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Math genius needed for calculating probability

STAR WARS: The Old Republic > English > General Discussion > Off-Topic
Math genius needed for calculating probability

Whojoo's Avatar


Whojoo
01.27.2014 , 03:30 AM | #1
Story:
There is a game with chances to win one out of 15 prizes. You need tickets to win a prize.
In total there are 1000 tickets in the game. You got 5 yourself.
The average of tickets per person is 3.5.

Question:
  1. What is the probablity if every ticket is only allowed to win one prize?
  2. What is the probablity if every person is only allowed to win one prize?


I know that for 1 prize it would be P = 1 / 200
Thing is, there are 15 prizes. How do you calculate this again?

Thanks in advance for the help
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JPryde's Avatar


JPryde
01.27.2014 , 04:01 AM | #2
Part 1 is rather easy:

The chance to win at least one is determined as 1-(chance to win none).

Chance to win none is
995/1000*994/999*993/998*992/997*991/996*
990/995*989/994*988/993*987/992*986/991*
985/990*984/989*983/988*982/987*981/986
= 985*984*983*982*981 / (1000*999*998*997*996)
~ 0.9270

Chance to win at least one ticket is therefore 1-0.9270 = 0.072925 if each ticket is allowed to win (and thus every ticket you do not own can be a winner).

Chance to win at least one price is 7.29% with 5 tickets and 15 draws.

Working on part 2, which is a lot more tricky.
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Whojoo's Avatar


Whojoo
01.27.2014 , 04:39 AM | #3
Quote: Originally Posted by JPryde View Post
Part 1 is rather easy:

The chance to win at least one is determined as 1-(chance to win none).

Chance to win none is
995/1000*994/999*993/998*992/997*991/996*
990/995*989/994*988/993*987/992*986/991*
985/990*984/989*983/988*982/987*981/986
= 985*984*983*982*981 / (1000*999*998*997*996)
~ 0.9270

Chance to win at least one ticket is therefore 1-0.9270 = 0.072925 if each ticket is allowed to win (and thus every ticket you do not own can be a winner).

Chance to win at least one price is 7.29% with 5 tickets and 15 draws.

Working on part 2, which is a lot more tricky.
I forgot that trick, damn so easy xD

Isn't part 2 the same? 1000 tickets with 3.5 for each person. Making 1000/3.5 possible winners which is ~286.
Followed by the calculation:
285 * 284 * 283 * 282 * 281 * 280 * 279 * 278 * 277 * 276 * 275 * 274 * 273 * 272 * 271
286 * 285 * 284 * 283 * 282 * 281 * 280 * 279 * 278 * 277 * 276 * 275 * 274 * 273 * 272
= ~0.95
P = 1 - 0.95 = ~0.05

Or am I forgetting something?
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JPryde's Avatar


JPryde
01.27.2014 , 04:50 AM | #4
No, part 2 unfortunately is tricky.

We got 15 winners to find.

First winner is easy... it is simply 1-995/1000 for the player (5 tickets) and if that fails, some other guy won, who had 3,5 tickets on average... being 1-x.

From second draw on, there is a chance, that a previous winner wins again, which would result in another draw. On Stage 1, we had 15 draws, flat and square. On Stage 2, we could have a lot more draws, cause each duplicate would result in another draw. But we do not know how many draws exactly it will be, we only know that there are roughly 285 players (1 + 1000/3.5) each holding 1-x tickets (avg. 3.5)

Your equation is easily to be falsified, as the chance to win on stage 2 obviously is higher than on stage 1, not lower. The chance to get drawn "directly" is as high as on stage 1... .but additionally each ticket can win on the "secondary" draw, when a winner wins again.... so the chance to win on stage 2 is higher than on stage 1... 5% cannot possibly be right.

Unfortunately, the possible variations of tickets on players are near endless... for example, 1000 tickets could be distributed as follows:
5 player, 712 on player 2 and 1 ticket each on player 3 through player 285 (this would be an average of 3,5035 still). That however would mean, that we got a really big chance of duplicates, as player 2 got over 70% chance to win the first ticket and still over 70% to get another ticket drawn of his.... this is easily more than if every player had indeed 3 or 4 tickets. (the avg is the same with distributing halve 3 and half 4 tickets, but the chance for a duplicate is a lot less.)
In fact we do not even know the numer of other players. The distribution of tickets is indeed endless. There doesn't need to be 285 players, you can have an avg of 3.5 with much less players too...

Therefore I wonder, what kind of excersise this is you got there, as it seems worthy of high class university stochastic. In computer science we did a lot of stochastic, but never had an exercise as complex as this.
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Whojoo's Avatar


Whojoo
01.27.2014 , 06:29 AM | #5
I see we're doing part 2 with different assumptions. I asume that once a player wins, all his tickets are removed from the ticket pool.

This is not for an exercise. I was bored and wanted to calculate my chances to win a beta key to some game
They allow you to have a total of 5 'tickets' and the widget also shows the amount of submitted tickets.
I asume, since it's a computer system deciding the winners, that a player's tickets are removed from the pool once the player has won something.
Ofcourse the probability is different once you assume that only the winning ticket is removed from the pool.

The 5% is very inacurate due to me rounding numbers here and there.
I do expect a number below your mentioned 7% since you're 1 out of ~286 players compared to you having 1 out of each 200 tickets (or me in this case).

PS. The 1000 and 3.5 are fictional numbers. In reality there are now 13k tickets in the pool with probably an average of 4.5-ish per person.
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JPryde's Avatar


JPryde
01.27.2014 , 06:47 AM | #6
Okay, those are obviously helpful infos... first of all, I wasn't aware of the 5 ticket limit and not of the "whole removal".

So every winning ticket is removing not 1, but 3,5 tickets on average from the draw pool.

That will make 1000, 996.5, 993, 989.5, 986, .... for the draw pool.
And still it is always 5 tickets less than that for the players tickets.... so 1-(995/1000 * 991.5/996.5 ...)

The fact that duplicate draws are not possible will only reduce the number of available tickets much faster... the end result on 15 draws will still be higher than on the first version of calculation.
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Whojoo's Avatar


Whojoo
01.27.2014 , 06:54 AM | #7
Well still have to calculate it with the real numbers. And since I'm already at a 7% chance with a 1000 tickets... I doubt I want to know my chances :P

But again, It was just an assumption from my side that tickets will be removed once the owner has won something. I can understand the difficulty in the equation if there's a chance of more than 15 draws. Kinda curious if you'd end up with a higher or lower chance, or nearly equal.

But I don't know the website's algorithm for selecting a winner. So we will never know who's assumption was the right one :P

Thanks for you help I can start calculating the probability that I will be on swtor this weekend :P Ofcourse with just my assumption :P
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JPryde
01.27.2014 , 06:55 AM | #8
Quote: Originally Posted by Whojoo View Post
Thanks for you help I can start calculating the probability that I will be on swtor this weekend :P Ofcourse with just my assumption :P
On SWTOR?
There is a beta key lottery running for SWTOR ? Where do I sign up
~~~ Macht Wächter ~~~
Vanjervalis Chain
Jhoira, Skarjis, Trântor, Ric-Xano, Sabri-torina, Tir-za, Shaina ...
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Whojoo's Avatar


Whojoo
01.27.2014 , 06:59 AM | #9
Quote: Originally Posted by JPryde View Post
On SWTOR?
There is a beta key lottery running for SWTOR ? Where do I sign up
haha :P not swtor ofcourse. But the lottery for the beta I want to join ends this thursday
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