## 21. Sunday Pie

On Sunday, as usual, Monsieur Dupont went for a walk. Meanwhile, his wife was trying in vain to decide whether he would remember to buy a pie and whether she would have to go to the bakery herself. In the end, she asked her son: “Do you know if your father intended to buy a pie?”

However, the child, predicting his father's behavior, is wrong one time out of three. Moreover, Madame Dupont believes that if the child is completely deprived of sweets, it will be twice as bad as if he gets two pies at once. How would you advise Madame Dupont to act: go and buy a pie herself regardless of what the son says, or go for the pie only if the boy says that his father did not intend to buy it?
21. Let the disappointment associated with buying two pies instead of one be evaluated by Madame Dupont as having a value of $\alpha$. Then, putting herself in her son's position, she should evaluate the disappointment caused by not buying any pies at all as having a value of $2 \alpha$. If she goes to the pastry shop herself, Madame Dupont risks experiencing disappointment of $\alpha$ with a probability of $1 / 2$ (since she cannot know in advance whether her husband will buy a pie or not, and considers the chances of these two events to be equal). On average, the undesirable effect of Madame Dupont's trip to the pastry shop is evaluated as $\alpha / 2$.

If she follows her son's advice, she will be wrong in one out of three cases. In this case, the undesirable effect of her mistake is equally likely to be evaluated as $\alpha$ and $2 \alpha$ (since the son's prediction is wrong one time out of three, both when the father does not buy a pie and when he does). Summing up the values $1 / 3 \cdot 1 / 2 \cdot \alpha$ and $1 / 3 \cdot 1 / 2 \cdot 2 \alpha$, we get the average value of the possible failure: $(1 / 6) \alpha + (1 / 3) \alpha = \alpha / 2$.

Thus, no matter what Madame Dupont does, the risk will be the same, both courses of action are exactly equivalent.