Hearthpwn

I needed to get a Bittertide Hydra with charge to win a game yesterday and it got me thinking about this question. I am not a smart man. please help if you know the answer.

We would need to know the total number of beast cards with special text that cost <5 mana, and then we need the total number of beasts with no text (or just keywords) that cost <5 mana.  If anyone has the information, that'd help.

Were you playing standard or wild?

standard

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Odds are pretty low.

The first card is the only one that has text (except a keyword), so this must be the Bittertide Hydra. First pick always has text I believe (can have a keyword as well, but must have text in addition to keyword).

The 2nd one has to be a Stonetusk Boar since this is the only beast with Charge keyword and no other text that costs less than 5. (Tundra Rhino has text so cannot be 2nd pick, even if it could you have to hero power for 2 to pick it, so you can't play a 10 mana zombeast from hand to win).

I can't be bothered counting  the number of beast with/without text costing 5 or less, but the probability would be

(1/(Numbers of beast costing 5 or less with text)) * (1/(number of beasts with no text or just a keyword costing 5 or less))

You only count neutral and hunter beasts.

Work out the answer and show your working...

I think there is a different number of beasts with text costing 5 or less and beasts with no text or keywords only costing 5 or less so I think richard's answer is not correct.

I think my formula is correct though (someone correct me if I am wrong).

EDIT: I am wrong ;)

It's a discover so you see  3 choices. And you are more likely to discover Hunter beasts (dunno if this applies, but if it does each hunter beast is 4 times more likely than non-hunter beasts).

Let me think some more...

Right, so assuming there is no 4x weighting for hunter beasts, chance to see a Bittertide Hydra first pick would be

1-((33/34) * (32/34) * (31/34))

1 - ((33/34) * (32/33) * (31/32)) [EDIT much after the fact: when you see a card in a discover option, it can't be chosen for the other picks, so the number in the denominator goes down by 1 each time]

which 1 minus is the chance to not see it. (You don't see the same choice twice).

If there is a weighting it's more complicated still.

You have to multiply it by the odds of getting a boar 2nd pick too.

DiamondDM13 didn't show his working, tut tut. Did you account for weighting and that you see 3 choices for each?

It's 50/50; you either get it or you don't. 

there are 37 beasts in standard with card text, and 19 with either blank text or keywords.  so if what mungo says is right the possibility would be (1/37) *(1/19)  sooooo it would be  a 0.00142% chance

posted that last comment without seeing what you guys had aswered already ( slow computer sorry)

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.

So... with weighting, it would be like having 4 copies of each hunter card in the card pool.

Since Bittertide Hydra and Stonetusk Boar are neutral, I think (will have to think more about this, could be wrong, if I am it is complicated more by how many hunter beasts you see in the discover option) you can just take the pool size to be 4*(hunter beasts in category) + (non-hunter beasts in category), call this 4H + N.

The choice to see a specific beast would be 1 - (((4H + N - 1)/(4H + N)) * ((4H + N - 2)/(4H + N)) * ((4H + N - 3)/(4H+N))

I think ;)

Do that for each category and multiply the 2 numbers together, multiply by 100 for a percentage.

I am on beer #4 so may be wrong.

Quote from AxletheBeast >>

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.

Quote from richard1234567891011 >>

Quote from mungo2613 >>

Right, so assuming there is no 4x weighting for hunter beasts, chance to see a Bittertide Hydra first pick would be

1-((33/34) * (32/34) * (31/34))

which 1 minus is the chance to not see it. (You don't see the same choice twice).

If there is a weighting it's more complicated still.

You have to multiply it by the odds of getting a boar 2nd pick too.

 Man you can choose 1 beast between 34 in the first choice and in the second choice 1 beast between 15. So the probability is (1/34)*(1/15)=0,19% if like he said in the post he wanted 2 specific beasts. 

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, Source

EDIT3:  Omitting Beasts of other classes (whoops).

Quote from Londipondi >>

Quote from AxletheBeast >>

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

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.