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That's not how probability works [0]. Say we assume a more realistic 40% success rate figure [1], given Jay's 2 trials: Probability of 1 success: (2 choose 1)
by gkop 13y ago
That's not how probability works [0]. Say we assume a more realistic 40% success rate figure [1], given Jay's 2 trials:
Probability of 1 success:
(2 choose 1)(0.4^1)(0.6^(2-1)) = 0.48
Probability of 2 successes:
(2 choose 2)(0.4^2)(0.6^(2-2)) = 0.16
0.48 + 0.16 gives us a 64% of at least one of Jay's restaurants succeeding.
So the odds were in fact in his favor.
[0] https://en.wikipedia.org/wiki/Binomial_distribution#Cumulative_distribution_function https://en.wikipedia.org/wiki/Binomial_distribution#Cumulati...
[1] http://www.businessweek.com/stories/2007-04-16/the-restaurant-failure-mythbusinessweek-business-news-stock-market-and-financial-advice http://www.businessweek.com/stories/2007-04-16/the-restauran...
- kennywinker 13y agoMy apologies! I'm a long way out from stats class now, but I've been doing some light A/B testing so I've been thinking about things from the perspective of statistical significance rather than simple probability. So with that in mind, only having two trials, can we really determine a deviation from the norm? The odds were in his favour (slightly) but not so much that we can say with any certainty that he or his tipless restaurants were the issue. I'm rusty on the math or I'd actually calculate statistical significance (p value?) of 2 tries 1 success vs two tries 0 successes. I suspect it's below 95% though.
- gkop 13y agoI agree we should not draw any conclusions from my arithmetic above. I just couldn't resist the opportunity to correct someone's math on HN :)
- Dylan16807 13y agoYou're still trying to draw conclusions from a 2-sample almost-even-odds outcome, which we really can't.