At a glance
라키 ranks #3,376,614 globally for subscribers (629), #2,508,514 globally for views (350K), 181 videos published.
Subscribers
629
Views
350,472
Videos
181
Charts
Subscribers, views & videos over time
Subscribers gained (weekly) · 6 mo
Views gained (weekly) · 6 mo
Engagement
Avg. views per video
1.93K
Total views ÷ total videos
Views per subscriber
557×
Total views ÷ subscribers
Subs per video
3
Subscribers ÷ total videos
Analytics tables
About
안녕하세욥!! 라키라는 채널을 운영중인 라키이빈다! 저는 종합게임을 할 거에요! 영상, 쇼츠 가림없이 모두 다 올릴 것입니다! 무슨 종합게임일까요? 로블록스, 쿠킹덤 입니다! 다른 게임이 될 수도 있습니다 여기는 뭐 하면 되냐면 구독 1초, 좋아요 1초, 알림설정 3초 총 5초만 쓰는걸 하면 됩니다 헷:) 제 시간은 소중해서 5초만 쓰라는 거냐? 그건 님이 아셔야죠 (크흠)
- Handle
- @라키의로블록스
- Joined
- February 3, 2024
More features coming
We're working on new analytics and tools. Stay tuned.
More features coming
We're working on new analytics and tools. Stay tuned.
Frequently asked questions
How accurate are these YouTube channel analytics?
Subscriber, view, and video counts come from YouTube's public API and refresh regularly. Growth and revenue figures are estimates built from tracked historical data, so they can differ slightly from what a channel owner sees privately in YouTube Studio.
How is a channel's global and country rank calculated?
Rank compares this channel's subscriber count and view count against every other channel Socialcounts.org tracks, both worldwide and within its home country.
How is the revenue estimate calculated?
It's a rough public estimate based on a $0.7–$2.0 CPM (revenue per 1,000 views) applied to view counts — not actual creator earnings, which depend on ad rates, audience geography, and monetization settings YouTube doesn't expose publicly.
Can I check analytics for any YouTube channel, not just my own?
Yes — search any public YouTube channel by name, handle, or channel URL. No login is required to see the basics.
Is this YouTube channel analytics checker free?
Yes. You get 10 free channel and video analytics lookups per rolling hour without an account — logging in, which is also free, removes that limit entirely.