determining a fair betting scheme

Question

Preface: I think this is interesting (and hopefully at least one other mathematician will agree!), but it's entirely possible that y'all will consider this too low-brow for MO. There isn't a completely definite answer, but I think there can probably be a near-consensus. If you're able, feel free to re-title if you think you have something more appropriate. Also, I have no idea what tag(s) to apply. Community-wiki?

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I'm trying to set up a winner-takes-all bet with four other friends about the order in which we end up getting married. So that nobody has to simultaneously shell out $100 and face the prospect of living alone for the rest of their life, it seems prudent to end the bet and determine the winner after 3 of us have gotten married. Here was my original scoring scheme:

123 > 132 > 213 > 231 > 312 > 321 > 12* > 13* > 1*2 > 1*3 > 21* > 23* > 2*1 > 2*3 > 31* > 32* > 3*1 > 3*2 > *12 > *13 > *21 > *23 > *31 > *32 > * 1 * > * 2 * > * 3 * > **1 > **2 > **3

[For example, the first ordering 123 > 132 means the following. Suppose I list 1: Aaron, 2: Ben, 3: Carlos. Then, I am better off if they get married in the order Aaron, Ben, Carlos than if they come out in order Aaron, Carlos, Ben.]

[Note that if you want to put asterisks around a number you need to include spaces, otherwise they'll disappear and the number will be italicized. Although I'm sure there's a way to avoid this.]

I made this based off of the naive initial assumption that getting the top three picks should beat anything else. But one friend pointed out that probably it should be that 12*>321 (for example), and so the question becomes:

• What are reasonable parameters for the ordering?
• How should they be prioritized?
• What ordering (if any) does this yield?

With regards to the third question, I wouldn't be at all surprised if some variant of Arrow Impossibility comes into play, but we'll cross that bridge when we get there...

Answer 1

I think it is easier if each person chooses an order for all 5 people (12345) but you still call it after only three people get married. To determine the winner, you play out the best possibility for 4th and 5th, and then you are rewarded 1 point for every relationship you get correctly.

In other words, you choose one person to be 1, and when you choose that person, you are saying that you believe he will beat 2-5. Every time you are right (1 before 2, 1 before 3, 1 before 4, 1 before 5) you get a point. Same with all other numbers.

So I label 12345. The actual occurrence is: 132. So,

13245 1 beats 3, 2, 4, 5 = four points 2 beats 4, 5 = two points 3 beats 4, 5 = two points 4 beats 5 = one point

Total = nine points.

That would beat 125, for example, which would get eight points, but it would tie with 124, which would get nine points. I think this relationship idea gets at the heart of the bet (who gets married before who).

The biggest issue with this is that you can easily get ties. There are 60 possible 3 number outcomes, 11 possible point scores. You could choose something like a lexical tie breaker, so 125>132. But you probably want to choose something more fun like, in the event of a tie, the winners take shots until only one man is left standing. Literally.

Answer 2

I get the following ranking. If I try to break it into broader divisions I get things in the order I don't like so I got the following:

1. Three guesses that are correct and in the right position.
2. Two guesses that are correct and in the right position.
3. Three correct guesses with one in the right position.
4. One correct guess in the right position and another correct guess not in the right position.
5. One correct guess in the right position
6. Three correct guesses with two in the right order none in the right position.
7. Two correct answers none in the right position in the right order
8. Two correct answers none in the right position in the wrong order.
9. One correct answer in the wrong position.
10. For tie breaking purposes if tied by the above criteria the set containing the minimal element not in both sets.
11. For tie breaking purposes if tied by the above criteria the set containing the minimal element not in the same position in both sets in the leftmost position. which I think gives the following order.

123 > 12x > 1x3 > x23 > 132 > 2x3 > 321 > 1x2 > 13x > x21> 32x > 2x3 > 1xx x2x> xx3 x13 >312 > 231 > x13 > x12 > 23x > 21x > 31x > 3x1 > x31 > x31 > x32 > x1x > xx1 > 2xx > > xx2 > 3xx > x3x.

Answer 3

Each player bets a permutation of ${1,...,5}$.

After three of you get married in order $x_1, x_2, x_3$, and while $y_a,y_b$ are not married yet, construct the set of two permutations $$W=\{(x_1, x_2, x_3, y_a, y_b), (x_1,x_2,x_3, y_b, y_a)\}.$$ The player who plays $\sigma$ gets a score $d(\sigma, W)=\min\{d(\sigma,\tau), \tau\in W\}$ where $d$ is distance for permutations, for instance $d(\sigma,\tau)$ the minimal number of transpositions needed to transform $\sigma$ to $\tau$.