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Data Interview Question

Value of Coin Flip Product

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Solution & Explanation

To solve the problem of calculating the expected value of the product of the number of heads and tails when a coin is flipped 100 times, we need to delve into the properties of the binomial distribution. Let's break it down step-by-step:

Understanding the Problem

  • Coin Tosses: We have 100 independent coin tosses.
  • Heads and Tails: Let HH be the number of heads and T=100HT = 100 - H be the number of tails.
  • Objective: Calculate E[H×T]E[H \times T], the expected value of the product of heads and tails.

Binomial Distribution Basics

  • Binomial Distribution: For a fair coin, HBinomial(n=100,p=0.5)H \sim \text{Binomial}(n=100, p=0.5).
  • Expected Value: E[H]=np=100×0.5=50E[H] = np = 100 \times 0.5 = 50.
  • Variance: Var(H)=np(1p)=100×0.5×0.5=25\text{Var}(H) = np(1-p) = 100 \times 0.5 \times 0.5 = 25.

Calculating E[H×T]E[H \times T]

  1. Expression: E[H×T]=E[H×(100H)]=E[100HH2]E[H \times T] = E[H \times (100 - H)] = E[100H - H^2].
  2. Using Linearity of Expectation:
    • E[100HH2]=100E[H]E[H2]E[100H - H^2] = 100E[H] - E[H^2].
  3. Finding E[H2]E[H^2]:
    • Var(H)=E[H2](E[H])2\text{Var}(H) = E[H^2] - (E[H])^2.
    • 25=E[H2]50225 = E[H^2] - 50^2.
    • E[H2]=25+2500=2525E[H^2] = 25 + 2500 = 2525.
  4. Substitute Back:
    • E[H×T]=100×502525=2475E[H \times T] = 100 \times 50 - 2525 = 2475.

Confidence Interval for E[H×T]E[H \times T]

  • Central Limit Theorem: For large nn, the distribution of HH is approximately normal.
  • Standard Error: SE=Var(H×T)/n\text{SE} = \sqrt{\text{Var}(H \times T)/n}.
  • Approximate Var(H×T)\text{Var}(H \times T):
    • Simplification using HH and TT as approximately normal gives Var(H×T)1002×0.5×0.5×(10.5)=2475\text{Var}(H \times T) \approx 100^2 \times 0.5 \times 0.5 \times (1 - 0.5) = 2475.
  • Confidence Interval Calculation:
    • 95% CI=[mean1.96×SE,mean+1.96×SE]95\% \text{ CI} = [\text{mean} - 1.96 \times \text{SE}, \text{mean} + 1.96 \times \text{SE}].
    • SE24751004.975\text{SE} \approx \sqrt{\frac{2475}{100}} \approx 4.975.
    • 95% CI=[24751.96×4.975,2475+1.96×4.975][2465.25,2484.75]95\% \text{ CI} = [2475 - 1.96 \times 4.975, 2475 + 1.96 \times 4.975] \approx [2465.25, 2484.75].

Conclusion

The expected value of the product of the number of heads and tails in 100 coin flips is 2475, with a 95% confidence interval approximately between 2465.25 and 2484.75. This reflects the variability and normal approximation of the binomial distribution at this scale.