Trap choices for Hunter

Nope, once mind controlled the minion is owned by the opposing player. It is working as intended.

TL;DR: Skip to the plot at the end of this answer, and to the simple approximation that follows.

(Aside: This question is probably more appropriate for mathSE than it is for Arqade. The math is quite involved.)

Caveat

I first want to make the obvious caveat: don't take these numbers too seriously. The probability that UtH+Buzzard is in the hunter's hand is entirely dependent on the previous cards played, and that player's playstyle. A skilled player would of course not rely on the numbers alone, but would have to take into account all previous turns and all previous cards played. It is complicated if not impossible to get actual numbers on this; no matter how much I take into account there will always be more to consider, and any calculation will be making some sort of assumption about the hunter's playstyle.

Example 1. On the hunter's turn 3, you have three 2/1's out, but he/she does not play UtH. More than likely he does not have UtH in his hand. Or maybe he is saving it, due to having both UtH and Buzzard in his hand. Regardless, the fact that he does not play UtH on this turn alters the probabilities in some way.

Example 2. On turn 4, the hunter plays down to 0 cards. Then you know of course that neither UtH nor Buzzard is in her hand. Thus from this point onward, you can no longer trust the probabilities in (for instance) MikeR's table; they are far too high, for they have not taken into account that the hunter received neither combo card all the way up through turn 4.

My point is that these probabilities (computed later in my answer) are merely a very rough estimation, and they do not take into account even half of the things they should. It also goes without saying that if at least one UtH or Buzzard has already been played (in combo or not in combo), the probabilities will be much lower.

Probabilities

The simplest case is that the deck contains no Tracking, and the player does not mulligan. In this case, MikeR is almost right. Suppose the hunter has so far seen x distinct cards. The probability that there is a starving buzzard and an Unleash the Hounds among these x cards (assuming 2 of each in the deck) is equal to the probability that there is a buzzard, plus the probability that there is an UtH, minus the probability that there is a buzzard or an UtH. This comes out to

[1 - (30-x)(29-x)/(30*29)] + [1 - (30-x)(29-x)/(30*29)]
  - [1 - (30-x)(29-x)(28-x)(27-x)/(30*29*28*27)]
= x(x-1)(x^2 - 113x + 3246)/657720

You can check this answer is correct by verifying that it gives 0, 0, 1, and 1 when you plug in x=0,1,29, and 30, respectively.

With Mulligan

Let s be the size of the starting hand, and suppose the hunter threw away k cards in the mulligan. Let x be the number of cards drawn so far, including the starting hand, but NOT including the cards thrown away. Assume that the player will never mulligan away either of UtH and Buzzard.

For each card thrown away in the mulligan, there is a chance of approximately (x-s)/(30-s) that that card is drawn sometime again in the x cards, and a chance of (30-x)/(30-s) that it isn't. (This would be exact, except that if one card is shuffled in among the x-s cards, the next card is more likely to be among the 30-s cards, as it cannot be in the same place as the first. Nevertheless, this difference should be negligible for x not too low or too high.) We can therefore approximate the effect of the mulligan quite well by saying that the number of cards x is increased by (30-x)/(30-s) for each card thrown away. Then the probability we obtain is

(x+r)(x+r-1)((x+r)^2 - 113(x+r) + 3246)/657720

where r = k*(30-x)/(30-s).

With Tracking

In almost all cases, the effect of Tracking will be equivalent to "draw 3 cards", for purposes of finding the combo. The only case where it is not equivalent is when BOTH (a) the hunter so far has neither of UtH and Buzzard, AND (b) both UtH and Buzzard are picked up among the 3 cards. Unfortunately, unlike with discards from Warlock cards like Soulfire and Doomguard, cards discarded from Tracking are not visible to the opponent, or else one could deal with these special cases just by watching the discard. But, erring on the side of overestimating the chances the hunter has the combo, I will assume that this case is negligible, and thus Tracking always increases the total number of cards drawn/seen by 3.

Actually, things get a bit complicated here. Supposing that the hunter will always play Tracking if he/she has it (to try and fish for the combo), the fact that the hunter has not played Tracking would have to be factored into the probability above, and it has not. On the other hand, if the hunter does play tracking, then that may decrease the chances that he/she has the combo, because he/she is probably fishing. However, I will sweep this objection under the rug and say it is covered by the large Caveat at the start of this answer. In general, it shouldn't be far off to assume that playing Tracking increases the value of x by three.

Summary:

In order to compute an approximate probability that the hunter has UtH+Buzzard in his/her hand, at any given moment, first compute or take note of

• x, the total number of cards drawn so far in the game by your opponent. Include the cards in the starting hand and cards drawn from Tracking or other card draw. Do not include the coin or cards drawn on the mulligan. An easy way to compute this is to subtract the opponent's remaining deck size from 30.

• s, the size of the hunter's starting hand, not including the coin.

• k, the number of cards thrown away on the mulligan.

• r, defined to be equal to k*(30-x)/(30-s).

Then compute the probability using the formula we found:

Probability = (x+r)(x+r-1)((x+r)^2 - 113(x+r) + 3246)/657720.

Here are particular cases of this formula.

Mulligan 0: ((x-1) x (x^2-113 x+3246))/657720
Start with 3 cards in hand, mulligan 1: ((13 x+15) (26 x+3) (338 x^2-38883 x+1137852))/87384843630
Start with 4 cards in hand, mulligan 1: ((5 x+6) (25 x+4) (625 x^2-71950 x+2107056))/60112450944
Start with 3 cards in hand, mulligan 2: ((5 x+12) (25 x+33) (625 x^2-73275 x+2186874))/69907874904
Start with 4 cards in hand, mulligan 2: ((2 x+5) (12 x+17) (12 x^2-1409 x+42117))/260904735
Start with 3 cards in hand, mulligan 3: ((4 x+15) (8 x+21) (16 x^2-1914 x+58329))/539412615
Start with 4 cards in hand, mulligan 3: ((23 x+64) (23 x+90) (529 x^2-63434 x+1937976))/300562254720
Start with 4 cards in hand, mulligan 4: ((11 x+47) (11 x+60) (121 x^2-14839 x+464034))/18785140920

Here's a plot of the probability:

Dotted lines indicate starting with 3 cards (going first), and solid lines indicate starting with 4 cards (going second). Red, Blue, Green, Orange, and Purple indicate 0,1,2,3, and 4 cards tossed in the mulligan, respectively.

As you can see, it turned out that starting with 3 versus 4 cards made hardly any difference. You can also see how the effect of the mulligan is very apparent in the early game, but becomes less significant later when you are more likely to redraw the cards you tossed away. (What appears to be error in the green, orange, and purple lines for low values x may actually be correct; note that x has to be at least equal to s for the probability to be meaningful, so the probability need not be 0 when x=0. If there is any error, it is due to the approximation in the mulligan, and it should be virtually insignificant.)

Was this worth the work? Probably not. The probability is reasonably close to linear, and since we are only looking for a general heuristic, we might as well just say the probability is x/30. If you use this formula, add (k/2) to x, where again k is the number of cards tossed in the mulligan. This is plenty of accuracy for the kind of thing you might need it for, and it has the large advantage of being easy to compute in your head.