$50 por cada camarón que metes 🦐
Published January 20, 2026
At a glance
Views
3,156,381
Likes
136,070
Comments
590
Performance vs. channel
All-time outlier score
Engagement rate
More features coming
We're working on new analytics and tools. Stay tuned.
About
- Published
- January 20, 2026
- Made for kids
- No
More features coming
We're working on new analytics and tools. Stay tuned.
How $50 por cada camarón que metes 🦐 is doing
It has 3.15M views in total. It ranks #2,296,160 among tracked YouTube videos by views.
Key facts about $50 por cada camarón que metes 🦐
- Total views
- 3.15M
- Total likes
- 136K
- Total comments
- 590
- Published
- January 20, 2026
- Global rank by views
- #2,296,160
Frequently asked questions
How many views does $50 por cada camarón que metes 🦐 have?
$50 por cada camarón que metes 🦐 has 3,156,381 views, 136,070 likes and 590 comments, ranking #2,296,160 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.