# Lindley's (1993) analysis of a version of Fisher's tea tasting lady example, using a mix of discrete and continuous priors

I am interested in explaining the version of Fisher's tea tasting lady example that is discussed in Lindley's (1993) 'The Analysis of Experimental Data: The Appreciation of Tea and Wine'. A similar question has been raised here: Reproducing a didactic example of Lindley (1993)

A lady claims to have the ability to taste whether milk or tea has been added first to a cup of tea. She is given 6 pairs of cups of tea. For each pair, one cup contains milk added first (followed by tea), whereas the other cup contains tea added first (followed by milk).

According to Lindley's Bayesian approach, the prior probability distribution (a mix of discrete and continuous) appropriate for tea is: $$0.8 \ \textrm{for} \ P = \frac{1}{2}$$ $$1.6(1 – P) \ \textrm{for} \ P > \frac{1}{2}$$

For 5R, 1W (i.e. 5 correct and 1 incorrect responses), the likelihood function is: $$P^5(1 - P)$$ For 6W (i.e. 6 correct responses), the likelihood function is: $$P^6$$

Lindley derives the following graph: Furthermore, he asserts that the value of the probability that P = 1/2 drops from 0.8 (prior) to 0.59 in the 5R, 1W case and 0.23 in the 6W case. I derived a different graph and computed that the value of the probability that P = 1/2 drops from 0.8 (prior) to 0.656 in the 5R, 1W case and 0.1093 in the 6W case. I was wondering about where I might have erred in my computations?

@Henry Thanks for your swift response! As requested, here are the calculations I used: $$P(P = \frac{1}{2} \,|\, 5R, 1W) = \left(\frac{P(5R, 1W \,|\, P = \frac{1}{2}) \times P(P = \frac{1}{2})}{P(5R, 1W)}\right) = \left(\left(P^5 \cdot (1 - P)\right) \times 0.8 \times \frac{1}{P(5R, 1W)}\right)$$

$$P(5R, 1W) = \int_{0}^{1} 0.8 \cdot P^5 \cdot (1 - P) \,dP$$ $$c_1 = \frac{1}{{P(5R, 1W)}} \approx 52.5$$

$$P(6R) = \int_{0}^{1} 0.8 \cdot P^6 \,dP$$

$$c_2 = \frac{1}{{P(6R)}} \approx 8.7497$$

$$P(P = \frac{1}{2} \,|\, 5R, 1W) \approx \left(\frac{1}{2}\right)^6 \times 0.8 \times 52.5 \approx 0.656$$

$$P(P = \frac{1}{2} \,|\, 6R) \approx \left(\frac{1}{2}\right)^6 \times 0.8 \times 8.7497 \approx 0.1093$$

• What were your calculations? I suspect your $0.656$ may come from ignoring the $0.8$ in the prior (if so I get $140/213 \approx 0.6573$). Though I may also have made an error: I get instead $112/185\approx 0.6054$ which is not quite $0.59$. I think the calculation should perhaps be $$\frac{\left. 0.8 P^5(1-P)\right|_{P=0.5}}{\left. 0.8 P^5(1-P)\right|_{P=0.5} + \int_{1/2}^1 1.6(1-P)P^5(1-P)\, dP}$$ Jun 5 at 8:29
• Meanwhile I get $\frac{112}{359} \approx 0.3120$ in the 6R (not 6W) case rather than your $0.1093$ or Lindley's $0.23$. My calculations are in an answer to the linked question Jun 6 at 8:59