way around working through each of the individual $1+5+15+35+70=126$ cases for Player A winning in five through nine games respectively.
The code below summarizes this counting.
probs = [.05*i for i in range(1,20)] for a in probs: for b in probs: pa = a / (1 + a) pb = b / (1 + b) r = (1  pa) * (1  pb) pAH = pa / (1  r) # probability a wins going first pBH = pb / (1  r) # probability b wins going first pAB = (1  pAH, pAH) cum_p = 0 # cumulative probability for s in ['1111', '01111', '001111', '0001111', '00001111']: tups = list(perm_unique(s)) for tup in tups: seq = [int(i) for i in list(tup)+['1']] p = pAB[seq[0]] for i in range(1, len(seq)): if seq[i1] == 0 and seq[i] == 0: p *= pBH elif seq[i1] == 1 and seq[i] == 0: p *= (1  pAH) elif seq[i1] == 0 and seq[i] == 1: p *= (1  pBH) elif seq[i1] == 1 and seq[i] == 1: p *= pAH cum_p += p print cum_p, print ""
My code makes use of some StackOverflow code to make unique permutations.
Posted on June 2, 2019
Updated on June 4, 2019
I made a few changes from the original post. I misread allowable order of operations (the equivalent of "parentheses are ok", which drastically increases solvable numbers). I decided to leave it in as an interesting twist on the problem.
I also left out the case of six small numbers (another misread), which I have since added.
Below are solutions to the Classic and Express problems posed on May 31's edition of the Riddler on fivethirtyeight.com, "Can You Win The Lotería?".
First, a brief solution to the Riddler Express. The question posed is: for $N=54$ total images, and $n=16$ images per player card, how often will the game end with one empty card of images?
Two things must be considered: how often do the two players have unique cards (a necessary condition), and if that is met, how often will an entire card be filled before any images on the second card are filled?
We must first consider the necessary condition that the players have unique images on their cards. It helps to think about image "selection" successively. The first player can pick their entire card without concern. Then the second player "picks" their images for the second card one by one. For $N=54$ and $n=16$, there are $54$ options, $5416=38$ of which are favorable, so probability the first image for the second card is not on the first card is $\frac{38}{54}$. The second image cannot be the first (and will not be the previous image), so there are now $53$ options, $37$ of which are favorable, yielding a probability of $\frac{37}{53}$. Continuing this logic, the probability the cards are unique is
$$ \frac{38}{54}\cdot\frac{37}{53}\cdot\frac{36}{52}\cdot...\cdot\frac{23}{39} \approx 0.001054.$$In general, the probability can be written as
$$ \displaystyle\frac{\binom{Nn}{n}}{\binom{N}{n}}.$$ Second, assuming the two cards are unique, how often will you win before the other player crosses a single image off their card? The extraneous cards are meaningless, akin to burn cards. WLOG, if you have images $1$$16$, your opponent has $17$$32$, then the favorable outcomes are all $16!$ arrangements of your cards, followed by all $16!$ arrangements of your opponents outcomes. The total number of arrangements is $32!$, the arrangements of all cards. The probability is then $$ \displaystyle\frac{16!\cdot 16!}{32!} \approx 1.664\times 10^{9}. $$In general, the probability can be written as
$$ \displaystyle\frac{(n!)^2}{(2n)!}. $$Note, there is no $N$ term in this expression. The simulation in the code below considers all $N$ cards to verify its invariance.
from ncr import ncr #n choose r import numpy as np def you_win(photo_order, n): """WLOG you have photos 0 to n1, opponent has n to 2n1 Check if all of your photos appear before any of your opponents""" n_count = 0 for i in photo_order: if i < n: n_count += 1 if n <= i < 2*n: return False if n_count == n: return True def main(): N = 54 # Total number of photos n = 16 # Number of photos per player num_sim = 10000 # Number of simulations # Simulate Unique count = 0 for _ in xrange(num_sim): list_1 = np.random.choice(N, n, replace=False) # Your photos list_2 = np.random.choice(N, n, replace=False) # Opponent's photos if set(list_1) & set(list_2): # Check for intersection count += 1 print "probability unique:", \ ncr(Nn, n) / float(ncr(N, n)), \ 1.0  count / float(num_sim) # Simulate You Win count = 0 for _ in xrange(num_sim): photo_order = np.arange(N) np.random.shuffle(photo_order) if you_win(photo_order, n): count += 1 print "probability you win:", 1./ncr(2 * n, n), count / float(num_sim) if __name__ == "__main__": main()
The remainder of this post will focus on the countdown problem, namely a brute force approach to counting all possible outcomes in a reasonable amount of time, follwed by a discussion of results. See code here.
The most fruitful combination is shown to be $5$, $6$, $8$, $9$, $75$, $100$, which yields $691$ different threedigit numbers. The least fruitful is shown to be $1$, $1$, $2$, $2$, $3$, $25$, which yields only $20$ different threedigit numbers.
Statistics for the various strategies (selecting 1 through 4 large numbers) are summarized, and it is shown that selecting two large numbers is the most advantageous, followed closely by selecting three large numbers.
Finally, a histogram of the threedigit numbers frequency among the $10,603$ combinations shows what is intuitive, at least in hindsight: smaller threedigit numbers, and multiples of $25$, $50$, and especially $100$ are the most common.
First we must consider all possible combinations of the small and large numbers. We will denote large numbers with $\Lone$, $\Ltwo$, $\Lthree$, and $\Lfour$ and small numbers with $\a$, $\b$, $\c$, $\d$ and $\e$. A player can selected between zero and four large numbers, and the small numbers may repeat once. There are $11$ possible scenarios, divided by double lines in the table below by strategy.
Pattern  Number of Combinations 

$\a\b\c\d\e\mathrm{f}$  $\binom{10}{6}$ 
$\Lone\a\a\b\b\c$  $\binom{4}{1}\cdot\binom{10}{2}\cdot\binom{8}{1}$ 
$\Lone\a\a\b\c\d$  $\binom{4}{1}\cdot\binom{10}{1}\cdot\binom{9}{3}$ 
$\Lone\a\b\c\d\e$  $\binom{4}{1}\cdot\binom{10}{5}$ 
$\Lone\Ltwo\a\a\b\b$  $\binom{4}{2}\cdot\binom{10}{2}$ 
$\Lone\Ltwo\a\a\b\c$  $\binom{4}{2}\cdot\binom{10}{2}\cdot\binom{8}{1}$ 
$\Lone\Ltwo\a\b\c\d$  $\binom{4}{2}\cdot\binom{10}{4}$ 
$\Lone\Ltwo\Lthree\a\a\b$  $\binom{4}{3}\cdot \binom{10}{1}\cdot \binom{9}{1}$ 
$\Lone\Ltwo\Lthree\a\b\c$  $\binom{4}{3}\cdot\binom{10}{3}$ 
$\Lone\Ltwo\Lthree\Lfour\a\a$  $\binom{4}{4}\cdot\binom{10}{1}$ 
$\Lone\Ltwo\Lthree\Lfour\a\b$  $\binom{4}{4}\cdot\binom{10}{2}$ 
These can be generated individually with the help of the Python itertools library. As a sample, consider $\Lone\Ltwo\a\a\b\c$.
import itertools L_array = [25., 50., 75., 100.] s_array = [float(i) for i in range(1, 11)] for s in itertools.combinations(s_array, 3): for L in itertools.combinations(L_array, 2): for s0 in s: loop_array = itertools.permutations(list(L) + list(s) + [s0])
This is probably the most complicated of the $11$ examples. Note the use of itertools.combinations vs itertools.permutations, the latter provides all arrangements and the former does not, which helps us reduce over counting. Over counting is not completely avoided however. The itertools library does not know that our two $\a$ values are repeats so we are over counting by a factor of $2$. This is not ideal, but the storage mechanism used (based on Python sets and dicts) eliminates error duplicate counting, so the only sacrifice is run time.
We will now take each permutation generated and loop through it's components left to right (i.e. consider the first number, then the first two numbers, then the first three, etc.). For example, given the permutation $100$, $50$, $10$, $2$, $7$, $4$, the first number "generated" is $100$. We then need to check all operations with the first two terms, $100+50$, $10050$, $100\times 50$, and $100\div 50$ which yields $150$, another hit, but three numbers, $50$, $5000$, and $2$ that are not hits. Those last three numbers are not hits, but cannot be ignored since additional terms may turn them into hits; for example $100\times 50\div 10\div 2=250$ is another hit.
Challenges also arise with the implementation of addition, subtraction, multiplication and division operations in this lefttoright paradigm. We have $5$ spaces in between the $6$ numbers for each operation, for $4^5=1024$ possible permutations of the operators. However, we must only consider those where left to right operation satisfies order of operations. For example, $+$ $$ $+$ $$ $$, $\div$ $\times$ $\div$ $\times$ $\div$, and $\times$ $\div$ $$ $$ $+$ are all acceptable but $+$ $\times$ $$ $$ $$ because placing addition before multiplication is the equivalent of adding parentheses. Using the above example, $100+50+10\times2$
Acceptable orders of operation can be found with the following code. This finds all lists of five operators where all multiplication and division occur before all addition and subtraction.
operator_list = [] # Indices for operators, 0,1,2,3 is *,/,+, for i4 in xrange(4 ** 5): cl = [i4 // 256 % 4, i4 // 64 % 4, i4 // 16 % 4, i4 // 4 % 4, i4 % 4] if good_order(cl): operator_list.append(cl) print len(operator_list) # 192
Proving that this covers every possible case is not simple, but it makes intuitive sense that when all permutations of the 6 numbers are considered, the $192$ operators considered are exhaustive.
We now have everything we need to search for every combination. Full implementation can be found here.
The combination $5$, $6$, $8$, $9$, $75$, $100$, yields $691$ different possible threedigit numbers, the most for any combination. The combination $1$, $1$, $2$, $2$, $3$, $25$, yields only $20$ possible combinations, the fewest. The top and bottom ten combinations are summarized in the tables below.
Number  Number of threedigit numbers created 

$5$, $6$, $8$, $9$, $75$, $100$  $691$ 
$5$, $7$, $8$, $9$, $75$, $100$  $652$ 
$6$, $7$, $9$, $10$, $75$, $100$  $650$ 
$6$, $8$, $9$, $10$, $75$, $100$  $648$ 
$3$, $8$, $9$, $10$, $75$, $100$  $648$ 
$2$, $5$, $6$, $9$, $75$, $100$  $643$ 
$5$, $6$, $8$, $9$, $50$, $100$  $641$ 
$5$, $6$, $7$, $9$, $75$, $100$  $640$ 
$2$, $5$, $8$, $9$, $75$, $100$  $638$ 
$4$, $7$, $9$, $10$, $75$, $100$  $636$ 
Number  Number of threedigit numbers created 

$1$, $1$, $2$, $2$, $3$, $25$  $20$ 
$1$, $1$, $2$, $2$, $4$, $25$  $21$ 
$1$, $1$, $2$, $3$, $3$, $25$  $25$ 
$1$, $1$, $3$, $4$, $4$, $25$  $34$ 
$1$, $1$, $2$, $2$, $6$, $25$  $35$ 
$1$, $1$, $2$, $2$, $9$, $25$  $35$ 
$1$, $1$, $2$, $2$, $5$, $25$  $35$ 
$1$, $1$, $2$, $2$, $7$, $25$  $35$ 
$1$, $1$, $2$, $2$, $4$, $50$  $36$ 
$1$, $1$, $2$, $4$, $4$, $25$  $36$ 
The table below shows that the best choice to pick two large numbers and four small numbers, but the strategy of picking three large is a close second. An outcome is defined as a threedigit number generated by a particular combination.
Pattern  Arrangements  Total Outcomes  Average # of Outcomes 

$\a\b\c\d\e\mathrm{f}$  $210$  $46,663$  $222.2$ 
$\Lone\a\a\b\b\c$  $1440$  $257,167$  $178.6$ 
$\Lone\a\a\b\c\d$  $3360$  $865,253$  $257.5$ 
$\Lone\a\b\c\d\e$  $1008$  $349,921$  $347.1$ 
One large  $253.5$  
$\Lone\Ltwo\a\a\b\b$  $270$  $58,221$  $215.6$ 
$\Lone\Ltwo\a\a\b\c$  $2160$  $694,966$  $321.7$ 
$\Lone\Ltwo\a\b\c\d$  $1260$  $574,520$  $456.0$ 
Two large  $\pmb{359.8}$  
$\Lone\Ltwo\Lthree\a\a\b$  $360$  $100,324$  $278.7$ 
$\Lone\Ltwo\Lthree\a\b\c$  $480$  $199,349$  $415.3$ 
Three large  $356.8$  
$\Lone\Ltwo\Lthree\Lfour\a\a$  $10$  $1,541$  $154.1$ 
$\Lone\Ltwo\Lthree\Lfour\a\b$  $45$  $10,619$  $236.0$ 
Four large  $221.1$ 
The histogram shows what we would intuitively expect, smaller numbers can be computed more often. There are also peaks at multiples of $25$, $50$, and especially $100$, another intuitive byproduct.
Posted on May 26, 2019
This is a solution to the problem posed on May 24's edition of the Riddler Classic on fivethirtyeight.com, "One Small Step For Man, One Giant Coin Flip For Mankind".
We will show that there are $12$ nontrivial solutions to the $3$astronaut problem with $4$ coins and $2$ nontrivial solutions to the $5$astronaut problem with $6$ coins.
The problem statement walks through the impossibility of using two coins, with $2^2=4$ outcomes, to determine a winner for three astronauts. The same can be said for three flips, which we will briefly discuss.
The key to attacking this problem for an $n$astronaut scenario is equally distributing outcomes amongst the first $n1$ astronauts. For the $3$astronaut, $3$flip case this translates to the following: among the $1$, $3$, $3$, and $1$ outcomes ($0$ heads, $1$ head, $2$ heads, and $3$ heads respectively), the first and second astronauts should have the same number each of $0$head, $1$head, $2$head, and $3$head outcomes. One possible assignment that meets this parameter is for the first and second astronauts to each have the distributions $0$, $1$, $1$, $0$, leaving the third astronaut with the distribution $1$, $1$, $1$, $1$. This is summarized in the table, below.
3 Heads 0 Tails  2 Heads 1 Tail  1 Head 2 Tails  0 Heads 3 Tails 


Astro #1  0  1  1  0 
Astro #2  0  1  1  0 
Astro #3  1  1  1  1 
Total  1  3  3  1 
This distribution, however, implies that the probability of flipping three heads and zero heads is both zero, a contradiction. All distributions with $3$ astronauts and $3$ coins yield similar contradictions.
There is no rigorous proof here to the claim that the first $n1$ astronauts must have the same distribution of outcomes, but the likelihood of them having different outcomes that independently add up to a very specific fraction is highly unlikely. Further work below will show it's analogous to polynomials with different integer coefficients having an identical real root between $0$ and $1$.
We broaden our horizons to a $4$coin solution, which has the $2^4=16$ outcomes of $1$, $4$, $6$, $4$, $1$, for $4$, $3$, $2$, $1$, and $0$ heads respectively. For three astronauts, we have significantly more ways to distribute the coin flips. For example, the first two astronauts could each be given the outcome distribution $0$, $2$, $3$, $2$, $0$, leaving $1$, $0$, $0$, $0$, $1$ for the third astronaut. This is summarized in the table, below.
4 Heads 0 Tails  3 Heads 1 Tail  2 Heads 2 Tails  1 Head 3 Tails  0 Heads 4 Tails 


Astro #1  0  2  3  2  0 
Astro #2  0  2  3  2  0 
Astro #3  1  0  0  0  1 
Total  1  4  6  4  1 
We now have to see if a solution exists with this distribution. Let $h$ be the probability of flipping a heads, and $t=1h$ be the probability of flipping a tails. We generate the probabilities of each flipping outcome: for $k$ heads in $n$ flips, a specific outcome (ignoring rearrangments) is $h^k t^{nk} = h^k (1h)^{nk}$. Using the current example from the above table, our probability polynomial for astronaut 1 (and equivalently astronaut 2) becomes
\begin{eqnarray} 2 h^3 (1h) + 3 h^2 (1h)^2 + 2 h (1h)^3 &=& \frac{1}{3} \\ h^4 + 2h^3 3 h^2 + 2h &=& \frac{1}{3} \end{eqnarray} $$ h = \frac{1}{6} \left( 3 \pm \sqrt{3 \left(4\sqrt{6} 9\right)} \right) \approx 0.242, 0.758 $$Therefore, we can state that one solution is with $h=.242$, the first astronaut getting assigned $2$, $3$, and $2$ each of the $3$head, $2$head, and $1$head outcomes, and the third astronaut being assigned the $4$head and $0$head outcomes.
As an aside, it makes sense to have symmetric solutions here (that is, they add up to $1$), since we picked symmetric coefficients $2$, $3$, and $2$. We expect an asymmetric pick like $1$, $3$, $2$ to not yield symmetric solutions.
Let's generalize further to find all $3$astronaut, $4$coin solutions. There are $4$ outcomes with $3$ heads, so astronauts one and two could each get $0$, $1$, or $2$. Similarly they could get $0$, $1$, $2$, or $3$ and $0$, $1$, or $2$ of the $2$head outcomes and $1$head outcomes respectively. We need to test all of these scenarios, which we accomplish using the following code.
import numpy as np import operator as op from functools import reduce from collections import namedtuple Solution = namedtuple('Solution', 'coefficients roots') def ncr(n, r): """Implementation of n choose r = n!/(r!*(nr)!)""" r = min(r, nr) numer = reduce(op.mul, range(n, nr, 1), 1) denom = reduce(op.mul, range(1, r+1), 1) return numer / denom def create_polys(num_coins): """Creates polynomial expansions in list form for n1 to 1 heads""" polys = [] for i in range(1, num_coins): poly = [] neg = (1)**i for j in range(0, i+1): poly.append(ncr(i, j) * neg) neg *= 1 for k in range(i+1, num_coins): poly.append(0) polys.append(poly) return polys def create_coeff_limits(num_coins, num_astronauts): """Determines the maximum allocation of head/tail outcomes based off the number of coins and astronauts""" coeffs = [] for i in range(1, num_coins): coeffs.append(ncr(num_coins, i) // (num_astronauts  1) + 1) return coeffs def is_valid_solution(root): """Checks if root is valid. Must be real and between 0 and 1.""" return np.isreal(root) and 0 < root < 1 def solve_poly(coeffs, polys, prob): """Generates and solves polynomials using np.roots coeffs: list of "coefficients", i.e. multipliers for the number of out head/tail outcomes polys: generated with create_polys prob: the negated probability 1/num_astronauts returns all valid solutions a list on named tuples """ sum_poly = [] valid_solutions = [] for degree in range(len(polys[0])): term = 0 for j in range(len(polys)): # for each polynomial term += coeffs[j] * polys[j][degree] sum_poly.append(term) sum_poly.append(prob) ans = np.roots(sum_poly) for root in ans: if is_valid_solution(root): valid_solutions.append(Solution(coeffs, root)) return valid_solutions def main(): num_astronauts = 3 num_coins = 6 prob = 1.0/num_astronauts polys = create_polys(num_coins) coeff_limits = create_coeff_limits(num_coins, num_astronauts) solutions = [] # Loop through all possible coefficient combinations coeff_dividers = [] for i in range(1, len(coeff_limits)): coeff_dividers.append(reduce(op.mul, coeff_limits[i:], 1)) # This helps us avoid recursion and nested for loops for i in range(reduce(op.mul, coeff_limits, 1)): coeffs = [] for j in range(len(coeff_dividers)): coeffs.append((i // coeff_dividers[j]) % coeff_limits[j]) coeffs.append(i % coeff_limits[1]) solutions += solve_poly(coeffs, polys, prob) for solution in solutions: print solution.coefficients, np.real(solution.roots) #strip 0.j print "Number of solutions:", len(solutions) if __name__ == '__main__': main()
This code, with num_astronauts=3 and num_coins=4 yields the following 12 solutions, where the five listed values correspond to distributions with $4$, $3$, $2$, $1$, and $0$ heads respectively.
$h$  Astronauts 1 & 2  Astronaut 3 

0.4505  0, 0, 3, 2, 0  1, 4, 0, 0, 1 
0.2861  0, 0, 3, 2, 0  1, 4, 0, 0, 1 
0.6051  0, 1, 3, 2, 0  1, 2, 0, 0, 1 
0.2578  0, 1, 3, 2, 0  1, 2, 0, 0, 1 
0.6967  0, 2, 2, 2, 0  1, 0, 2, 0, 1 
0.3033  0, 2, 2, 2, 0  1, 0, 2, 0, 1 
0.7139  0, 2, 3, 0, 0  1, 0, 0, 4, 1 
0.5495  0, 2, 3, 0, 0  1, 0, 0, 4, 1 
0.7422  0, 2, 3, 1, 0  1, 0, 0, 2, 1 
0.3949  0, 2, 3, 1, 0  1, 0, 0, 2, 1 
0.7579  0, 2, 3, 2, 0  1, 0, 0, 0, 1 
0.2421  0, 2, 3, 2, 0  1, 0, 0, 0, 1 
The problem is easily generalized with the above code to higher numbers of astronauts. There are two solutions for $5$ astronauts and $6$ coins, where the seven listed values correspond to distributions with $6$, $5$, $4$, $3$, $2$, $1$, and $0$ heads respectively.
$h$  Astronauts 1, 2, 3, & 4  Astronaut 5 

0.4326  0, 1, 3, 5, 3, 1, 0  1, 2, 3, 0, 3, 2, 1 
0.5674  0, 1, 3, 5, 3, 1, 0  1, 2, 3, 0, 3, 2, 1 
$h$  Astronauts 1, 2, 3, & 4  Astronaut 5 

0.4662  0, 0, 3, 8, 8, 5, 1, 0  1, 7, 9, 3, 3, 1, 3, 1 
0.3636  0, 0, 3, 8, 8, 5, 1, 0  1, 7, 9, 3, 3, 1, 3, 1 
0.4536  0, 0, 4, 7, 8, 5, 1, 0  1, 7, 5, 7, 3, 1, 3, 1 
0.3863  0, 0, 4, 7, 8, 5, 1, 0  1, 7, 5, 7, 3, 1, 3, 1 
0.5215  0, 0, 4, 8, 8, 5, 1, 0  1, 7, 5, 3, 3, 1, 3, 1 
0.3468  0, 0, 4, 8, 8, 5, 1, 0  1, 7, 5, 3, 3, 1, 3, 1 
0.527  0, 0, 5, 7, 8, 5, 1, 0  1, 7, 1, 7, 3, 1, 3, 1 
0.359  0, 0, 5, 7, 8, 5, 1, 0  1, 7, 1, 7, 3, 1, 3, 1 
0.5343  0, 0, 5, 8, 7, 5, 1, 0  1, 7, 1, 3, 7, 1, 3, 1 
0.3845  0, 0, 5, 8, 7, 5, 1, 0  1, 7, 1, 3, 7, 1, 3, 1 
0.543  0, 0, 5, 8, 8, 4, 1, 0  1, 7, 1, 3, 3, 5, 3, 1 
0.4207  0, 0, 5, 8, 8, 4, 1, 0  1, 7, 1, 3, 3, 5, 3, 1 
0.5513  0, 0, 5, 8, 8, 5, 0, 0  1, 7, 1, 3, 3, 1, 7, 1 
0.4487  0, 0, 5, 8, 8, 5, 0, 0  1, 7, 1, 3, 3, 1, 7, 1 
0.5742  0, 0, 5, 8, 8, 5, 1, 0  1, 7, 1, 3, 3, 1, 3, 1 
0.3363  0, 0, 5, 8, 8, 5, 1, 0  1, 7, 1, 3, 3, 1, 3, 1 
0.4564  0, 1, 2, 8, 8, 5, 1, 0  1, 3, 13, 3, 3, 1, 3, 1 
0.3758  0, 1, 2, 8, 8, 5, 1, 0  1, 3, 13, 3, 3, 1, 3, 1 
0.5278  0, 1, 3, 8, 8, 5, 1, 0  1, 3, 9, 3, 3, 1, 3, 1 
0.3532  0, 1, 3, 8, 8, 5, 1, 0  1, 3, 9, 3, 3, 1, 3, 1 
0.5376  0, 1, 4, 7, 8, 5, 1, 0  1, 3, 5, 7, 3, 1, 3, 1 
0.3673  0, 1, 4, 7, 8, 5, 1, 0  1, 3, 5, 7, 3, 1, 3, 1 
0.5519  0, 1, 4, 8, 7, 5, 1, 0  1, 3, 5, 3, 7, 1, 3, 1 
0.396  0, 1, 4, 8, 7, 5, 1, 0  1, 3, 5, 3, 7, 1, 3, 1 
0.5674  0, 1, 4, 8, 8, 4, 1, 0  1, 3, 5, 3, 3, 5, 3, 1 
0.4326  0, 1, 4, 8, 8, 4, 1, 0  1, 3, 5, 3, 3, 5, 3, 1 
0.5793  0, 1, 4, 8, 8, 5, 0, 0  1, 3, 5, 3, 3, 1, 7, 1 
0.457  0, 1, 4, 8, 8, 5, 0, 0  1, 3, 5, 3, 3, 1, 7, 1 
0.5995  0, 1, 4, 8, 8, 5, 1, 0  1, 3, 5, 3, 3, 1, 3, 1 
0.3409  0, 1, 4, 8, 8, 5, 1, 0  1, 3, 5, 3, 3, 1, 3, 1 
0.5566  0, 1, 5, 6, 8, 5, 1, 0  1, 3, 1, 11, 3, 1, 3, 1 
0.3849  0, 1, 5, 6, 8, 5, 1, 0  1, 3, 1, 11, 3, 1, 3, 1 
0.5831  0, 1, 5, 7, 7, 5, 1, 0  1, 3, 1, 7, 7, 1, 3, 1 
0.4169  0, 1, 5, 7, 7, 5, 1, 0  1, 3, 1, 7, 7, 1, 3, 1 
0.604  0, 1, 5, 7, 8, 4, 1, 0  1, 3, 1, 7, 3, 5, 3, 1 
0.4481  0, 1, 5, 7, 8, 4, 1, 0  1, 3, 1, 7, 3, 5, 3, 1 
0.6155  0, 1, 5, 7, 8, 5, 0, 0  1, 3, 1, 7, 3, 1, 7, 1 
0.4657  0, 1, 5, 7, 8, 5, 0, 0  1, 3, 1, 7, 3, 1, 7, 1 
0.6287  0, 1, 5, 7, 8, 5, 1, 0  1, 3, 1, 7, 3, 1, 3, 1 
0.351  0, 1, 5, 7, 8, 5, 1, 0  1, 3, 1, 7, 3, 1, 3, 1 
0.6151  0, 1, 5, 8, 6, 5, 1, 0  1, 3, 1, 3, 11, 1, 3, 1 
0.4434  0, 1, 5, 8, 6, 5, 1, 0  1, 3, 1, 3, 11, 1, 3, 1 
0.6137  0, 1, 5, 8, 7, 4, 0, 0  1, 3, 1, 3, 7, 5, 7, 1 
0.5464  0, 1, 5, 8, 7, 4, 0, 0  1, 3, 1, 3, 7, 5, 7, 1 
0.6327  0, 1, 5, 8, 7, 4, 1, 0  1, 3, 1, 3, 7, 5, 3, 1 
0.4624  0, 1, 5, 8, 7, 4, 1, 0  1, 3, 1, 3, 7, 5, 3, 1 
0.641  0, 1, 5, 8, 7, 5, 0, 0  1, 3, 1, 3, 7, 1, 7, 1 
0.473  0, 1, 5, 8, 7, 5, 0, 0  1, 3, 1, 3, 7, 1, 7, 1 
0.649  0, 1, 5, 8, 7, 5, 1, 0  1, 3, 1, 3, 7, 1, 3, 1 
0.3713  0, 1, 5, 8, 7, 5, 1, 0  1, 3, 1, 3, 7, 1, 3, 1 
0.6242  0, 1, 5, 8, 8, 2, 1, 0  1, 3, 1, 3, 3, 13, 3, 1 
0.5436  0, 1, 5, 8, 8, 2, 1, 0  1, 3, 1, 3, 3, 13, 3, 1 
0.6364  0, 1, 5, 8, 8, 3, 0, 0  1, 3, 1, 3, 3, 9, 7, 1 
0.5338  0, 1, 5, 8, 8, 3, 0, 0  1, 3, 1, 3, 3, 9, 7, 1 
0.6468  0, 1, 5, 8, 8, 3, 1, 0  1, 3, 1, 3, 3, 9, 3, 1 
0.4722  0, 1, 5, 8, 8, 3, 1, 0  1, 3, 1, 3, 3, 9, 3, 1 
0.6532  0, 1, 5, 8, 8, 4, 0, 0  1, 3, 1, 3, 3, 5, 7, 1 
0.4785  0, 1, 5, 8, 8, 4, 0, 0  1, 3, 1, 3, 3, 5, 7, 1 
0.6591  0, 1, 5, 8, 8, 4, 1, 0  1, 3, 1, 3, 3, 5, 3, 1 
0.4005  0, 1, 5, 8, 8, 4, 1, 0  1, 3, 1, 3, 3, 5, 3, 1 
0.6637  0, 1, 5, 8, 8, 5, 0, 0  1, 3, 1, 3, 3, 1, 7, 1 
0.4258  0, 1, 5, 8, 8, 5, 0, 0  1, 3, 1, 3, 3, 1, 7, 1 
0.6678  0, 1, 5, 8, 8, 5, 1, 0  1, 3, 1, 3, 3, 1, 3, 1 
0.3322  0, 1, 5, 8, 8, 5, 1, 0  1, 3, 1, 3, 3, 1, 3, 1 
The code below is an bruteforce solution to the Riddler Express problem about soccer team selection. The smallest solution found was with jersey numbers $1$, $5$, $6$, $8$, $9$, $10$, $12$, $15$, $18$, $20$, and $24$.
import itertools def brute_soccer(team_list): average_score = sum([1./i for i in team_list]) if abs(average_score2.0) < 1e8: print team_list, average_score for i in itertools.combinations(range(1, 25), 11): brute_soccer(i)