Problem 267: Billionaire

Project Euler Problem 267 Solution

EulerSolve provides an optimized solution for Project Euler Problem 267, Billionaire, with C++, Python, Java, and a step-by-step mathematical explanation.

Problem Summary A gambler starts with capital $1$. In each of $n$ fair tosses, the gambler bets the same fixed fraction $f$ of the current capital. A win triples the stake portion, so total capital is multiplied by $(1+2f)$; a loss destroys the stake portion, so total capital is multiplied by $(1-f)$. For the original Project Euler problem, $n=1000$ and the target is $T=10^9$. The C++ solution keeps both quantities as parameters, so the same derivation works for any $n\ge 1$ and $T>1$. Mathematical Approach 1. Wealth After Exactly $k$ Wins If exactly $k$ of the $n$ tosses are wins, then the order of wins and losses does not matter, because multiplication is commutative. The final wealth is therefore $$W_k(f)=(1+2f)^k(1-f)^{n-k}.$$ So for each possible win count $k$, the problem is a one-variable optimization over the admissible range $0\le f<1$. 2. Why the Code Optimizes the Logarithm Directly maximizing $W_k(f)$ is equivalent to maximizing its logarithm, because $\ln(x)$ is strictly increasing on $x>0$. The code therefore works with $$g_k(f)=\ln W_k(f)=k\ln(1+2f)+(n-k)\ln(1-f).$$ This removes huge intermediate numbers and turns the product into a sum. It also makes the derivative easy to analyze. 3....

Detailed mathematical approach

Problem Summary

A gambler starts with capital $1$. In each of $n$ fair tosses, the gambler bets the same fixed fraction $f$ of the current capital. A win triples the stake portion, so total capital is multiplied by $(1+2f)$; a loss destroys the stake portion, so total capital is multiplied by $(1-f)$.

For the original Project Euler problem, $n=1000$ and the target is $T=10^9$. The C++ solution keeps both quantities as parameters, so the same derivation works for any $n\ge 1$ and $T>1$.

Mathematical Approach

1. Wealth After Exactly $k$ Wins

If exactly $k$ of the $n$ tosses are wins, then the order of wins and losses does not matter, because multiplication is commutative. The final wealth is therefore

$$W_k(f)=(1+2f)^k(1-f)^{n-k}.$$

So for each possible win count $k$, the problem is a one-variable optimization over the admissible range $0\le f<1$.

2. Why the Code Optimizes the Logarithm

Directly maximizing $W_k(f)$ is equivalent to maximizing its logarithm, because $\ln(x)$ is strictly increasing on $x>0$. The code therefore works with

$$g_k(f)=\ln W_k(f)=k\ln(1+2f)+(n-k)\ln(1-f).$$

This removes huge intermediate numbers and turns the product into a sum. It also makes the derivative easy to analyze.

3. Best Fixed Fraction for a Given $k$

Differentiating gives

$$g_k'(f)=\frac{2k}{1+2f}-\frac{n-k}{1-f}.$$

Setting the derivative to zero yields the critical point

$$f_*=\frac{3k-n}{2n}.$$

The second derivative is

$$g_k''(f)=-\frac{4k}{(1+2f)^2}-\frac{n-k}{(1-f)^2}<0,$$

so $g_k$ is concave and any interior critical point is automatically the global maximum.

This gives exactly the three branches used in minimum_wins_needed:

1. If $k=n$, then every toss is a win, so $W_n(f)=(1+2f)^n$ is increasing and the optimum is the boundary limit $f\to1^{-}$, giving $\max g_n = n\ln 3$.

2. If $f_*\le 0$, then the concave maximum lies to the left of the admissible interval, so the best admissible point is $f=0$, with $W_k(0)=1$ and therefore $g_k(0)=0$.

3. If $0<f_*<1$, then the optimum is achieved at that interior point and the code evaluates $g_k(f_*)$ directly.

4. Turning Optimization into a Threshold in $k$

For a fixed admissible $f$, replacing one loss by one win multiplies wealth by

$$\frac{1+2f}{1-f}>1 \qquad (0<f<1).$$

So for every fixed $f$, the quantity $W_k(f)$ is nondecreasing in $k$, and strictly increasing when $f>0$. Taking the maximum over all admissible $f$ preserves this monotonicity, so $\max_f g_k(f)$ also increases with $k$.

That means there is a clean threshold $k_{\min}$: the smallest number of wins for which the target becomes reachable, namely

$$k_{\min}=\min\{k:\max_f g_k(f)\ge \ln T\}.$$

Once this threshold is found, every outcome with at least $k_{\min}$ wins is successful, and every outcome below it is impossible to rescue with any fixed fraction.

5. The Probability Is a Binomial Tail

Because the tosses are fair, the number of wins $K$ follows

$$K\sim\mathrm{Binomial}(n,1/2).$$

Therefore the answer is

$$P_{\mathrm{succ}}=\Pr[K\ge k_{\min}]=\sum_{k=k_{\min}}^{n}\binom{n}{k}2^{-n}.$$

The program does not compute factorials. Instead it starts with

$$P_0=2^{-n}$$

and uses the recurrence

$$P_{k+1}=P_k\frac{n-k}{k+1}.$$

This is the standard ratio between consecutive binomial masses and is much more stable numerically.

6. Worked Checkpoint: $n=2,\;T=2$

The code contains this test and expects the answer $1/4$. Let us derive it exactly.

If $k=1$, then

$$f_*=\frac{3\cdot1-2}{2\cdot2}=\frac14.$$

At this optimum,

$$W_1\!\left(\frac14\right)=\left(1+\frac12\right)\left(1-\frac14\right)=\frac32\cdot\frac34=\frac98<2.$$

So one win is not enough.

If $k=2$, then every toss is a win and the optimum is the boundary $f\to1^{-}$, where

$$W_2(f)\to 3^2=9>2.$$

Hence $k_{\min}=2$. The success probability is just the probability of two wins in two fair tosses:

$$P_{\mathrm{succ}}=\binom22 2^{-2}=\frac14.$$

7. Worked Checkpoint: $n=4,\;T=4$

The second internal test expects $5/16$. Again we can derive it by hand.

If $k=2$, then

$$f_*=\frac{3\cdot2-4}{2\cdot4}=\frac14,$$

$$W_2\!\left(\frac14\right)=\left(\frac32\right)^2\left(\frac34\right)^2=\frac{81}{64}<4.$$

Thus two wins are still insufficient.

If $k=3$, then

$$f_*=\frac{3\cdot3-4}{2\cdot4}=\frac58,$$

and

$$W_3\!\left(\frac58\right)=\left(1+\frac{10}{8}\right)^3\left(1-\frac58\right)=\left(\frac94\right)^3\frac38=\frac{2187}{512}>4.$$

Therefore $k_{\min}=3$. The success probability is the binomial tail

$$\Pr[K\ge3]=\binom43 2^{-4}+\binom44 2^{-4}=\frac{4+1}{16}=\frac{5}{16}.$$

How the Code Works

parse_arguments reads three command-line controls: --tosses=<n>, --target=<T>, and --skip-checkpoints.

minimum_wins_needed scans $k=0,1,\dots,n$. For each $k$, it computes the best possible log-wealth according to the three-case rule above. The first $k$ whose best log-wealth reaches $\ln T$ is returned.

binomial_tail_probability builds the probabilities $P_k=\Pr[K=k]$ from $P_0=2^{-n}$ using the binomial ratio recurrence, then sums from $k_{\min}$ to $n$.

solve_probability is just the composition of those two steps. The two checkpoints in run_checkpoints verify that both the threshold search and the tail summation are working correctly before the main Euler instance is evaluated.

Complexity Analysis

The threshold search over $k$ is $O(n)$. Building and summing the binomial probabilities is also $O(n)$. The implementation stores a probability vector of length $n+1$, so the memory usage is $O(n)$.

A streaming implementation could reduce the memory to $O(1)$, because only the current probability is needed, but the vector version is simpler to read and perfectly adequate for $n=1000$.

Footnotes and References

Problem page: https://projecteuler.net/problem=267
Kelly criterion intuition: Wikipedia - Kelly criterion
Binomial distribution: Wikipedia - Binomial distribution

Problem 267 source code

C++

#include <cmath>
#include <cstdint>
#include <iomanip>
#include <iostream>
#include <string>
#include <vector>

namespace {

struct Options {
    int tosses = 1000;
    long double target = 1.0e9L;
    bool run_checkpoints = true;
};

bool parse_int_after_prefix(const std::string& arg, const std::string& prefix, int& value) {
    if (arg.rfind(prefix, 0U) != 0U) {
        return false;
    }
    const std::string tail = arg.substr(prefix.size());
    if (tail.empty()) {
        return false;
    }
    int parsed = 0;
    for (char c : tail) {
        if (c < '0' || c > '9') {
            return false;
        }
        parsed = parsed * 10 + static_cast<int>(c - '0');
    }
    value = parsed;
    return true;
}

bool parse_ld_after_prefix(const std::string& arg, const std::string& prefix, long double& value) {
    if (arg.rfind(prefix, 0U) != 0U) {
        return false;
    }
    const std::string tail = arg.substr(prefix.size());
    if (tail.empty()) {
        return false;
    }
    char* end_ptr = nullptr;
    value = std::strtold(tail.c_str(), &end_ptr);
    return end_ptr != nullptr && *end_ptr == '\0';
}

bool parse_arguments(int argc, char** argv, Options& options) {
    for (int i = 1; i < argc; ++i) {
        const std::string arg(argv[i]);
        if (arg == "--skip-checkpoints") {
            options.run_checkpoints = false;
            continue;
        }
        if (parse_int_after_prefix(arg, "--tosses=", options.tosses) ||
            parse_ld_after_prefix(arg, "--target=", options.target)) {
            continue;
        }
        std::cerr << "Unknown argument: " << arg << '\n';
        return false;
    }
    return options.tosses >= 1 && options.target > 1.0L;
}

long double binomial_tail_probability(const int n, const int k_min) {
    if (k_min <= 0) {
        return 1.0L;
    }
    if (k_min > n) {
        return 0.0L;
    }

    std::vector<long double> prob(static_cast<std::size_t>(n + 1), 0.0L);
    prob[0] = std::ldexp(1.0L, -n);  // 2^-n
    for (int k = 0; k < n; ++k) {
        prob[static_cast<std::size_t>(k + 1)] =
            prob[static_cast<std::size_t>(k)] * static_cast<long double>(n - k) /
            static_cast<long double>(k + 1);
    }

    long double sum = 0.0L;
    for (int k = k_min; k <= n; ++k) {
        sum += prob[static_cast<std::size_t>(k)];
    }
    return sum;
}

int minimum_wins_needed(const int n, const long double target) {
    const long double ln_target = std::log(target);
    for (int k = 0; k <= n; ++k) {
        long double best_gain = -1.0e300L;
        if (k == n) {
            best_gain = static_cast<long double>(n) * std::log(3.0L);
        } else {
            const long double f_star =
                (3.0L * static_cast<long double>(k) - static_cast<long double>(n)) /
                (2.0L * static_cast<long double>(n));
            if (f_star > 0.0L && f_star < 1.0L) {
                best_gain =
                    static_cast<long double>(k) * std::log(1.0L + 2.0L * f_star) +
                    static_cast<long double>(n - k) * std::log(1.0L - f_star);
            } else if (f_star <= 0.0L) {
                best_gain = 0.0L;
            }
        }
        if (best_gain >= ln_target) {
            return k;
        }
    }
    return n + 1;
}

long double solve_probability(const int tosses, const long double target) {
    const int k_min = minimum_wins_needed(tosses, target);
    return binomial_tail_probability(tosses, k_min);
}

bool run_checkpoints() {
    // For two tosses and target 2.0, requiring exactly two wins gives probability 1/4.
    {
        const long double p = solve_probability(2, 2.0L);
        if (std::fabsl(p - 0.25L) > 1e-15L) {
            std::cerr << "Checkpoint failed for tosses=2 target=2.0" << '\n';
            return false;
        }
    }
    // For four tosses and target 4.0, at least three wins are required at optimum.
    {
        const long double p = solve_probability(4, 4.0L);
        const long double expected = 5.0L / 16.0L;
        if (std::fabsl(p - expected) > 1e-15L) {
            std::cerr << "Checkpoint failed for tosses=4 target=4.0" << '\n';
            return false;
        }
    }
    return true;
}

}  // namespace

int main(int argc, char** argv) {
    Options options;
    if (!parse_arguments(argc, argv, options)) {
        return 1;
    }
    if (options.run_checkpoints && !run_checkpoints()) {
        return 2;
    }
    const long double ans = solve_probability(options.tosses, options.target);
    std::cout << std::fixed << std::setprecision(12) << static_cast<double>(ans) << '\n';
    return 0;
}

Python

import math

def binomial_tail_probability(n, k_min):
    if k_min <= 0:
        return 1.0
    if k_min > n:
        return 0.0
        
    prob = [0.0] * (n + 1)
    # math.ldexp is available in math package but pow(2.0, -n) is also fine
    prob[0] = math.ldexp(1.0, -n)
    
    for k in range(n):
        prob[k + 1] = prob[k] * (n - k) / (k + 1)
        
    return sum(prob[k_min : n + 1])

def minimum_wins_needed(n, target):
    ln_target = math.log(target)
    for k in range(n + 1):
        best_gain = -1.0e300
        if k == n:
            best_gain = n * math.log(3.0)
        else:
            f_star = (3.0 * k - n) / (2.0 * n)
            if 0.0 < f_star < 1.0:
                best_gain = k * math.log(1.0 + 2.0 * f_star) + (n - k) * math.log(1.0 - f_star)
            elif f_star <= 0.0:
                best_gain = 0.0
                
        if best_gain >= ln_target:
            return k
            
    return n + 1

def solve_probability(tosses, target):
    k_min = minimum_wins_needed(tosses, target)
    return binomial_tail_probability(tosses, k_min)

def solve():
    ans = solve_probability(1000, 1.0e9)
    return f"{ans:.12f}"

if __name__ == '__main__':
    print(solve())

Java

import java.util.Locale;

public class Euler267 {
    static double binomialTailProbability(int n, int kMin) {
        if (kMin <= 0)
            return 1.0;
        if (kMin > n)
            return 0.0;

        double[] prob = new double[n + 1];
        prob[0] = Math.pow(2.0, -n);

        for (int k = 0; k < n; k++) {
            prob[k + 1] = prob[k] * (n - k) / (k + 1);
        }

        double sum = 0.0;
        for (int k = kMin; k <= n; k++) {
            sum += prob[k];
        }
        return sum;
    }

    static int minimumWinsNeeded(int n, double target) {
        double lnTarget = Math.log(target);
        for (int k = 0; k <= n; k++) {
            double bestGain = -1.0e300;
            if (k == n) {
                bestGain = n * Math.log(3.0);
            } else {
                double fStar = (3.0 * k - n) / (2.0 * n);
                if (fStar > 0.0 && fStar < 1.0) {
                    bestGain = k * Math.log(1.0 + 2.0 * fStar) + (n - k) * Math.log(1.0 - fStar);
                } else if (fStar <= 0.0) {
                    bestGain = 0.0;
                }
            }
            if (bestGain >= lnTarget) {
                return k;
            }
        }
        return n + 1;
    }

    static double solveProbability(int tosses, double target) {
        int kMin = minimumWinsNeeded(tosses, target);
        return binomialTailProbability(tosses, kMin);
    }

    public static String solve() {
        double ans = solveProbability(1000, 1.0e9);
        return String.format(Locale.US, "%.12f", ans);
    }

    public static void main(String[] args) {
        System.out.println(solve());
    }
}