1991issue C051-4
How equal independent stakes change the odds of a complete loss
A two-thirds success chance still leaves a one-third probability of losing the entire stake in a single position. Holding that success-failure-pair fixed, equal-share-allocation turns complete-loss and overall-loss-probability into a discrete count across a chosen number of independent stakes.
- A stylized trial with success chance two thirds and failure chance one third still leaves a one-third probability of losing the entire stake when capital is concentrated in one position.
- A finite starting purse cannot rely on the long-run two-thirds success rate that an unlimited purse split into many equal independent amounts would eventually realize.
- Five equal independent positions lower the complete-loss probability to 0.0041 and the overall-loss-probability to 0.21, so the complementary chance of a net gain is 0.79.
- Ten equal independent positions reduce the all-failure probability to 0.000017 and the overall-loss-probability to 0.077, so the chance of a win or a break-even outcome is 0.923.
A favorable pair can still empty a single stake
A selection process can still have a higher chance of success than of failure and produce sizable losses if the entire purse is committed to one outcome.
A stylized independent trial with success chance two thirds and failure chance one third still leaves a one-third probability of losing the entire stake when capital is concentrated in a single position.
Those two assumed per-position chances form a success-failure-pair, and they must add to one.
A finite purse cannot wait for the long-run average
An unlimited purse split into many equal independent amounts would eventually realize the two-thirds success rate, but a finite starting capital cannot rely on that long-run average.
Equal-share-allocation is the division of that finite starting purse into a set number of equal independent stakes.
Five equal independent positions
Under the same two-thirds and one-third pair, splitting the stake into five equal independent positions lowers the probability of losing every position to 0.0041.
That complete-loss case is the single outcome in which every allocated position fails.
For five independent positions, the combinations in which failures outnumber successes sum to an overall-loss-probability of 0.21, so the complementary probability of a net gain is 0.79.
Enumerating paths with a binomial count
Independent position outcomes can be enumerated with the binomial expansion of failure plus success raised to the number of positions, where those two probabilities sum to one.
The binomial-probability-model is a discrete count of independent successes and failures across a fixed number of equal positions.
For five positions the expansion coefficients are 1, 5, 10, 10, 5, and 1, and the first three terms cover five failures, four failures with one success, and three failures with two successes.
The same count at ten positions, or any other number
Expanding the same model to ten independent positions reduces the all-failure probability to 0.000017 and the overall-loss-probability to 0.077, so the chance of a win or a break-even outcome is 0.923.
The same combinatorial procedure is not limited to five or ten names: changing the number of positions or the success and failure probabilities produces the matching discrete distribution.
All readings on this track · 7 readings
- 1989Auditing price motifs against binomial chance
- 1991How equal independent stakes change the odds of a complete loss
- 1991Binomial counts for unrelated position construction
- 1996Log-change regression and binomial outlier clusters as an evaluation pipeline
- 1996Constructing a log-change stationarity screen with regression or binomial tests
- 1998Binomial baselines for discount-rate change timing
- 2002Trade-count horizon for equity-curve survival