July 24, 2006
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Speed Dating Odds
Given this very, very hypothetical situation:There are 17 men and 17 women at a speed dating event. At the end
of the night, each gets to write down their top two dates. What is the
probability of a match being made?To simplify the problem, assume each person only has one choice, and
that guy #1 chooses girl #1, etc… What are the chances that a girl
has picked the guy who chose her? The prob that she hasn’t picked him
is 16/17. So the odds that all 17 girls miss is (16/17)^17 = .357. So
the odds of at least one match is pretty good already: 1-.357 = 64.3%.
Of course, it improves when you allow choice #2. In that
case the odds of a single girl missing her pick drops to 15/17. So 17
girls all missing = (15/17)^17 = .119. Odds of at least one match:
1-.119 = 88.1%.
Corrected by altoz: the above paragraph holds true if you allow either gender 2 choices, but not both genders. Allowing both genders increases the chances to 98.6%, as per his comment below. Hey, 98.6 is the temperature of the human body. Coincidence? I think not…If you assume the guys and girls are at least somewhat socially savvy
and able to sense a mutual connection, then the odds go up
dramatically. Start booking your weekends at the church now…The big question is, if you eventually marry your choice #2, do you ever tell your spouse they were #2???
Comments (19)
Not to quibble, but I don’t know if your model’s very fair to the individual — a participant isn’t concerned about whether the group as a whole results as a match, they want to know if they will walk home with a match.
So, their specific odds are much smaller.
The chance of her missing one of the two guys that picked her is just straight 1 – (15/17), right? Granted, I think a 12% chance is better than most people would fare at any bars or parties now that I think about it…
You are right about the odds of an individual, whose odds are only 2/17 = 12%. Those are better odds than my experience in bars!
This problem was studied by Gale and Shapley in the American Mathematical Monthly (1962). A brief synopsis is available here. The takeaway is that if you ask each man to rank-list the 17 women, and you ask each woman to rank-list the 17 men, then there are a number of ways you can create “stable” matches. By stable, I mean that there is no man and woman that could simultaneously defect from their assigned matches, pair off together, and both be happier than they were in the algorithm’s output. A variant of this algorithm is used to match residency programs with first-year medical residents.
Wow, changed, does that mean you’ve hit on 17 wymin or more at bars at one time?!?! Dang, my hero!
throw in some game theory and this, too
you know, i just redid this and the probabilities aren’t correct.
the probability that the guy chosen by the girl at #1 has her down at #1 or #2 is 2/17. the probability that the guy chosen by the girl at #2 has her down at #1 or #2 is also 2/17.
Thus, the probability of a match (at #1 or #2) for a given girl is 1-(15/17)*(15/17) = 22%
The probability of 0 matches is
((15/17)^2)^17 = 1.4%
Thus, the probability of a match is 98.6%
The expected value (where if person A chose B and C and both B and C had A down would count as 2 matches) is something like:
(1-(15/17)^2)*17 = 3.76
So we should have expected 3.76 couples assuming a completely random distribution. BTW, I got to this conclusion because I was trying to run some simulations on this.
Thanks, I knew I could count on you guys to figure this out!
you guys are NERDS!!
i’m so impressed.
“husband!? what was all that one in a million talk?”
-Lloyd, Dumb and Dumber
No matter what simulations you run or probability statistics or theoretical framework…..
I WAS THE TOTAL MAC-DADDY THAT NIGHT!!!! MUHahahhahahahaha!!!!!
Your simulations will have to figure in only 16 guys and 15 wymin (-2 for the two wymin who matched with me). Oh and then, you’d have to take away 1 choice from all the wymin, since it’ll be a constant (meaning all 17 wymin put my name as #1)….
that sounds pretty competitive of you, jwkcheng. a little testosterone-heavy from working out recently?
! b
James, you forgot to figure in:
1) The wymin will put you down for both #1 and #2. After a taste of the JWK-Master, who could settle?
2) The men will put you down for #1 and #2 also.
So 2 of the wymin will luck out (although for both to really luck out, you’d all have to move to Utah). And that leaves 31 broken-hearted singles!
ybrfdy: It’s not a competition when you know what the outcome is going to be before it actually happens!
Zoiks!!! >;^P
changed: So true so true! My heart goes out to the 31 broken-hearted singles, and even more so to Wymin #2…. so close yet so far. =( sadness.
Maybe I should retire now from speed-dating as to keep the hope alive for all of you Kraft cheese slices out there!
hmmm…altoz beat me to it with regards to the probabilities, and he said it much more elegantly than i ever could. i look forward to jwkcheng’s post on this.
hey changed, i can’t believe you sat and thought through/calculated all of this. you nerd! =D please say hello to the the missus for me.
I’m clueless about the probability stuff, but the answer to that last “big question” is “no.”
(hanna)
oh geez…this post started off so fun for me, then i came upon the word problem-esque scenario/probability math stuff, and skimming others’ responses…i got all panicky, head-rush, sweaty palms all over again…
just like the last math class i ever took: algebra II/trig.
but, kudos to all who get it….
LoL. You should write a little book of all of this quantitative and graphed-out nerdy wisdom on dating. i’d buy it
hello – remember to take into account personalities…