Hide and seek game

Question

$A$ and $B$ go to the Senate to play a game of Hide-and-Seek. There are $100$ rooms in the Senate, and $B$ picks one of them and hides there till the game ends.

$A$, at the beginning of every turn, picks one room and searches in it. Since he is human and thus fallible, he only has a $60\%$ chance of finding $B$ if she is in fact hiding in the room. If he fails to find her in the room, $B$'s score increases by one, and the next turn starts, whereupon $A$ must pick another room to check (he can also check the same room again). $B$ remains in the room she initially chose.

Given that $A$ and $B$ play with perfect strategies, what is the expected value of $B$'s score at the end of the game?

I tried with $1$ room. There is only one hiding place for $B$. So the expected score of $$B = \sum_{n=1}^{\infty}0.4^{n-1}0.6(n-1)=\frac{2}{3}.$$

Answer 1

NOTE: This solves a different problem. In this calculation it is assumed that A searches randomly. That is sub-optimal. It is clearly better to go through all the rooms before repeating. Perhaps the calculation still has some value so I'll leave it up.

Let E be the expected score (= Expected number of rooms searched - 1). Look at the first search, it can have one of three (mutually exclusive) outcomes:

Outcome A: A finds B! Probability = $\frac {1}{100} \frac{3}{5}$ And in this case E is 0.

Outcome B: A looks in the right room but fails to find B. Probability = $\frac {1}{100} \frac{2}{5}$ And in this case E goes to E + 1.

Outcome C: A looks in the wrong room. Probability = $\frac {99}{100}$ And in this case E also goes to E + 1.

Combining these, we get $$E = \frac {1}{100} \frac{3}{5} (0) \;\;+\;\;\frac {1}{100} \frac{2}{5} (E + 1) + \frac {99}{100} (E + 1) $$ This is easily solved to yield $E= \frac{497}{3}$

Answer 2

$A$ is better off searching a new room rather than an old room.
Suppose $A$ searched all hundred rooms (even after finding $B$).
They have a $3/5$ chance of finding $B$ during this sweep, and if they do it is after searching $50.5$ rooms on average. So $$E=\frac35(49.5)+\frac25(100+E)\\ \frac35E=29.7+40\\ E=116\frac16$$

Answer 3

You forgot to include that the answer must be accurate to .1 decimal.

I believe you are over complicating the question. There are 100 possible rooms for player B to hide in. You are trying to find the odds of not only player A locating player B, but also with the factor that player A only has a 60% chance of finding player B, if in fact they have chosen the correct room.

Based on the question, they are playing "A" game of hide and seek, thus meaning one game.

The odds of player A finding player B in 100 rooms with only an expected accuracy of locating player B in the correct room of 60%, would mean that player B has the odds of obtaining a score of 166.7 if you are rounding up.

If you choose to not round up, then it would be 166.6, or more accurately 166.666667

Simply 100 rooms divided by 60% accuracy.

Edit tried it out and was not the correct answer, sorry but now this is bugging me. Interested in what it is.