Symmetry in the Binomial Model

Why Binomial Symmetry Is Useful

In a binomial experiment with success probability one half, a count of $k$ successes is just as likely as a count of $n-k$ successes. This gives a useful way to check exact probabilities and to turn some upper-tail calculations into lower-tail calculations. The reflection is around $n/2$, so it matters whether $n$ is even or odd and whether an event includes its boundary.

The symmetry is not a general property of every binomial distribution. It depends on the probability of success being exactly one half, as well as on the binomial model's fixed number of independent trials with a common success probability. Before applying it, state the model and check these conditions.

Definition: Let $X$ count the successes in $n$ independent Bernoulli trials, each with success probability $p=1/2$. Then $X\sim\operatorname{Bin}(n,1/2)$, with support $0,1,\ldots,n$. Its probability mass function is $$ P(X=k)=\binom{n}{k}\left(\frac12\right)^k\left(\frac12\right)^{n-k} =\frac{\binom{n}{k}}{2^n}, \qquad k=0,1,\ldots,n. $$ For every supported count $k$, the reflected count $n-k$ has the same probability: $$ P(X=k)=P(X=n-k). $$

How the Reflection Works

The reflection sends a count $k$ to $n-k$. For example, with $n=8$, a count of 2 is paired with a count of 6; with $n=7$, a count of 2 is paired with a count of 5. When $n$ is even, the middle count $n/2$ reflects to itself. When $n$ is odd, there is no single middle integer: the two central counts, $(n-1)/2$ and $(n+1)/2$, reflect to each other.

The same reflection applies to sets of counts, but it reverses their order. An inclusive lower tail through $k$ is paired with an inclusive upper tail starting at $n-k$. These inclusive boundaries are important: the reflected event is not obtained by simply reusing $k$ as the upper-tail threshold.

P(X\leq k)=P(X\geq n-k), \qquad P(X\geq r)=P(X\leq n-r),

These identities apply when the thresholds are integers in the support. A more general way to use the rule is to reflect every count in an event. For instance, the event $a\leq X\leq b$ becomes $n-b\leq X\leq n-a$. Thus intervals with reflected endpoints have equal probabilities. If an interval maps to itself, its probability is unchanged by reflection; this fact alone does not imply that its probability is one half.

Why the Formula Holds

There are two equivalent justifications: the binomial mass formula and a pairing of trial outcomes. The mass formula makes the role of $p=1/2$ explicit. For an integer $k$ between 0 and $n$,

P(X=k)=\binom{n}{k}\frac{1}{2^n}.

The binomial coefficient counts the patterns with exactly $k$ successes among $n$ trials. Choosing which $k$ trials are successes is equivalent in number to choosing which $n-k$ trials are failures, so $\binom{n}{k}=\binom{n}{n-k}$. Consequently,

P(X=k) =\frac{\binom{n}{k}}{2^n} =\frac{\binom{n}{n-k}}{2^n} =P(X=n-k).

To obtain the tail identity, sum the equal point probabilities over the reflected counts. The counts $0,1,\ldots,k$ reflect to $n,n-1,\ldots,n-k$, which are exactly the counts from $n-k$ through $n$. Therefore $P(X\leq k)=P(X\geq n-k)$. The same one-to-one pairing proves the corresponding identity for an upper tail.

When the result does not apply: If $p\neq 1/2$, the probabilities at $k$ and $n-k$ are generally unequal. The reflection argument also requires the binomial setup: a fixed number of independent trials with the same success probability. Do not apply the identities to a count merely because it is an integer between 0 and $n$.

Using Symmetry in Calculations

Worked Example: Matching Exact Counts and a Lower Tail

As a hypothetical model, suppose eight independent pieces are inspected, and each has probability $1/2$ of being flagged for a specified condition. Let $X$ be the number flagged. Find $P(X=2)$, $P(X=6)$, and $P(X\leq2)$.

Model and method. The trial count is fixed at $n=8$, the trials are independent, and each flag probability is $p=1/2$. Thus $X\sim\operatorname{Bin}(8,1/2)$. Use the binomial mass formula for the exact counts and sum the masses for the lower tail. No interest rate or mortality assumption is part of this probability model.

Exact count of 2. There are $\binom{8}{2}=8\cdot7/(2\cdot1)=28$ ways to select the two flagged pieces, out of $2^8=256$ equally likely patterns. Hence

P(X=2)=\frac{\binom{8}{2}}{2^8} =\frac{28}{256} =\frac{7}{64} =0.109375.

Exact count of 6. By reflection, 6 successes pair with $8-6=2$ successes. Directly, $\binom{8}{6}=28$, so

P(X=6)=\frac{\binom{8}{6}}{256} =\frac{28}{256} =\frac{7}{64} =0.109375.

Lower tail through 2. The counts 0, 1, and 2 have respectively $\binom{8}{0}=1$, $\binom{8}{1}=8$, and $\binom{8}{2}=28$ patterns. Therefore

P(X\leq2)=\frac{1+8+28}{256} =\frac{37}{256} =0.14453125.

The reflected-tail identity gives the check $P(X\leq2)=P(X\geq6)$. The upper-tail counts 6, 7, and 8 have $28+8+1=37$ patterns, confirming the same result. Rounded to four decimal places, the lower-tail probability is 0.1445. This calculation avoids the mistake of pairing $X\leq2$ with $X\geq2$; the correct reflected threshold is $8-2=6$.

Worked Example: Reflecting an Upper-Tail Event

In a hypothetical sequence of seven independent trials, each trial results in a success with probability $1/2$. Let $X$ be the number of successes. Find the probability of at least five successes.

Model and method. Here $X\sim\operatorname{Bin}(7,1/2)$. The event $X\geq5$ reflects to $X\leq7-5=2$. Calculate the smaller lower tail rather than adding upper-tail terms.

P(X\geq5)=P(X\leq2) =\frac{\binom{7}{0}+\binom{7}{1}+\binom{7}{2}}{2^7}.

The three coefficients are $\binom{7}{0}=1$, $\binom{7}{1}=7$, and $\binom{7}{2}=7\cdot6/(2\cdot1)=21$. Their sum is $1+7+21=29$, and $2^7=128$. Thus

P(X\geq5)=\frac{29}{128}=0.2265625.

Rounded to four decimal places, the probability is 0.2266. A boundary error would be to reflect the event to $X\leq3$, perhaps by subtracting 5 from 7 and then adding one. But the counts 5, 6, and 7 pair with 2, 1, and 0, respectively, so the reflected event is $X\leq2$, including exactly 2.

Worked Example: Comparing Reflected Intervals

Suppose a hypothetical batch contains ten items, with each item independently having probability $1/2$ of meeting a specified test criterion. Let $X$ count the items meeting it. Find the probability that either 2 through 4 items meet the criterion or 6 through 8 items meet it.

Model and method. The count satisfies $X\sim\operatorname{Bin}(10,1/2)$. Reflecting an interval $a\leq X\leq b$ gives $10-b\leq X\leq10-a$. Thus 2 through 4 reflects to 6 through 8; the two events are disjoint, so their probabilities can be added.

First calculate the probability of the lower interval. The required coefficients are $\binom{10}{2}=45$, $\binom{10}{3}=120$, and $\binom{10}{4}=210$. Their sum is $45+120+210=375$. Since there are $2^{10}=1024$ equally likely trial patterns,

P(2\leq X\leq4)=\frac{375}{1024}.

By symmetry, $P(6\leq X\leq8)=375/1024$ as well. The intervals do not overlap, so

P\big((2\leq X\leq4)\text{ or }(6\leq X\leq8)\big) =\frac{375}{1024}+\frac{375}{1024} =\frac{750}{1024} =\frac{375}{512} =0.732421875.

Rounded to four decimal places, the requested probability is 0.7324. The method avoids recalculating three upper-interval coefficients. It also avoids treating the two interval events as one continuous interval: counts 5 are not included.

Common Mistakes and Exam Traps

The most important check is whether $p=1/2$. For a general binomial model, the mass at $k$ is $\binom{n}{k}p^k(1-p)^{n-k}$. Although the binomial coefficients at $k$ and $n-k$ agree, the remaining factors need not.

For example, consider $X\sim\operatorname{Bin}(4,0.6)$, which is not a fair-trial model. A mistaken use of symmetry would assert $P(X=1)=P(X=3)$. Direct calculation gives

P(X=1)=\binom{4}{1}(0.6)(0.4)^3=4(0.6)(0.064)=0.1536,

whereas

P(X=3)=\binom{4}{3}(0.6)^3(0.4)=4(0.216)(0.4)=0.3456.

These values are different. The correction is to use symmetry only after checking that the success probability is one half. Also track inclusive boundaries when reflecting events: for a lower tail $X\leq k$, the matching upper tail is $X\geq n-k$, not $X\geq n-k+1$. A reliable check is to reflect the endpoint counts themselves; $k$ must pair with $n-k$.

Key takeaway: Under $X\sim\operatorname{Bin}(n,1/2)$, counts $k$ and $n-k$ have equal probabilities. Reflecting an event means replacing each count $j$ by $n-j$; for inclusive tails, $P(X\leq k)=P(X\geq n-k)$. Verify the model and preserve the endpoint.

Check Your Understanding

For each problem, use the binomial assumptions stated and apply symmetry where it helps. Probabilities are rounded to four decimal places when needed.

A hypothetical set of six independent trials has success probability $1/2$. Find $P(X=1)$ and use symmetry to identify a different count with the same probability. (Answer: $P(X=1)=P(X=5)=6/64=0.0938$.)
For $X\sim\operatorname{Bin}(9,1/2)$, find $P(X\geq7)$ by reflecting the event and calculating the resulting tail. (Answer: $P(X\geq7)=P(X\leq2)=46/512=23/256=0.0898$.)
For $X\sim\operatorname{Bin}(8,1/2)$, calculate $P(3\leq X\leq5)$ by adding the exact probabilities. (Answer: $(56+70+56)/256=182/256=91/128=0.7109$.)
A hypothetical process has twelve independent trials, each with success probability $1/2$. Find $P(X\leq4)$ using the binomial coefficients. (Answer: $P(X\leq4)=P(X\geq8)=(1+12+66+220+495)/4096=794/4096=397/2048=0.1938$.)
For $X\sim\operatorname{Bin}(5,1/2)$, find the probability that either all trials succeed or none succeed. Explain why the two outcomes have equal probability. (Answer: $P(X=5\text{ or }X=0)=1/32+1/32=1/16=0.0625$, since 0 and 5 are reflected counts.)

What You'll Learn