Binomial Probability Recursion

Why Recursion Helps

A binomial probability table can require several separate calculations if each probability is evaluated from the binomial formula. When adjacent probabilities are related by a simple ratio, one accurately calculated probability can instead be used to generate the others. This is especially useful when a problem asks for several count probabilities or when the table itself provides a check on the calculation.

The recursion is not a different probability model: it is an algebraic consequence of the binomial probability formula. It applies only when the count has a binomial distribution with fixed parameters, and the recurrence must stop at the endpoints of the distribution’s support.

Definition: Suppose $N$ counts successes in $n$ independent trials, each with the same success probability $p$. Then $N\sim\operatorname{Bin}(n,p)$, where $q=1-p$, and $$ P(N=k)=\binom{n}{k}p^kq^{n-k},\qquad k=0,1,\ldots,n. $$ The support is the integer set $\{0,1,\ldots,n\}$. For an integer $k$ outside this set, $P(N=k)=0$.

The Consecutive-Probability Ratio

Write $b_k=P(N=k)$. For $0\leq k<n$, the ratio of the next probability to the current one is

\frac{b_{k+1}}{b_k} =\frac{\binom{n}{k+1}p^{k+1}q^{n-k-1}}{\binom{n}{k}p^kq^{n-k}} =\frac{n-k}{k+1}\frac{p}{q}.

This formula assumes $0<p<1$, so that both $p$ and $q$ are positive. It is valid only for $k=0,\ldots,n-1$, where both probabilities are within the support. Rearranging gives the forward recursion:

b_{k+1}=b_k\frac{n-k}{k+1}\frac{p}{q}.

Why the Recursion Holds

The ratio follows by dividing the binomial formula at $k+1$ by the formula at $k$. The powers simplify to $p/q$, while the ratio of the binomial coefficients simplifies as follows:

\frac{\binom{n}{k+1}}{\binom{n}{k}} =\frac{n!}{(k+1)!(n-k-1)!}\frac{k!(n-k)!}{n!} =\frac{n-k}{k+1}.

Combining those two simplifications gives the stated ratio. The derivation also explains why each step depends on the current index $k$: the number of remaining trials changes as the count increases, while the denominator $k+1$ comes from the next binomial coefficient.

To build a table, calculate one probability directly—often $b_0=q^n$—then apply the recurrence repeatedly. If a different probability is easier to calculate, the table can be built upward from it or downward using the reciprocal ratio. Whichever anchor is chosen, all generated values must stay within the support.

Confirm the model.
Check that there are $n$ fixed, independent trials with the same success probability $p$, and set $q=1-p$.

Choose an anchor.
Calculate one supported probability directly, using $b_0=q^n$ when convenient or another binomial probability when it is simpler.

Apply the ratio within the support.
For each step from $k$ to $k+1$, multiply by $\frac{n-k}{k+1}\frac{p}{q}$. Do not continue past $k=n$.

Check the result.
Confirm that probabilities are nonnegative, the table has no positive entries outside $0,\ldots,n$, and the entries sum to 1 apart from any rounding.

Worked Example: Building a Full Table from Zero Claims

A hypothetical group contains five independent policies. Each policy has probability $0.20$ of producing a claim during the year. Let $N$ be the number of policies that produce a claim. Build the probability table recursively and find the probability of between one and three claims, inclusive.

Method and assumptions. Each policy is one trial, the number of trials is $n=5$, and the common claim probability is $p=0.20$. Independence and the common probability give $N\sim\operatorname{Bin}(5,0.20)$. Thus $q=1-p=0.80$, and the support is $0,1,2,3,4,5$.

Anchor. The probability of zero claims is $b_0=q^5=0.80^5=0.32768$. For each next entry, multiply by $\frac{5-k}{k+1}\frac{0.20}{0.80} =\frac{5-k}{k+1}(0.25)$.

Recursion. For $k=0$, the multiplier is $\frac{5}{1}(0.25)=1.25$, so $b_1=0.32768(1.25)=0.40960$. For $k=1$, the multiplier is $\frac{4}{2}(0.25)=0.50$, so $b_2=0.40960(0.50)=0.20480$. For $k=2$, the multiplier is $\frac{3}{3}(0.25)=0.25$, so $b_3=0.20480(0.25)=0.05120$. For $k=3$, the multiplier is $\frac{2}{4}(0.25)=0.125$, so $b_4=0.05120(0.125)=0.00640$. For $k=4$, the multiplier is $\frac{1}{5}(0.25)=0.05$, so $b_5=0.00640(0.05)=0.00032$.

The requested probability is $P(1\leq N\leq3)=b_1+b_2+b_3=0.40960+0.20480+0.05120=0.66560$. As a table check, the six entries sum to $0.32768+0.40960+0.20480+0.05120+0.00640+0.00032=1.00000$. The answer is $0.6656$ to four decimal places. The method avoids separately evaluating four binomial coefficients, while the total checks that no supported count was omitted.

Support and Boundary Checks

The recurrence generates probabilities for supported counts only. For a binomial count with $n$ trials, $n+1$ probabilities belong in the table, from $b_0$ through $b_n$. There is no next step from $b_n$: the formula for a transition to $b_{n+1}$ is not part of the recurrence, and $P(N=n+1)=0$. A count below zero is likewise impossible.

When $0<p<1$, every supported probability is positive. At the special parameter values $p=0$ or $p=1$, the ratio formula is not usable because it divides by $q$ or $p$ in one direction. Handle these cases directly: if $p=0$, then $P(N=0)=1$; if $p=1$, then $P(N=n)=1$. These are degenerate distributions, not ordinary cases for the ratio recursion.

Worked Example: Using an Exact Fraction and Checking the Upper Tail

A hypothetical inspection process checks six independent components. Each component has probability $1/3$ of being defective. Let $N$ count defective components. Build the table using exact fractions and find $P(N\geq4)$.

Model and anchor. Here $n=6$, $p=1/3$, and $q=2/3$, so $N\sim\operatorname{Bin}(6,1/3)$ with support $0,\ldots,6$. Start with $b_0=(2/3)^6=64/729$. The multiplier from $b_k$ to $b_{k+1}$ is $\frac{6-k}{k+1}\frac{1}{2}$.

For $k=0$, the multiplier is $3$, giving $b_1=192/729$. For $k=1$, it is $\frac{5}{4}$, giving $b_2=240/729$. For $k=2$, it is $\frac{2}{3}$, giving $b_3=160/729$. For $k=3$, it is $\frac{3}{8}$, giving $b_4=60/729$. For $k=4$, it is $\frac{1}{5}$, giving $b_5=12/729$. For $k=5$, it is $\frac{1}{12}$, giving $b_6=1/729$.

The requested tail includes exactly the supported counts 4, 5, and 6: $P(N\geq4)=b_4+b_5+b_6=(60+12+1)/729=73/729=0.100137\ldots$. Rounded to four decimal places, the answer is $0.1001$. The support check is essential: there are seven entries, indexed from 0 to 6, and their numerators sum to $64+192+240+160+60+12+1=729$, so the table sums to 1.

Ratios and the Most Likely Count

The same ratio can show where the probabilities increase and decrease. Since $b_k>0$ for every supported $k$ when $0<p<1$, $b_{k+1}>b_k$ exactly when the ratio is greater than 1. The ratio decreases as $k$ increases: its numerator $n-k$ decreases and its denominator $k+1$ increases. Consequently, once the probabilities stop increasing, they cannot begin increasing again.

b_{k+1}\geq b_k \quad\Longleftrightarrow\quad \frac{n-k}{k+1}\frac{p}{q}\geq1 \quad\Longleftrightarrow\quad (n+1)p\geq k+1.

This comparison identifies the mode, meaning a supported count with the largest probability. If $(n+1)p$ is not an integer, the unique mode is $\lfloor(n+1)p\rfloor$. If $(n+1)p$ is an integer, there are two modes, $(n+1)p-1$ and $(n+1)p$, because the ratio between those adjacent probabilities equals 1.

Worked Example: Locating the Most Likely Claim Count

A hypothetical pool has ten independent policies, each with probability $0.30$ of producing a claim in the year. Let $N$ be the claim count. Find its most likely value, and use recursion to compare the probabilities at that value and the next larger count.

Model. The assumptions give $N\sim\operatorname{Bin}(10,0.30)$, with $q=0.70$ and support $0,\ldots,10$. To check the location of the maximum, examine the consecutive ratios around 3. From 1 to 2, the multiplier is $\frac{9}{2}\frac{3}{7}=\frac{27}{14}>1$, and from 2 to 3 it is $\frac{8}{3}\frac{3}{7}=\frac{8}{7}>1$. From 3 to 4 it is $\frac{7}{4}\frac{3}{7}=\frac{3}{4}<1$. Because the ratios decrease with $k$, the probabilities increase up to 3 and decrease after 3. Thus the most likely count is 3.

For a numerical check, calculate the anchor $b_2$ directly: $b_2=\binom{10}{2}(0.30)^2(0.70)^8 =45(0.09)(0.05764801)=0.2334744405$. Then $b_3=b_2(8/7)=0.266827932$, and $b_4=b_3(3/4)=0.200120949$. To four decimal places, these probabilities are $0.2335$, $0.2668$, and $0.2001$, respectively. They confirm the increase to 3 and decrease after it. The comparison avoids assuming that the mode is simply the mean; the mode is determined by the adjacent-probability ratios.

Common Mistakes

Reversing the ratio. The forward multiplier is $\frac{n-k}{k+1}\frac{p}{q}$, not $\frac{k+1}{n-k}\frac{q}{p}$. The latter is the reciprocal and is appropriate only when stepping backward.
Using the wrong index. The transition from $b_k$ to $b_{k+1}$ uses $n-k$ and $k+1$. Substituting $k+1$ for $k$ throughout shifts the recursion by one count.
Confusing the support with an unbounded count. A binomial count cannot exceed $n$. Do not generate or include a positive probability for $n+1$.
Forgetting to check the starting value. A recursion propagates an error in its anchor. Confirm the model parameters and compute the anchor from the binomial formula before generating the table.
Applying the ratio at a degenerate parameter. If $p=0$ or $p=1$, use the point-mass distribution directly instead of dividing by zero.

Key takeaway: For $N\sim\operatorname{Bin}(n,p)$ with $0<p<1$, adjacent probabilities satisfy $P(N=k+1)=P(N=k)\frac{n-k}{k+1}\frac{p}{1-p}$ for $k=0,\ldots,n-1$. Anchor the recursion carefully, remain within the support, and check that the probabilities sum to 1.

Check Your Understanding

For each problem, identify the binomial parameters before applying a ratio or calculating a probability. Answers are given to four decimal places where appropriate.

A hypothetical set of four independent devices each fails during a test with probability $0.20$. If $N$ is the number that fail, use the binomial formula to anchor at $P(N=1)$, then use the ratio to find $P(N=2)$. (Answer: $P(N=2)=0.1536$)
A hypothetical team tests seven independent sensors, each with probability $0.50$ of detecting a fault. What is $P(N=7)$, and how many supported probabilities belong in the full table? (Answer: $P(N=7)=1/128=0.0078$; 8 probabilities)
A hypothetical portfolio has eight independent contracts, each with probability $0.25$ of a claim. Starting from $P(N=2)$, use the ratio to find $P(N=3)$. (Answer: $P(N=3)=1701/8192=0.2076$)
A hypothetical group contains nine independent risks, each with claim probability $0.40$. Determine the modal claim count using consecutive-probability ratios, including whether the mode is unique. (Answer: Modes are 3 and 4; they tie because $(9+1)(0.40)=4$)
A hypothetical process has three independent trials, each with success probability $0.20$. What is the probability of more than three successes, and why should that count not appear as a positive entry in a recursively generated table? (Answer: $0$; the support ends at 3)

What You'll Learn