At any given point in the bomb defusing process, Ilsa has cut ~i~ wires, completely defused ~a~ bombs, and partially defused ~b~ bombs. A partially-defused bomb has exactly one of its wires cut.

We can do dynamic programming where ~DP[a][b]~ is the probability that there are ~a~ completely-defused bombs and ~b~ partially-defused bombs after Ilsa has cut ~2a+b~ wires. Necessarily, the sum of all DP values for the same number of wires cut must equal ~1~. Note that all numbers of cut wires ~i~ are equally likely before Ethan walks into the room.

We start with ~DP[0][0] = 1~, and with all other DP values initialized to ~0~. When ~0~ wires have been cut, it must be the case that no bombs are either completely or partially defused. We can then iteratively compute the DP value for all states where ~1~ wire is cut, then ~2~ wires, and so on as follows:

For a given number of wires ~i~, such that we've already computed all DP values that correspond to ~i~ wires being cut, we can compute the values for ~i+1~ wires being cut by iterating over all possible states ~(a, b)~ such that ~2a+b = i~. For each such state, we compute the probability ~p~ that the next wire cut will completely defuse a partially-defused bomb, and the probability ~q~ that the next wire cut will be the first wire cut on a new bomb:

• There are ~b~ wires that would complete a partially-defused bomb if cut, and ~2N-i~ uncut wires. Therefore, ~p = \frac{b}{2N-i}~.
• ~p~ and ~q~ must sum to ~1~. Therefore, ~q = 1-p~.

Then, we update two of the states for ~i+1~ wires as follows:

• With probability ~p~, there will be one more completely-defused bomb and one fewer partially-defused bomb. Therefore, we can increase ~DP[a+1][b-1]~ by ~DP[a][b] \times p~.
• With probability ~q~, there will be one more partially-defused bomb. Therefore, we can increase ~DP[a][b+1]~ by ~DP[a][b] \times q~.

Let ~U~ be the random variable of the number of uncut wires (and let ~u~ be a specific number of uncut wires). The answer we actually want to compute is the expected number of uncut wires, given that we know that there are ~K~ completely-defused bombs:

$$\displaystyle E[U \mid a = K] = 1 \times P(U = 1 \mid a = K) + \dots + 2N \times P(U = 2N \mid a = K)$$

Let ~S[K]~ be the sum of ~DP[K][b]~ over all possible values of ~b~. We can then state that:

$$\displaystyle P(U = u \mid a = K) = \frac{DP[K][2N-2K-u]}{S[K]}$$

The probability of being in a given state ~(a, b)~ that corresponds to ~u~ uncut wires is proportional to ~DP[a][b]~. Since all values of ~u~ are equally likely without any information, the probability of being in state ~(a, b)~ is simply ~\frac{DP[a][b]}{S[K]}~ after you know that ~a = K~.

Computing the table of DP values takes ~\mathcal O(N^2)~ time, and combining them together takes ~\mathcal O(N)~ time, so the total time complexity is ~\mathcal O(N^2)~.

Taking the second sample case as an example, there are three states in which Ethan might walk in on Ilsa: ~(0, 0)~, ~(0, 1)~, and ~(0, 2)~, since ~a~ is known to be ~0~. It must be the case that Ilsa has ~2~, ~3~, or ~4~ wires remaining since with any fewer wires remaining there would have been at least ~1~ completely-defused bomb. All of these values of ~u~ are equally likely when Ethan walks into the room, but when he observes that no bombs have been completely defused, the chance that exactly ~2~ wires remain uncut is decreased since there's a chance that one bomb would have been completely defused by that point.

The chance that ~2~ cut wires belong to the same bomb (out of ~2~ total bombs) is ~\frac 1 3~, so the chance of having ~2~ wires uncut and ~0~ completely-defused bombs is only ~\frac 2 3~ the chance of having ~3~ wires uncut or ~4~ wires uncut. Observe that ~DP[0][2] = \frac 2 3~, while ~DP[0][0] = DP[0][1] = 1~.

$$\displaystyle \begin{align*} P(U = 4 \mid a = 0) &= \frac{DP[0][0]}{S[K]} = \frac{1}{8/3} = \frac 3 8 \\ P(U = 3 \mid a = 0) &= \frac{DP[0][1]}{S[K]} = \frac{1}{8/3} = \frac 3 8 \\ P(U = 2 \mid a = 0) &= \frac{DP[0][2]}{S[K]} = \frac{2/3}{8/3} = \frac 2 8 \end{align*}$$

Dropping terms that equal ~0~, the answer we want to compute is:

$$\displaystyle \begin{align*} E[U \mid a = 2] &= 2 \times P(U = 2 \mid a = 0) + 3 \times P(U = 3 \mid a = 0) + 4 \times P(U = 4 \mid a = 0) \\ &= \left(2 \times \frac 2 8\right) + \left(3 \times \frac 3 8\right) + \left(4 \times \frac 3 8\right) \\ &= \frac{25}{8} = 3.125 \end{align*}$$