Here is another hide-and-seek game but played with an inverse handicap. If the last case would have given a clue to the seeker 75% of the time, this time, it would help the hider 75% of the time on one location, i.e., the house. So the situation is
- There are two options to hide – behind bushes, behind the house.
- If the person hides behind the house, and the opponent searches the house, there is a 75% chance the seeker will not spot the hider (camouflage effective 75% of the time.
- If the person hides behind the bush, and the opponent searches it, it is normal, and no camouflage works.
We have seen similar puzzles before. Suppose there was no complication of camouflage; the players should mix their choices randomly at equal probabilities (0.5).
How should one search?
Let p be the probability that the seeker should search behind the house. The best value for p is such a way that the hider finds no particular incentive to choose one place over the other. In other words, the p makes the payoffs for the hider to hide behind bush and house equal.
Payoff to hide behind house = Payoff to hide behind bushes.
Pay off for hide behind house = p[1 x 0.75 + 0 x 0.25] + (1-p) x 1 = 0.75p + 1 – p
Pay off for hiding behind bushes = p x 1 + (1-p) x 0 = p
If the hider hides behind the house and the seeker searches the house, there is a 75% probability that she will gain one point and a 25% chance of zero. On the other hand, if the hider is behind the house and the seeker searches the bush, the hider gets a full score (one).
Solving for p,
0.75p + 1 – p = p
p = 0.8
How should one hide?
Let q be the probability that the hider must hide behind the house. Following the logic as before,
Payoff to search behind house = Payoff to search behind bushes.
Payoff to search behind house = q[0 x 0.75 + 1 x 0.25] + (1-q) x 0 = 0.25q
Payoff to search behind bushes = q x 0 + (1-q) x 1 = 1 – q
Solving for q,
0.25q = 1 – q
q = 0.8
Reference
Camouflaged Hide and Seek: William Spaniel
