Given a success rate of 1% and 100 tries....

  • Context: High School 
  • Thread starter Thread starter llatosz
  • Start date Start date
  • Tags Tags
    Probabilities Rate
Click For Summary

Discussion Overview

The discussion revolves around calculating the probabilities of achieving successes in a scenario where there is a 1% success rate over 100 attempts. Participants explore the probability of no successes, one success, and two successes, while also considering the implications for a video game context where players seek to obtain items with low drop rates.

Discussion Character

  • Mathematical reasoning
  • Exploratory
  • Debate/contested

Main Points Raised

  • One participant calculates the probability of no successes as 0.99^100, suggesting a 36.6% chance.
  • Another participant questions the probability of achieving one success, proposing different interpretations of the calculation, including combinations of successes and failures.
  • There is a discussion about the number of combinations for achieving one or two successes, with some participants noting the complexity of calculating these probabilities.
  • One participant mentions that the probability of exactly two wins involves counting the number of ways to achieve those wins, leading to a discussion about permutations and combinations.
  • Another participant introduces the binomial coefficient formula for calculating combinations, which is acknowledged and discussed further.
  • Some participants express confusion over the calculations and seek clarification on the probabilities associated with multiple successes.
  • There is a mention of the probability of zero successes approximating e^-1 for large n, which is noted as an interesting point by participants.

Areas of Agreement / Disagreement

Participants express varying interpretations of the calculations, leading to some disagreements about the correct approach to determining probabilities. While there is some consensus on the general method of using combinations and permutations, specific numerical results and interpretations remain contested.

Contextual Notes

Participants highlight the importance of understanding the definitions of success and failure probabilities, as well as the need to account for combinations when calculating multiple successes. Some calculations are noted to be incorrect or incomplete, but no consensus is reached on the final probabilities.

Who May Find This Useful

This discussion may be useful for individuals interested in probability theory, particularly in contexts involving low success rates and multiple trials, such as gaming or statistical modeling.

llatosz
Messages
62
Reaction score
9
What is the probability of no success? 1 success? 2 successes?

No success: 0.99^100 = 36.6% chance?

1 success: (0.99^99) + 0.01 = 37.97% chance? or (0.99^99) * (0.01)^1 = 0.369% chance ? You would think it would be fairly likely

2 successes: ? -Depends on formula for 1 success

Can someone show me how to calculate? This is to show a friend how you won't be guaranteed to get an item with a 1% drop rate if you buy 100 drops for a video game, but I want to explore it a bit further.
 
Physics news on Phys.org
What would you say the chances of success on throw 1 and failure on every following throw?
And the chance of fail at throw 1, success at throw 2 and fail on the rest?
Etc.

You would think it would be fairly likely
WHY do you think that? There's a reason!
 
Last edited:
Merlin3189 said:
What would you say the chances of success on throw 1 and failure on every following throw?
And the chance of fail at throw 1, success at throw 2 and fail on the rest?
Etc.

Yes that is kinda what I'm asking lol. I'm aware that for 1 success there are 100 different combinations, and that for 2 successes there are like 998 or 999 combinations.

So the chance of success on throw 1 is 1%, and failure on every following throw is .99^99 = 36.97%, so the chance of 1 success would be 37.97% like I said?
Would this be the chance of 1 success out of 100?
 
You would think it would be fairly likely
You'd think that, because you have 100 chances of getting it, despite the low chance each time.

If you state it non-mathematically, I think it is easier:
What's the chance of winning on one particular throw? 0.01
What's the chance of winning on one particular throw and losing on the other 99 throws? 0.01 x 0.9999
Whats the chance of winning on ANY particular throw out of the 100 throws and losing on the other 99? 100 times previous answer.

P(1 success) is p1q(100-1) x (combinations: 1 from 100)
Which comes to 0.370 (3sf) or 0.3697 (4sf)

For 2 wins, you are going to say p2q(100-2)
and that is, say, P(WWLLLL...) , but it is also P(LLLWLLWLLL...) or P(WLLLLLLLLLWLLLL...) etc.
So you need to count how many ways you can get 2 wins out of 100 throws, 100C2 and add up all those probabilities.
Or more simply, just multiply P(2 specific wins) by 100C2
 
Merlin3189 said:
P(1 success) is p1q(100-1) x (combinations: 1 from 100)
Which comes to 0.370 (3sf) or 0.3697 (4sf)

For 2 wins, you are going to say p2q(100-2)
and that is, say, P(WWLLLL...) , but it is also P(LLLWLLWLLL...) or P(WLLLLLLLLLWLLLL...) etc.
So you need to count how many ways you can get 2 wins out of 100 throws, 100C2 and add up all those probabilities.
Or more simply, just multiply P(2 specific wins) by 100C2

Ah okay! Thanks! So p is probability of success and q is probability of fail from what I infer.
So, going forward "1 win" just means any 1 win, so,

P(1 win) = (0.01) * (0.99)99 * 100combos = 0.3697 = 36.97%
But then
P(2 wins) = (0.01)2 * (0.99)98 * (99*99)combos = 0.366 = 36.6% ... which doesn't make sense.

Also allow me to check my understanding if i may:
P(0 win) + P(1 win) + P(2 wins) + P(3 wins) ~=~ 0.95 to 1 because that covers nearly all possibilities since 4+ wins at 1% win rate should be incredibly rare
 
note the probability of zero successes is

##\binom{n}{0} (\frac{1}{n})^0 (1-\frac{1}{n})^n = (1-\frac{1}{n})^n \approx e^{-1}## for large n
 
llatosz said:
Ah okay! Thanks! So p is probability of success and q is probability of fail from what I infer.
Oops! Sorry that slipped through without my noticing. Yes. And q = 1-p
So, going forward "1 win" just means any 1 win, so,

P(1 win) = (0.01) * (0.99)99 * 100combos = 0.3697 = 36.97%
But then
P(2 wins) = (0.01)2 * (0.99)98 * (99*99)combos = 0.366 = 36.6% ... which doesn't make sense.

Also allow me to check my understanding if i may:
P(0 win) + P(1 win) + P(2 wins) + P(3 wins) ~=~ 0.95 to 1 because that covers nearly all possibilities since 4+ wins at 1% win rate should be incredibly rare
0.3660 + 0.3697 + 0.1849 = 0.9206 so, yes, there's not much chance left now
But your calculation for 2 combinations is not right.
P(2 specific wins) = 0.012 x 0.9998 is ok, but any 2 from 100 combinations is not 99 * 99
There are 100 positions for one win and 99 remaining positions for the other win, = 9900 permutations
But that counts every pair of wins twice - eg. WWLLL... or WWLLL... / WLLWLLL... or WLLWLLL... etc
So the number of combinations (when we ignore the order of selection) is 9900 / 2 = 4950
So P(exactly 2 wins in any positions) = 0.012 x 0.9998 x 4950 = 0.000037346 x 4950 = 0.1849

When you get on to 3 wins, there are 100 x 99 x 98 permutations, but each set of 3 wins gets counted 6 times (3 x 2 x 1)
and for 4 wins, there are 100 x 99 x 98 x 97 permutations, with each set of 4 wins getting counted 24 times (4 x 3 x 2 x 1 )

So the C(N,r) formula = ## \frac {N!} {(N-r)! \times r!}##
so, eg. ## C(100,4) = \frac {100!} { (100 - 4 )! \times 4!} = \frac {100!} {96! \times 4!} = \frac {100.99.98.97} {4.3.2.1} = 3,921,225 ##
 
  • Like
Likes   Reactions: llatosz
StoneTemplePython said:
note the probability of zero successes is

##\binom{n}{0} (\frac{1}{n})^0 (1-\frac{1}{n})^n = (1-\frac{1}{n})^n \approx e^{-1}## for large n

Oh wow, youre right, that's pretty neat!

Merlin3189 said:
There are 100 positions for one win and 99 remaining positions for the other win, = 9900 permutations
But that counts every pair of wins twice - eg. WWLLL... or WWLLL... / WLLWLLL... or WLLWLLL... etc
So the number of combinations (when we ignore the order of selection) is 9900 / 2 = 4950

Oh okay, this totally makes sense now!
And yes, I am getting the same result as you for 3 wins too! Makes sense, thank you! :D
Now I can formulate some numbers for my friend's loot drop gambling in a video game we play! There's an item he wants with a 1% drop rate and his idea was that he can buy 100 chances ($1 each) to guarantee himself the item lol.

Thank you very much for all your help, you are kind and it is fully appreciated :)
 
You're welcome.
So what is your conclusion, or your friend's? Does he think 100 tickets is enough to be certain of winning?
I was a little surprised when I worked out how many tickets I'd need, to have more chance of winning than not winning.
 
  • #10
With a 1% chance on each try, the chance of not winning on anyone try is 99%. The chance of not winning on either of 2 tries is 99% of 99%, or 0.99^2, which is 98.01%. For 100 tries, the chance of not winning at least once is 0.99^100, which is 0.366032, or 36.6032%. The chance of winning at least once is 100% minus the chance of not winning at least once, which for 100 tries with 1% chance of winning on each try, is 100% - 36.6032%, which is 63.3968%.
 

Similar threads

High School Probability in a dice game
  • · Replies 41 ·
2
Replies
41
Views
6K
High School How to calculate this conditional probability?
  • · Replies 7 ·
Replies
7
Views
1K
Undergrad Understanding more about probability? (beginner here)
  • · Replies 10 ·
Replies
10
Views
2K
High School Monty Hall Paradox: Why Does It Matter What Host Knows?
  • · Replies 12 ·
Replies
12
Views
3K
Undergrad Need a quick double-check on probability formula for 100-sided die
  • · Replies 5 ·
Replies
5
Views
3K
Undergrad What is the Probability of Winning in a Raffle Ticket Draw with Replacement?
  • · Replies 12 ·
Replies
12
Views
6K
Graduate For a Given a Probability of Success, How Many Successes in a Sample?
  • · Replies 3 ·
Replies
3
Views
3K
High School Baccarat math and beating the edge
  • · Replies 3 ·
Replies
3
Views
3K
High School The Paradox of 1 – 1 + 1 – 1 + 1 – 1 + …
  • · Replies 5 ·
Replies
5
Views
3K
High School Generating Stats and Probabilities
  • · Replies 2 ·
Replies
2
Views
2K