Unfair Coin Detection
You suspect a coin might be unfair. Specifically, you hypothesize that it could be unfair such that the probability of getting Heads, P(H), is only 0.01. You flip this coin 10 times and observe 10 consecutive Heads.
Assuming your prior belief that the coin is this specific type of "unfair" (P(H)=0.01) is 50%, and your prior belief that it's a "fair" coin (P(H)=0.5) is also 50%, what is the probability that the coin is indeed this "unfair" type, given the evidence of 10 Heads?
Related Concepts
Hint
- Define your hypotheses:
- H₁: Coin is unfair (P(Heads) = 0.01). Call this U.
- H₂: Coin is fair (P(Heads) = 0.5). Call this F.
- Define the evidence E: Observing 10 Heads in 10 flips.
- What are your prior probabilities P(U) and P(F)? (Given as 0.5 each).
- Calculate the likelihood of the evidence under each hypothesis:
- P(E | U): If the coin is unfair (P(H)=0.01), what's the probability of 10 Heads? (0.01)¹⁰.
- P(E | F): If the coin is fair (P(H)=0.5), what's the probability of 10 Heads? (0.5)¹⁰.
- Calculate the total probability of the evidence P(E) using the law of total probability:
P(E) = P(E | U)P(U) + P(E | F)P(F) - Apply Bayes' Theorem to find P(U | E).
Think about which hypothesis makes the evidence (10 heads) more plausible.
Explanation: Unfair Coin Detection
You have a coin. You're 50/50 on whether it's fair (Heads 50% of the time) or if it's a very specific "unfair" coin (Heads only 1% of the time). You flip it 10 times and get 10 Heads!
Now, what's the chance it's that specific "unfair" coin (P(H)=0.01)?
- If it WAS the "unfair" P(H)=0.01 coin: Getting 10 Heads in a row would be incredibly rare (0.01 multiplied by itself 10 times – a tiny number!).
- If it WAS a "fair" P(H)=0.5 coin: Getting 10 Heads is still unlikely (0.5 multiplied by itself 10 times), but it's MUCH more likely than with the P(H)=0.01 coin.
Since the evidence (10 Heads) is vastly more probable if the coin is fair compared to if it's this specific P(H)=0.01 unfair coin, our belief will shift heavily towards the coin not being this P(H)=0.01 type of unfair. It's far more likely to be fair (or unfair in a way that favors heads, but that's not the hypothesis we're testing here).
We will use Bayes' Theorem to determine the probability that the coin is the specified type of "unfair" given the evidence of 10 consecutive heads.
Let the hypotheses be:
- U: The coin is unfair, with P(Heads) = P(H|U) = 0.01.
- F: The coin is fair, with P(Heads) = P(H|F) = 0.5.
Let E be the evidence: 10 Heads in 10 flips.
We want to find P(U | E).
1. Define Prior Probabilities
We are given neutral priors for the two hypotheses:
P(U) = 0.5(Prior probability that the coin is this specific type of unfair)P(F) = 0.5(Prior probability that the coin is fair)
Note: P(U) + P(F) = 1, as these are the only two possibilities considered for the coin's state.
2. Calculate the Likelihood of the Evidence under each Hypothesis
The probability of getting 10 Heads in 10 independent flips is P(H)10.
- Likelihood of 10 Heads if the coin is unfair (P(H)=0.01):
P(E | U) = (0.01)10 = 10-20
(This is an extremely small number: 0.00000000000000000001) - Likelihood of 10 Heads if the coin is fair (P(H)=0.5):
P(E | F) = (0.5)10 = 1 / 210 = 1 / 1024 ≈ 0.0009765625
3. Calculate the Total Probability of the Evidence (P(E))
Using the Law of Total Probability:
P(E) = P(E | U) × P(U) + P(E | F) × P(F)
P(E) = (10-20 × 0.5) + ((1/1024) × 0.5)
P(E) = (0.5 × 10-20) + (0.0009765625 × 0.5)
P(E) ≈ (0.5 × 10-20) + 0.00048828125
Since 0.5 × 10-20 is incredibly small compared to 0.00048828125, P(E) is overwhelmingly dominated by the second term.
P(E) ≈ 0.00048828125
(Using the provided answer's approximation for P(10 heads) if P(U) & P(F) were 0.5: P(E) ≈ 0.00049. Let's use the more precise value for now.)
4. Apply Bayes' Theorem to Find P(U | E)
Bayes' Theorem states:
P(U | E) = [P(E | U) × P(U)] / P(E)
P(U | E) = [(10-20) × 0.5] / 0.00048828125
P(U | E) = (0.5 × 10-20) / 0.00048828125
P(U | E) ≈ 1.024 × 10-17
This is an extremely small number, practically zero.
Final Result & Conclusion
The probability that the coin is the specific type of "unfair" (P(Heads)=0.01), given that 10 heads were observed in 10 flips, is:
P(Unfair with P(H)=0.01 | 10 Heads) ≈ 1.024 × 10-17 ≈ 0
Conclusion: The evidence of 10 consecutive heads makes it extremely unlikely that the coin is unfair in the way specified (i.e., having P(Heads) = 0.01). The evidence strongly supports the alternative hypothesis that the coin is fair (or, more broadly, that P(Heads) is much higher than 0.01).
If we were to calculate P(F | E):
P(F | E) = [P(E | F) × P(F)] / P(E)
P(F | E) = [ (1/1024) × 0.5 ] / 0.00048828125
P(F | E) = 0.00048828125 / 0.00048828125 = 1 (approximately, due to the dominance of this term)
So, given the evidence, we become almost certain that the coin is not the P(H)=0.01 type of unfair, and instead, it is far more likely to be fair (or unfair in a way that favors heads).
Impact of Evidence: This problem powerfully illustrates how overwhelming evidence can drastically shift our beliefs. Even with an initial 50/50 assumption, observing 10 heads makes the hypothesis that P(H)=0.01 virtually impossible. The likelihood of the evidence under the "unfair P(H)=0.01" hypothesis is so minuscule compared to the likelihood under the "fair" hypothesis that the posterior probability for "unfair P(H)=0.01" becomes negligible.
Important Consideration: The definition of "unfair" is critical here. If "unfair" could mean P(H)=0.99, then 10 heads would make that type of unfairness very likely. The question specifically tests the P(H)=0.01 unfairness.
Hypothesize: How would your conclusion change if your prior belief P(Unfair with P(H)=0.01) was much higher, say 0.99, and P(Fair) was 0.01? Would 10 heads still make you believe the coin is fair?