The cheapest ticket and the cheapest route are not the same purchase.

An evaluation has just ended in a breach, and three doors are open. They are usually presented as one door at three prices, and they are not.

• A reset re-runs the rule set that has just closed the account. It is normally discounted, and on some products it does not begin from a clean account: part of the drawdown allowance is already spent when the reset starts.
• A fresh account re-runs that same rule set from the beginning, at the full ticket price.
• A switch buys an attempt under a different rule set, at whatever that one costs, and runs a mechanic your failure says much less about.

This site already prices one route. The cost of a funded account is the fee divided by the pass probability, and the expected number of purchases is one divided by that probability; the arithmetic is set out here and is not repeated. But that prices a ticket. It does not choose between tickets, and choosing needs the same quantity computed for all three, because the doors differ in the fee and in the probability, and nothing makes the cheaper fee arrive attached to the better odds.

So fix the comparison object once: expected total spend until one funded account, computed for each door, smallest wins. Two doors then compare by the ratio of their expected spends, which is the ratio of their fees divided by the ratio of their probabilities. Setting that ratio to one gives the whole of the price argument in a single line.

Take the reset when its discount is deeper than the probability it costs you. Not when it is cheap — when it is cheaper by more than it is worse.

Everything that follows is the two terms of that comparison. What does a partly spent allowance actually do to the probability, and what does the failure you are holding tell you about which door you are standing at? The first has a clean answer. The second has an uncomfortable one.

A reset that begins from a clean account is simply a discounted fresh attempt, and the choice between them is only about price. A reset that begins with part of the allowance already consumed is a different object: the same target, less room, and therefore a different probability. Pricing that needs one honest map from how much of the allowance is gone to what happens to the pass probability, and the map has to be derived rather than asserted.

The one used here is not invented for this page. It is the classical ruin problem in its diffusion form — Feller, chapter XIV, cited in full below. Treat the attempt as a drifted random walk that ends when equity has fallen the allowance below its starting point or risen the target above it; the probability of reaching the target first then has a closed form. Nothing about the drift is assumed. It is recovered from the pass probability the reader states, so your own number goes in and the haircut comes out. The idealisation’s limits are set out at the end, and they are real.

The rising line never keeps up with the axis underneath it, and that is the finding. Surrender 40% of the allowance and you give up 26.9% of the pass probability — 67% of the share you lost, not all of it. Across the whole range drawn, the haircut never reaches 83% of the allowance surrendered. Room is worth less than its share.

The line does bend, but gently, and the measurement is worth more here than the adjective: the last ten points of carryover cost 11.8% of pass probability against 5.8% for the first ten. A factor of two, then, not an order of magnitude.

Both facts have the same mechanical cause. The allowance a carried floor removes is the part furthest from where you are standing: room you only ever reach on paths that have already gone badly, and from deep in the hole the same adverse drift that put you there makes reaching the target unlikely anyway. So a reset takes the least valuable room first. What it leaves you is the room that stops ordinary noise from ending the attempt, and that is the room that was doing the work — which is also why the last slice costs more than the first.

A carried floor is real and it is priced below its own size. At the fees resets are usually offered at, it is not what decides the question.

It does eventually decide it. A 40% discount stops covering its own haircut once the reset starts with 54.6% of the allowance gone; a 20% discount runs out at 31.1%. Which of those you are being offered is a fact about one product’s terms, and it is the fact worth reading before the fee is.

The serious objection to a reset was never the fee and is not the floor. It is that a reset re-runs the mechanic that just beat you: if the mechanic was the cause rather than the luck, repeating it repeats the cause. That objection has force only if the mechanic was the cause — and the single failure in your hand is the entire evidence base for deciding that.

So price the evidence rather than assuming it. Take two possibilities: this rule set suits how you trade, or it does not, and the two carry different pass probabilities. A failure is one observation. How strongly it argues for the second possibility is the ratio of how often each kind of trader fails — the unsuited failure rate over the suited one. That ratio multiplies the odds, and the odds are what accumulate across attempts.

With the worked example’s pair, one failure carries 0.46 bits and moves an even prior to 58%. Reaching 90% takes seven consecutive failures.

The reason is worth stating plainly, because it generalises past this page. Both kinds of trader fail most of the time. An observation that is likely under both hypotheses discriminates between neither, and failure is exactly such an observation in a market where the base rate of failure is high under every hypothesis you might hold. Evidence is scarce precisely where failure is normal.

Widen the gap and it barely helps. Even the steepest pair drawn above — a trader who passes half the time against one who almost never does — still needs four consecutive failures to reach the same confidence, because even the suited trader fails half the time. Narrow the gap and it becomes hopeless: the flattest pair needs more attempts than most traders will ever buy.

That is the uncomfortable answer, and it is the honest one. The failure that prompts the decision is not enough to settle the decision. What is left is the prior — what you knew about that mechanic against how you trade before you bought anything — and the prior is not evidence you acquired by failing. It is evidence you either had or did not.

The two terms are now in hand, so the comparison can be drawn whole. Two things vary and both are observable before you pay: how deep the reset’s discount is, and how much more passable the other rule set is for the way you trade. Everything else — the prices, the two probabilities, the belief, the carryover — is held at the worked example.

The dashed vertical is the smaller question. Left of it the reset’s discount does not cover its haircut and a fresh account is the better of the two same-mechanic routes; right of it the reset is. It sits at 21.1% off, which is a shallower discount than most resets are priced at — which is the arithmetic behind the observation that resets usually do win the ticket comparison.

The wine line is the question that matters. At the worked example’s discount, the other rule set has to be 157% as passable before the dearer ticket becomes the cheaper route. That is not a nudge. That is a different rule set behaving very differently against the same trader.

Rule sets do differ by that much; the floor mechanic alone moves pass probability substantially across real rule sets, and that page measures it. The point of the map is not that switching never pays. It is that switching has to clear a bar set by the reset’s discount, and the deeper that discount, the higher the bar — which is why the frontier rises to the right. A cheap reset makes a switch harder to justify, not easier.

The worked example is a worked example. The prices are whatever you are being offered, the carryover is whatever that product’s terms say, and the two pass probabilities are yours. Below, all of it moves.

One readout in that panel deserves its own paragraph, because it corrects something this site publishes.

The last cell is the one-rate answer: the fee divided by a single pass probability, which is the formula elsewhere on this site and the formula almost everyone uses. Compare it with the fresh-account cell beside it. They disagree, always in the same direction, and often by a lot: on the worked example’s inputs the one-rate answer sits 33% below the correct figure, $1,768 against $2,648.

The cause is not an error in the formula. It is that the formula needs a pass probability, and a trader who does not know which kind of trader they are does not have one — they have a distribution over two. Expected purchases is then the average of one-divided-by-each-probability, not one divided by the average probability. Those are different numbers, the second is always the smaller, and the gap widens the less sure you are. A trader who is certain of their own pass rate loses nothing by using the short formula. A trader who has just failed and is asking which door to take is, by construction, not that trader.

Applied by someone who does not know their own pass probability — which is everyone at the moment this decision is made — fee divided by pass probability understates the bill. On the inputs above, by 33%.

The floor map is an idealisation. Constant drift and variance, one floor, one target, no daily limit, no consistency clause, no session boundary, no minimum trading days. Real rule sets carry several clauses, and the binding one is often not the floor at all. What the idealisation is for is a defensible shape for the carryover cost; the sensitivity to its geometry is stated beside the figure, and the level moves with it.

Two kinds of trader is a caricature. Suitability to a mechanic is continuous. Two points are the smallest model in which the evidence question can be asked at all, and a continuum would change the numbers without changing the finding, which rests on failure being common under every hypothesis rather than on there being exactly two.

If a rule set is genuinely impossible for you, none of this applies. Should the unsuited pass probability be truly zero rather than merely low, expected spend is not large but infinite, and no comparison of expectations means anything. The right object then is a budget, not an expectation. The panel will not let that probability reach zero, for exactly this reason.

This prices attempts, not the prize. Nothing here says whether a funded account is worth the expected spend, or what surviving the funded phase costs. That comparison lives on the fee page linked below.

No firm is named and no reset terms are transcribed. Whether a particular product’s reset returns a clean account or a partly spent one is a per-product fact that has to be read in that product’s own rules, and it varies. The glossary defines the terms.

A phrase that does not discriminate

The fee page argues that when the underlying pass probability is the problem, the correct response is usually not a reset — its own answer is a smaller risk size, a fourth door this page does not price. That conclusion is not what is at issue here. What is worth separating out is the reason given along the way: a cheap reset does not fix bad odds.

Taken alone, that reason does not discriminate among the three doors this page compares. None of them changes your pass probability by itself: a fresh account does not fix bad odds either, and neither does a switch unless the other mechanic genuinely suits you better. So the sentence is true of a reset, but it is equally true of rebuying and switching — options the fee page is not arguing against. The argument against a reset that does survive here is narrower and harder: a reset re-runs a mechanic you may be unsuited to, and the evidence section above is the reason that argument is harder than it looks, because one failure barely establishes that you are unsuited to anything.

This narrows the reason rather than rebutting the page it comes from. The fee page’s actual remedy — a smaller risk size — sits outside the reset/rebuy/switch comparison entirely, and nothing here disputes it. What this page adds is that, among the three doors it does compare, the reset usually wins on price alone, and the case against it has to be made on the mechanic, on a prior formed before the failure, and against a bar the map above draws explicitly.

Is a reset always the cheapest of the three routes?

No, and the condition is exact. A reset beats a fresh account when its discount is deeper than the share of pass probability it costs you, which on the worked example means any discount past 21.1%. Below that the cheaper ticket is the dearer route. Most resets are discounted more steeply than that, which is why the reset usually wins the price comparison — but a reset that starts from a heavily consumed allowance can cross the line even at a substantial discount.

How many failures does it take before switching is justified by the evidence alone?

More than most traders will buy. With the pair drawn in the evidence figure, reaching 90% confidence that the rule set is the problem takes seven consecutive failures from an even prior; with a narrower gap between the two kinds of trader it takes 18. That is not an argument against ever switching. It is an argument that the case for switching has to come from what you knew about the mechanic beforehand, not from the breach that prompted the question.

Does a discount code change any of this?

It changes the choice between routes even though it does not change the number of attempts. A coupon shaves the size of each purchase and not the count, which is why it is a weak lever on the total. But the comparison between two routes is a ratio of fees against a ratio of probabilities, so a coupon that applies to one route and not to another moves that ratio directly, and can flip the ordering without touching anyone’s pass probability.

Why is the expected number of purchases larger than one divided by my average pass probability?

Because you are averaging the wrong quantity. If your pass probability is uncertain, the expected number of attempts is the average of one-divided-by-each-possible-probability, and that is always at least one divided by the average probability. The low-probability case contributes an enormous number of attempts and drags the average up, while it contributes only a small amount to the average probability. On the worked example the short formula lands 33% below the correct total.