Deriving Go Or No-Go Thresholds from Payoffs

Why a Probability Threshold Depends on Payoffs

A probability-of-success estimate does not, by itself, determine whether a development program should proceed. The same estimated probability can support different decisions when the development cost, potential return, failure outcome, or value of stopping differs. A go/no-go threshold makes that relationship explicit: it is the success probability at which proceeding and stopping have equal expected value, given specified payoffs.

The threshold is not a universal probability standard. It is derived for a particular decision, with payoffs expressed on a consistent basis. For example, an estimated probability of success of 30% may be sufficient for a program with a relatively low cost and a large success-contingent return, but insufficient for a costly program with a modest return.

Definition: Let $p$ be the probability of the defined development success outcome. Let $V_s$ and $V_f$ be the values, at the decision point, of the success and failure outcomes before subtracting the proceed cost $C$. Let $S$ be the value of stopping at that point. The proceed cost is incurred if the program proceeds, regardless of outcome. All values must be measured on the same monetary basis and at the same decision point.

This setup has two available actions, proceed and stop, and two possible outcomes after proceeding, success and failure. The probability $p$ must refer to the success event that determines which payoff applies. That event should be defined clearly for the decision; a probability for a different endpoint or a different stage of development may not be suitable. The earlier material on calibration diagnostics for probability-of-success models is relevant because a threshold comparison is only as useful as the probability supplied to it.

Expected Payoffs and the Threshold

Under the specified model, the expected value of proceeding is the probability-weighted average of the success and failure values, less the cost paid to proceed. Stopping yields $S$. The decision rule follows by comparing those two quantities.

EV(\text{proceed}) = pV_s + (1-p)V_f - C

EV(\text{stop}) = S

Proceeding is preferred by the expected-value criterion when $EV(\text{proceed}) > S$. The break-even threshold is found by setting the two expected values equal and solving for $p$.

$$ pV_s + (1-p)V_f - C = S $$

$$ p(V_s - V_f) = S + C - V_f $$

Threshold result: If $V_s > V_f$, the break-even probability is $$ p^* = \frac{S+C-V_f}{V_s-V_f}. $$ For $p>p^*$, proceeding has the higher expected value; for $p<p^*$, stopping has the higher expected value. At $p=p^*$, the two actions have equal expected value. The threshold must be interpreted in light of the feasible range $0\leq p\leq1$.

The condition $V_s>V_f$ means that success is more valuable than failure. It is what makes the expected value of proceeding increase with $p$: its slope with respect to $p$ is $V_s-V_f$. If the two outcome values are equal, changing the success probability does not change the proceed payoff; there is no probability threshold to solve for. If $V_s<V_f$, the expected value decreases as $p$ increases, so the usual interpretation that a higher success probability favors proceeding does not apply.

A threshold above 1 means that proceeding is not preferred for any feasible probability, while a threshold below 0 means proceeding is preferred for every feasible probability under this payoff model. These are informative outcomes, not calculation errors: they indicate that the cost and alternative value are so large or small relative to the outcome values that changing $p$ across its possible range cannot reverse the decision.

Worked Applications

Worked Example: A Simple Success-Contingent Return

Suppose a hypothetical development decision has a proceed cost of $C=\$18$ million. Success yields $V_s=\$90$ million, failure yields $V_f=\$0$, and stopping yields $S=\$0$. What is the break-even probability, and should the program proceed if the estimated success probability is 24%?

The success value exceeds the failure value, so the threshold formula applies. Substituting the values, in millions of dollars:

p^* = \frac{S+C-V_f}{V_s-V_f} = \frac{0+18-0}{90-0} = \frac{18}{90} = 0.2000

The break-even probability is 20.00%. At $p=0.24$, the expected value of proceeding is:

EV(\text{proceed}) = 0.24(90)+0.76(0)-18 =21.6-18 =\$3.6\text{ million}

The expected value of stopping is \$0 million. Since \$3.6 million is greater than \$0 million, proceeding has the higher expected value under these assumptions. This conclusion compares expected values; it does not mean that success is certain or that the program will realize a \$3.6 million payoff.

Worked Example: A Valuable Stopping Alternative and a Failure Loss

Consider a hypothetical program with a proceed cost of $C=\$12$ million. Conditional on proceeding, the success value is $V_s=\$52$ million and the failure value is $V_f=-\$3$ million. Stopping instead has value $S=\$4$ million. Find the threshold and compare the actions when $p=0.40$. The outcome values include any outcome-contingent costs or consequences; the \$12 million proceed cost is separate.

Here $V_s>V_f$, so:

p^* = \frac{S+C-V_f}{V_s-V_f} = \frac{4+12-(-3)}{52-(-3)} = \frac{19}{55} = 0.3455

The threshold is approximately 34.55%. A probability of 40.00% is above that threshold. Direct calculation verifies the comparison:

EV(\text{proceed}) = 0.40(52)+0.60(-3)-12 =20.8-1.8-12 =\$7.0\text{ million}

Stopping yields \$4 million, so proceeding has an expected-value advantage of $\$7.0-\$4.0=\$3.0$ million. Omitting the stopping value or treating failure as worth zero would change the decision calculation and could produce a different threshold.

Worked Example: A Threshold Above One

Suppose a hypothetical program has $C=\$25$ million, $V_s=\$30$ million, $V_f=\$5$ million, and $S=\$12$ million. Determine whether any feasible success probability makes proceeding preferable.

The break-even calculation gives:

p^* = \frac{12+25-5}{30-5} = \frac{32}{25} = 1.280

A probability cannot exceed 1, so the threshold is not attainable. To check the result at the most favorable feasible probability, set $p=1$:

EV(\text{proceed}\mid p=1)=1(30)+0(5)-25 =\$5\text{ million}

Even if success were certain, proceeding would yield an expected value of \$5 million, below the \$12 million value of stopping. Therefore, no feasible success probability makes proceeding preferable under these payoffs.

Worked Example: Excluding a Sunk Cost

A hypothetical program has already incurred \$8 million in development costs. A further \$10 million must be spent to proceed from the current decision point. If the program succeeds, its value at this point before the further cost is \$50 million; if it fails, the value is \$0. Stopping is worth \$0. If the success probability is 30%, should the program proceed? The past \$8 million cannot be recovered under either action.

The threshold uses the further cost, not the amount already spent:

p^* = \frac{0+10-0}{50-0} = 0.2000

At $p=0.30$, the expected value from the current decision point is:

EV(\text{proceed}) = 0.30(50)+0.70(0)-10 =15-10 =\$5\text{ million}

Stopping is worth \$0 million, so proceeding has the higher expected value. The \$8 million already spent is not part of this comparison because it is incurred regardless of the current action. Including it would give a misleading threshold of $(10+8)/50=0.3600$, which answers a different question by charging the current decision for an unrecoverable past cost.

Assumptions, Checks, and Interpretation

The algebra is simple, but a defensible threshold requires a coherent decision model. In particular, the calculation assumes that the two outcome states cover the relevant possibilities, $p$ is the probability of the defined success state, and the payoffs apply to those states. It also assumes that expected value is the criterion used to rank actions. For monetary values, that is a risk-neutral criterion: it compares probability-weighted monetary outcomes without adding a separate adjustment for risk preferences or constraints.

All values need to be expressed at the same decision point. If success or failure values arise later, they must be made comparable to present values consistently. Costs that will be paid only after a particular outcome belong in that outcome's value; a cost paid whenever the program proceeds belongs in $C$. A cost should not appear in both places. Costs already incurred and identical under proceed and stop are sunk for this decision and cancel from the comparison. A recoverable asset or other value obtained by stopping should instead be included in $S$.

Several quick checks help identify errors:

Check the direction. If $V_s>V_f$, increasing $p$ must increase $EV(\text{proceed})$. A calculation implying the opposite usually contains a sign or payoff-order error.
Check the boundary cases. At $p=0$, proceeding is worth $V_f-C$; at $p=1$, it is worth $V_s-C$. Compare both with $S$ to see whether a threshold inside the probability range is plausible.
Check units and timing. The cost, stop value, and outcome values must use compatible monetary units and a common valuation date.
Check what success means. A probability for one endpoint or development stage cannot automatically be substituted for the probability associated with the payoff states in this decision.
Check for double counting. Make sure the proceed cost is not already embedded in $V_s$ or $V_f$, and do not charge the current decision for a sunk cost that does not vary by action.

The threshold is a concise alternative to calculating both expected values separately, but the two approaches are algebraically equivalent. Direct expected-value calculation can be easier to audit when payoffs are complex; solving for $p^*$ shows exactly how high the success probability must be for proceeding to match stopping. A decision tree can organize decisions with multiple stages or branches, whereas the formula here summarizes one proceed-versus-stop comparison with two outcome states. A fixed probability rule may be easy to apply across programs, but it does not generally encode their differing costs, returns, and stopping values.

A numerical threshold is not automatically a recommendation. Its usefulness depends on whether the success probability and payoffs are relevant and credible, and whether expected monetary value is suitable for the decision maker. If the threshold is close to the estimated probability, modest changes in inputs may reverse the preferred action; report the inputs and the comparison, rather than presenting the threshold as a universal biological or regulatory standard.

Common pitfall: “The probability of success exceeds 20%, so the program should proceed” is incomplete reasoning if 20% came from a different program’s payoffs. Derive the threshold from the current proceed cost, success and failure values, and stopping value; then compare it with the probability for the success outcome defined in this decision.

Check Your Understanding

For each question, state the relevant comparison. Worked solutions follow in the same box.

A program costs \$15 million to proceed. Success is worth \$60 million, failure is worth \$0, and stopping is worth \$0. Find the threshold and state the preferred action at $p=0.20$.
A program has $C=\$8$ million, $V_s=\$28$ million, $V_f=\$4$ million, and $S=\$6$ million. Find the threshold and the expected-value advantage of proceeding at $p=0.50$.
A program has $C=\$20$ million, $V_s=\$25$ million, $V_f=\$0$, and $S=\$10$ million. Can proceeding be preferred at any feasible probability? Explain using the threshold or a boundary check.
Why should a cost already incurred, which cannot be recovered under either action, ordinarily be excluded from the current threshold calculation?

Solutions. 1. $p^*=(0+15-0)/(60-0)=0.2500$. Since $0.20<0.2500$, stopping is preferred. Specifically, proceeding is worth $0.20(60)-15=-\$3$ million, compared with \$0 from stopping. 2. $p^*=(6+8-4)/(28-4)=10/24=0.4167$. At $p=0.50$, proceeding is worth $0.50(28)+0.50(4)-8=\$8$ million, while stopping is worth \$6 million; the expected-value advantage is \$2 million. 3. $p^*=(10+20-0)/(25-0)=1.200$, above 1. At $p=1$, proceeding is worth $25-20=\$5$ million, still less than the \$10 million stopping value, so no feasible probability makes proceeding preferable. 4. The past cost is the same under both current actions, so it does not change their payoff difference. Including it would penalize the current proceed choice for an expenditure that cannot be avoided or recovered by stopping now.