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.

Listen Morty, I hate to break it to you, but what people calls "love" is just a chemical reaction that compels animals to breed. It hits hard, Morty, then it slowly fades, leaving you stranded in a failing marriage. I did it. Your parents are gonna do it. Break the cycle, Morty. Rise above. Focus on ZOMBEAST science.

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))

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).

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.

Roughly 0,01442307692307692307692307692308% chances, in standard. Only 1 option for the first part and 1 option for the second part to work, so that are roughly the odds. (Rhino only shows up on first option, so the only possible option to match with Hydra with charge is Boar)

I came up with 39 Text Beasts and 16 plain Beast from what I have experienced that they count as Text and Plain, it could not be entirely accurate.

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

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.

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

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.

Yeah, that's correct if it is unweighted. I overcomplicated things by trying to not include duplicates, but there is only one you are interested in so it doesn't matter. I am doing beermath atm.

I'm pretty sure if hunter beasts are weighted 4x more likely like other discover options it is more complicated.

EDIT: No, I am right, seeing 3 beasts makes it much more likely to see the one you want - it's not just a straight 1 in 34 chance.

Chance is 1 - (not seeing it in a pick from 3)

Which is what I posted assuming no weighting for hunter beasts, i.e.

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

EDIT2: Disregard that, you need to reduce the denominator as well as the numerator since the pool gets smaller as you remove cards from it. (so, 1 - ((33/34) * (32/33) * (31/32))).

*Posted after Maehlice calculated it correctly at the end of page 2 of this thread.

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.

Minion (6)Ability (23)Playable Hero (1)Were you playing standard or wild?

standard

Listen Morty, I hate to break it to you, but what people calls "love" is just a chemical reaction that compels animals to breed. It hits hard, Morty, then it slowly fades, leaving you stranded in a failing marriage. I did it. Your parents are gonna do it. Break the cycle, Morty. Rise above. Focus on ZOMBEAST science.

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.

Roughly 0,01442307692307692307692307692308% chances, in standard. Only 1 option for the first part and 1 option for the second part to work, so that are roughly the odds. (Rhino only shows up on first option, so the only possible option to match with Hydra with charge is Boar)

I came up with 39 Text Beasts and 16 plain Beast from what I have experienced that they count as Text and Plain, it could not be entirely accurate.

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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

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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.

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~~Yeah, that's correct if it is unweighted. I overcomplicated things by trying to not include duplicates, but there is only one you are interested in so it doesn't matter. I am doing beermath atm.~~~~I'm pretty sure if hunter beasts are weighted 4x more likely like other discover options it is more complicated.~~~~1 - ((33/34) * (32/34) * (31/34))~~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.