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replied 1939d
| attacker catching up from a given deficit is analogous to a Gambler's
Ruin problem. Suppose a gambler with unlimited credit starts at a deficit and plays potentially an
infinite
1 / 30
replied 1939d
| number of trials to try to reach breakeven. We can calculate the probability he ever
reaches breakeven, or that an attacker ever catches up with the honest chain, as follows
replied 1939d
| [8]:

![](

Given our assumption that p > q, the probability drops exponentially as the number of blocks the
attacker has to catch up with increases.
replied 1939d
| With the odds against him, if he doesn't make a lucky
lunge forward early on, his chances become vanishingly small as he falls further behind.

We now consider how long the recipient
replied 1939d
| of a new transaction needs to wait before being
sufficiently certain the sender can't change the transaction. We assume the sender is an attacker
who wants to make the recipient
replied 1939d
| believe he paid him for a while, then switch it to pay back to
himself after some time has passed. The receiver will be alerted when that happens, but the
sender hopes it will be too
replied 1939d
| late.

The receiver generates a new key pair and gives the public key to the sender shortly before
signing. This prevents the sender from preparing a chain of blocks ahead of time by
replied 1939d
| working on
it continuously until he is lucky enough to get far enough ahead, then executing the transaction at
that moment. Once the transaction is sent, the dishonest sender starts
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| working in secret on a
parallel chain containing an alternate version of his transaction.

The recipient waits until the transaction has been added to a block and z blocks have
replied 1939d
| been
linked after it. He doesn't know the exact amount of progress the attacker has made, but
assuming the honest blocks took the average expected time per block, the attacker's
replied 1939d
| potential
progress will be a Poisson distribution with expected value:

![](

To get the probability the attacker could still catch up now, we
replied 1939d
| multiply the Poisson density for
each amount of progress he could have made by the probability he could catch up from that point:

![](

Rearranging
replied 1939d
| to avoid summing the infinite tail of the distribution...

![](

Converting to C code...

```
#include <math.h>
double AttackerSuccessProbability(doub
replied 1939d
|le q, int z)
{
double p = 1.0 - q;
double lambda = z * (q / p);
double sum = 1.0;
int i, k;
for (k = 0; k <= z; k++)
{
double poisson = exp(-lambda);
for (i = 1; i <= k; i++)
replied 1939d
| poisson *= lambda / i;
sum -= poisson * (1 - pow(q / p, z - k));
}
return sum;
}
```

Running some results, we can see the probability drop off exponentially with z.

```
q=0.1
z=
replied 1939d
|0 P=1.0000000
z=1 P=0.2045873
z=2 P=0.0509779
z=3 P=0.0131722
z=4 P=0.0034552
z=5 P=0.0009137
z=6 P=0.0002428
z=7 P=0.0000647
z=8 P=0.0000173
z=9 P=0.0000046
z=10 P=0.0000012
q=0.3
z=
1
1
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|0 P=1.0000000
z=5 P=0.1773523
z=10 P=0.0416605
z=15 P=0.0101008
z=20 P=0.0024804
z=25 P=0.0006132
z=30 P=0.0001522
z=35 P=0.0000379
z=40 P=0.0000095
z=45 P=0.0000024
z=50 P=0.0000006
1
1
replied 1939d
|```

Solving for P less than 0.1%...

```
P < 0.001
q=0.10 z=5
q=0.15 z=8
q=0.20 z=11
q=0.25 z=15
q=0.30 z=24
q=0.35 z=41
q=0.40 z=89
q=0.45 z=340
```

**12. Conclusion**
We have
2
2
replied 1939d
| proposed a system for electronic transactions without relying on trust. We started with
the usual framework of coins made from digital signatures, which provides strong control
3
2
replied 1939d
| of
ownership, but is incomplete without a way to prevent double-spending. To solve this, we
proposed a peer-to-peer network using proof-of-work to record a public history of
1
1
replied 1939d
| transactions
that quickly becomes computationally impractical for an attacker to change if honest nodes
control a majority of CPU power. The network is robust in its unstructured
1
1
replied 1939d
| simplicity. Nodes
work all at once with little coordination. They do not need to be identified, since messages are
not routed to any particular place and only need to be delivered on
1
replied 1939d
| a best effort basis. Nodes can
leave and rejoin the network at will, accepting the proof-of-work chain as proof of what
happened while they were gone. They vote with their CPU power,
1
1
replied 1939d
| expressing their acceptance of
valid blocks by working on extending them and rejecting invalid blocks by refusing to work on
them. Any needed rules and incentives can be enforced
1
1
replied 1939d
| with this consensus mechanism.

**References**
[1] W. Dai, "b-money," http://www.weidai.com/bmoney.txt, 1998.
[2] H. Massias, X.S. Avila, and J.-J. Quisquater, "Design of a secure
1
1
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| timestamping service with minimal
trust requirements," In 20th Symposium on Information Theory in the Benelux, May 1999.
[3] S. Haber, W.S. Stornetta, "How to time-stamp a digital
1
1
replied 1939d
| document," In Journal of Cryptology, vol 3, no
2, pages 99-111, 1991.
[4] D. Bayer, S. Haber, W.S. Stornetta, "Improving the efficiency and reliability of digital time-stamping,"
In
2
1
replied 1939d
| Sequences II: Methods in Communication, Security and Computer Science, pages 329-334, 1993.
[5] S. Haber, W.S. Stornetta, "Secure names for bit-strings," In Proceedings of the 4th
replied 1939d
| ACM Conference
on Computer and Communications Security, pages 28-35, April 1997.
[6] A. Back, "Hashcash - a denial of service counter-measure,"
http://www.hashcash.org/papers/hashcas
1
1
replied 1939d
|h.pdf, 2002.
[7] R.C. Merkle, "Protocols for public key cryptosystems," In Proc. 1980 Symposium on Security and
Privacy, IEEE Computer Society, pages 122-133, April 1980.
[8] W.
1
replied 1939d
| Feller, "An introduction to probability theory and its applications," 1957.
1
1


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