Before the second intermission we need stats help
April 28, 2018 6:16 PM Subscribe
We know that Buffalo, Montreal and Carolina have the top 3 NHL draft picks, but not the ordering. We also know that they had, in order, 18.5%, 9.5% and 3.0% chance of winning the first draft pick. Given that we know one of them did, what is each team's chance of getting the first draft pick?
I'm arguing with a friend based on the stats here, which show each team's chance of getting the first pick and of one of the top three picks. We are arguing between Buffalo having 18.5/(18.5+9.5+3.0) and 18.5/49.4 (where 49.4 is the chance they had to win one of the top three).
I'm arguing with a friend based on the stats here, which show each team's chance of getting the first pick and of one of the top three picks. We are arguing between Buffalo having 18.5/(18.5+9.5+3.0) and 18.5/49.4 (where 49.4 is the chance they had to win one of the top three).
I've seen so much questionable math on Sabres fan sites that I don't even know where to start. I am reasonably certain that Buffalo has the best likelihood of winning. Lol. As an honatary Buffalonian, though, I'm also reasonably certain that Buffalo will not win.
posted by kevinbelt at 6:52 PM on April 28, 2018 [1 favorite]
posted by kevinbelt at 6:52 PM on April 28, 2018 [1 favorite]
My house mathematician has worked this out - he says:
Buffalo: 36.34% chance of getting the first pick
Montreal: 34% chance
Carolina: 29.66% chance
posted by LobsterMitten at 6:56 PM on April 28, 2018 [1 favorite]
Buffalo: 36.34% chance of getting the first pick
Montreal: 34% chance
Carolina: 29.66% chance
posted by LobsterMitten at 6:56 PM on April 28, 2018 [1 favorite]
It's 18.5/(18.5+9.5+3.0), or 59.7%, and so on. You can't use the 49.4 number because it's conditional on the chance they had to win top 3, but not the chance for the other teams to have the other two.
posted by hermanubis at 6:58 PM on April 28, 2018
posted by hermanubis at 6:58 PM on April 28, 2018
More of a writeup from him:
This is a nice question.
The second suggestion isn't correct -- because you can't just use one team's odds in isolation from the other teams. This is a reciprocals problem, you're taking a sum of reciprocals, and that makes calculations more difficult.
The first suggestion is a good thought, although also incorrect.
In order to figure out the answer, you have to know how they actually determine the top three. A natural way to determine it would be to give each team some number (varying for the diff teams) of ping-pong balls, mix them all together in a bin, and draw a ball from the bin at random. That's the #1 team. Then they remove all of that team's balls from the pool, and repeat the process to get the #2 team, and so on. If they're doing it that way, we can figure out the answer, but I think it will be more complicated than the first suggestion.
[...]
Ok, I initially thought first suggestion would be close to correct, but that turns out to be wrong, because the odds of each of the top 3 teams being picked to be in the top 3 are so low to begin with that once they're there, their odds are more equal than you would think.
Do the pingpong ball process -- there are 100 pingpong balls in a bin, they're assigned to each team in proportion to that team's odds, so Buffalo has 18.5 pingpong balls and so on. All the remaining ping pong balls out of the 100 are representing the rest of the league, i.e. not one of these three teams.
There are 6 possible permutations for how the top 3 will turn out (BMC,BCM,CBM,CMB,MCB,MBC). So you work out the odds for each scenario. The first pick influences the odds of the second and third picks, so those odds will be different in each scenario. For example, Carolina has a 3/100 chance of being picked first -- if they are picked first, then Buffalo's odds of being picked second go to 18.5/97. But if the reverse scenario happened, Buffalo was picked first, then Carolina's odds of being picked second would go to 3/81.5. The numerator for each team will stay constant but the denominators will change.
For example, to get the odds of the BMC scenario, you multiply: (probability that B is picked first) x (probability that M is picked second given that B is first) by (probability that C is picked third given the other two in their positions).
In other words, BMC = P(B1)*P(M2 given B1)*P(C3 given M2 and B1)
Do that for all six scenarios, sum the six to get the total denominator (D).
Then BMC/D is the probability that the BMC scenario has occurred, given that we know Buffalo, Montreal, and Carolina are the top three.
Since there are so many other pingpong balls, the relative change in denominators is not that much, so the odds of each of the six scenarios are more similar than you would think. If you were only putting in pingpong balls for the three teams, the denominators would change a lot more relative to each other, so it's important that we remember to put in all the balls representing the rest of the league.
posted by LobsterMitten at 7:31 PM on April 28, 2018 [3 favorites]
This is a nice question.
The second suggestion isn't correct -- because you can't just use one team's odds in isolation from the other teams. This is a reciprocals problem, you're taking a sum of reciprocals, and that makes calculations more difficult.
The first suggestion is a good thought, although also incorrect.
In order to figure out the answer, you have to know how they actually determine the top three. A natural way to determine it would be to give each team some number (varying for the diff teams) of ping-pong balls, mix them all together in a bin, and draw a ball from the bin at random. That's the #1 team. Then they remove all of that team's balls from the pool, and repeat the process to get the #2 team, and so on. If they're doing it that way, we can figure out the answer, but I think it will be more complicated than the first suggestion.
[...]
Ok, I initially thought first suggestion would be close to correct, but that turns out to be wrong, because the odds of each of the top 3 teams being picked to be in the top 3 are so low to begin with that once they're there, their odds are more equal than you would think.
Do the pingpong ball process -- there are 100 pingpong balls in a bin, they're assigned to each team in proportion to that team's odds, so Buffalo has 18.5 pingpong balls and so on. All the remaining ping pong balls out of the 100 are representing the rest of the league, i.e. not one of these three teams.
There are 6 possible permutations for how the top 3 will turn out (BMC,BCM,CBM,CMB,MCB,MBC). So you work out the odds for each scenario. The first pick influences the odds of the second and third picks, so those odds will be different in each scenario. For example, Carolina has a 3/100 chance of being picked first -- if they are picked first, then Buffalo's odds of being picked second go to 18.5/97. But if the reverse scenario happened, Buffalo was picked first, then Carolina's odds of being picked second would go to 3/81.5. The numerator for each team will stay constant but the denominators will change.
For example, to get the odds of the BMC scenario, you multiply: (probability that B is picked first) x (probability that M is picked second given that B is first) by (probability that C is picked third given the other two in their positions).
In other words, BMC = P(B1)*P(M2 given B1)*P(C3 given M2 and B1)
Do that for all six scenarios, sum the six to get the total denominator (D).
Then BMC/D is the probability that the BMC scenario has occurred, given that we know Buffalo, Montreal, and Carolina are the top three.
Since there are so many other pingpong balls, the relative change in denominators is not that much, so the odds of each of the six scenarios are more similar than you would think. If you were only putting in pingpong balls for the three teams, the denominators would change a lot more relative to each other, so it's important that we remember to put in all the balls representing the rest of the league.
posted by LobsterMitten at 7:31 PM on April 28, 2018 [3 favorites]
LobsterMItten that is mind blowing, thanks for posting the write up! That's really counterintuitive, Buffalo has way higher odds, how can they not be more likely to get the first pick?
I would have bet a lot of money that option one was exactly correct even after your first post. Reading the write up I then got what I'd missed 'in theory' but I still didn't get that the effect could be so big, I had to run a simulation to convince myself this was really going on.
It might have been more obvious as a problem if there were only two top slots, Buffalo and Carolina, out of all the teams. It is unlikely that Carolina is in the top two so they got really lucky on one pick. Once you know they got lucky, whether they got lucky on the first pick or the second pick is almost fifty-fifty.
posted by mark k at 10:29 AM on April 29, 2018 [1 favorite]
I would have bet a lot of money that option one was exactly correct even after your first post. Reading the write up I then got what I'd missed 'in theory' but I still didn't get that the effect could be so big, I had to run a simulation to convince myself this was really going on.
It might have been more obvious as a problem if there were only two top slots, Buffalo and Carolina, out of all the teams. It is unlikely that Carolina is in the top two so they got really lucky on one pick. Once you know they got lucky, whether they got lucky on the first pick or the second pick is almost fifty-fifty.
posted by mark k at 10:29 AM on April 29, 2018 [1 favorite]
This thread is closed to new comments.
posted by beaning at 6:49 PM on April 28, 2018