Post History
#7: Post edited
- 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
- 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
- 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
- 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
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
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
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.