Mathematics

I'm playing a computer game in which you can spend 100 coins to be given a 10% chance to upgrade a weapon, or spend 200 coins for a 20% chance, or 300 coins for a 30% chance and so on, up to 1000 coins for a 100% chance.

My theory is that if you only have 300 coins to spend, it would be better to do 3 attempts since 10% + 10% + 10% = 30% and you have the added bonus of having a chance to spend less coins if it succeeds before the 3rd try.

Not sure if my theory is correct because a 100% chance is not the same as 10 10% chances. Also after getting failures, the odds remain the same. But I don't have the logic/math knowledge to have an answer and explanation.

Note:
For the purpose of the question, let's assume the game is somehow computing true random numbers and therefore able to give true 10%-100% odds
Upgrading weapons is something done very very often. Let's say the average player wants to upgrade ~1,000 times.

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Comments on Are 3 10% chances better than one 30% chance (when penalized by a variable for failures)?

Are 3 10% chances better than one 30% chance (when penalized by a variable for failures)?

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