What are the odds of winning the pokemon lottery?

Obviously, the odds of winning a Master Ball is $\frac{1}{65536}$ . From now on, we will focus on the odds of winning a Rare Candy.

We partition the cases of winning a Rare Candy into five:

In all of the following case, let $x$ denotes the number which we want to replace so that it does not match the trainer ID.

Case $1$ : $xbcde$

We partition case $1$ into subcases $1(a)$ and $1(b)$ .

Subcase $1(a)$ : $xbcde$ where $\overline{bcde}$ ranges from $0000$ to $5535$ .

Obviously, $x$ can only range from $0$ to $6$ . Therefore, there are $7\times 5536= 38752$ IDs of this type. If we are to replace $x$ by another number ( so as to create IDs which differ in the first digit), we only have $6$ choices as replacement.

Subcase $1(b)$ : $xbcde$ where $\overline{bcde}$ ranges from $5536$ to $9999$ .

Notice that $x$ can only range from $0$ to $5$ . If $x>5$ , then $xabcd$ would be out of range. Therefore, there are $6\times 4464= 26784$ IDs of this type. If we are to replace $x$ by another number ( so as to create IDs which differ in the first digit), we only have $5$ choices as replacement.

It is obvious that the two subcases above partition the set of $65536$ possible trainer ID. Now, we attempt to calculate average number of loto IDs which match every single digit of the trainer ID except the first by considering the number of such loto IDs generated by each element in the set of $65536$ trainer IDs and then taking the average. Note that for the $38752$ trainer IDs in subcase $1$ , if we replace the first digit of the trainer ID, there are $6$ possible replacements. In subcase $2$ , if we replace the first digit of the trainer ID, there are $5$ possible replacements. Therefore, the average number of loto IDs generated by each of the $65536$ trainer IDs for case $1$ is $\frac{38752\times 6 + 26784\times 5}{65536}$ . To calculate the probability of case $1$ appearing for any of the $65536$ trainer IDs, we simply take $\frac{38752\times 6 + 26784\times 5}{65536^2}$ .

From case $2$ onwards, I will only list the subcases and probabilities since the reasoning is similar to that of case $1$ .

Case $2$ : $axcde$

Subcase $2(a)$ : $axcde$ where $a$ ranges from $0$ to $5$ while $\overline{cde}$ ranges from $000$ to $999$ .

Notice that $x$ can take values from $0$ to $9$ .Therefore, there are $6\times 10 \times 1000= 60000$ IDs of this type. For replacement of the second digit, we have $9$ possible replacements.

Subcase $2(b)$ : $axcde$ where $a=6$ while $\overline{cde}$ ranges from $000$ to $535$ .

Notice that $x$ can take values from $0$ to $5$ .Therefore, there are $1\times 6 \times 536= 3216$ IDs of this type. For replacement of the second digit, we have $9$ possible replacements.

Subcase $2(c)$ : $axcde$ where $a=6$ while $\overline{cde}$ ranges from $536$ to $999$ .

Notice that $x$ can take values from $0$ to $4$ .Therefore , there are $1\times 5 \times 464= 2320$ IDs of this type. For replacement of the second digit, we have $9$ possible replacements.

Combining all the subcases, the probability of case $2$ appearing is $\frac{60000\times 9 + 3216\times 5 +2320\times 4 }{65536^2}$ .

Case $3$ : $abxde$

Subcase $3(a)$ : $abxde$ where $\overline{ab}$ ranges from $00$ to $64$ while $\overline{de}$ ranges from $00$ to $99$ .

Notice that $x$ can take values from $0$ to $9$ .Therefore, there are $65\times 10 \times 100= 65000$ IDs of this type. For replacement of the third digit, we have $9$ possible replacements.

Subcase $3(b)$ : $abxde$ where $\overline{ab}=65$ while $\overline{de}$ ranges from $00$ to $35$ .

Notice that $x$ can take values from $0$ to $5$ .Therefore, there are $1\times 6 \times 36= 216$ IDs of this type. For replacement of the third digit, we have $5$ possible replacements.

Subcase $3(c)$ : $abxde$ where $\overline{ab}=65$ while $\overline{de}$ ranges from $36$ to $99$ .

Notice that $x$ can take values from $0$ to $4$ .Therefore, there are $1\times 5 \times 64= 320$ IDs of this type. For replacement of the third digit, we have $4$ possible replacements.

Combining all the subcases, the probability of case $3$ appearing is $\frac{65000\times 9 + 216\times 5 +320\times 4 }{65536^2}$ .

Case $4$ : $abcxe$

Subcase $4(a)$ : $abcxe$ where $\overline{abc}$ ranges from $000$ to $654$ while $\overline{e}$ ranges from $0$ to $9$ .

Notice that $x$ can take values from $0$ to $9$ .Therefore, there are $655\times 10 \times 10= 65500$ IDs of this type. For replacement of the fourth digit, we have $9$ possible replacements.

Subcase $4(b)$ : $abcxe$ where $\overline{abc}=655$ while $\overline{e}$ ranges from $0$ to $5$ .

Notice that $x$ can take values from $0$ to $3$ .Therefore, there are $1\times 6 \times 4= 24$ IDs of this type. For replacement of the fourth digit, we have $3$ possible replacements.

Subcase $4(c)$ : $abcxe$ where $\overline{abc}=655$ while $\overline{e}$ ranges from $6$ to $9$ .

Notice that $x$ can take values from $0$ to $2$ .Therefore, there are $1\times 3 \times 4= 12$ IDs of this type. For replacement of the fourth digit, we have $2$ possible replacements.

Combining all the subcases, the probability of case $4$ appearing is $\frac{65500\times 9 + 24\times 3 +12\times 2 }{65536^2}$ .

Case $5$ : $abcdx$

Subcase $5(a)$ : $abcdx$ where $\overline{abcd}$ ranges from $0000$ to $6552$ while $\overline{e}$ ranges from $0$ to $9$ .

Notice that $x$ can take values from $0$ to $9$ .Therefore, there are $6553\times 10= 65530$ IDs of this type. For replacement of the fifth digit, we have $9$ possible replacements.

Subcase $5(b)$ : $abcdx$ where $\overline{abcd}=6553$ while $\overline{e}$ ranges from $0$ to $5$ .

Notice that $x$ can take values from $0$ to $5$ .Therefore, there are $1\times 6= 6$ IDs of this type. For replacement of the fifth digit, we have $5$ possible replacements.

Combining all the subcases, the probability of case $5$ appearing is $\frac{65530\times 9 + 6\times 5 }{65536^2}$ .

We are finally getting there! To calculate the probability of winning a Master Ball or Rare Candy in the first go, we simply add up the probability of winning the Master Ball and the probability of five cases above. This yields the desired answer.