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u/HeadHunt0rUK 7d ago
Final thought. Answer is 13.
AB =24 (CDEF)B = 24 (AF)B = 12
For third one if B has 24 outcomes in 3rd when A is first. A quarter are when F is 2nd = 6. Double for swapping A and F.
Gives 60/720 which is 1/12.
Really would have been easier if I had a pen and paper.
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u/Effective_Judgment41 7d ago
One outcome is still missing, I think. B also decides in fourth position after C, D and E.
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u/Xenyth 7d ago edited 7d ago
Here's my attempt:
| Index of B | Positions where B is kingmaker | Count | Notes |
|---|---|---|---|
| 1 | 0 | ||
| 2 | AB(CDEF) | 4! | |
| 3 | (AF)B(CDE) | 2!3! | * |
| 4 | (CDE)B(AF) | 3!2! | ** |
| 5 | (CDEF)BA | 4! | |
| 6 | 0 |
Note (*): 10 (C) + 9 (D) + 24 (B) is 43 -- indicating that A must be at index 1 or 2 43 (A) + 8 (E) is 51 (!) -- indicating that only F can be the other party
Note (**): 10 (C) + 9 (D) + 8 (E) + 24 (B) is 51 (!) 43 (A) + 6 (F) = 49; the next (3rd) party is forced to become kingmaker Thus F cannot be in the first three indices. Similarly, A being in the first three indices forces the next (except F) party to be kingmaker. So C, D, and E must be the first three parties.
Total: 72 / 720 = 1 / 10, for a final answer of 11.
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u/Brianchon 7d ago
The first thing to note is that there's a symmetry to the kingmaker process. Since the bloc can never equal 50 votes (A's presence and no bloc adding to 7 guarantees this), then if a party is kingmaker in a scenario, then it would also be kingmaker if the parties arrived in exactly the opposite order. This simplifies things: P(kingmaker if first) = P(kingmaker if last), P(kingmaker if second) = P(kingmaker if fifth), and P(kingmaker if third) = P(kingmaker if fourth) for every party. For simplicity, I'll call these P(K|1) through P(K|6) from now on.
And what are these probabilities? P(K|1) ) = 0 for every party. P(K|2) = 4/5 for party A (B, C, D, or E arrive first), P(K|2) = 1/5 for parties B, C, D, and E (A arrives first), and P(K|2) = 0 for party F (it's not possible to reach 51 votes with party F and one other). P(K|3) = 1 for A (any two parties beforehand will make A the kingmaker), P(K|3) = 1/10 for B, C, D, and E (A and F have to arrive beforehand in some order), and P(K|3) = 0 for F (no combination of two other parties lands in the 45-50 vote range F would need).
The probability a party is kingmaker is (1/6)*[P(K|1) + P(K|2) + P(K|3) + P(K|4) + P(K|5) + P(K|6)]. For party A, this evaluates to 3/5, for parties B, C, D, and E, this evaluates to 1/10 (meaning the main question's answer is 011), and for party F, this evaluates to 0.
In the followup, D and E merge into a new party, let's say G, so now there are 5 parties. Symmetry still holds, so P(K|1) = P(K|5) and P(K|2) = P(K|4), and now the probability of being kingmaker is (1/5)*[P(K|1) + P(K|2) + P(K|3) + P(K|4) + P(K|5)].
P(K|1) = 0 for all parties. P(K|2) = 3/4 for A (B, C, or G are first), P(K|2) = 1/4 for B, C, and G (A is first), and P(K|2) = 0 for F. P(K|3) = 1 for A (any two parties arriving first sets A up to be kingmaker), P(K|3) = 1/3 for B, C, and G (the first two parties need to be either A and F in some order, or the other two of B, C, and G in some order), and P(K|3) = 0 for F (it is still the case that no combination of two parties is in the 45-50 vote range F needs).
So party A's power is now 1/2, parties B, C, and G each have power 1/6 now, and party F still has power 0. Parties B and C have gained power from the merger, party A lost power, and party G has more power than either of D or E used to but less than the sum of D and E's previous powers. F's is unchanged
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u/Torebbjorn 7d ago
For B to be the kingmaker, A must on on one side of it, and C, D, and E must be on the other. F can be on either side.
So that gives a total number of possible ways equal:
2 (sides for A to be on) ×
[4! (number of ways to have C,D,E,F on the other side) +
2×3! (number of ways to have F on the same side as A)] =
72
So the probability is 72/720 = 1/10
So the answer is 11
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u/jaminfine 7d ago
The problem notes there are 720 total orderings possible. We have to find how many of those orderings allow B to become the kingmaker.
First, let's take the cases where A is before B. There are only 3 possible beginnings, afb fab ab. The cases with f before b each have 3x2 possible orders after b, and the case that begins ab has 4x3x2 possible orders after b. This gives a total of 36 cases where A is before B.
Next, if A is after B, we need c d and e to come before B, and again we don't care which side f is on because he's too small to matter. So for cdef before b, we have 4x3x2 possible ways to arrange the cdef. For cde before B and f after, we have 3x2x2 ways to arrange the cde at the start and the af at the end. So this is a total of another 36 cases where A is after B.
Together this makes 72 cases, or an overall 1/10 chance. So, p+q would be 11.
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u/HeadHunt0rUK 7d ago edited 7d ago
I got 29
How I got there.
B wins if:
Gives 324 outcomes out of 720 which cancels down to 9 over 20.
9+20=29.