Jackson Floyd

2022-03-27

Confidence of a failure

A joint of steel pipe (casing) has a 1% failure rate. 400 joints of casing are in a typical well. How may wells can I drill or joints can I run, before I am 90% confident of having at least one casing failure? You can leave out the well to simplify the problem: Casing has a 1% failure rate. How many joints of casing can I have before I am 90% confident in having a failure? Thanks for re teaching me. I think it is called gambler's ruin.

A joint of steel pipe (casing) has a 1% failure rate. 400 joints of casing are in a typical well. How may wells can I drill or joints can I run, before I am 90% confident of having at least one casing failure? You can leave out the well to simplify the problem: Casing has a 1% failure rate. How many joints of casing can I have before I am 90% confident in having a failure? Thanks for re teaching me. I think it is called gambler's ruin.

Esteban Sloan

Beginner2022-03-28Added 21 answers

This is a problem using the geometric distribution. Distributions with that name are defined in different ways.

My definition is that X is the number of the trial on which we see the first casing failure.

Then the PDF of X is$f\left(k\right)=P(X=k)=(1-p){p}^{k-1}$ , for $k=1,2,\cdots$ . Also, the CDF is $F\left(k\right)=P(X\le k)=\sum _{i=1}^{k}f\left(i\right).$

We can use R statistical software to make a table of the PDF and CDF for your problem and then print out the first few and the last few rows of the table:

$>x=1:230;\text{}pdf=.01\cdot {.99}^{x-1};cdf=cum\sum \left(pdf\right)$

$>TAB=cb\in d(x,pdf,cdf)$

$>head\left(TAB\right);tail\left(TAB\right)$

$$\begin{array}{|cccc|}\hline & x& pdf& cdf\\ [1,]& 1& 0.01000000& 0.01000000\\ [2,]& 2& 0.00990000& 0.01990000\\ [3,]& 3& 0.00980100& 0.02970100\\ [4,]& 4& 0.00970299& 0.03940399\\ [5,]& 5& 0.00960596& 0.04900995\\ [6,]& 6& 0.00950990& 0.05851985\\ \hline\end{array}$$

$$\begin{array}{|cccc|}\hline & x& pdf& cdf\\ [225,]& 225& 0.001052649& 0.8957877\\ [226,]& 226& 0.001042123& 0.8968299\\ [227,]& 227& 0.001031701& 0.8978616\\ [228,]& 228& 0.001021384& 0.8988830\\ [229,]& 229& 0.001011170& 0.8998941\\ [230,]& 230& 0.001001059& 0.9008952\\ \hline\end{array}$$

It seems that it will take 230 fittings to be 90% sure of a casing failure. I will leave it to you to figure out how to sum the geometric series to get the CDF. Then you can solve to find the smallest k such that$F\left(k\right)\ge .9.$

Note: The version of the geometric distribution programmed into R is for the random variable Y which counts the trials up to, but not including, the first casing failure. Thus in R, qgeom(.9, .01) returns 229.

My definition is that X is the number of the trial on which we see the first casing failure.

Then the PDF of X is

We can use R statistical software to make a table of the PDF and CDF for your problem and then print out the first few and the last few rows of the table:

It seems that it will take 230 fittings to be 90% sure of a casing failure. I will leave it to you to figure out how to sum the geometric series to get the CDF. Then you can solve to find the smallest k such that

Note: The version of the geometric distribution programmed into R is for the random variable Y which counts the trials up to, but not including, the first casing failure. Thus in R, qgeom(.9, .01) returns 229.

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