<p>I'm messing around with bitcoins a bit, and I'm now looking at the confirmations. In several places I read that 6 is the "normal" figure of confirmations needed to say a transactions has actually and irreversibly occurred. I also read <a href="https://bitcoin.stackexchange.com/questions/1170/why-is-6-the-number-of-confirms-that-is-considered-secure">this excellent Q</a>, which explains a lot more. </p>

<p>I still wonder about one thing though: in the case of bad intentions, how big is the chance of double spending in the case of 0, 1, 2, 3, 4, 5 and 6 confirmations expressed in percentages?</p>

<p>I'm after a little table which should look something like this:</p>

<pre><code>confirmations | chance of double spending (all figures are made up)
-----------------------------------------
0             | 10%
1             | 4%
2             | 1%
3             | 0.5%
4             | 0.2%
5             | 0.05%
6             | 0.001%
</code></pre>

CC BY-SA 3.0
2014-03-22T19:16:38.977Z
It depends almost entirely on how much hashing power is available to the attacker.
Nate Eldredge
CC BY-SA 3.0
2014-03-22T20:26:19.057Z
I think it is `hash-power^confirmations` where `hash-power` is percent of network hashing you have. Note how for 0 confirmations on any hashing power you have 100% chance to double spend. This of course assumes network is perfect.
John T
CC BY-SA 3.0
2014-03-22T21:11:54.683Z
@NateEldredge - Alright, so let's say we're dealing with an attacker with a fairly large, to very large hashing power (I personally have no idea what large in this sense is..). What are reasonable numbers in that case? I'm not even after exact numbers, some kind of estimation would already be good, to get an idea of the chances of double spending.
kramer65
CC BY-SA 3.0
2014-03-23T01:28:03.617Z
The original bitcoin paper says something on this regard, doesn't it? Besides, there's a white paper by Meni Rosenfeld expanding on this, as well as another analysis of how much security is reduced by the shorter block times in Litecoin and its offspring vs the 10 mins in Bitcoin (actually, not much)
Joe Pineda
CC BY-SA 3.0
2014-03-23T01:33:01.840Z
Oh, both Rosenfeld's paper and the relevant part of Satoshi's are linked to from the question you reference!
Joe Pineda
CC BY-SA 3.0
2014-03-24T11:53:26.060Z
It's really impossible to assign numbers to these odds. You can only list odds when you depend solely on luck, but luck is only one of the several factors that are relevant here.
Steven Roose
<confirmations><doublespend><zero-confirmation>
How big is the chance of double spending for 0, 1, 2, 3, 4, 5 and 6 confirmations?
kramer65
# How big is the chance of double spending for 0, 1, 2, 3, 4, 5 and 6 confirmations?

(Asked by: username_0 on 2014-03-22)

I'm messing around with bitcoins a bit, and I'm now looking at the confirmations. In several places I read that 6 is the "normal" figure of confirmations needed to say a transactions has actually and irreversibly occurred. I also read [this excellent Q](https://bitcoin.stackexchange.com/questions/1170/why-is-6-the-number-of-confirms-that-is-considered-secure), which explains a lot more.

I still wonder about one thing though: in the case of bad intentions, how big is the chance of double spending in the case of 0, 1, 2, 3, 4, 5 and 6 confirmations expressed in percentages?

I'm after a little table which should look something like this:

```
confirmations | chance of double spending (all figures are made up)
-----------------------------------------
0             | 10%
1             | 4%
2             | 1%
3             | 0.5%
4             | 0.2%
5             | 0.05%
6             | 0.001%

```

**Tags:** confirmations, doublespend, zero-confirmation

> **Comments:**
> **username_1**: It depends almost entirely on how much hashing power is available to the attacker.  
> **username_2**: I think it is `hash-power^confirmations` where `hash-power` is percent of network hashing you have. Note how for 0 confirmations on any hashing power you have 100% chance to double spend. This of course assumes network is perfect.  
> **username_0**: @username_1 - Alright, so let's say we're dealing with an attacker with a fairly large, to very large hashing power (I personally have no idea what large in this sense is..). What are reasonable numbers in that case? I'm not even after exact numbers, some kind of estimation would already be good, to get an idea of the chances of double spending.  
> **username_3**: The original bitcoin paper says something on this regard, doesn't it? Besides, there's a white paper by Meni Rosenfeld expanding on this, as well as another analysis of how much security is reduced by the shorter block times in Litecoin and its offspring vs the 10 mins in Bitcoin (actually, not much)  
> **username_3**: Oh, both Rosenfeld's paper and the relevant part of Satoshi's are linked to from the question you reference!  
> **username_4**: It's really impossible to assign numbers to these odds. You can only list odds when you depend solely on luck, but luck is only one of the several factors that are relevant here.  

---

## 0 answers

