Jufo81 Joined: May 23, 2010
• Posts: 344
March 30th, 2018 at 5:14:43 AM permalink
Hi,
this probability question arised from the latest episode of TV show Survivor, a show which I am sure everyone knows about. The latest episode can been seen here within US: https://www.cbs.com/shows/survivor/video/pCyKg5imLvLv6Bb3N_DwV_3KQL4BAMrN/survivor-ghost-island-fate-is-the-homie/

In the episode there were two original tribes (tribe = group of people playing together): Orange with 6 players and Purple with 9 players. These 15 players were randomly divided into three new tribes: A, B and C so that each tribe consisted of 5 members. The draw was conducted by contestants blindly picking buffs in turn, there were 5 buffs of each kind A,B and C.

What happened in the draw was that the Orange minority tribe with 6 members ended up being split 2-2-2 into the three new tribes, meaning that each tribe A,B,C ended up with 2 Orange and 3 Purple contestants. Can anyone calculate the probability of this outcome, because an uneven split seems to be much more likely?

To put the question into simpler from: You can think of there being 9 Purple balls and 6 Orange balls. The balls are randomly labeled A,B,C so that there are 5 of each A,B,C. What are the odds that exactly 2 Orange balls get label A, B and C? Last edited by: Jufo81 on Mar 30, 2018
ChesterDog Joined: Jul 26, 2010
• Posts: 964
March 30th, 2018 at 7:59:56 AM permalink
Quote: Jufo81

...What happened in the draw was that the Orange minority tribe with 6 members ended up being split 2-2-2 into the three new tribes, meaning that each tribe A,B,C ended up with 2 Orange and 3 Purple contestants. Can anyone calculate the probability of this outcome, because an uneven split seems to be much more likely?... I get 0.12397 as the probability. And I find that 2 Orange with 3 Purple in each new tribe is the most likely outcome. I, too, expected an uneven distribution more likely.

By the way, the number of ways for 2 Oranges in each of the first two tribes is combin(6,2) * combin(4,2) = 90.
ThatDonGuy Joined: Jun 22, 2011
• Posts: 4926
March 30th, 2018 at 8:42:21 AM permalink
Maybe I'm reading the problem wrong, but I get about 1/5 - actually, exactly 200/1001.

There are (15)C(6) (i.e. COMBIN(15,6)) = 5005 ways to draw 6 balls from the 15. There are (5)C(2) = 10 pairs of As that can be drawn, (5)C(2) pairs of Bs, and (5)C(2) pairs of Cs, so the probability = (10 x 10 x 10) / 5005 = 200 / 1001.
MidwestAP Joined: Feb 19, 2012
• Posts: 1262
March 30th, 2018 at 9:08:45 AM permalink
I get the same result at ThatDonGuy. There are 10 ways that exactly two orange end up in any group.

OOPPP
OPOPP
OPPOP
OPPPO
POOPP
POPOP
POPPO
PPOOP
PPOPO
PPPOO

The chances that any one of these combinations is drawn for the first group is 10[(6*5*9*8*7)/(15*14*13*12*11)]=60/143=0.41958

Given that the first group met the requirements, that chances that any one of the combination is drawn for the second group is 10[(4*3*6*5*4)/(10*9*8*7*6)]=10/21=0.47619

The third group, by default has to meet the requirements.

Therefore, multiplying these two probabilities above gives us 200/1001= 19.98%
ChesterDog Joined: Jul 26, 2010
• Posts: 964
March 30th, 2018 at 9:23:40 AM permalink
Quote: MidwestAP

I get the same result at ThatDonGuy. There are 10 ways that exactly two orange end up in any group.

OOPPP
OPOPP
OPPOP
OPPPO
POOPP
POPOP
POPPO
PPOOP
PPOPO
PPPOO

The chances that any one of these combinations is drawn for the first group is 10[(6*5*9*8*7)/(15*14*13*12*11)]=60/143=0.41958

Given that the first group met the requirements, that chances that any one of the combination is drawn for the second group is 10[(4*3*6*5*4)/(10*9*8*7*6)]=10/21=0.47619

The third group, by default has to meet the requirements.

Therefore, multiplying these two probabilities above gives us 200/1001= 19.98%

Thanks for the explanation. I now agree with you and Don.
Wizard Joined: Oct 14, 2009
• Posts: 23301
March 30th, 2018 at 10:45:41 AM permalink
Quote: MidwestAP

Therefore, multiplying these two probabilities above gives us 200/1001= 19.98%

I agree. Here is my math: =(COMBIN(6,2)*COMBIN(9,3)/COMBIN(15,5))*(COMBIN(4,2)*COMBIN(6,3)/COMBIN(10,5))

Where combin(x,y)=x!/((y!*(x-y)!)
Where x! = 1*2*3*...*x

I haven't seen the episode yet. I tend to lose interest as the large-breasted women get voted out, which was the case last week.
It's not whether you win or lose; it's whether or not you had a good bet.
Jufo81 Joined: May 23, 2010
• Posts: 344
March 30th, 2018 at 11:19:13 AM permalink
Quote: ChesterDog

I get 0.12397 as the probability. And I find that 2 Orange with 3 Purple in each new tribe is the most likely outcome. I, too, expected an uneven distribution more likely.

By the way, the number of ways for 2 Oranges in each of the first two tribes is combin(6,2) * combin(4,2) = 90.

Hmm I got the same result as ChesterDog: 0.12397. All the possible ways to split the 6 Orange balls into three groups of five are {5,1,0},{4,2,0},{4,1,1},{3,2,1},{3,3,0},{2,2,2}, and out of these {3,2,1} is by far the most likely, not {2,2,2}.

If we start with {5,1,0} it can be ordered in 6 ways ({5,1,0},{5,0,1},{1,0,5},{1,5,0},{0,1,5},{0,5,1}) each of which is equally likely. If we label the orange balls a...f there are six ways to put them in {5,1,0} so the total number of these combinations is 6*6 = 36.

Similarily the total number of combinations for the other cases:
{4,2,0}: 6*(6)C(2) = 90
{4,1,1}: 3*6*5 = 90
{3,2,1}: 6*6*(5)C(2) = 360
{3,3,0}: 3*(6)C(3) = 60
{2,2,2}: (6)C(2)*(4)C(2) = 90 (<- what ChesterDog also wrote)

So in total there are 36 + 90 + 90 + 360 + 60 + 90 = 726 combinations and thus the probability for {2,2,2} split is 90/726 = 0.12397.

If this is wrong and the answer is 19.98% why is this so?
ThatDonGuy Joined: Jun 22, 2011
• Posts: 4926
March 30th, 2018 at 1:09:08 PM permalink
Quote: Jufo81

So in total there are 36 + 90 + 90 + 360 + 60 + 90 = 726 combinations and thus the probability for {2,2,2} split is 90/726 = 0.12397.

If this is wrong and the answer is 19.98% why is this so?

You are not counting the number of ways the 9 purple balls can be put into the remaining positions.
For example, each of the 36 {5,1,0} distributions of orange balls has (9)C(4) = 126 {0,1,5} distributions of purple balls.

For each distribution of orange balls, the total number of (orange ball distributions) x (purple ball distributions) is:
5,1,0: 36 x 126 = 4536
4,2,0: 90 x 504 = 45360
4,1,1: 90 x 630 = 56700
3,2,1: 360 x 1260 = 453600
3,3,0: 60 x 756 = 45360
2,2,2: 90 x 1680 = 151200
The ratio of 2,2,2 distributions to the total = 151,200 / 756,756 = 200 / 1001.
Also notice that 3,2,1 is the most likely distribution.
Jufo81 Joined: May 23, 2010
• Posts: 344
March 30th, 2018 at 1:12:31 PM permalink
Quote: ThatDonGuy

You are not counting the number of ways the 9 purple balls can be put into the remaining positions.
For example, each of the 36 {5,1,0} distributions of orange balls has (9)C(4) = 126 {0,1,5} distributions of purple balls.

For each distribution of orange balls, the total number of (orange ball distributions) x (purple ball distributions) is:
5,1,0: 36 x 126 = 4536
4,2,0: 90 x 504 = 45360
4,1,1: 90 x 630 = 56700
3,2,1: 360 x 1260 = 453600
3,3,0: 60 x 756 = 45360
2,2,2: 90 x 1680 = 151200
The ratio of 2,2,2 distributions to the total = 151,200 / 756,756 = 200 / 1001.
Also notice that 3,2,1 is the most likely distribution.

Thanks for this, I get it now!