How to calculate this probability?

[Hrus]

How to calculate this probability?

Somebody asked me to calculate probability for this:
While gambling amulets with high-level character, he got Tal-Rasha's Amulet and Highlord's Wrath for 10M gold.

10M/63K=158 gambles
odds for set: 1/1000
odds for TR amu between set amus:1/45
odds for uniques: 1/2000
odds for HW between u. amus: 5/75

RESULT?
(mine was 0.000182~, but I am not sure if I did it right)

 

[krischan]

Your numbers are correct. Wearing Gheed's Fortune and an Edge bow will reduce gambling costs by up to 30%.

What does the number 0.000182 mean ?

The conclusion you can draw is that you have to spend an average of about 2 billion of gold to gamble Tal's amulet and 1.3 billion for Highlord's, when wearing an edge bow and Gheeds' with a perfect vendor price reduction.

 

[Hrus]

0.000182 (or 0.0182%) should be the final probability of gambling both amus for 10M gold. I didn't write how I calculated it, because I wanted others to have some nice mathematic exercise. And I am almost sure I didn't calculate it right.

 

[krischan]

If something happens with a chance of X (X being a number from 0 to 1), the chance for it to happen at least once after N tries is 1 - (1 - X)^N

 

[Hrus]

If something happens with a chance of X (X being a number from 0 to 1), the chance for it to happen at least once after N tries is 1 - (1 - X)^N

Yep. I just did the same for Highlord and Tal Rasha's amu and multiply both probabilities. But I am not sure if I can do this that way because the events are in relation. If the highlords are gambled, Tal's amu can't be gambled in the same try.

 

[krischan]

That will have an insificant influence. The chance for getting Tal's amu will smaller by just 1/2000 - the chance for a unique amulet spoiling the roll for succeeding with the roll for Tal's amu: 1999/2000 * 1/45000 = 0.0000222111 instead of 0.0000222222.

 

[Arbedark]

1/1000 x 1/45 = 1/45000 per roll of getting a TR.

1/2000 x 5/75 = 1/30000 per roll of getting a highlords.

158 Gambles = 156 "unsuccessful" rolls. and 2 "Successful" rolls.

Argh GOD I hate stats!

Roll order has an affect on this.

Let U = Unsuccessful roll.
Let TR = Tal Rasha
Let HL = Highlords

156 U, TR, HL
155 U, TR, HL, U

Etc etc. Lots of different possibilities. Add up the probability of each of these different events happening and you will have final probability.

Edit: Stupid me.

Chance = 158 x 156 x Probability of NOT getting TR or HL x Probability of getting HL x probability of getting TR

158 x 156 x 1/30000 x 1/45000 x 0.9999(4 s.f)

= 1.8256763470314403292181069958848e to the -5



Arb

 

[krischan]

I don't know what you actually calculated here. :scratch:

Chance for getting HL Wrath with 158 rolls = 1 - (1 - 1/30000)^158 = 0.00525

Any roll resulting in a unique amulet (1/2000 chance) prevents getting Tal's amu, so the chance for that amulet with 158 rolls is 1 - (1 - 1999/2000 * 1/45000)^158 = 0.00350

 

[Arbedark]

I calculated the probability of getting both Highlords and Tal Rasha's amulet from 158 rolls.

Ignore everything before "Edit: Stupid me" as that is mostly incorrect.





Arb

 

[krischan]

OK, now I think I understand it. It seems you approximated (1-X)^N = 1-N*X which is rather accurate with X being positive and much smaller than 1, being the case here.

OK, enough smartass talk. Question answered :lol:

 

[pncwd]

OK, now I think I understand it. It seems you approximated (1-X)^N = 1-N*X which is rather accurate with X being positive and much smaller than 1, being the case here.

OK, enough smartass talk. Question answered :lol:

I love math and statistics, but when I got out of high school x number years ago I left all that stuff behind. But just reading you guys arguments (not bickering type the math type) I just want to go back and learn it all over again. LOL

you guys rock, what do you do for a living, human calculators. hehe, j/k

basically it boils down to luck I think that is the easiest way to put it. If it is anything other than luck I want you guys to join me in a casino somewhere. that way I will know when to play and when not to play.

:thumbsup: :thumbsup: :thumbsup:

 

[krischan]

I studied physics and knowing that (1+-X)^N ~= 1+-N*X with |X|<<1 is first semester math

 

[Warrior of Light]

I have nothing to do on the work, so the answer is:

(1-(1-(1/45)*(1/1000))^158)* (1-(1-(5/75)*(1/2000))^158) = 1/54314 (0.0000184114)

 

[Arbedark]

I'm studying math from tomorrow at uni.

I used a longer method than Krischan used, and if someone cares to evaluate my answer without "e" to the negative 5, then we'll see if the answers match up (when i did it quickly was using scientific calc on computer, there will also be a rounding error due to me using 0.9999 to 4 significant figures instead of the exact value).

I've probably got it wrong since its been over a year since i've done probability in maths. And i've probably used assumptions which are incorrect.




Arb

 

[Hrus]

I did it exactly as Warrior of Light, but I did write one zero less in my post. Stupid me.

 

[lemonD]

Under the assumpt that the chance a ring be a set is independent of the chance it be an unique:

HL, chance to be Hilord: 0.0005*5/75
TR, chance to be Tal's: 0.001/45
NT, chance of getting nothing: 1-HL-TR+HL*TR since HL, TR are independent
nHL, chance not be Hilord: 1-HL
nTR, chance not be Tal's: 1-TR


Chance of getting exact 1 Hilord and 1 Tal's: HL*TR*NT^156*158*157 = 1.82163e-5
Chance of getting at least 1 Hilord and at least 1 Tal's = 1-nHL^158-nTR^158+NT^158 = 1.84114e-5


I think im right

 

[Hrus]

This seems right lemonD. :thumbsup:
Nice first post. :clap: