# Talk:Abyssal bludgeon

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buy limit is 8 but wiki is dumb so i dont know how to edit.

no i dont have a source someone in cc figured it out

## Expected Sire kills for full bludgeon

The following message is copied from User talk:90.252.148.164#Full bludgeon chance from Sire

Hey there, you just changed the Abyssal bludgeon page to state that 480 is the expected number of Sire kills to get the full bludgeon. What did you base this number off? From what I can tell, the previous number of 618 kills was based on the following:

• There's a 1/100 chance to get an unsired per kill.
• Every unsired has a 62/128 chance to give any bludgeon piece, without any duplicates being possible before completing a bludgeon.
• Effectively, this is a 1/2 chance to get a bludgeon piece for each unsired. Conveniently, that's the same as a coin toss.
• StackExchange to the rescue (because I'm too lazy myself to work this all out formally): https://math.stackexchange.com/a/3378200/191172 says you should expect to need 6 coin tosses (i.e. unsireds) until your third time getting heads on the coin, following $E_{n}=n*E_{1}$ .
• Now going back to the actual $P$ robability: 1/206 to get a bludgeon piece (1/100 * 62/128 = 1/206.45), which means $E_{1}=1/P=206$ .
• Filling in $n=3$ for three required pieces, gives an $E$ xpected number of Sire kills of $E_{3}=3*206=618$ (or more accurately, $3*206.45=619$ )

I'm not sure what formula you followed to get an approximate of 480, but going by this, I'm quite convinced the old number of 618 was correct. I'd love to see your calculations though! Joeytje50 talk 22:15, 22 March 2021 (UTC)

Sorry i didn't know how to reply to messages so i guess ill edit this one haha. Sorry my maths is probably bad, i went on that bludgeon was a 1/2 chance from an unsired to make it easier for myself, and i got that after 5 unsireds you have a 16/32 of all outcomes to get 3 bludgeon pieces, in other words 50% chance. and then using the wiki dry calculator it said by 467 kills you have a 50% chance of having 5 unsireds of more. but as a bludgeon piece isn't 1/2 its slightly lower i rounded it up to 480. If this is all wrong my bad, not my strong point, the whole 16/32 chance thing could be wrong, however even if it was 6 unsireds on average to get a bludgeon then the kills it would take on average is more like 570, as you don't just times the drop rate by 6 for average probability. I guess check out the wiki dry calculator page, might explain it better. Sorry for wasting your time if i've got the wrong end of the stick on this one. —The preceding unsigned comment was added by 90.252.148.164 (talk) on 17:20, 23 March 2021 (UTC).

The following message is copied from User talk:138.199.29.134#Bludgeon expected kills

Hey there, I saw you slightly changed the amount of expected kills to receive the full bludgeon. You mention the formula ${\frac {p*r}{1-p}}$ in your edit summary, which supposedly results in 616.4 kills for the full bludgeon. Filling in the values you mention in your summary, this would be ${\frac {\left[{\frac {1}{100}}*{\frac {62}{128}}\right]*3}{1-\left[{\frac {1}{100}}*{\frac {62}{128}}\right]}}\simeq {\frac {1}{68.5}}$ which is clearly not what you meant.

In an attempt to reproduce your result though, I was able to figure out that the formula you must have used would be $P={\frac {p}{r*(1-p)}}$ or $E_{r}={\frac {r*(1-p)}{p}}$ , which would give $E_{3}=3*{\frac {1-\left[{\frac {1}{100}}*{\frac {62}{128}}\right]}{\left[{\frac {1}{100}}*{\frac {62}{128}}\right]}}\simeq 616.4$ .

However, understanding which formula you used doesn't mean I understand quite where this formula is coming from. Filling this formula in for something that's easier to understand intuitively, such as a coin flip (50/50 chance) would imply that the chance to get three (non-consecutive) heads in a series of coin flips, if your formula is correct, would be $E_{3}=3*{\frac {1-{\frac {1}{2}}}{\frac {1}{2}}}=3*{\frac {\frac {1}{2}}{\frac {1}{2}}}=3$ . I think we both can see clearly that the expected number of coin tosses to get three heads is not three coin tosses. So there must be something wrong in your generalized formula. I'm not quite sure how you were able to arrive on a formula that involves ${\frac {p}{1-p}}$ in this situation, but I'd be more than happy to read your reasoning and perhaps improve the page.

For now, your edit has been undone by someone else. I don't quite agree with the edit summary in that undo edit, but I do think there must be some mistake in your formula, so for now I think the value on the page should remain at 619. Please enlighten me if I am making a mistake with my calculations. See this talk page message I've left someone else yesterday, if you want to read through the logic behind an expected value of 619 kills. Joeytje50 talk 20:04, 23 March 2021 (UTC)

Hi Joey, the distribution of the number of kills to get the bludgeon is negative binomial. The properties of that distribution can readily be found online, such as wikipedia. Since people don't like fractions they clearly don't want the mean so I've put in the median. This isn't in closed form but can be numerically calculated. Please let me know if you feel like anything is missing. Thanks.

Just noticed you mentioned this on your talk page after I undid the edit on the bludgeon page. I might be misunderstanding the statistics though, so if you get a chance can you put the calculations you're doing on Talk:Abyssal bludgeon? I think having the discussion there is ultimately going to be better than having it spread around multiple IP talk pages, and I've asked Joey to throw his info there as well :). Sorry if I was a bit hasty on undoing the change you made, but there does seem to be some confusion between multiple people on the real rate and the math needed to figure it out. BigDiesel2m (talk) 20:38, 23 March 2021 (UTC)

Please continue this discussion below this line

Just to clarify,

Hi, we want 3 successes where each has probability 62/12800. This is a textbook examples of the negative binomial distribution with parameters p=1-62/12800 and r=3. The properties of the distribution can be found quickly by googling. The mean is p*r/(1-p)+r, telling the average number of kills which need not be integral. The mode is floor(p*(r-1)/1-p)+r, telling the most likely number of kills. The median can be calculated numerically, it says how many kills most will have to endure. We can calculate other quantiles to get a confidence interval. It all just immediately follows, so please don't revert.

Apologies, the formulas I've given is for the number of kills without an unsired which gives a bludgeon piece. The actual answer is of course 3 higher. In your example where p=1/2, you're finding the expected number of tails. The expected number of tosses is 3+that. Good catch. (NOte, I've updated the formulas above.. now they're correct for what we want..)138.199.29.134

Ah, yes that sounds much more reasonable. I've edited the page now to include both the mean and the mode, so people can choose which one they find more relevant. It's funny to see that the two routes to get to the expected number of kills of 619 seem to be quite different approaches, but in the end come down to the same number. Anyway, thanks for sharing your thoughts, and thanks for this clarification! Joeytje50 talk 21:42, 23 March 2021 (UTC)

I was expecting to have to explain how to calculate the median since it's not obvious like the mean. For what it's worth, the python code below will return it. Notice you must have the distribution.

from scipy.stats import nbinom
r,p=3,62/12800
nbinom.mean(r,p)+r
nbinom.median(r, p)+r


—The preceding unsigned comment was added by 138.199.29.134 (talk) on 22:43, 23 March 2021 (UTC).

Thanks, that'll be useful in case anyone is unsure, or wants to calculate this for something else. The way I verified the number 552 was using our 'dry calc', which gives a 50% chance to get 3 or more bludgeon pieces if you fill in 62/12800, 552 and 3 in the input boxes. However a script like that doesn't just verify but also calculates, is probably useful to have listed here. Thanks!
PS: To sign, please use four tildes, that way the tildes will be substituted by a link to your user info, as well as the time of your comment. Sadly, the two tildes you inserted did not work. Joeytje50 talk 23:13, 23 March 2021 (UTC)
Hi Joey, thank you for the pointers and fixing up my formatting. Just for future reference, using the Dry calc just worked by coincidence there and won't work in general. This answer again requires thinking about probability: the dry calc uses a binomial distribution but this is about a negative binomial distribution. (To be technical for a second, the dryness calc only works for that as the probability approaches zero, i.e. because the bludgeon chance is low.) For example, if you run the python code above for r,p=123,73/75 and then feed the result into the dry calc, you won't get 50%. They're just measuring different things. 138.199.29.134 23:58, 23 March 2021 (UTC)