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2/135 is failed flights/total flights. Over the life of the shuttle[0], there were 355 astronauts who flew, 14 died. So for every ~25 astronauts who flew on th

2/135 is failed flights/total flights. Over the life of the shuttle[0], there were 355 astronauts who flew, 14 died. So for every ~25 astronauts who flew on the shuttle, 1 died.. 0) https://www.space.com/12376-nasa-space-shuttle-program-facts-statistics.html https://www.space.com/12376-nasa-space-shuttle-program-facts...

ebbv 9y ago
Oh my lord you guys that is not how probabilities work. You can’t look at what actually happened and then say “Well that was the chance.” If I flip a coin 20 times and happen to end up with a 15:5 split that doesn’t mean the chance of heads is 3:1. Also the context of that 1 in 12 in the quote is important. They’re saying that was the chance on the first flight in retrospect. The shuttle and it’s supporting facilities and processes didn’t stay static, they improved over time.
- craftyguy 9y ago
  I didn't say it was a probability. Quite literally, if you look at how many people flew and how many of those died, 1 in 25 died. There was a 100% 'chance' of 1 in 25 dying on the shuttle because that's what actually happened. There's no probability there.
- drb91 9y ago
  Wait, so a ratio by itself is meaningless without showing how the ratio was derived? If only we could verify the 1/12!
- whatshisface 9y ago
  If statistics didn't converge to probabilities in the high-n limit then science would be impossible. In the 15:5 coin example, the conclusion is skewed by a massive prior that coins are fair. When we're talking about things like space shuttle launches, it's not reasonable to expect our understanding of the system to be so good that we can boldly forge through with estimates that disagree with results.
- joshuamorton 9y ago
  >Oh my lord you guys that is not how probabilities work. You can’t look at what actually happened and then say “Well that was the chance.” If I flip a coin 20 times and happen to end up with a 15:5 split that doesn’t mean the chance of heads is 3:1. Erm, over a large sample size, yes this is exactly how things work. Sure 15/5 may be too small, but 150/50 isn't. That's well after the point where its reasonable to believe that your coin is rigged. As the other user mentioned, if you don't have the prior that the coin is fair, a 15/5 outcome may (and does) indeed imply that heads are more likely than tails. In fact, 15/5 would imply a less than 1% chance of a fair coin, given uniform priors. (Beta[15, 5] -> 1st percentile is 50.175)
  - pps43 9y ago
    But in this case the samples are small. If you flip a coin once and got tails, what is the probability of heads?
    - joshuamorton 9y ago
      That's not quite the same question as what I'm answering. I'm saying "given you flip a coin once and it comes up heads, what is the probability that its a fair coin" (or previously, given you flipped a coin 20 times, and it came up 15/5, what is the likelyhood it is a fair coin). And with a single flip, you can't actually answer that question (BetaDist[1, 0] is undefined). Now, once you've answered that question, you can integrate to get a probability that the next coin is heads, but with a single flip you can't even answer the first question. You need to have gotten heads and tails each at least once.
      - isostatic 9y ago
        Flip a coin 50 times and get 50 heads, it's almost impossible it's a fair coin. Does your BetaDist function work in that case?
        joshuamorton 9y ago
        No, the beta distribution BetaDist[a, b] is only defined for a, b strictly > 0. When a or b is very large and the other is zero, you can treat it as 1 and get a decent approximation, but it'll be a an estimate that biases towards the center.
- grondilu 9y ago
  > If I flip a coin 20 times and happen to end up with a 15:5 split that doesn’t mean the chance of heads is 3:1. Isn't it the best you can say, though? 20 is a small number but when it's all you have, there's nothing you can do about it. Of course we know the "real" probability is a 1/2 because we know how coin tosses work, but if we didn't know it and coin tosses were black boxes, we'd have to go with this 3:1 chance.
  - hikarudo 9y ago
    The chance of heads being 3:1 is the maximum likelihood estimate, which indeed is the best we can do if we don't know anything else. But usually we do have an idea of reasonable values.
- quickben 9y ago
  I believe he is right though. You are considering future probability chance, and are right, but he is considering past average. In this case his factual outcome trumps your potential forecast.
TangoTrotFox 9y ago
Take an example of why this number is probably not as meaningful as you might think. We have only one astronaut. And this astronaut flies on a ship 1000 times. And then on the 1000th time it blows up. Well we had 1 astronaut that flew and he died. So it would be fair to say that 100% of astronauts that flew on our imaginary ship died on it, but it would lack much of any meaning. Accounting for repeats, there were 833 individuals that flew on the shuttle and 14 died. The ratio there is about 60:1 which is predictably comparable to the ratio of successful:failed flights.
delbel 9y ago
The Challenger Crew somehow died and also they are actually still alive. This is probably due to a timeline shift, for example there is no "Jiffy" peanut butter in this timeline, aka Mandela effect. Unfortunately we do not know what is causing this yet. https://1.bp.blogspot.com/-8NO8Dkx_yY8/V5loXQBvWkI/AAAAAAAIaRk/LZurjZJp5wcYYjwVEVixFJ39mMiU-83lQCLcB/s640/13654201_1288270541184295_8806315802829802667_n.jpg https://1.bp.blogspot.com/-8NO8Dkx_yY8/V5loXQBvWkI/AAAAAAAIa...
jacquesm 9y ago
That's one very twisted way of computing the chance of dying on the shuttle. If you launched the same crew over and over again they would all have died. If you used 'fresh' people for every launch then the chance according to your computation would have been lower. And yet, through all of that the underlying machinery would be identical and that is the thing we're trying to evaluate here. So even though you can compute that particular number it is not a very useful number to calculate. A better way would be to look at the 2/135 number to determine vehicle safety and to use the number of astronauts on board to get to a generic figure and for each individual astronaut to use the 2/135 multiplied by the number of flights they took to figure out their chances. Of course these are still very low numbers and the error bars will therefore be large.

ebbv 9y ago

Oh my lord you guys that is not how probabilities work. You can’t look at what actually happened and then say “Well that was the chance.” If I flip a coin 20 times and happen to end up with a 15:5 split that doesn’t mean the chance of heads is 3:1. Also the context of that 1 in 12 in the quote is important. They’re saying that was the chance on the first flight in retrospect. The shuttle and it’s supporting facilities and processes didn’t stay static, they improved over time.

craftyguy 9y ago
I didn't say it was a probability. Quite literally, if you look at how many people flew and how many of those died, 1 in 25 died. There was a 100% 'chance' of 1 in 25 dying on the shuttle because that's what actually happened. There's no probability there.
drb91 9y ago
Wait, so a ratio by itself is meaningless without showing how the ratio was derived? If only we could verify the 1/12!
whatshisface 9y ago
If statistics didn't converge to probabilities in the high-n limit then science would be impossible. In the 15:5 coin example, the conclusion is skewed by a massive prior that coins are fair. When we're talking about things like space shuttle launches, it's not reasonable to expect our understanding of the system to be so good that we can boldly forge through with estimates that disagree with results.
joshuamorton 9y ago
>Oh my lord you guys that is not how probabilities work. You can’t look at what actually happened and then say “Well that was the chance.” If I flip a coin 20 times and happen to end up with a 15:5 split that doesn’t mean the chance of heads is 3:1. Erm, over a large sample size, yes this is exactly how things work. Sure 15/5 may be too small, but 150/50 isn't. That's well after the point where its reasonable to believe that your coin is rigged. As the other user mentioned, if you don't have the prior that the coin is fair, a 15/5 outcome may (and does) indeed imply that heads are more likely than tails. In fact, 15/5 would imply a less than 1% chance of a fair coin, given uniform priors. (Beta[15, 5] -> 1st percentile is 50.175)
- pps43 9y ago
  But in this case the samples are small. If you flip a coin once and got tails, what is the probability of heads?
  - joshuamorton 9y ago
    That's not quite the same question as what I'm answering. I'm saying "given you flip a coin once and it comes up heads, what is the probability that its a fair coin" (or previously, given you flipped a coin 20 times, and it came up 15/5, what is the likelyhood it is a fair coin). And with a single flip, you can't actually answer that question (BetaDist[1, 0] is undefined). Now, once you've answered that question, you can integrate to get a probability that the next coin is heads, but with a single flip you can't even answer the first question. You need to have gotten heads and tails each at least once.
    - isostatic 9y ago
      Flip a coin 50 times and get 50 heads, it's almost impossible it's a fair coin. Does your BetaDist function work in that case?
      - joshuamorton 9y ago
        No, the beta distribution BetaDist[a, b] is only defined for a, b strictly > 0. When a or b is very large and the other is zero, you can treat it as 1 and get a decent approximation, but it'll be a an estimate that biases towards the center.
grondilu 9y ago
> If I flip a coin 20 times and happen to end up with a 15:5 split that doesn’t mean the chance of heads is 3:1. Isn't it the best you can say, though? 20 is a small number but when it's all you have, there's nothing you can do about it. Of course we know the "real" probability is a 1/2 because we know how coin tosses work, but if we didn't know it and coin tosses were black boxes, we'd have to go with this 3:1 chance.
- hikarudo 9y ago
  The chance of heads being 3:1 is the maximum likelihood estimate, which indeed is the best we can do if we don't know anything else. But usually we do have an idea of reasonable values.
quickben 9y ago
I believe he is right though. You are considering future probability chance, and are right, but he is considering past average. In this case his factual outcome trumps your potential forecast.

TangoTrotFox 9y ago

Take an example of why this number is probably not as meaningful as you might think. We have only one astronaut. And this astronaut flies on a ship 1000 times. And then on the 1000th time it blows up. Well we had 1 astronaut that flew and he died. So it would be fair to say that 100% of astronauts that flew on our imaginary ship died on it, but it would lack much of any meaning. Accounting for repeats, there were 833 individuals that flew on the shuttle and 14 died. The ratio there is about 60:1 which is predictably comparable to the ratio of successful:failed flights.

delbel 9y ago

The Challenger Crew somehow died and also they are actually still alive. This is probably due to a timeline shift, for example there is no "Jiffy" peanut butter in this timeline, aka Mandela effect. Unfortunately we do not know what is causing this yet. https://1.bp.blogspot.com/-8NO8Dkx_yY8/V5loXQBvWkI/AAAAAAAIaRk/LZurjZJp5wcYYjwVEVixFJ39mMiU-83lQCLcB/s640/13654201_1288270541184295_8806315802829802667_n.jpg https://1.bp.blogspot.com/-8NO8Dkx_yY8/V5loXQBvWkI/AAAAAAAIa...

jacquesm 9y ago

That's one very twisted way of computing the chance of dying on the shuttle. If you launched the same crew over and over again they would all have died. If you used 'fresh' people for every launch then the chance according to your computation would have been lower. And yet, through all of that the underlying machinery would be identical and that is the thing we're trying to evaluate here. So even though you can compute that particular number it is not a very useful number to calculate. A better way would be to look at the 2/135 number to determine vehicle safety and to use the number of astronauts on board to get to a generic figure and for each individual astronaut to use the 2/135 multiplied by the number of flights they took to figure out their chances. Of course these are still very low numbers and the error bars will therefore be large.