5

Playing around the Starving Buzzard and Unleash the Hound combo is probably the most important factor in a match against a hunter. Letting them draw too many cards and clear the board with the hounds can easily lose you a game. But playing too few minions in fear of UtH can be equally dangerous, especially for decks that rely on having many minions on the board.

Until now I mostly went with gut feeling on the decision to play around UtH or limit the number of minions on my side. But I'd like to put my decisions on a more solid base here.

What is the actual probability that a hunter has Unleash the Hounds and Starving Buzzard on the hand at any point in time? It should be possible to calculate that, taking into account the mulligan phase, tracking and any other card draw. The result would obviously depend on the number of rounds played, or more exactly on the number of cards drawn by the hunter.

  • 1
    This might be impossible to determine, given that we have NO IDEA what the hunter wants from tracking, and we cannot see what they've discarded! We also have no clue what they want to mulligan for, as they don't know what kind of deck we're playing, and we don't know if they're mulliganning for UTH, traps, bow, or kill commands. If you're willing to state some assumptions (deck size 26 because of tracking, ignore mulligan phase, 2 UTH 2 starving buzzards) this question may be answerable. – PunDefeated Aug 21 '14 at 17:37
  • @PunDefeated I think the assumption that they are looking for the combo in the mulligan phase and with tracking are reasonable. I don't want to explicitly restrict it any further in case an answerer has a better argument for different restrictions. – Mad Scientist Aug 21 '14 at 17:39
  • If I play hunter into control warrior, there's almost no way I'm looking for UTH Buzzard, but I guess that's a reasonable assumption for most cases. Tracking still makes me nervous, there are the cases where both parts, if not 3 of the 4 combo cards can show up. This all being said, probability isn't my strong suit, but maybe I can try to figure it out with the 26-card-deck-because-of-tracking assumption before someone smarter can figure out the real math :) – PunDefeated Aug 21 '14 at 17:44
  • In what way do you expect to get answered here? In percentage? Or in x out of y? – AtlasEU Aug 21 '14 at 19:22
  • 6
    Yes. If you're wondering if they have it, the answer is yes. – Studoku Aug 21 '14 at 20:17
11

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:

enter image description here

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.

  • I love this so much – Dustin Mar 3 '15 at 17:01
  • @Dustin Glad you liked it! – 6005 Mar 3 '15 at 20:55
2

Since the combo consists of two different cards, but each card has two identical copies, it's easiest to break this down into a two-part question:

1) What is the chance that he has UTH?
2) Assuming that he has UTH, what is the chance that he has Buzzard?

Let x be the number of cards drawn in total. That means 30-x cards remain in the deck. The chance that neither copy of UTH is in his hand is: (30-x)*(29-x)/(30*29)

Therefore, the chance that he has at least one copy of UTH is: 1-((30-x)*(29-x)/(30*29))

Assuming that UTH is in his hand, there are 29 cards left that we care about, 29-x of which are still in the deck. The chance that neither copy of buzzard is in his hand is: (29-x)*(28-x)/(29*28)

Therefore, the chance that he has at least one copy of buzzard is: 1-((29-x)*(28-x)/(29*28))

Multiplying those two results together gives the answer.

Plugging in possible values for x:

4.    6.60%
5.    9.94%
6.   13.77%
7.   18.03%
8.   22.64%
9.   27.52%
10.   32.60%
11.   37.82%
12.   43.11%
13.   48.42%
14.   53.69%
15.   58.86%
16.   63.89%
17.   68.73%
18.   73.34%
19.   77.67%
20.   81.71%
21.   85.40%
22.   88.72%
23.   91.66%
24.   94.17%
25.   96.26%
26.   97.89%
27.   99.07%
28.   99.77%
29.  100.00%
30.  100.00%

Edit: correcting typo (I had referred to UTH instead of buzzard in the second equation)

Also, Jason has a valid point regarding mulligan affecting the percentage. However, it doesn't affect it in the ways implied so far.

In the simplest case throws back everything. Since the cards are being reshuffled into the deck before redraw, it is as if the opening draw never happened. Therefore, you can use the table above just as it is written.

Where the mulligan actually affects the percentages is if the hunter keeps cards. In the simplest case, assume the hunter keeps only UTH and Buzzard and throws back everything else. If he keeps 1, you know with 100% that he has either UTH or buzzard, and can use 1-((29-x)*(28-x)/(29*28)) to determine his chance of having the full combo. (I did not include the percentage for that but they can be easily computed) Note that x would include the card kept but not the cards thrown back.

The problem is that you don't know what algorithm the hunter uses. (Maybe he also keeps tracking, or hyena, or always keeps at least 1 card no matter what) Now it's like poker, where the raw percentages are a good starting point, but cannot be solely relied on. You have to balance probability with intuition.

  • 2
    Don't forget the differences in going first vs going second (3 cards versus 4) and mulligans specifically looking for UTH, which can give you 8 attempts to get the card in your opening hand. – au revoir Aug 21 '14 at 19:51
  • I based the percentages on number of cards drawn instead of turn number to allow for those differences. However, no player can act before drawing their 4th card anyway so I didn't include x=3. You are right that mulligans do affect the percentages and I am editing my answer to reflect that. – Mike R Aug 21 '14 at 22:38
  • Good answer. I'm fairly sure your probabilities are slightly (significantly?) off due to the fact that if the hunter has two UtHs in the first x cards, the formula (28-x)(27-x)/(28*27) should be used instead of (29-x)(28-x)/(29*28) for the chance of not having either starving buzzard. – 6005 Aug 25 '14 at 22:28
  • One way to see your formula is slightly off: plug in x=1. You should get 0, because there is no chance of getting both the buzzard and the UtH in only one card. yet your formula gives a nonzero answer. – 6005 Aug 25 '14 at 22:54
  • Also, plugging in x=30 gives .997537 instead of the correct 1.0. – 6005 Aug 25 '14 at 23:04
0

You want to calculate the Hypergeometric Distribution based on your wish for 2 different cards out of 4 out 30.

You can get the correct answer here: http://stattrek.com/online-calculator/hypergeometric.aspx

-1

Deck = 30 cards.

Going first = 27 cards.

Draw to 3 mana(cost of UTH) = 24 cards.

Assuming 2 UTH cards in the deck = 2/24 which is 8% chance of them being able to cast UTH with no other buffs.

Needing 5 mana with starving buzzard = 4/22 cards 18% chance assuming they have 2 starving buzzards and 2 UTH in their deck.

If they use Tracking I guess you can do the math, but it gets more complicated. Obviously if they go second it's slightly different and the later in the game the higher the chance of this happening.

  • 1
    dont forget mulligan, if you really need that combo u can throw away other cards. – SSilicon Aug 21 '14 at 18:10
  • 1
    That's why I specified going first. The mulligan really doesn't change the percentage because your hand is selected from the same group of cards that you threw away. For example, if you throw away all 3 cards, there is a possibility that you will draw the same 3 cards again. – FoxMcCloud Aug 21 '14 at 18:22
  • 1
    @ChaseC - The mulligan changes the initial start percentage, because you have the ability to see 6 or 8 cards before the game starts. – au revoir Aug 21 '14 at 20:22
  • This is incorrect when talking about percentages when you put back the same cards, and DRAW FROM THE CARDS YOU PUT BACK. The mulligan phase is important, but doesn't change the fact that you still draw from the cards you put back in your deck. – FoxMcCloud Aug 22 '14 at 12:51
  • @ChaseC Yes and no. Blizzard claims (though I have not tried to check) that you can't draw the cards you put back on the mulligan, though of course you can draw the cards you put back later on. So there is some chance that mulliganing will end up increasing the probability (if the cards get shuffled in near the bottom of the deck), and some chance that it will not change the probability (if the cards get shuffled in near the top of the deck). – 6005 Aug 25 '14 at 22:42

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