Probability problem using Stirling's formula

Question

A full deck of 52 cards is divided into half at random. Using Stirling’s formula, estimate the probability that each half contains the same number of red and black cards.

Answer 1

We can ignore the suits and focus on the suit colours. Since there are only two colours, red and black, we have a binary situation. The probability p=½ that a card in the divided pack would be either a red or a black. So p=q=½ and the binomial expression is:

(p+q)²⁶=1, so, expanding:

p²⁶+26p²⁵q+325p²⁴q²+...+²⁶C_rp^26-rq^r+...+325p²q²⁴+26pq²⁵+q²⁶.

Since p=q=½, ²⁶C_rp^26-rq^r=²⁶C_r/2²⁶.

Each term in the expansion represents a probability. p²⁶ or q²⁶ is the probability that all cards in the divided pack will be the same colour. ²⁶C₁₃/2²⁶ represents the probability that there are equal numbers of black and red cards. This evaluates to 10400600/67108864=0.155 or 15.5% approximately.

We now need to use Stirling's approximation to see how it compares with the actual probability.

Stirling's formula is:

n!≈√(2πn)(n/e)ⁿ and (2n!)≈√(4πn)(2n/e)²ⁿ.

²ⁿC_n=(2n!)/(n!)²≈2√(πn)(2n/e)²ⁿ/(2πn(n/e)²ⁿ)=2²ⁿ/√(πn)).

Therefore ²ⁿC_n≈2²ⁿ/√(πn))=10501063 compared to the actual value 10400600.

And ²ⁿC_n/2²ⁿ≈1/√(πn), and when n=13, this evaluates to about 0.156 or 15.6%.