Timeline for Can John Gottman predict which couples will divorce with extremely high accuracy?
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
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| S Apr 19 at 16:02 | history | bounty ended | CommunityBot | ||
| S Apr 19 at 16:02 | history | notice removed | CommunityBot | ||
| Apr 12 at 15:53 | comment | added | Evargalo | The second block quote is a very convoluted way of explaining what a p-value is without saying it's a p-value... | |
| Apr 11 at 20:06 | comment | added | Ray Butterworth | Without knowing the actual predictions, it's impossible to analyze the results. Even knowing how many divorces were predicted, whether successfully or not, would make a big difference. But as it stands, the 88.4% accurate "all stay together" prediction would score 100% accuracy for staying together and 0% accuracy for divorcing. One might expect their 90% accurate prediction to be almost as unreliable when broken down like that. | |
| Apr 11 at 17:22 | comment | added | Ray Butterworth | @EvanCarroll says "I wish I could bounty a comment.". A bigger problem is that even if it were an actual Answer, it might be disqualified and deleted for being purely theoretical or as original research. (See: FAQ: What are theoretical answers?). I think an acceptable answer would have to contain a link to some authoritative source saying the same thing. | |
| S Apr 11 at 14:15 | history | bounty started | Evan Carroll | ||
| S Apr 11 at 14:15 | history | notice added | Evan Carroll | Draw attention | |
| S Oct 3, 2019 at 23:01 | history | bounty ended | CommunityBot | ||
| S Oct 3, 2019 at 23:01 | history | notice removed | CommunityBot | ||
| Sep 29, 2019 at 11:41 | comment | added | naive | One cannot only look at accuracy and evaluate predictions. You should enquire about false positives and false negatives. Look up about precision and recall. Especially when the data is imbalanced ( less number of divorcees ), predicting all remain together gives very good accuracy. Look up Henry's comment. But, the recall will be zero in this case and will mean that the model is no good. | |
| S Sep 25, 2019 at 21:50 | history | bounty started | Dacromir | ||
| S Sep 25, 2019 at 21:50 | history | notice added | Dacromir | Authoritative reference needed | |
| Sep 13, 2019 at 22:37 | comment | added | Henry | Without meeting the 147 couples, I could make individual predictions about whether they divorce in four years which would would turn out to be 88.4% accurate when 17 of them had divorced: all I would have to do is say each couple would stay together. It does not make 90% look so impressive | |
| Sep 13, 2019 at 21:00 | history | tweeted | twitter.com/StackSkeptic/status/1172615949108830208 | ||
| Sep 13, 2019 at 16:28 | comment | added | Nat |
I tried playing with the numbers, but couldn't reproduce. If anyone else wants to try, it's (7/56)*(6/55)*(5/54)*(4/53)*(3/52)*(2/51)*(1/50) * bin(56,7) * (0.125 ^ 7) * ((1 - 0.125) ^ (56 - 7)) for the full calculation (and deleting the latter part of the query is equivalent to assuming that he was given the fact that it'd be 7 total).
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| Sep 13, 2019 at 15:25 | history | edited | Laurel |
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| Sep 13, 2019 at 15:17 | comment | added | Nat | "For the Gottman, Katz and Hooven study, where Gottman et. al. picked out all seven divorced couples out of 56, the probability is approximately .000000000384 or 3.84×10-9.": Do they explain that calculation? I mean, if we assume he knew it'd be 7 divorces in 56, then that's (7/56)*(6/55)*(5/54)*(4/53)*(3/52)*(2/51)*(1/50), which is ~4.31*10^-9. Or if we don't assume that he knew it'd actually be 7, then with a binomial distribution and 12.5% odds, it'd be ~15.9% chance of getting 7, for ~6.87*10^-10. So where'd they get "3.84×10-9" from? | |
| Sep 13, 2019 at 10:55 | comment | added | kutschkem | The FAQ seems odd as @Fifth_H0r5eman mentioned, but I also think it sounds plausible to be able to predict who will divorce in four years (not who will divorce EVER). After all, four years is not THAT long, and most issues are probably already present years before the actual divorce. | |
| Sep 13, 2019 at 10:33 | comment | added | Fifth_H0r5eman | ... 1 x 10-19 times. That is the number point one (0.1) with 18 zeroes in front of the number one ... - That is possibly the worst written description of a small standard form I've seen. Also, worth noting that if he just guessed all couples would stay together, in the US he'd have ~60% success rate. If you're actually looking into problems in a marriage, it doesn't seem that outrageous that 94% is possible. Picking balls out of a bag is random, social relation problems really aren't, we're very good at detecting them... | |
| Sep 13, 2019 at 8:05 | history | asked | Dacromir | CC BY-SA 4.0 |