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#9: Post undeleted by user avatar dustytrash‭ · 2021-01-05T21:40:46Z (almost 4 years ago)
#8: Post deleted by user avatar dustytrash‭ · 2021-01-05T21:39:17Z (almost 4 years ago)
#7: Post edited by user avatar dustytrash‭ · 2021-01-05T20:42:35Z (almost 4 years ago)
clarified note
  • 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.
  • There is no penalty for failure, only that you lose the coins spent.
  • 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.
  • Upgrading weapons is something done very very often. Let's say the average player wants to upgrade ~1,000 times.
  • 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.
  • There is no penalty for failure, only that you lose the coins spent.
  • 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.
#6: Post edited by user avatar dustytrash‭ · 2021-01-05T20:33:43Z (almost 4 years ago)
  • 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.
  • There is no penalty for failure, only that you lose the coins spent.
  • 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.
  • 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.
  • There is no penalty for failure, only that you lose the coins spent.
  • 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.
  • Upgrading weapons is something done very very often. Let's say the average player wants to upgrade ~1,000 times.
#5: Post edited by user avatar dustytrash‭ · 2021-01-05T20:15:42Z (almost 4 years ago)
clarity / language used
  • 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.
  • There is no penalty for failure, only that you lose the coins spent.
  • My theory is that since `10% + 10% + 10% = 30%`, it would be better to do 3 sets of 10%, because your odds at the beginning and end are the same, 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.
  • 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.
  • There is no penalty for failure, only that you lose the coins spent.
  • 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.
#4: Post edited by user avatar dustytrash‭ · 2021-01-05T20:11:20Z (almost 4 years ago)
  • 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.
  • There is no penalty for failure, only that you lose the coins spent.
  • My theory is that since `10% + 10% + 10% = 30%`, it would be better to do 3 sets of 10%, because your odds at the beginning and end are the same, 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 thoery 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.
  • 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.
  • There is no penalty for failure, only that you lose the coins spent.
  • My theory is that since `10% + 10% + 10% = 30%`, it would be better to do 3 sets of 10%, because your odds at the beginning and end are the same, 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.
#3: Post edited by user avatar dustytrash‭ · 2021-01-05T20:10:57Z (almost 4 years ago)
  • 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.
  • There is no penalty for failure, only that you lose the coins spent.
  • My theory is that since `10% + 10% + 10% = 30%`, it would be better to do 3 sets of 10%, because your odds at the beginning and end are the same, 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 thoery 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.
  • 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.
  • There is no penalty for failure, only that you lose the coins spent.
  • My theory is that since `10% + 10% + 10% = 30%`, it would be better to do 3 sets of 10%, because your odds at the beginning and end are the same, 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 thoery 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.
#2: Post edited by user avatar dustytrash‭ · 2021-01-05T20:10:25Z (almost 4 years ago)
  • Are 3 10% chances better than one 30% chance?
  • Are 3 10% chances better than one 30% chance (when penalized by a variable for failures)?
#1: Initial revision by user avatar dustytrash‭ · 2021-01-05T20:09:42Z (almost 4 years ago)
Are 3 10% chances better than one 30% chance?
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.

There is no penalty for failure, only that you lose the coins spent.

My theory is that since `10% + 10% + 10% = 30%`, it would be better to do 3 sets of 10%, because your odds at the beginning and end are the same, 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 thoery 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.