<p>The following holds for a generalized case on the required time for having <code>m</code> blocks with a confidence of <code>p 100\%</code>. The time <code>X</code> between consecutive blocks is exponential distributed, i.e. <code>X~Exp[k]</code> with <code>k=1/10</code>.
The sum of <code>m</code> independent and identical exponential random variables follows the Erlang distribution, that means for <code>X_i~Exp[k]</code>,  <code>X_1+X_2+...+X_m~Erlang[m,k]</code>. The required time for having <code>m</code> blocks with a confidence of <code>p 100%</code> is given by the <code>p</code>-quantile of the <code>Erlang[m,k]</code> distribution. In your case <code>m=6</code> and <code>p=0.99</code>. The quantile <code>Q_Erlang[m,k](p)</code> does not have a closed form expression. However, it is given in terms of the inverse regularized gamma function which can be evaluated numerically.</p>

<p>Plugging in the values leads to <code>Q_Erlang[6,1/10](0.99)=131.085</code>, which means that in <code>99%</code> of all cases <code>6</code> blocks will be mined in less than <code>131</code> minutes.</p>

<p><a href="https://i.sstatic.net/qJ05t.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/qJ05t.png" alt="enter image description here"></a></p>

CC BY-SA 3.0
2017-04-06T03:57:42.697Z
Thank you for your detailed answer. It looks like the answer is much more advanced than my level :)
Is there any formula for Q_Erlang that I can use to calculate values for different parameters on calculator?
I mean how can I calculate the value of Q_Erlang[6,1/10](0.99) on a calculator?
DigitalPerson
CC BY-SA 3.0
2017-04-06T07:04:23.760Z
@Digi An online source for the evaluation of the quantiles is https://www.wolframalpha.com/input/?i=Quantile%5BErlangDistribution%5B6,+1%2F10%5D,+0.99%5D
stat_facts
CC BY-SA 3.0
2017-04-05T17:13:13.040Z
2017-04-05T22:10:08.877Z
2017-04-05T22:10:08.877Z
stat_facts
<p><strong>TL;DR: Same result as <a href="https://bitcoin.stackexchange.com/a/52598/5406">stat_facts</a>, but another way of explaining it. ;)</strong></p>

<p>Discovery of blocks is a <a href="https://bitcoin.stackexchange.com/q/22978/5406">Poisson process</a>.</p>

<p>The probability that <code>x</code> blocks are found in the time that we expected <code>λ</code> blocks to be found is:</p>

<p><a href="https://i.sstatic.net/yJ0GT.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/yJ0GT.gif" alt="enter image description here"></a></p>

<p>Now, you have specified that you want to have six blocks found with a confidence of 99%.</p>

<p>The sum of probabilities of all possible outcomes always is 1. This means that we can express the probability of all outcomes where <em>at least</em> six blocks have been found as "100% minus the probability of having found less than six blocks".</p>

<pre><code>p(x≥6|λ) = 1 - p(5|λ) - p(4|λ) - p(3|λ) - p(2|λ) - p(1|λ) - p(0|λ) = 0.99
</code></pre>

<p>Plugging the <code>x</code> values into the above formula we get:</p>

<p><a href="https://i.sstatic.net/J5PGX.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/J5PGX.gif" alt="enter image description here"></a></p>

<p>The resulting probability curve looks like this (y-axis is probability and x-axis is λ):</p>

<p><a href="https://i.sstatic.net/HsICm.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/HsICm.png" alt="enter image description here"></a></p>

<p>Finally, you were asking for <code>p(x≥6|λ) = 0.99</code> which turns out to be at about <code>λ ≈ 13.1085</code>. Since the expected time for one block is 10 minutes, this means that in 131 minutes for 99% of the cases at least six blocks will have been found.</p>

CC BY-SA 3.0
2017-04-06T14:19:52.527Z
2017-04-06T14:19:52.527Z
2017-04-13T12:47:33.743Z
Murch
<p>I had this question in an exam but I was not very sure how to solve it:</p>

<p>Suppose Bob the merchant wants to have a policy that orders will ship within x minutes after receipt of payment. What value of x should Bob choose so that with 99% confidence 6 blocks will be found within x minutes</p>

<p>Do you have in idea about the formula that we have to use to find the answer?</p>

CC BY-SA 3.0
2017-04-05T05:48:19.727Z
It's a [Poisson process](https://www.probabilitycourse.com/chapter11/11_1_2_basic_concepts_of_the_poisson_process.php).
Nate Eldredge
<confirmations>
how long does it take to get 6 confirmations with confidence of 99%
Digi
# how long does it take to get 6 confirmations with confidence of 99%

(Asked by: username_0 on 2017-04-05)

I had this question in an exam but I was not very sure how to solve it:

Suppose Bob the merchant wants to have a policy that orders will ship within x minutes after receipt of payment. What value of x should Bob choose so that with 99% confidence 6 blocks will be found within x minutes

Do you have in idea about the formula that we have to use to find the answer?

**Tags:** confirmations

> **Comments:**
> **username_1**: It's a [Poisson process](https://www.probabilitycourse.com/chapter11/11_1_2_basic_concepts_of_the_poisson_process.php).  

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## 2 answers

### Answer by username_2 on 2017-04-05. Score: 2

The following holds for a generalized case on the required time for having `m` blocks with a confidence of `p 100\%`. The time `X` between consecutive blocks is exponential distributed, i.e. `X~Exp[k]` with `k=1/10`.
The sum of `m` independent and identical exponential random variables follows the Erlang distribution, that means for `X_i~Exp[k]`, `X_1+X_2+...+X_m~Erlang[m,k]`. The required time for having `m` blocks with a confidence of `p 100%` is given by the `p`-quantile of the `Erlang[m,k]` distribution. In your case `m=6` and `p=0.99`. The quantile `Q_Erlang[m,k](p)` does not have a closed form expression. However, it is given in terms of the inverse regularized gamma function which can be evaluated numerically.

Plugging in the values leads to `Q_Erlang[6,1/10](0.99)=131.085`, which means that in `99%` of all cases `6` blocks will be mined in less than `131` minutes.

[![enter image description here](https://i.sstatic.net/qJ05t.png)](https://i.sstatic.net/qJ05t.png)

> **Comments:**
> **username_3**: Thank you for your detailed answer. It looks like the answer is much more advanced than my level :) Is there any formula for Q_Erlang that I can use to calculate values for different parameters on calculator? I mean how can I calculate the value of Q_Erlang[6,1/10](0.99) on a calculator?  
> **username_2**: @username_0 An online source for the evaluation of the quantiles is https://www.wolframalpha.com/input/?i=Quantile%5BErlangDistribution%5B6,+1%2F10%5D,+0.99%5D  

---

### Answer by username_4 on 2017-04-06. Score: 0

**TL;DR: Same result as [stat\_facts](https://bitcoin.stackexchange.com/a/52598/5406), but another way of explaining it. ;)**

Discovery of blocks is a [Poisson process](https://bitcoin.stackexchange.com/q/22978/5406).

The probability that `x` blocks are found in the time that we expected `λ` blocks to be found is:

[![enter image description here](https://i.sstatic.net/yJ0GT.gif)](https://i.sstatic.net/yJ0GT.gif)

Now, you have specified that you want to have six blocks found with a confidence of 99%.

The sum of probabilities of all possible outcomes always is 1. This means that we can express the probability of all outcomes where *at least* six blocks have been found as "100% minus the probability of having found less than six blocks".

```
p(x≥6|λ) = 1 - p(5|λ) - p(4|λ) - p(3|λ) - p(2|λ) - p(1|λ) - p(0|λ) = 0.99

```

Plugging the `x` values into the above formula we get:

[![enter image description here](https://i.sstatic.net/J5PGX.gif)](https://i.sstatic.net/J5PGX.gif)

The resulting probability curve looks like this (y-axis is probability and x-axis is λ):

[![enter image description here](https://i.sstatic.net/HsICm.png)](https://i.sstatic.net/HsICm.png)

Finally, you were asking for `p(x≥6|λ) = 0.99` which turns out to be at about `λ ≈ 13.1085`. Since the expected time for one block is 10 minutes, this means that in 131 minutes for 99% of the cases at least six blocks will have been found.


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