Talk:Uzi

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Drop chance source?

Where does this 1/80000 information come from? — Preceding unsigned comment added by 50.152.227.94

Source code. --0icke0 (talk) 10:40, 8 October 2013 (UTC)

Another failed fix?

Can anyone confirm the drop rate is actually fixed? I've been farming Angry Trappers for quite a long time, but with no result. — Preceding unsigned comment added by 5.167.248.46 at 12:28, 26 October 2013 (UTC)

My 2 Uzis immediately after the update and third one a bit later confirm that. Barhandar (talk) 12:36, 26 October 2013 (UTC)
Thank you. I've just got the damn thing. Now - it's pumpkin time! — Preceding unsigned comment added by 46.146.73.224 at 14:09, 26 October 2013 (UTC)

Help

I playing on mobile, having fun and im like "oh, i should get an Uzi, I dont have one yet" but long time of farming no result anyone got any tips for me? plz help meee 172.249.169.192 20:58, 14 September 2017 (UTC)

Well since mobile still has 1/80000 chance to get it to drop dont expect yourself to get one — Preceding unsigned comment added by 112.207.203.26 at 06:03, November 3, 2017 (UTC)
I understand your frustration, as I have tried farming a Uzi on Mobile version Mobile as well. However, you should go to the forums to discuss game-related questions or ask for help with farming. —ϟAwesome_Diamondsϟtalk 12:00, 3 November 2017 (UTC)
Mobile's drop chance is 1/200. Basically, with each Angry Trapper defeated, a 200-sided dice is rolled, and if you get a "1", you get an Uzi. You could end up with an Uzi after the first Angry Trapper you defeat, or after the 1 millionth. – Ferretwings (talk) 01:05, 8 November 2017 (UTC)

High Velocity Bullets

Visually, high velocity bullets shoot faster on the Uzi and other bullet velocity increasing guns. But I have always wondered if they are affected at all? Are they? Uygfjmhgvh (talk) 08:49, 16 April 2019 (UTC)

All weapons with projectiles have different Velocity values, ammo shares this property. The Uzi has a high base Velocity, and this value is added to the already high one of High Velocity Bullets, so naturally the Velocity of an Uzi and High Velocity Bullets is higher than with other guns. Bame66 (talk) 16:39, 16 April 2019 (UTC)
Thank you. Uygfjmhgvh (talk) 12:49, 6 May 2019 (UTC)

Vandalized

Just managed to catch the beginning of a chain of vandalizations, you can see it in the history (until it gets purged at least).... i signed up on google to try and revert it, but find i have no clue how to actually revert to the pre-vandalized page (i can find it in logs, but dont know how to revert properly). Figured posting here might draw an admins attention faster. — Preceding unsigned comment added by Thornstromb (talk • contribs) at 00:31, 20 October 2020 (UTC)

I have managed to undo the edit. I'm no wiki expert, but usually what I do when trying to undo many consecutive edits is to press "edit" on the revision directly before the chain of vandalism, then save without making changes. That saves the page as the selected revision, essentially allowing reversion to a certain point in history. Hope that helps, though I don't know if that's the only way.--2057clones (talk) 00:48, 20 October 2020 (UTC)
He is now blocked, thanks. --Dinoxel (talk) 00:57, 20 October 2020 (UTC)

Drop chance explanation

Explanation for the tip "One would need to defeat 230 Angry Trappers for a 90% chance of obtaining an Uzi.":

This is a Bernoulli process, a sequence of statistically independent Bernoulli trials. Each trial is described as "Kill an Angry Trapper and see if it drops an Uzi". "Success" is if it does drop one, "failure" is if it doesn't. [math]\displaystyle{ p }[/math] is the probability of success. As of Desktop 1.4.4.7, [math]\displaystyle{ p = 1\% = 0.01 }[/math].

The stochastic variable [math]\displaystyle{ X }[/math] is the number of successes (number of dropped Uzis) and [math]\displaystyle{ n }[/math] is the number of trials (number of killed Angry Trappers). [math]\displaystyle{ X }[/math] is binomially distributed ([math]\displaystyle{ X \sim B(n;p) }[/math]), the probability mass function is [math]\displaystyle{ P(X=k) = B(n;p;k) = \binom{n}{k} \times p^k \times (1-p)^{n-k} }[/math] with [math]\displaystyle{ k }[/math] as a given number of successes.

How many trials are needed at least so that the probability of at least 1 success ([math]\displaystyle{ P(X \ge 1) }[/math]) is at least 90% ([math]\displaystyle{ \ge 0.9 }[/math])?

[math]\displaystyle{ \begin{align} P(X \ge 1) & \ge 0.9 \\ 1 - P(X=0) & \ge 0.9 \\ P(X=0) & \le 0.1 \\ B(n;0.01;0) & \le 0.1 \\ \binom{n}{0} \times 0.01^0 \times (1-0.01)^{n-0} & \le 0.1 \\ 0.99^n & \le 0.1 \\ n \times \log 0.99 & \le \log 0.1 \\ n & \ge 229.1052883 \end{align} }[/math]

At least 230 Angry Trappers need to be killed for the probability of obtaining at least 1 Uzi to be at least 90%.

Also see Talk:Armored Skeleton#Armor polish drop? for a similar, more detailed explanation.

--Rye Greenwood (talk) 20:51, 8 November 2022 (UTC)