Meta · Published 2026-06-03 · 4 min read
The maths behind fishing for a winning campaign
I am trying a new format and I am nervous — I do not know how interesting you find maths. If the post lands I will publish more like it.
What is "fishing for a catch"?
A common UA strategy. The buyer launches three to five campaigns with an identical or similar setup. A few days later they look at the results and find the campaign that caught on — the one that showed a positive ROAS. The rest are switched off.
The logic: the Meta, TikTok or Google algorithm ramps each campaign up differently, and the buyer's job is to find the one that got lucky with a good audience.
A thought experiment
Let us test that strategy with maths. Take an app that definitely does not make money, and see whether the buyer can still "find a catch".
If you never did probability theory at university, do not worry. The frightening formulas can be skipped; go straight to the conclusion.
The app's parameters
- LTV per paying user = $32
- Conversion to paying = 10%
- CPI = $4
- True ROAS = 80%
Note that the app is definitely loss-making. Each install brings $3.20 on average and costs $4. The buyer launches five campaigns at $50 each. Question: what is the probability that at least one shows a ROAS of 100% or more?
The model
Step 1. The coin. Every install is a coin toss. With probability 10% the user pays $32. With probability 90% they pay nothing.
Step 2. How many tosses. A $50 budget at a $4 CPI is 12 installs. That is 12 coin tosses.
Step 3. What counts as success. For the campaign to show a ROAS of 100% or more we need at least 2 payers out of 12. Two times $32 is $64 of revenue against $50 of spend, which is a ROAS of 128%.
Step 4. What is the probability? The probability of getting 2 or more payers out of 12 with a 10% chance each is computed through the binomial distribution. The result: 34%.
So every third campaign will show a ROAS of 100% or more. That is simply how randomness works.
Step 5. Five campaigns. The probability that at least one of the five shows a positive result: 1 − (1 − 0.34)^5 = 86%.
The formula, which you may safely skip
The binomial distribution gives the probability of getting exactly K successes out of N attempts: P(K of N) = C(N,K) × p^K × (1−p)^(N−K), where C(N,K) = N! / (K! × (N−K)!) is the number of combinations.
For our case: P(0 payers of 12) = 0.9^12 = 0.28. P(1 payer of 12) = 12 × 0.1 × 0.9^11 = 0.38. P(0 or 1) = 0.28 + 0.38 = 0.66. P(2 or more) = 1 − 0.66 = 0.34.
For several campaigns: P(at least one shows ROAS ≥ 100%) = 1 − (1 − P(one))^M, where M is the number of campaigns.
How the probability depends on budget
Spend $50 per campaign gives 12 installs and needs 2 or more payers. One campaign: 34%. At least 1 of 5: 86%.
Spend $150 gives 37 installs and needs 5 or more payers. One campaign: 31%. At least 1 of 5: 84%.
Spend $300 gives 75 installs and needs 10 or more payers. One campaign: 22%. At least 1 of 5: 71%.
Spend $1,000 gives 250 installs and needs 32 or more payers. One campaign: 9%. At least 1 of 5: 36%.
What came out of it
At $50 per campaign the chance of seeing a positive ROAS on at least one of five campaigns is 86%. You will almost certainly see a positive ROAS.
Even at a more or less normal spend of $300 the chance is 71%.
But note: the probability of a false positive on a single campaign at $300 is only 22%. By launching one campaign instead of five, the buyer cuts the chance of fooling themselves from 71% to 22% — in other words, gets three times the reliability out of the same judgement.
In short: better to launch one campaign and watch it than to launch five and watch all of them.