Hearthpwn

My weighting theory was debunked, hunter cards are no more likely than neutral beasts, I was overcomplicating things like I always do ;)

Beasts from other classes don't appear though, so just neutrals and hunter cards. Assuming Maehlice did the adding up of beasts with and without text (or just keywords) correctly the answer they gave looks correct. So a combinatorics approach should give the same result.

Quote from mungo2613 >>

My weighting theory was debunked, hunter cards are no more likely than neutral beasts, I was overcomplicating things like I always do ;)

Beasts from other classes don't appear though, so just neutrals and hunter cards. Assuming Maehlice did the adding up of beasts with and without text (or just keywords) correctly the answer they gave looks correct. So a combinatorics approach should give the same result.

EDIT: Was using wrong numbers apparently

Ok, Maehlice counted 39 and 17 not 34 and 15 though, do your figures agree in that case?

Apparently the picture posted does not include beasts from KFT (not checked myself, just about to start beer #7).

You're right, I should've paid closer attention. Updated and numbers now agree.

Well that's a relief. Case closed!

Quote from mungo2613 >>

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. 

 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.

Quote from mungo2613 >>

Quote from mungo2613 >>

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. 

 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 was wrong to disagree with this, you were correct, the chances to see a specific beast is just 1/(num beasts) since there is no weighting for hunter beasts.

Seeing 3 cards makes no odds since not seeing the specific card has odds

1 - ((N-1)/N * (N-2)/(N-1) * (N-3)/(N-2))

= N/N - ((N-1)(N-2)(N-3)/(N(N-1)(N-2))

= N/N - ((N-1)(N-2)(N-3)/(N(N-1)(N-2))

= N/N - (N-3)/N

= (N-(N-3))/N

= 3/N

so the probability is the same.

Algebra wins the day and proved me wrong ;) Well played

Yeah, I missed the factor of 3 missing from his argument.

But it does turn out that you don't need to do the 1 - (don't see the beast) calculation because you can just multiply the odds of seeing a single beast by 3.

Quote from DiamondDM13 >>

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.

So anyone has a correct list of the two sets of choices BTW? I didn't find one anywhere. 

I have been working on this. And I can give you an answer:
First, we need to consider two facts:

• In Standard, we have 56 eligible beasts (36 neutral and 20 hunter beasts)
• Hunter beasts are 4 times more likely to appear than neutral beasts

With these facts, we can determine the probability of getting a Neutral or Hunter minion

For the first pick, we have 8 possible scenarios: all 3 beast options are neutral (N,N,N), the first 2 beast are neutral and the third one is hunter (N,N,H), ..., etc

I calculated the probabilties in this public spreadsheet if you want to check my calculations:

https://docs.google.com/spreadsheets/d/1on0EkZivCYn2BHsCc6Sh1LGQPgZqG6Pt0MnR2pUfrKg/edit?usp=sharing

The probability of getting a neutral minion can be calculated adding all the probabilities where at least 1 neutral minion can be picked: 68.80%.  Then if we want to know the probability of picking an specific neutral minion, we need to divide this value by 36 = 1.94%

Now, we need to calculate the probability of picking a Neutral beast in the second pick given the fact we already picked a neutral minion in the first pick:
Following the same table and considering we have now 35 out of 55 neutral beasts, we have:

If we sum all the scenarios where where at least 1 neutral minion can be picked, we have: 67.95%, and dividing by 55 = 1.91%

The probability of picking a second specific neutral minion is 1.91%

So the probabilty of picking two specific neutral beasts (Bittertide Hydra + Stonetusk Boar) is: 1.94% * 1.91% = 0.04%

Please advise if you see any mistake in my calculations or my method.

Hunter beasts are not more likely to appear than normal beasts, Maehlice and mysticsnake already came up with the correct answer.

EDIT: Citation for the hunter beasts not appearing more often than neutral ones is in post #35 on page 2 of this thread.

you're right. Also I didn't consider some cases in my calculations. I guess I won't sleep tonight until find out the answer

Read post #39 and get some sleep ;)

EDIT: And using my algebra from post 58 makes it simpler still (no need to do the 1-(don't see a specific beast) calculation, replace it with 3/(num beasts per selection), i.e.

3/39 * 3/17 = 9/663 = 3/221 (same answer mysticsnake came up with via combinatoric argument)

= 0.01357 ish

i.e. 1.357%

EDIT2: You don't really even need to do any algebra to see that if you pick 3 beasts from a pool of N with no duplicates being shown means that the chance of getting a specific one is 3/N.

This is because you see 3 beasts and you do not see (N-3) of them. The chance of your specific beast showing up in the selection is therefore 3/N. You need to do the 1-(not seeing a specific beast) calculation (or use a combinatoric argument which is basically stuff involving numbers from Pascal's Triangle) if more than one beast is an acceptable pick though I think.

disregard this message