On (American) Survivor S38E4,\302\240an intersting scenaio arose. There\302\240was initially\302\2402 teams of 6 (blue team) and 9 persons (yellow team).
Then they decided to mix it up. Each player was randomly assigned a buff, (red, green or orange =3 colored buffs *5).= 3 teams of 5.What was interesting is 5 out of the original 6 blue team are now in the new red team. 5 of the original yellow team are in the new green team and 4 of the original yellow team are in the new orange team
What is the chances of that happening?
Perhaps the easy way to look at it is, what is the probability that you pick five greens and they are all original blues, and pick five reds and they are all original yellows.Method 1:The total number of set partitions of n distinct items into e sets of length L, with n = e*L, is n!/e!/(L!)^e. For 15 into 3x5, [e=3,L=5] that is 126,126 unique team combinations.
(I think it was vv who furnished this formula for a related problem. Don't ask me how he derived it). Stop press:I found this
req := f(m) = binomial(m*n-1, n-1)*f(m-1); rsol := rsolve({req, f(0) = 1}, f(m))eval(rsol, {m = 3, n = 5})and then he went to this closed form solutionfmk := factorial(m*n)/(factorial(m)*factorial(n)^m); eval(fmk, [m = 3, n = 5])So there would be at least 1 instance of all blue ending up on one team, and all yellow on another. And the complement. So 2/%Method 2 (old maths profs soln):
I think this is a problem of sampling without replacement, which uses the hypergeometric distribution.
OK you have initially 15 people with 6 blue and 9 yellow. And now take the simpler case of one buff. Call getting an original blue a success S and not getting a blue a failure F. Say we want to know the probability of getting 1 original blue in the new red. Means the other four are original yellow. P(1) = 5C1 * (6/15)^1 * (9/15)^4 = 0.2592 and now I think I\342\200\231ve oversimplified it. Yes, the binomial distribution is sampling with replacement. The correct answer is 5C1 * (6/15) * (9/14) * (8/13) * (7/12) * (6/11) = 0.2517.
I think the answer to the whole problem is the number of combinations possible of 1 S and 4 F in the new red, and 5 S and 0 F in the new green, and 0 S and 5 F in the new orange buff, which should be 15!/((1! * 4!) * (5! * 0!) * (0! * 5!)) = 1.30767E12/(24*120*120) = 3783780.
This has to be multiplied by the probability that the first person picked was blue, the next four yellow, the next five blue, the next zero yellow, the next zero blue and the last five yellow. This probability works out to be the same regardless of the order in which these choices are arranged, and is (6/15) * (9/14)*(8/13)*(7/12)*(6/11) * (5/10)*(4/9)*(3/8)*(2/7)*(1/6) * (5/5)*(4/4)*3/3)*(2/2)*(1/1) = 6/15 * (9!/5!)/(14!/10!) * 5!/(10!/5!) * 1 = 0.4*0.1259*0.003968 = 0.0001998.There are 15! Possible arrangements of the 15 players. How many of these have the number of original blue and yellow players in the buffs chosen at random? You have to distribute 6 original blues and 9 original yellows into 15 slots. There are 6 ways of putting an original blue into the first slot of the red buff and 9!/5! ways of filling the remaining five spots with original yellows and then there are 5! ways of placing the remaining five original blues in the red and 5! ways of putting the remaining five original yellows in the five yellow slots, and if you divide this by 15! You get the above probability which as I said is exactly the same if you get the same numbers in the buffs in a different order. However, there are 5C1 ways of choosing which of the five blue buff are the original blue, but the other two buffs, there is only one way each.
So the answer should in fact be 5*0.0001998 = 0.000999. Plausible, but it seems a bit small.Which is correct, and is there a Maple solution?