Correction of error in the calculation of the probability of deleting…

notebooks/Fuzzer.ipynb

@@ -863,12 +863,43 @@
     "     ], \"9 ** 0.5\")"
    ]
   },
+  {
+   "attachments": {},
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": []
+  },
+  {
+   "attachments": {},
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "The chance is actually higher than you may think. If you remove `/` (the root of all files), for instance, your entire file system will be gone. If you remove `.` (the current folder), all the files in the current directory will be gone. \n",
+    "\n",
+    "The probability of generating a string that is exactly 1 character long is 1/101, this is because the length of the string is determined by calling random.randrange(0, max_length + 1), where the default value of max_length is 100. Per the description given of random.randrange, that should return a random number in [0, 99 + 1). So, we end up with the inclusive range [0, 100] where there are 101 values in the interval.\n",
+    "\n",
+    "For `/` or `.` to be produced, you need a string length of 1 (chance: 1 out of 101) and one of these two characters (chance: 2 out of 32)."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "1/101 * 2/32"
+   ]
+  },
   {
    "attachments": {},
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "The chance is actually higher than you may think. If you remove `/` (the root of all files), for instance, your entire file system will be gone. If you remove `~` (your home directory), all your files will be gone. If you remove `.` (the current folder), all the files in the current directory will be gone. For any of these to be produced, you need a string length of 1 (chance: 1 out of 100) and one of these three characters (chance: 3 out of 32), which indeed is a chance of about one in a thousand."
+    "The above code block precludes the possiblity of removing `~` (your home directory), this is because the probability of generating the character '~' is not 1/32; it is 0/32. The characters are created by calling chr(random.randrange(char_start, char_start + char_range)), where the default value of char_start is 32 and the default value of char_range is 32. The documentation for chr reads, \"[r]eturn the string representing a character whose Unicode code point is the integer i.\" The Unicode code point for '~' is 126 and therefore, not in the interval [32, 64). \n",
+    "\n",
+    "If the code were to be changed so that char_range = 95 then the probability of obtaining the character '~' would be 1/94 , thus resulting in the probability of the event of deleting all files being equal to 0.000332\n",
+    "\n",
+    "And all your files in the home directory will be gone"
    ]
   },
   {
@@ -877,14 +908,27 @@
    "metadata": {},
    "outputs": [],
    "source": [
-    "1/100 * 3/32"
+    "3/94 * 1/94 * 99/101"
    ]
   },
   {
+   "attachments": {},
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "However, we can actually deal with any string as long as the _second_ character is a space – after all, `rm -fr / WHATEVER` will first deal with `/`, and only then with whatever follows. The chances for the first character are 3 out of 32, for the space 1 out of 32, so we're more at 1 out of 300:"
+    "However, we can actually deal with any string as long as the _second_ character is a space – after all, `rm -fr / WHATEVER` will first deal with `/`, and only then with whatever follows. The chances for the first character are 2 out of 32 as the code block above only allows for the probability of obtaining a `/` or a `.` but not a `~`.\n",
+    "\n",
+    "For the space the probability is 1 out of 32.\n",
+    "\n",
+    "We have to include the term for the probability of obtaining at least 2 characters which is required for the scenario of obtaining a space as the second character. This probability is 99/101 because it is calculated as (1 - probabilty of obtaining a single character or no character at all), so it is equal to 1-(2/101).\n",
+    "\n",
+    "Therefore, the probability calculation for the event of deleting all files in the case of having a space for the second character is:\n",
+    "\n",
+    "[probability of obtaining '/' or '. ' followed by a space] = [the probability of obtaining either the '/' character or the '. ' character] * [the probability of obtaining space] * [Probability of getting at least 2 characters] = 0.001914\n",
+    "\n",
+    "\n",
+    "\n",
+    "Diagram of probability of obtaining at least 2 characters."
    ]
   },
   {
@@ -893,7 +937,7 @@
    "metadata": {},
    "outputs": [],
    "source": [
-    "3/32 * 1/32"
+    "2/32 * 1/32 * 99/101"
    ]
   },
   {