Conditional entropy to predict a outcome

I have given the dataset of three different customers: $$A,B,C$$ visiting a restaurant which sells $$4$$ different kind of dishes: $$B1, B2, B3, B4$$.

The table shows past orders of each of the three customers (o$$1 \to$$ o$$10$$) where o$$1$$ means first order and o$$2$$ means second order and so on. How can conditional probability help us here to predict the next order for each of the following customers?

customer/order: o$$10$$ o$$9$$ o$$8$$ o$$7$$ o$$6$$ o$$5$$ o$$4$$ o$$3$$ o$$2$$ o$$1$$

$$A$$: $$B1$$ $$B2$$ $$B2$$ $$B2$$ $$B2$$ $$B2$$ $$B1$$ $$B1$$ $$B1$$ $$B1$$

$$B$$: $$B3$$ $$B2$$ $$B4$$ $$B1$$ $$B3$$ $$B2$$ $$B1$$ $$B3$$ $$B2$$ $$B4$$

$$C$$: $$B4$$ $$B2$$ $$B1$$ $$B2$$ $$B4$$ $$B3$$ $$B4$$ $$B2$$ $$B4$$ $$B2$$

Let $$B_{j, i}$$ denote the dish $$B_j$$ at the $$i$$th order.

Applying Markov chain model per customer, we can compute conditional probabilites as below.

For A, $$P(B_{1,i+1}|B_{1,i}) = \frac{P(B_{1,i+1}, B_{1, i})}{P(B1_i)} = \frac{3 / 9}{4 / 9} \\ P(B_{2,i+1}|B_{1,i}) = \frac{1 / 9}{4 / 9}\\ P(B_{3,i+1}|B_{1,i}) = P(B_{4,i+1}|B_{1,i}) = 0$$ For B, $$P(B_{1,i+1}|B_{3,i}) = 2/2 = 1$$, $$P(B_{2,i+1}|B_{3,i}) = P(B_{3,i+1}|B_{3,i}) = P(B_{4,i+1}|B_{3,i}) = 0$$

For C, $$P(B_{2,i+1}|B_{4,i}) = 2/3$$, $$P(B_{3,i+1}|B_{4,i}) = 1/3$$, $$P(B_{1,i+1}|B_{4,i}) = P(B_{4,i+1}|B_{4,i}) = 0$$

Conditional entropy for A is \begin{align} H(B_{*, i+1}|B_{*, i}) & = - \sum_{j, k} P(B_{j,(i+1)}, B_{k,i}) \log \frac{P(B_{j,(i+1)}, B_{k,i})}{P(B_{k,i})} \\ & = - \left( Q(B_{1, i+1}|B_{1, i}) + Q(B_{2, i+1}|B_{1, i}) + Q(B_{1, i+1}|B_{2, i}) + Q(B_{2, i+1}|B_{2, i}) \right) \\ & = - \left( \frac{3}{9}\log\frac{3}{4} + \frac{1}{9}\log\frac{1}{4} + \frac{1}{9}\log\frac{1}{5} + \frac{4}{9}\log\frac{4}{5} \right)\\ & \approx 0.53 \end{align} , where $$Q(B_{j,(i+1)}, B_{k,i}) = P(B_{j,(i+1)}, B_{k,i}) \log \frac{P(B_{j,(i+1)}, B_{k,i})}{P(B_{k,i})}$$

For B, \begin{align} H(B_{*, i+1}|B_{*, i}) & = - \left( \frac{1}{9}\log\frac{1}{2} + \frac{1}{9}\log\frac{1}{2} + \frac{3}{9}\log\frac{3}{3} + \frac{2}{9}\log\frac{2}{2} + \frac{2}{9}\log\frac{2}{2} \right) \\ & \approx 0.15 \end{align} For C, \begin{align} H(B_{*, i+1}|B_{*, i}) & = - \left( \frac{1}{9}\log\frac{1}{1} + \frac{1}{9}\log\frac{1}{4} + \frac{3}{9}\log\frac{3}{4} + \frac{1}{9}\log\frac{1}{1} + \frac{2}{9}\log\frac{2}{3} + \frac{1}{9}\log\frac{1}{3} \right)\\ & \approx 0.46 \end{align}

• Thank you so much. Could you help me relate all these probabilities to their conditional entropies? – Amit Kattal Apr 24 at 15:46
• Thank you so much. Can you help me one last time. How can we most accurately predict the next order from each of the people using record of other participants using mutual information? – Amit Kattal Apr 25 at 5:36