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Raikou 100% IV chance

Has anyone calculated the chances of encountering a perfect Raikou and successfully catching it at the same time? Assuming you have an average of 10 balls per try with great curve throw for each ball thrown with no misses. Golden razz is used for each throw. The badge for catching the pokemon type is also gold. What are the odds?

Asked by AsianKetchup7 years 8 months ago

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Answers

1/216 chance of encountering one.

Chance of catching varies on how good your throws are.

Edited the question for further details.

Encountered once and I was given 6 balls. Luckily I managed to catch it on the fifth ball.

This is that RNG BS. I never get below 11 balls ever and still have yet to run into a perfect Legendary. Only 3 all together over 91%. A 91% Lugia I caught. A 96% Lugia that fled, and a 96% Raikou I caught. No other Legendaries and I have over 70 candies for all of them so far too.

The number of balls has nothing to do with Mon iv.

I know, it has everything with how much you contributed in defeating the Raid Boss. Crazy how lots can leech of you and always end up with the better IV Pokemon anyways.

3 out of a possible 14 balls are due to your own personal contribution. If you are on a minority team in the raid, your team doesn't have control, and you rejoin the battle instead of being a leech in the lobby, you get 6 balls. Or if it happened before they gave out 6 balls, he could have done +1 damage and only gotten 6 balls.

Still not contributing much. I've literally been in Raids where it was me and one other person in a 10-12 man Raid and we still did the most damage overall and did we get a shot at a perfect or above 91%, nope. We did the most damage too, got more balls but got RNG'd with some BS. Twice actually in level 5 Raids .

The chance of encountering one is 1/216, the chance of capturing it with those params is 70%. Multiply the two and you get ~1/308.

Glug - You are the only one that really answered the question. Good job! I understand the 1/216 part (iv range is 10 to 15 for each of H/A/D, so 6 possibilities for each of the three ivs, so 1/6 x 1/6 x 1/6 = 1/216). But it don't understand how you got the 70%. Please elucidate, Master Glug.

Edit: Pork Chop might have answered the question, too - with the question later being edited to provide the specific parameters regarding the balls, etc.