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There are $N$ different sized targets, numbered $1, 2,\dots, N$. A blindfolded shooter shoots towards random directions. Because target sizes are different, the hit probabilities of each target are different, say $p_1, p_2,\dots ,p_N$. A bullet hole is left on the target each time a target is hit. The shooter does not know whether a shot hit a target or not.

Note that a shoot could be "void", i.e., $p_1+p_2+\cdots+p_N\le1$. One shot at most hits one target.

The shooter keeps shooting until $X$ out of the $N$, $X<N$, targets have bullet holes. ( each of the $X$ targets is hit at least once). Then the shooter is told to stop.

The question is: at the end of the game, what is the probability that target $k$ has NOT been hit, $1\leq k\leq N$.

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Try $N=2,3,4$, and $X=1,...,N-1$ to see if there is a pattern. – Michael Jan 21 at 15:57
Not sure if this helps, but the answer in non-closed form is $\sum_f \prod_{i=1}^X \frac{p_{f(i)}}{\sum_{j = 1}^N p_j - \sum_{j=1}^{i-1} p_{f(j)}}$, where the sum extends over all injective mappings $f \colon \{1,\dots X\} \to \{1, \dots, N\} - \{k\}$. – David Jan 21 at 16:27
@David could you explain what $p_{f(j)}$ is? – AYang Jan 21 at 16:31
For a given injective mapping $f$, it's $p_i$ for $i = f(j)$. – David Jan 21 at 16:32
that's a tough question! What company was that? Clearly the result is equal to $(1-p_k)^X\sum_{l=0}^\infty (1-p_k)^lP(\text{"you take $X+l$ shots"})$ but that probability seems to complex to calculate.. I would be really interested in reading the solution! – Ant Jan 21 at 16:46
up vote 4 down vote accepted

We may as well suppose the game continues until all targets have been hit (which will happen eventually if all $p_j > 0$; we may as well remove any targets that have $p_j = 0$).

For each subset $S$ of $\{1,\ldots, N\}$, let $p_S = \sum_{s \in S} p_s$ be the probability that a shot hits a member of $S$, and let $a_{i,t}(S)$ be the probability that target $i$ is one of the first $t$ targets in set $S$ to be hit. You want $1 - a_{i,X}(\{1,\ldots,N\})$. Of course $a_{i,t}(S) = 0$ if $i \notin S$, and we also take it to be $0$ if $t = 0$. Otherwise, conditioning on the first target in $S$ to be hit, $$ a_{i,t}(S) = \dfrac{p_i}{p_S} + \sum_{j \in S \backslash \{i\}} \dfrac{p_j}{p_S} a_{i,t-1}(S \backslash \{j\}) $$ Now I claim that $$ a_{i,t}(S) = \sum_{T \subseteq S \backslash \{i\}} c(t,|T|,|S|) \frac{p_i}{p_{T \cup \{i\}}} $$ for some constants $c(t,m,n)$, $0 \le m \le n-1$. I will prove this by induction on $t$. In the case $t=1$ we have $a_{i,1}(S) = p_i/p_S$, so $c(1,m,n) = 1$ if $m = n-1$, $0$ otherwise.

If $t >1$, we have (with $|S|=n$): $$ \eqalign{a_{i,t}(S) &= \dfrac{p_i}{p_S} + \sum_{j \in S \backslash \{i\}} \dfrac{p_j}{p_S} \sum_{T \subseteq S \backslash \{i,j\}} c(t-1, |T|,n-1) \dfrac{p_i}{p_{T \cup \{i\}}}\cr &= \dfrac{p_i}{p_S} + \sum_{m=0}^{n-2} c(t-1,m,n-1) \sum_{T \subseteq S \backslash \{i\}: |T| = m} \sum_{j \in S \backslash (T \cup \{i\})} \dfrac{p_j p_i}{p_S p_{T \cup \{i\}}}\cr &= \dfrac{p_i}{p_S} + \sum_{m=0}^{n-2} c(t-1,m,n-1) \sum_{T \subseteq S \backslash \{i\}: |T| = m} \dfrac{p_{S \backslash (T \cup \{i\})} p_i}{p_S p_{T \cup \{i\}}}\cr &= \dfrac{p_i}{p_S} + \sum_{m=0}^{n-2} c(t-1,m,n-1) \sum_{T \subseteq S \backslash \{i\}: |T| = m} \dfrac{(p_S - p_{T \cup \{i\}}) p_i}{p_S p_{T \cup \{i\}}}\cr &= \dfrac{p_i}{p_S} + \sum_{m=0}^{n-2} c(t-1,m,n-1) \sum_{T \subseteq S \backslash \{i\}: |T| = m} \left(\dfrac{p_i}{p_{T \cup \{i\}}} - \dfrac{p_i}{p_S}\right)\cr &= \sum_{T \subseteq S \backslash \{i\}} c(t,|T|,n) \dfrac{p_i}{p_{T \cup \{i\}}} }$$ where $c(t, m,n) = c(t-1, m,n-1)$ if $m < n-1$ while $$c(t, n-1, n) = 1 - \sum_{m=0}^{n-2} {n-1 \choose m} c(t-1,m,n-1)$$

Hmm: it looks like $$ c(t, m, n) = \cases{1 & for $t=n,m=0$\cr (-1)^{n+m+t} {m-1 \choose {t+m-n}} & $ n-m \le t \le n$\cr 0 & otherwise\cr}$$ There ought to be an inclusion-exclusion proof for this.

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Thank you for your input. I'm studying this solution. Could you help me to understand how you obtain $a_{i,t}(S) = \dfrac{p_i}{p_S} + \sum_{j \in S \backslash \{i\}} \dfrac{p_j}{p_S} a_{i,t-1}(S \backslash \{j\})$ ? – AYang Jan 22 at 21:41
Let $J$ be the first target in set $S$ to be hit. $J = j$ with probability $p_j/p_S$, for each $j \in S$. If $J = i$ then target $i$ is one of the first $t$ targets in $S$ to be hit. If $J = j \ne i$, then for $i$ to be one of the first $t$ targets in $S$ to be hit what you need is that $i$ is one of the next $t-1$ targets in $S \backslash \{j\}$ to be hit, i.e. in effect you start over with $t$ replaced by $t-1$ and $S$ replaced by $S \backslash \{j\}$. – Robert Israel Jan 22 at 22:24
In the derivation of $t>1$ case, the LHS should be $a_{i,t}(S)$ instead of $a_{i,1}(S)$, right? – AYang Jan 24 at 4:21
Thanks for catching the typo. I edited. – Robert Israel Jan 24 at 7:17
Thank you Robert. Now I understand your approach until the very last step -- the second case ($n-m\leq t\leq n$) of value of $c(t,m.n)$. Could you shed a light here? – AYang Jan 25 at 4:13

At the risk of shooting myself in the foot, here's how I'd approach it:

The key principle is that we can ignore any action that doesn't produce a tangible result. In particular, this includes missing a target, so we can start by normalizing the probabilities: write $\tilde{p_i} =\frac{p_i}{\sum_{1\leq k\leq N}p_k}$ so that we can work with values that have $\sum\tilde{p_i}=1$.

But from here, it should be clear that the process is just drawing without replacement - after our trial, we'll have some set of $X$ items, and the probability that those items are $i_0, i_1, \ldots, i_X$ is $\tilde{p}_{i_0}\cdot\tilde{p}_{i_1}\cdots\tilde{p}_{i_X}$. So the probability that target $k$ is hit is just $\dfrac{\sum_{\{S:\left|S\right|=X \wedge k\in S\}} \mathbb{P}_S}{\sum_{\{S: \left|S\right|=X\}} \mathbb{P}_S}$, where the lower sum is over all $X$-element subsets $S$ of $1\ldots N$ and the upper sum is over all those subsets with $k\in S$, and $\mathbb{P}_S=\prod_{i\in S}\tilde{p}_i$ (and of course, the probability that $k$ isn't hit is just the complement of this value).

EDIT: as pointed out by Nate Eldredge in a comment below, the formula can't simplify to one independent of the other probabilities; let the below serve as a cautionary tale about making assumptions!

This formula itself should offer further simplification, I'm reasonably sure, but that will take a little bit of chewing. (I strongly suspect that everything else is just a red herring and the final probability is a simple expression in $\tilde{p}_k$ and $\tilde{q}_k=1-\tilde{p}_k$, but I need to work through the details and have my ducks in a row before I'm certain of that.)

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It has to depend on $X$ since, if $X=N$ the probability is $0$. – David Kleiman Jan 21 at 18:53
@DavidKleiman Yep - I realized that after my initial post (which guessed that it was uniformly $\tilde{p}_k$) and made a quick edit. – Steven Stadnicki Jan 21 at 18:58
It can't just be a function of $\tilde{p}_k$. Consider $N=3$, $X=2$ and $p_1 = 0.01$. If we have $p_2 = 0.01$ and $p_3 = 0.98$ then the probability of hitting target 1 is about $1/2$. If we have $p_2 = p_3 = 0.495$ then the probability of hitting target 1 is much smaller. – Nate Eldredge Jan 21 at 19:18
@NateEldredge Oooh, that's a great example. Very good catch; thank you! I was missing the 'diagonal' terms in my mental products - turns out that, e.g., $p_1p_2+p_1p_3+p_2p_3$ isn't quite the same as $(p_1+p_2+p_3)^2$... :P – Steven Stadnicki Jan 21 at 19:43
I should have stated that $X<N$ – AYang Jan 21 at 21:31

I don't think it's easy. Say we have p1 = p2 = 0.49, and p3 .. p22 = 0.01. For the first two objects, the chance of being the first object is 0.49, the chance of being first or second is a bit above 0,98, and then it gets close to 1 very, very quickly.

For the other objects, the chance of being the first object is 0.01, the chance of being one of the first two is about 0.03, and to be among the first 2 + n the chance is about n/20.

If the probabilities are less extreme then I think it gets quite difficult, for example 20 items with 0.03 ≤ pk ≤ 0.07.

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Guys please criticize the following answer to my own question.

Let variable $L$ be the number of shots by the end of the game. Then, clearly, the probability of the $k$-th target survives is $(1-p_k)^L$.

Then the real question is how to determine $L$ or the $E(L)$, the expectation of $L$.

Let $q_i=1-p_i$ be the probability of missing a target for one single shot.

With $L$ shots, the probability that the $i$-th target has been hit at least once is $1-q_i^L$. Then given a set of targets $S$, where $k\notin S$ and $|S|=X$, the probability that all targets in $S$ have been hit at least once, let's call it $S$ completed, is that \begin{equation} P_{\mbox{S completed}} = \prod_{i\in S} (1-q_i^L) \end{equation}

Note that there are $C_{N-1}^{X}$ target set $S$ satisfying $k\notin S$ and $|S|=X$, forming a set $\tilde{S} = \{S \mid k\notin S \mbox{ and }|S|=X \}$. As said, $|\tilde{S}|=C_{N-1}^{X}$.

Because the game is over when $L$ shots have been made, then the probability that at least one $S\in\tilde{S}$ completed equals to 1 (I'm not sure if I'm correct by saying this). Or \begin{equation} \sum_{S\in\tilde{S}} P_{\mbox{S completed}} = 1 \end{equation}

With all of this, we have \begin{equation} \sum_{S\in\tilde{S}} \prod_{i\in S} (1-(1-p_i)^L) = 1 \end{equation}

Knowing $p_i$, $1\leq i \leq N$, $L$ can be obtained using computation tools like Matlab or R. (I haven't verified the solution with the tools yet)

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Your first clearly statement is wrong. Suppose $x = n$, then your result should be $0$, but your clearly result is nonzero. – DanielV Jan 21 at 21:12
@DanielV thank you for pointing this out. $X$ should be always less than $N$, i.e., $X<N$. I should have clarified this in the original post. – AYang Jan 21 at 21:18
In logic, from $A \implies B$ it follows that for any addition assumption $C$, that $(A \text{ and } C) \implies B$. It doesn't matter if you add the assumption $x < n$, the clearly statement is still false. You can add the assumption that it is tuesday, you can add the assumption that there are no odd perfect numbers, it doesn't matter. The statement is still false. Adding assumptions does not change conclusions. Removing assumptions can make a false statement into undefined, changing assumptions can change a false statement into true, but adding assumptions never changes conclusions. – DanielV Jan 21 at 21:30
Then I don't get why $(1-p_k)^L$ does not hold for the $X<N$ case. For each shot, the probability of missing target $k$ is $(1-p_k)$, for $L$ consecutive shots, wouldn't the probability of still missing it be $(1-p_k)^L$ ? What I'm missing here? – AYang Jan 21 at 21:38
"By the end of the game". The game only ends when certain conditions are met. All possible sequences are not end game sequences. – DanielV Jan 21 at 21:46

I think the right way to think about this is to realize that the game is over if and only if you've hit $X$ unique targets, so the relevant game states are all subsets of $[1, N]$ of size $X$ - the ones that contain $k$ are states where you've hit target k before the game is over.

Looking at pure combinations, the number of states where you've hit k is:

$$n_{k} = {X - 1 \choose N - 1}$$

Because you're selecting the first one in the set to be $k$ and selecting $X - 1$ from the remaining $N - 1$ combinations. The number of combinations that do not include $k$ are:

$$n_{!k} = {X \choose N - 1}$$

So in the limit where all targets are equally likely, the answer would be:

$$\mathbf{p}_{!k} = \frac{n_{!k}}{n_{k} + n_{!k}}$$

(Note that the denominator is the total number of end game states.)

When the probabilities of hitting targets is different, I think you need to use the same approach, but using the approach from this question to calculate the sum of the product of all the relevant subsets.

Using the notation from the Wikipedia article on elementary symmetric polynomials, and calling $\mathbf{T}_{!k} = \{p_i\}$ where $i \in [1, N]$ and $i \neq k$:

$$P_{k} = p_{k} \cdot e_{X-1}(\mathbf{T}_{!k})$$ $$P_{!k} = e_{X}(\mathbf{T}_{!k})$$

And the answer is: $$\mathbf{p}_{!k} = \frac{P_{!k}}{P_{k} + P_{!k}}$$

Edit To verify that this is the correct answer, I wrote a python script that runs an empirical test by actually randomly playing the game a number of times and counting the number of games in which $k$ was hit:

from __future__ import division

from copy import copy
from itertools import combinations
from operator import mul
from functools import reduce, partial

import numpy as np

# Empirical approach
def play_game(t_list, k, X):
    if k >= len(t_list):
        raise ValueError("k cannot be greater than N")

    if X >= len(t_list):
        raise ValueError("X must be <= N")

    targ_probs = np.array(t_list) / sum(np.array(t_list))
    targ_inds = np.arange(0, len(targ_probs))
    targ_set = set()

    while True:
        c_target = np.random.choice(targ_inds, p=targ_probs)

        if c_target == k:
            return False


        if len(targ_set) >= X:
            return True

def empirical_probability(n_games, t_list, k, X):
    game_outcomes = [play_game(t_list, k, X) for ii in range(n_games)]
    return sum(game_outcomes) / n_games

# Analytical approach
def elementary_symmetric_polynomial(m, t_set):
    return sum(map(partial(reduce, mul), combinations(t_set, m)))

def analytical_probability(t_list, k, X):
    t_k = t_list[k]
    t_set = np.array(t_list)
    t_set = t_set[[ii for ii in range(0, len(t_list)) if ii != k]]

    # All subsets of size X not containing k
    p_nk = elementary_symmetric_polynomial(X, t_set)

    # All subsets of size X - 1 not containing k
    p_k = t_k * elementary_symmetric_polynomial(X-1, t_set)

    return p_nk / (p_k + p_nk)

if __name__ == "__main__":
    n_games = 5000

    # Randomly generate a game set
    N = 6
    t_list = np.random.random((N,))
    t_list /= sum(t_list)

    empirical = [['X = '] + [str(X) for X in range(2, N-1)]]
    analytical = copy(empirical)
    for k in range(0, len(t_list)):
        ana_answers = ['k = {}'.format(k)]
        emp_answers = ['k = {}'.format(k)]
        for X in range(2, N-1):
            ana_answers.append(round(analytical_probability(t_list, k, X), 3))
            emp_answers.append(round(empirical_probability(n_games, t_list, k, X), 3))


    print("Empirical answers:")
    for row in empirical:
        print('\t'.join(str(x) for x in row))

    print("Analytical answers:")
    for row in analytical:
        print('\t'.join(str(x) for x in row))

Choosing $N = 6$ and running the game 5000 times, here is one run of results:

p = [ 0.3305, 0.0350, 0.04995, 0.0831, 0.1503, 0.3509]
Empirical answers:
X =     2   3   4
k = 0   0.373   0.153   0.045
k = 1   0.916   0.833   0.721
k = 2   0.892   0.779   0.591
k = 3   0.813   0.642   0.407
k = 4   0.671   0.432   0.199
k = 5   0.348   0.141   0.041
Analytical answers:
X =     2   3   4
k = 0   0.397   0.218   0.108
k = 1   0.908   0.811   0.649
k = 2   0.871   0.741   0.542
k = 3   0.792   0.608   0.374
k = 4   0.652   0.416   0.225
k = 5   0.38    0.207   0.102

I think these differences are well within the noise of the empirical measurement. The algorithm for the elementary symmetric polynomial calculation came from this question.

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thank you for your input! I'm studying your approach. Looking at your measurement, one observation is that the noise increases while X increases. – AYang Jan 22 at 17:29
Is your approach subject to $p_1+p_2+\cdots+p_N=1$ ? – AYang Jan 22 at 19:36
@AYang There is no constraint on $\{p_i\}$, because of the normalization condition in the denominator - you normalize "probability of reaching an end-game state that contains k" against "probability that you've reached an end game state". – Paul Jan 22 at 19:54
By the way, as a general rule during job interviews, I find that counting relevant states is often the key insight to problems like this. I think human brains work better with populations than probabilities. – Paul Jan 22 at 20:07
In your python code, you used np.random.choice to mimic the shots. The second parameter of the "choice" function requires all probability sums to 1. However, in real game case, a shot could be void. – AYang Jan 22 at 20:18

The question is obviously too difficult to be answered analytically, at least in a job interview. What it's not so difficult is to design an algorithm to evaluate the answer numerically. (Perhaps that was the idea in the interview?)

An equivalent formulation: we have $N$ balls with weights $p_1, p_2 \cdots p_N$. We extract the balls without replacement, with probabilities equal to their normalized weights. We want $P_{j,t}$, the probability that, after $t$ extrations, ball $j$ has not been yet extracted.

In particular, $P_{j,0}=1$, $P_{j,N}=0$,$P_{j,1}=1-p_j$,

To get the idea, suppose $t=3$, $j=N$. Then to get $P_{j,t}$ we need to sum over all the probabilities of the length $3$ extraction starting subsequences that don't include $j=N$. For example, (assuming normalized initial weighs) the subsequence $(1,2,3\cdots )$ would have probability

$$ p_1 \frac{p_2}{1-p_1} \frac{p_3}{1-p_1-p_2} $$ and so on.

Then $${P_{j,t}= \sum_{\sigma \in {\mathbb P}_{N,t,j} } \prod_{i=1}^t \frac{p_{\sigma_i}}{\left(1- \sum_{k=1}^{i-1} p_{\sigma_k} \right)}}$$

where ${\mathbb P}_{N,t,j} $ denotes all the permutations of $t$ elements, taken from the set $\{1, 2 \cdots N \}$ with element $j$ excluded.

The sum over such permutations can be concisely computed by recursion. For example, in Java:

public class Balls {

    // not necessarily normalized
    final static double prob[] = { 0.3, 0.2, 0.2, 0.15, 0.1, 0.05 };
    final static int N = prob.length;

     * Computes probabiity that element j (j=0..N-1, same index as in prob array)
     * does not appear in the first pos positions

     * computeProb(element, 0) = 1
     * computeProb(element, 1) = 1-p_j 
     * computeProb(element, N) = 0 
    public static double computeProb(int j, int pos) {
        double[] p0 = removeAndNormalize(prob, j, false);
        return allPermutations(p0, pos);

    static double allPermutations(double p[], int depth) {
        double ac = 0;
        if( depth <= 0 ) return 1;
        for( int i = 0; i < p.length; i++ ) 
            ac += p[i] * allPermutations(removeAndNormalize(p, i, true), depth - 1);
        return ac;

    // normalizes probabities in place
    static void normalize(double[] list) {
        double ac = 0;
        for( double px : list )         ac += px;
        for( int i = 0; i < list.length; i++ )          list[i] /= ac;

    // removes element of array, and optionally normalizes 
    // original (untouched) must be already normalized 
    static double[] removeAndNormalize(double p[], int index, boolean normalize) {
        double[] q = new double[p.length - 1];
        for( int i = 0, j = 0; i < q.length; i++, j++ ) {
            if( i == index ) j++;
            q[i] = normalize ? p[j] / (1 - p[index]) : p[j];
        return q;

For example, here's some results for $N=6$

prob=[0.3, 0.2, 0.2, 0.15, 0.1, 0.05]
t=0  1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 
t=1  0.70000 0.80000 0.80000 0.85000 0.90000 0.95000 
t=2  0.44794 0.59624 0.59624 0.68615 0.78423 0.88919 
t=3  0.24803 0.39457 0.39457 0.50573 0.64561 0.81148 
t=4  0.10545 0.20779 0.20779 0.30642 0.46950 0.70305 
t=5  0.02477 0.06330 0.06330 0.11122 0.21732 0.52009 

prob=[60.0, 5.0, 4.0, 3.0, 2.0, 0.1]
t=0  1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 
t=1  0.19028 0.93252 0.94602 0.95951 0.97301 0.99865 
t=2  0.02777 0.63673 0.70856 0.78082 0.85349 0.99264 
t=3  0.00277 0.34721 0.43412 0.54870 0.68413 0.98307 
t=4  0.00014 0.12570 0.18124 0.27503 0.45035 0.96754 
t=5  0.00000 0.00568 0.00942 0.01701 0.03533 0.93256

In IdeOne, so you can play with it.

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What do the columns in your answer mean? If it's X, starting with 1, why isn't the first column not equal to $1-p_t$? Since $\sum_i p_i = 1$ in your example, you always hit exactly one target in the first round. Also, I would expect the $X = 6$ column to be all 1.0 – Paul Jan 22 at 14:07
@Paul Each column correspond to an element index ($j$ in my formulation, $k$ in the original question). $X$ is $t$ in my formulation (rows). see that the row $t=1$ gives $1-p_i$ (normalized). The row $t=6$ (trivially one) is not displayed – leonbloy Jan 22 at 14:20
I meant that row $t=6$ is trivially zero. (the probability that some element has not appeared after $N$ extractions is, of course, zero) – leonbloy Jan 22 at 14:30
Ah, I was reading it wrong. That makes sense. Of course, I would challenge your notion that the question is too difficult to be answered analytically in a job interview setting. My answer below is analytical and something you could come up with during a job interview. I added code to prove that it's the correct result. – Paul Jan 22 at 15:13

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