IV Consensus Guide: Nakamoto consensus: the double-spend race, the arithmetic of confirmations, and the deployed most-work rule 4 Playing catch-up 6 What the numbers say

5 Waiting for confirmations

The merchant’s defence is patience: ship only after the paying transaction has $n$ confirmations (Definition 2.3). This section computes exactly how much that patience buys in the paper’s first-seen model. The paper’s model of the attack timeline has one convention worth surfacing before the theorem.

Remark 5.1 (The pre-mined block).

The paper assumes the attacker banks one block before the race is scored: “we assume one block was pre-mined by the attacker before commencing the attack” (§4). The attacker forks from the public leaf at the moment the merchant’s transaction is broadcast, and a patient attacker times the broadcast to coincide with a private block he has just found but not announced. At the moment the honest chain reaches $n$ confirmations, the attacker who has found $m$ further blocks therefore holds $m+1$ , and the deficit walk of Definition 4.3 starts at

z\;=\;n-m-1.

The convention is a modelling choice, not a law: an attacker with no banked block starts at $z=n-m$ , and every probability below shrinks accordingly. It is also, as Theorem 5.2 shows, precisely the choice that makes the final formula collapse into a symmetric race.

By Proposition 3.5, the number $m$ of attacker blocks found while the honest network finds its $n$ is negative binomial. Averaging the catch-up probability of Theorem 4.4 over $m$ gives the central result for the specification and Rosenfeld first-seen tie rule. Section 8.1 explains the additional success opportunity supplied by Zebra’s tie-break. The model is not an exact deployed rollback probability.

Theorem 5.2 (Double-spend success probability; Rosenfeld eq. (1)).

Let the merchant wait for $n\geq 1$ confirmations, under Assumptions 4.1 and 4.2 and the convention of Remark 5.1, with a private branch accepted only after it becomes strictly heavier. The probability that the attacker’s branch eventually overtakes the public chain is $r(q,n)=1$ if $q\geq p$ , and otherwise

r(q,n)\;=\;1-\sum_{m=0}^{n-1}\binom{m+n-1}{m}\left(p^{n}q^{m}-p^{m}q^{n}\right% )\;=\;2\sum_{m=0}^{n-1}\binom{m+n-1}{m}p^{m}q^{n}.

(2)

Proof.

Conditioning on $m$ and applying Theorem 4.4 at $z=n-m-1$ ,

r\;=\;\sum_{m=0}^{\infty}P(m)\,a_{n-m-1}\;=\;\underbrace{\sum_{m=0}^{n-1}P(m)% \left(\frac{q}{p}\right)^{\!n-m}}_{\text{behind: must catch up}}\;+\;% \underbrace{\sum_{m=n}^{\infty}P(m)}_{\text{already ahead}},

valid for $q<p$ ; when $q\geq p$ every $a_{n-m-1}=1$ and $r=\sum_{m}P(m)=1$ immediately. For the first sum, each term transforms exactly:

\binom{m+n-1}{m}p^{n}q^{m}\left(\frac{q}{p}\right)^{\!n-m}\;=\;\binom{m+n-1}{m% }p^{m}q^{n},

which by Lemma 3.6 is the probability that the attacker’s $n$ -th block arrives while the honest network holds $m$ ; summed over $m\leq n-1$ , the first sum is $\mathrm{Pr}[A_{n}]$ , the probability that the attacker wins the first-to- $n$ race. The second sum is $\mathrm{Pr}[m\geq n]=1-\mathrm{Pr}[H_{n}]=\mathrm{Pr}[A_{n}]$ by the same lemma. Hence $r=2\mathrm{Pr}[A_{n}]$ , the right-hand form of (2); the middle form follows from $\mathrm{Pr}[A_{n}]=1-\mathrm{Pr}[H_{n}]$ applied to one of the two copies. The printed upper limit in the paper’s equation (1) is $n$ rather than $n-1$ ; the $m=n$ term is identically zero, so the forms agree. □

Remark 5.3 (The race doubling).

The proof exposes a structure the paper’s algebra passes through silently: under the one-pre-mined-block convention, the catch-up half — the honest chain reaches $n$ first and the attacker later overtakes — has exactly the same probability as the already-ahead half, in which the attacker reaches $n$ first. A first-to- $n$ winner may have fallen behind and recovered along the way; neither half asserts otherwise. Thus the double-spend probability is twice the chance of winning a first-to- $n$ race. The identity also absorbs the boundary case: at $q=p$ each race half is exactly $\tfrac{1}{2}$ , giving $r=1$ with no case split, continuously with the $q>p$ regime. The algebra here is the paper’s; reading its two sums as the two halves of one race, and the observation that the pre-mine convention is what makes them equal, are this volume’s presentational additions (verified by script over random $(q,n)$ ; the two sums agree to floating-point precision).

Example 5.4 (A full race, exactly).

Take $q=1/5$ (a twenty-percent attacker) and $n=6$ . The deficit starts at $z=5-m$ , the catch-up factor is $(q/p)^{6-m}=4^{-(6-m)}$ , and every quantity below is an exact rational, printed to six decimals (script-computed):

$m$	$P(m)$	$a_{5-m}=4^{-(6-m)}$	product
$0$	$0.262144$	$0.000244$	$0.000064$
$1$	$0.314573$	$0.000977$	$0.000307$
$2$	$0.220201$	$0.003906$	$0.000860$
$3$	$0.117441$	$0.015625$	$0.001835$
$4$	$0.052848$	$0.062500$	$0.003303$
$5$	$0.021139$	$0.250000$	$0.005285$
catch-up half (sum of products)			$0.011654$
already-ahead half, $\mathrm{Pr}[m\geq 6]$			$0.011654$
$r(0.2,\,6)$			$0.023308$

The two halves agree to every printed digit — they are equal exactly, by Remark 5.3 — and the closed form (2) evaluates to the same $0.023308$ . A merchant facing a possible twenty-percent adversary who ships after six confirmations is assigned a $2.33\%$ strict-overtake probability by this model.

Remark 5.5 (Satoshi’s approximation).

The whitepaper’s §11 computes the same quantity with one approximation and two offsetting conventions: the attacker’s progress is approximated as Poisson with mean $\lambda=nq/p$ (the law $\mathrm{Pr}[k]=e^{-\lambda}\lambda^{k}/k!$ on the nonnegative integers); no block is pre-mined, but success is scored at breakeven rather than strictly ahead, and those two conventions cancel exactly — the whitepaper’s catch-up factor for $m$ attacker blocks is $(q/p)^{n-m}$ , term for term the factor of Remark 5.1. The Poisson approximation is the entire gap. In each of the following script-computed comparisons it pushes the estimate down: at $q=10\%$ , $n=6$ , the whitepaper’s formula gives $0.0243\%$ against the exact $0.0591\%$ — an understatement by a factor of about $2.4$ ; at $q=30\%$ , $n=6$ : $13.211\%$ against $15.645\%$ ; at $q=10\%$ , $n=10$ : $0.0001\%$ against $0.0008\%$ . The direction is not uniform over all $(q,n)$ ; the reason to prefer the negative binomial model is that it is exact under the stated assumptions, not that the approximation is always one-sided.

$q$	$n{=}1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$
$2\%$	$4.000$	$0.237$	$0.016$	$0.0011$	$8\!\cdot\!10^{-5}$	$<10^{-5}$
$5\%$	$10.000$	$1.450$	$0.232$	$0.039$	$0.0066$	$0.0012$	$2.1\!\cdot\!10^{-4}$	$4\!\cdot\!10^{-5}$	$<10^{-5}$
$10\%$	$20.000$	$5.600$	$1.712$	$0.546$	$0.178$	$0.059$	$0.020$	$0.0067$	$0.0023$	$0.0008$
$20\%$	$40.000$	$20.800$	$11.584$	$6.669$	$3.916$	$2.331$	$1.401$	$0.848$	$0.516$	$0.316$
$30\%$	$60.000$	$43.200$	$32.616$	$25.207$	$19.762$	$15.645$	$12.475$	$10.003$	$8.055$	$6.511$
$40\%$	$80.000$	$70.400$	$63.488$	$57.958$	$53.314$	$49.300$	$45.769$	$42.621$	$39.787$	$37.218$
$45\%$	$90.000$	$85.050$	$81.375$	$78.342$	$75.716$	$73.375$	$71.252$	$69.299$	$67.487$	$65.793$
$48\%$	$96.000$	$94.003$	$92.508$	$91.264$	$90.177$	$89.201$	$88.307$	$87.478$	$86.703$	$85.972$
$50\%$	$100$ for every $n$

Table 1: The double-spend success probability

r(q,n)

, in percent, computed by script from Theorem 5.2. On the rows the paper’s Table 1 also prints (

q\in\{2,10,20,30,40,48,50\}\%

), the script reproduces its values cell-for-cell; the

5\%

and

45\%

rows are this volume’s additions.

Refer to caption — Figure 3: The success probability $r(q,n)$ against the attacker’s hashrate share $q$ , for $n\in\{1,2,4,6,8,10\}$ confirmations, logarithmic vertical scale clipped at $10^{-5}$ ; coordinates computed by script from Theorem 5.2. Every curve reaches $1$ at $q=\tfrac{1}{2}$ (dashed) and stays there: beyond the majority line, confirmations are worthless. Below it, each added confirmation drops $r$ by a roughly constant factor — the curves are near-equally spaced in log scale — and the factor improves as $q$ falls.