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# What are my chances of finding a Ditto with 6 perfect IVs in the Friend Safari?

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Or a 5 IVs and 4 IVs on that topic. Perfect is an IV of 31.

All FS (I'll be using that term for Friend Safari) Pokemon have two perfect IVs, so that's 1/1 if you want a 2 IV Ditto.

There is a 1/32 chance you can get any IV from 0 to 31. So if you want a 3 IV Ditto, there is a 1/32 chance.

Same with 4 IVs, you just multiply the chance by 32 since there are 32 new possible IVs, or 1/1024.

You keep going with 5 IVs and 6IVs to get 1/32768 and 1/1048576 respectively.

Comparing this with a regular wild Ditto you can catch in the Pokemon Village, that has a 1/1073741824 (more than 1/one trillion) chance to have perfect IVs.

tl;dr Simple Chart:
2 IV FS Pokemon: 1/1
3 IV FS Pokemon: 1/32
4 IV FS Pokemon: 1/1024
5 IV FS Pokemon: 1/32768
6 IV FS Pokemon: 1/1048576
0 IV Normal Pokemon: 31/32
1 IV Normal Pokemon: 1/32
2 IV Normal Pokemon: 1/1024
3 IV Normal Pokemon: 1/32768
4 IV Normal Pokemon: 1/1048576
5 IV Normal Pokemon: 1/33554432
6 IV Normal Pokemon: 1/1073741824

Source: Simple Calculations and Experience with your torture questions before

by
edited by
In case you're wondering what the actual fraction for a 0 IV Pokemon is (unsimplified) it's 1040187392 / 1073741824. This adds up to 96.875%.
Just going to post that this is inaccurate.

1,073,741,824 is just over 1 billion (not trillion).

Second, the calculations you posted for all of your calculations assume that if you are looking for a 2 IV Pokemon there are only 2 stats. There are 15 possible combinations of 2 IVs existing. That means that there are 13,852,815 possible combinations out of the 1,073,741,824

So about 1 in 78 to get a 2 IV Pokemon.

Now you just need to remember that these are the "binary" (maxed or not maxed) possibilities.
1 outcome for zero IVs
6 outcomes for 1 IV
15 outcomes for 2 IVs
20 outcomes for 3 IVs
15 for 4 IVs
6 for 5 IVs
1 for 6 IVs

Aside from the 6 IVs scenario, you just need to use

z=x*31^y

where
x = the number of binary combinations for desired IVs
y = 6  - number of maxed IVs desired.

You're chances are z/32^6

The guy above me, Le Scraf, the chance for a 0 IV Pokemon is only

1 in 1,073,741,824

...