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

[llatosz]

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.

 

[Merlin3189]

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!

 

[llatosz]

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?

 

[Merlin3189]

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

 

[llatosz]

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 * 100_combos = 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

 

[StoneTemplePython]

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

 

[Merlin3189]

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 * 100_combos = 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. $C(100,4) = \frac {100!} { (100 - 4 )! \times 4!} = \frac {100!} {96! \times 4!} = \frac {100.99.98.97} {4.3.2.1} = 3,921,225$

 

[llatosz]

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!

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 :)

 

[Merlin3189]

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.

 

[sysprog]

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%.