Why did you need a Bittertide Hydra specifically? Just for the 8 attack? I imagine the chances of getting a combined 8+ attack with charge off Build-A-Beast is much, MUCH higher than specifically getting Bittertide Hydra and a charge beast.

I had 6 mana left and my opponent had lethal on board

I'm more confused now. You had 6 mana after hitting the Hero Power button? You needed an 8-attack with Charge? Why didn't you have 8 mana after hitting the button?

It doesn't matter if you just meant the question for knowledge's sake though I suppose.

Blizzard have always said class cards have 4x more chance to appear in a discover.

That's why Stonehill Defender is awesome in Paladin.

Anyway, I corrected my reply where I agreed with you, picking out of 3 assuming no weighting is much higher chance than 1/34 for seeing a hydra.

Maehlice seems to be on the right track but I don't think he's accounting for not being able to see the same card twice in a discover option. (Hence my formula where instead of cubing you reduce the number by 1 each time, I think that isn't valid though, we may need to consider the scenario where you see 0, 1, 2 or 3 hunter beasts separately). That involves making a probability tree, not hard but hard to convey in ASCII format ;)

Yeah we don't know, so the right thing to do is calculate both probabilities (weighted and not weighted), then we have all bases covered.

Maehlice is nearly there I think. I still think we may have to split the cases based on how many hunter beasts we see in the discover if they are weighted.

We definitely have to consider how many hunter beasts we see if it is weighted towards hunter beasts.

Each hunter beast you see effectively removes 4 from the pool of beasts you can see for the next pick instead of 1 removed if you see a non-hunter beast.

So a tree diagram will need to be drawn up, tedious :/

They're not complicated though and we just need to separate the cases where we see 0, 1 or 2 hunter beasts (3 would not count since both beasts we are interested in are neutral).

OK, so you need to factor in whether 0, 1, 2 or 3 hunter beasts are seen in the discover options.

Now I'm trying to decide if the order it decides whether it's a hunter beast or not is important (I don't think it is at this beer level). Pretty sure we need to take into account how likely it is to be offered 0, 1, 2 or 3 though.

In any case man if it is weighted the caclulation is simplier than your. You simply have to add 4 to hunter's beasts instead of 1. (because he wanted 2 neutral) So there are 14 hunter's beasts in the first choice and and 1 in the second choice. So you can simply add 3 (4-1) for each hunter's beasts. So you have to add 42 in the first choice and 3 in the second. 34+42=76 15+3=18 so (1/76) *(1/18)=0,073% and this is your probability to find the Hydra with charge if hunter's beasts have a 4x chance to be found.

I don't think it's as simple as that, since once you see 1 hunter beast as an option, the probability for the 2nd and 3rd options to be hunter beasts also goes down by more than if the first beast was not a hunter beast.

EDIT: Basically, the more hunter beasts offered in the discover the more likely it is for the remaining cards to be non-hunter beasts since the available pool is reduced 4 times as much for each hunter card shown in the pick.

Build-a-Beast allows the player to create a Zombeast through a Discover-like interface. The player is presented with 3 Beast minions twice to choose from. The first choice will always be from Beasts with card text, whereas the second choice will always be from Beasts with either only keywords or no card text at all.^{[1]} A minion card is then generated for the player with the combined mana costs, Attack, Health and card text of the two Beasts. The Zombeast will have the card art and quotes of the first minion chosen.^{[2]}^{[3]}

Minions with Battlecry and Enrage count as being in the first pool of cards, since they have additional text besides the keyword to specify what their effect does (for example, "Battlecry: Deal 1 damage", or "Enrage: +2 Attack", instead of simply "Battlecry" or "Enrage").^{[4]}^{[5]}

Hunter Beasts and neutral Beasts have an equal chance of being offered.^{[1]}

The player will never be offered a Beast that costs more than 5 mana, since this could potentially create a Zombeast that costs more than 10 mana and is thus impossible to play under normal circumstances.^{[6]}

True! But not the point. Build-A-Beast doesn't Discover, it just has the same interface. Discover effects specifically add the card to your hand. It if doesn't SAY Discover and put the chosen card in your hand, it's not Discover and doesn't necessarily follow the same rules.

In this case I checked Hearthstone Gamepedia and found cited notes on this: it is not like Discover; there is an equal chance to pull class and non-class minions off Build-A-Beast.

OK, looks good then. That is the answer. This thread escalated rather quickly :)

EDIT: Windows calculator is annoying, if you paste something you have to press = for it to evaluate it fully, lol. That was confusing me for about the last half hour.

Note to self: next time download Python for use as a calculator.

It's a good job that just 1 beast from each pick was required, otherwise we could have been arguing about it for days ;) [It's not difficult if X beasts are acceptable from pick 1 and Y are acceptable from pick 2 though. Gets a bit more complicated if what you see in your first pick makes options in the 2nd pick viable].

I think we all learnt something today anyway (I learnt not to try and do probability while drinking but hope I provided some hints for people to reach the correct answer).

EDIT: In before someone asks, if I build a Zombeast, and I need a either the classic Bittertide Hydra/Stonetusk Boar combo or the alternative Vicious Fledgling/Stonetusk Boar combo but I'd then need the Vicious Fledgling to get windfury into either +1/+1 or +3 attack, what are the odds for that?

EDIT: In before someone asks, if I build a Zombeast, and I need a either the classic Bittertide Hydra/Stonetusk Boar combo or the alternative Vicious Fledgling/Stonetusk Boar combo but I'd then need the Vicious Fledgling to get windfury into either +1/+1 or +3 attack, what are the odds for that?

Chance to get any one particular adaptation is 1 - (9/10) * (8/9) * (7/8) = 30 %

Chance to get any of two adaptations is 1 - (8/10) * (7/9) * (6/8) = 53.333 %

So to get a Stonetusk Fledgling and then roll Windfury into +1/+1 or +3 attack is 0.01357466 * 0.3 * 0.53333 = 0.21719456 %

EDIT: Why did I think there were only 6 adaptations? Corrected ...

Good, Good my minions... Today, the probability to get a specific monster. Tomorrow THE WORLD!

(Laughs maniacally)

It only occurs to me now that one of us should have chimed in with a Mr. Meeseeks joke somewhere in here (especially after the guy did the Rick one) since you literally asked people to come and help you come up with an answer. =P

This is one of the most productive threads I've ever seen in all of my years on forums. xD

At 5-mana or less, there are 20 Hunter Beasts and 36 Neutral Beasts

Since Class cards are 4x more likely to appear, the total pool is essentially out of (20*4) + 36 = 116.

The odds of NOT getting one of the two desired beasts on the first step is (114/116) * (113/115) * (112/114) = 94.872563 %

Therefore the odds of succeeding in the 1st step is 5.127436 %

And, the odds of NOT getting the other desired beast on the second step is (114/115) * (113/114) * (112/113) = 97.391304 %

Therefore, the odds of succeeding in the 2nd step is 2.608695 %

The odds of succeeding at both is 0.05127436 * 0.02608695 =

0.133759 %EDIT: Forgot to consider duplicates.EDIT2: Sources or duplicates: Source,Source,SourceEDIT3: Omitting Beasts of other classes (whoops)."Nerf Paper," said Rock.

Blizzard have always said class cards have 4x more chance to appear in a discover.

That's why Stonehill Defender is awesome in Paladin.

Anyway, I corrected my reply where I agreed with you, picking out of 3 assuming no weighting is much higher chance than 1/34 for seeing a hydra.

Maehlice seems to be on the right track but I don't think he's accounting for not being able to see the same card twice in a discover option. (Hence my formula where instead of cubing you reduce the number by 1 each time, I think that isn't valid though, we may need to consider the scenario where you see 0, 1, 2 or 3 hunter beasts separately). That involves making a probability tree, not hard but hard to convey in ASCII format ;)

Yeah we don't know, so the right thing to do is calculate both probabilities (weighted and not weighted), then we have all bases covered.

Maehlice is nearly there I think. I still think we may have to split the cases based on how many hunter beasts we see in the discover if they are weighted.

We definitely have to consider how many hunter beasts we see if it is weighted towards hunter beasts.

Each hunter beast you see effectively removes 4 from the pool of beasts you can see for the next pick instead of 1 removed if you see a non-hunter beast.

So a tree diagram will need to be drawn up, tedious :/

They're not complicated though and we just need to separate the cases where we see 0, 1 or 2 hunter beasts (3 would not count since both beasts we are interested in are neutral).

https://www.mathsisfun.com/data/probability-tree-diagrams.html

"Nerf Paper," said Rock.

OK, so you need to factor in whether 0, 1, 2 or 3 hunter beasts are seen in the discover options.

Now I'm trying to decide if the order it decides whether it's a hunter beast or not is important (I don't think it is at this beer level). Pretty sure we need to take into account how likely it is to be offered 0, 1, 2 or 3 though.

From https://hearthstone.gamepedia.com/Deathstalker_Rexxar:

My previous answer was way off. Recalculating ...

"Nerf Paper," said Rock.

Ok, equal chances for hunter and non-hunter beasts makes it a lot simpler.

So it's just (1-(not bittertide hydra)) * (1-(not stonetusk boar)) then...

Bittertide Hydra can only be chosen during the first Discover set, which is out of 39 cards.

Stonetusk Boar can only be chosen during the second Discover set, which is out of 17 cards.

Odds of succeeding the first step is 1 - ( 38/39 * 37/38 * 36/37 ) = 7.692307 %

Odds of succeeding the second step is 1 - ( 16/17 * 15/16 * 14/15 ) = 17.647058 %

Odds of succeeding both steps is 0.07692307 * 0.17647058 =

1.357466 %"Nerf Paper," said Rock.

"Nerf Paper," said Rock.

OK, looks good then. That is the answer. This thread escalated rather quickly :)

EDIT: Windows calculator is annoying, if you paste something you have to press = for it to evaluate it fully, lol. That was confusing me for about the last half hour.

Note to self: next time download Python for use as a calculator.

Good, Good my minions... Today, the probability to get a specific monster. Tomorrow THE WORLD!

(Laughs maniacally)

It's a good job that just 1 beast from each pick was required, otherwise we could have been arguing about it for days ;) [It's not difficult if X beasts are acceptable from pick 1 and Y are acceptable from pick 2 though. Gets a bit more complicated if what you see in your first pick makes options in the 2nd pick viable].

I think we all learnt something today anyway (I learnt not to try and do probability while drinking but hope I provided some hints for people to reach the correct answer).

EDIT: In before someone asks, if I build a Zombeast, and I need a either the classic Bittertide Hydra/Stonetusk Boar combo or the alternative Vicious Fledgling/Stonetusk Boar combo but I'd then need the Vicious Fledgling to get windfury into either +1/+1 or +3 attack, what are the odds for that?

Chance to get any one particular adaptation is 1 - (9/10) * (8/9) * (7/8) = 30 %

Chance to get any of two adaptations is 1 - (8/10) * (7/9) * (6/8) = 53.333 %

So to get a Stonetusk Fledgling and then roll Windfury into +1/+1 or +3 attack is 0.01357466 * 0.3 * 0.53333 =

0.21719456 %EDIT: Why did I think there were only 6 adaptations? Corrected ...

"Nerf Paper," said Rock.

Bittertide Hydra/Stonetusk Boar would work too though.

So it's same answer as before + (1-(same answer as before)) * your new answer

I think (beer #6 now).

EDIT: There are 10 adaptations not 6 anyway :P