Communities

Writing
Writing
Codidact Meta
Codidact Meta
The Great Outdoors
The Great Outdoors
Photography & Video
Photography & Video
Scientific Speculation
Scientific Speculation
Cooking
Cooking
Electrical Engineering
Electrical Engineering
Judaism
Judaism
Languages & Linguistics
Languages & Linguistics
Software Development
Software Development
Mathematics
Mathematics
Christianity
Christianity
Code Golf
Code Golf
Music
Music
Physics
Physics
Linux Systems
Linux Systems
Power Users
Power Users
Tabletop RPGs
Tabletop RPGs
Community Proposals
Community Proposals
tag:snake search within a tag
answers:0 unanswered questions
user:xxxx search by author id
score:0.5 posts with 0.5+ score
"snake oil" exact phrase
votes:4 posts with 4+ votes
created:<1w created < 1 week ago
post_type:xxxx type of post
Search help
Notifications
Mark all as read See all your notifications »
Users Search
Help Dashboard Sign In Sign Up
Q&A Meta
Q&A
Posts Tags Edits
Ask Question

Post History

81%
+7 −0
Q&A Are 3 10% chances better than one 30% chance (when penalized by a variable for failures)?

3 answers  ·  posted 3y ago by dustytrash‭  ·  last activity 3y ago by Grove‭

Question logic basic odds
#9: Post undeleted by user avatar dustytrash‭ · 2021-01-05T21:40:46Z (over 3 years ago)
#8: Post deleted by user avatar dustytrash‭ · 2021-01-05T21:39:17Z (over 3 years ago)
#7: Post edited by user avatar dustytrash‭ · 2021-01-05T20:42:35Z (over 3 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 (over 3 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 (over 3 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 (over 3 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 (over 3 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 (over 3 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 (over 3 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.
logic basic odds