Annie and Becky are playing hide and seek. Annie wants to hide, and it’s Becky’s job to find out. Annie has two choices – hide behind a house or the bushes. But there is a problem: if she hides behind the hose, there is a 75% probability that the surveillance system will catch her and a light will lit in the front. How often should Annie try to hide behind the house?
We have seen similar puzzles before. Suppose there was no complication due to light. In that case, intuition suggests that Annie and Becky should mix their choices randomly at equal probabilities.
Probabilities for searching
Becky’s strategy is designed to make both choices equally attractive to Annie. Let p be the probability that Becky search the house. Then, for Annie, the payoffs much match.
Hide behind house = Hide behind bushes.
Pay off for hide behind house = 0 x 0.75 + 0.25 [0 x p + 1 x (1-p)].
Pay off for hiding behind bushes = 1 x p + 0 x (1-p)
Explanation for the first equation. If Annie hides behind the house, there is a 75% probability that the light will lit and Becky will go and find her (or Annie ends up with 0). On the other hand, inside the 25% case, where no light is lit, there is a p chance Becky will go for the house and a (1-p) chance Becky will miss.
Solving for p,
0.25 – 0.25 p = p
p = 0.2
Probabilities for hiding
Let q be the probability for Annie to hide behind the house. Note that Becky will need to apply probabilities only when there is no light in front of the house. Annie’s strategy is to make payoffs for Becky equal (given no light).
The chance that Annie hides behind the house given no light is a conditional probability, and we must apply Bayes’ rule.
P(HH|NL) = P(NL|HH) x P(HH) / [ P(NL|HH) x P(HH) + P(NL|HB) x P(HB)]
= 0.25q /[0.25q + (1-q)]
Similarly, The probability Annie hides behind the bushes given no light is:
P(HB|NL) = P(NL|HB) x P(HB) / [P(NL|HB) x P(HB) + P(NL|HH) x P(HH)]
= (1-q) /[(1-q) + 0.25q]
Now, back to payoffs:
Payoff for Search house | no light = Payoff for Seach bushes | no light
Payoff for Search house | no light = 1 x [0.25q /[0.25q + (1-q)]] + 0 x (1-q) /[(1-q) + 0.25q]
Payoff for Search bushes | no light = 0 x [0.25q /[0.25q + (1-q)]] + 1 x (1-q) /[(1-q) + 0.25q]
0.25 q /[0.25q + (1-q)]] = (1-q) /[(1-q) + 0.25q]
0.25 q = (1-q)
q = 1/1.25 = 0.8
This is counterintuitive; Annie will hide behind the house 80% of the time, knowing the light will be lit 75% of those occasions.
Reference
Expert-Level Hide and Seek: William Spaniel
