Contra quien peleó? @IbaiLlanos
Published December 22, 2024
At a glance
Views
150,216
Likes
14,762
Comments
1,147
Performance vs. channel
All-time outlier score
Engagement rate
More features coming
We're working on new analytics and tools. Stay tuned.
About
- Published
- December 22, 2024
- Made for kids
- No
More features coming
We're working on new analytics and tools. Stay tuned.
How Contra quien peleó? @IbaiLlanos is doing
It has 150K views in total. It ranks #10,874,231 among tracked YouTube videos by views.
Key facts about Contra quien peleó? @IbaiLlanos
- Total views
- 150K
- Total likes
- 14.7K
- Total comments
- 1.14K
- Published
- December 22, 2024
- Global rank by views
- #10,874,231
Frequently asked questions
How many views does Contra quien peleó? @IbaiLlanos have?
Contra quien peleó? @IbaiLlanos has 150,216 views, 14,762 likes and 1,147 comments, ranking #10,874,231 among tracked YouTube videos by views.
How accurate are these YouTube video analytics?
View, like, and comment 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 the creator sees privately in YouTube Studio.
What is the outlier score?
How this video's views compare to the channel's usual Shorts or long-form video. A 2x score means double the usual views for that channel, 0.5x means half.
What is engagement rate and how is it calculated?
Likes and comments per view, compared to this channel's typical engagement rate — shown as above or below what's usual for the channel.
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 count — not actual creator earnings, which depend on ad rates, audience geography, and monetization settings YouTube doesn't expose publicly.
Is this YouTube video analytics checker free?
Yes. You get 10 free channel and video analytics lookups per 24 hours without an account — logging in, which is also free, removes that limit entirely.