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, 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%
  • It depends almost entirely on how much hashing power is available to the attacker. Mar 22, 2014 at 19:16
  • 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.
    – John T
    Mar 22, 2014 at 20:26
  • @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
    Mar 22, 2014 at 21:11
  • 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
    Mar 23, 2014 at 1:28
  • Oh, both Rosenfeld's paper and the relevant part of Satoshi's are linked to from the question you reference!
    – Joe Pineda
    Mar 23, 2014 at 1:33


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