Jacksonville Jaguars Fan Forums

Quote:I read a topic on nfl.com and 32 out 40 WRs failed to score double digit touchdowns the following season after posting their first double digit touchdown season.


Fortunately both of our receivers scored 10 or more times so I'm hoping we fall in the 20%.


After reading it again, he didn't even include Allen Robinson.

 

So if there is a 20% chance of 1 player repeating with 10+ TDs and the Jags had 2 players do it in 2015, then there is actually a 40% chance that 1 will have 10+ TDs this year and 4% chance that BOTH will have 10+ TD's. 

 

Because TD's have so much variance, it would be more interesting to see how consistent those players were for the following 2 years +/- from their first 10+ TD season.

 

One thing that would be interesting would be how often for on teams where a player fails to reach 10+ TDs the 2nd time, how often another player increases their total to 10+ TDs. As in say, neither get 10+ TD's, but now J. Thomas does.

They good.

Quote:So if there is a 20% chance of 1 player repeating with 10+ TDs and the Jags had 2 players do it in 2015, then there is actually a 40% chance that 1 will have 10+ TDs this year and 4% chance that BOTH will have 10+ TD's. 

 

Because TD's have so much variance, it would be more interesting to see how consistent those players were for the following 2 years +/- from their first 10+ TD season.

 

One thing that would be interesting would be how often for on teams where a player fails to reach 10+ TDs the 2nd time, how often another player increases their total to 10+ TDs. As in say, neither get 10+ TD's, but now J. Thomas does.

 


I may be wrong, but I don't think that's how that works.

I think both of those guys would easily trade 10 TDs for 10 wins.  They have a direct incentive to want less wear and tear on their bodies.  Having Yeldon or Ivory 'stealing' their TDs will be good for them as it opens the passing game up by moving the safeties. It also prolongs their career and greatly improves their chances at a second big contract.

There's a 91% chance that the 20% is wrong.

Quote:So if there is a 20% chance of 1 player repeating with 10+ TDs and the Jags had 2 players do it in 2015, then there is actually a 40% chance that 1 will have 10+ TDs this year and 4% chance that BOTH will have 10+ TD's.

If you really believe this, then I have some excellent games of chance that I wish to engage you in.

Quote:If you really believe this, then I have some excellent games of chance that I wish to engage you in.

 

Quote:I may be wrong, but I don't think that's how that works.

 

OP said that Hurns and Robinson have a 20% chance of repeating. I was just explaining with simple statistics what the probability would be for one of them or both of them, because it would be different.

 

Now this is only basic statistics. It is the equivalent of what are the chances of getting head on a coin flip. Given football is not simple statistics because of some other variables, I could not calculate that. I wasn't give any advanced information as to what happens if both players are on the same team, etc. Not to mention 40 times is not a high number of repetitions, if you want to disregard those factors, you would need a sample size probably way in the hundreds. 

Quote:

Well done!

Quote:OP said that Hurns and Robinson have a 20% chance of repeating. I was just explaining with simple statistics what the probability would be for one of them or both of them, because it would be different.

Okay, thanks for the explanation. So, the chance for either of them to repeat is 20%, and thus the chance for one or the other to repeat is 40%. Got it.


So here's the game we're going to play. I'm going to toss a coin twice. The chance that the first toss will be heads is 50%, right? So the chance that one of the tosses will be heads is 100%. It's guaranteed. Given that, you'd be fine with offering me, say, million to one odds against me tossing tails twice in a row, yes?


Like you said, basic statistics.

Quote:Okay, thanks for the explanation. So, the chance for either of them to repeat is 20%, and thus the chance for one or the other to repeat is 40%. Got it.


So here's the game we're going to play. I'm going to toss a coin twice. The chance that the first toss will be heads is 50%, right? So the chance that one of the tosses will be heads is 100%. It's guaranteed. Given that, you'd be fine with offering me, say, million to one odds against me tossing tails twice in a row, yes?


Like you said, basic statistics.

 

I was doing the math wrong which happens when you do math post call. Since 4% of the time, they would both do it at the same time. So the chances either of them having a 10+ TD season is 36%.

 

For your situation:

The chances for you to win would be for both tosses to be tails. Chances for 1 toss to be tails is 50% (or .5). That means the chances for tails to occur twice is  0.5 x 0.5 or 0.25. So the chances for heads to turn up once would be 1 - 0.25 = 0.75 or 75%. For a third toss, it then becomes 1 - (0.5 x 0.5 x 0.5) = 1 - (0.125) = 0.875 or 87.5 %. 

Quote:I was doing the math wrong which happens when you do math post call. Since 4% of the time, they would both do it at the same time. So the chances either of them having a 10+ TD season is 36%.

I love questions like this. It's been so long since I took statistics, I've forgotten how to do it. I had to look it up.


Bottom line, you are correct, if the odds of one of them making it is 20%, then since there are two of them, the odds of at least one of them making it is 36%.


But I'm not sure I agree with the way you came up with the answer. The way I understand it, if the odds of either one of them not making it is 80%, then the odds of both of them not making it is .8 x .8 = .64. Which means the odds of at least one of them making it is 36%.

Quote:I was doing the math wrong which happens when you do math post call. Since 4% of the time, they would both do it at the same time. So the chances either of them having a 10+ TD season is 36%.

Better. The easier way to handle these cases is normally to invert the calculation. The chance of them both *failing* to do it is 0.8*0.8 = 0.64. Which gives, as you say, 36% chance of one or the other or both doing it. Your problem is that you originally counted that 4% twice.
 

Quote:For your situation:
The chances for you to win would be for both tosses to be tails. Chances for 1 toss to be tails is 50% (or .5). That means the chances for tails to occur twice is  0.5 x 0.5 or 0.25. So the chances for heads to turn up once would be 1 - 0.25 = 0.75 or 75%. For a third toss, it then becomes 1 - (0.5 x 0.5 x 0.5) = 1 - (0.125) = 0.875 or 87.5 %.

Exactly. You can't just add probabilities the way you originally did. The coin toss example makes that obvious (and the fact that it's 50-50 makes things even easier...)

Quote:Better. The easier way to handle these cases is normally to invert the calculation. The chance of them both *failing* to do it is 0.8*0.8 = 0.64. Which gives, as you say, 36% chance of one or the other or both doing it. Your problem is that you originally counted that 4% twice.

 


Exactly. You can't just add probabilities the way you originally did. The coin toss example makes that obvious (and the fact that it's 50-50 makes things even easier...)

 

Yeah, I made the calculation that for them happening together, but then forgot that for the either/or calculation, the events are not mutually exclusive. 

I think we need a separate math forum.


This thread is giving me a headache.

id rather have double digit wins anyway

Quote:I think we need a separate math forum.


This thread is giving me a headache.

Stick with things you know, like 91%, and work up from there.