Illinois Football Recruiting Thread

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#101      
The article said he had almost 372 snaps so definitely got some playing time but pretty limited stats from all those snaps.

That's a good chunk of snaps. Not much sticking out. Was a solid tackler. Not sure if he was playing hurt but he was really bad the final three games of the season.
 
#102      
looks like he provides depth, but not a starter type if Jacas goes pro
 
#106      
Great updates from coach. My only 2 cents is that our Rb's aren't over the top great. There is no Pat Bryant playing running back. We would be playing with gasoline if we did have a PB in the back field. It would help Luke to next level.
We really don't know how good our RB room is with a suspect OL under producing. Bringing in a 5 star wouldn't guarantee improvement.
 
#107      
Boy, wonder who’s who.

So, theoretically, if one is at 95%, do you mean that the second is at a 65-70% or that the odds of getting both at 65-70%? Because of it it’s the latter….

To solve for the odds of the second event (let's call it Event Gabe), we can use the formula for the probability of two independent events happening together. The probability of both events X and Gabe occurring (P(C ∩ Gabe)) is related to the probabilities of X and Gabe as follows:

\[
P(X \cap Gabe) = P( ) \times P(Gabe \mid X)
\]

where:

- \( P(X) \) is the probability of Event X (95% or 0.95),
- \( P(Gabe \mid X) \) is the probability of Event Gabe happening given that Event X has occurred.

You are given that the probability of both events happening together is between 65% and 70% (or 0.65 to 0.70). So, we have the equation:

\[
P(X \cap Gabe) = P(C) \times P(Gabe) = 0.65 \text{ to } 0.70
\]

Now, solve for \( P(Gabe) \):

\[
P(Gabe) = \frac{P(X \cap Gabe)}{P(X)}
\]

Substitute the values:

\[
P(Gabe) = \frac{0.65}{0.95} \approx 0.684 \quad \text{to} \quad P(Gabe) = \frac{0.70}{0.95} \approx 0.737
\]

So, the probability of Event Gabe occurring is approximately between **68.4% and 73.7%**.
You've got the right idea, but let's tweak it slightly. If Event X has a 95% probability (0.95), and the combined probability of both Event X and Event Gabe happening is between 65% and 70%, then we're looking to find the probability of Event Gabe occurring given that Event X has already occurred.

The formula for the probability of two independent events happening together is:

P(X \cap Gabe) = P(X) \times P(Gabe)

Given:
P(X) = 0.95
P(X ∩ Gabe is) between 0.65 and 0.70

We can find P(Gabe) by rearranging the formula:

P(Gabe) = P(X ∩ cap Gabe)
____________
P(X)


Substitute the values:

P(Gabe) = 0.65
____________ ≈ approx 0.684
0.95

P(Gabe) = 0.70
____________ ≈ approx 0.737
0.95


So, the probability of Event Gabe occurring given that Event X has occurred is approximately between **68.4% and 73.7%**.

Keep in mind, probabilities are inherently uncertain, but this should give you a solid range to work with!
 
#108      
You've got the right idea, but let's tweak it slightly. If Event X has a 95% probability (0.95), and the combined probability of both Event X and Event Gabe happening is between 65% and 70%, then we're looking to find the probability of Event Gabe occurring given that Event X has already occurred.

The formula for the probability of two independent events happening together is:

P(X \cap Gabe) = P(X) \times P(Gabe)

Given:
P(X) = 0.95
P(X ∩ Gabe is) between 0.65 and 0.70

We can find P(Gabe) by rearranging the formula:

P(Gabe) = P(X ∩ cap Gabe)
____________
P(X)


Substitute the values:

P(Gabe) = 0.65
____________ ≈ approx 0.684
0.95

P(Gabe) = 0.70
____________ ≈ approx 0.737
0.95


So, the probability of Event Gabe occurring given that Event X has occurred is approximately between **68.4% and 73.7%**.

Keep in mind, probabilities are inherently uncertain, but this should give you a solid range to work with!
200-71.gif
 
#109      
Sorry but we/I don't tweak much on this board I/we don't know what that means...dumb it down please...
 
#114      
We need to close the deal. Would be a huge in-state pickup.

He said Bielema was his favorite coach. That Trimble is always on him. Talked about how Rusk would be a 5th year senior when he was a freshman. Said he was at practice today and sat in on the tight end meeting with Coach Discher. I can’t imagine not getting him after listening to that interview. It’s the most positive I’ve ever heard a prospective recruit. He gave a funny answer too why he (and college coaches) likes Kittle more than Kelce as a tight end.
 
#117      
On first quote … 5 star running back … Reached out to us … Bret staying loyal to his guys …

Second quote … Can confirm … Lot of guys already ALL IN … Someone turn on the DMX … 😎😎😎

Third quote … Couple of those former guys are dying to come back here … They left and realized how good they had it here … Willing to come back for $0 NIL … We run a professional program here and focus on development … Other programs … Not so much …
Gotta wonder how many of those calls are coming from West Lafayette
 
#118      
Um, that would be very difficult to turn down a 5 star player because we are too deep at a position. With injuries that our RB's seem to get over the years, Feagin two years in a row, Laughery had injury issues for a couple of years, and this year might be the 1st year McCray was healthy. I just kinda cringe when hearing that we are turning down 5 star players.
I understand what you are saying but we don't know what sort of baggage that 5* is bringing with him. Given the togetherness of the team and what it took to build it, it's far more important to keep your culture intact and bring in players that fit it than to bring talented players who don't.
 
#119      
On first quote … 5 star running back … Reached out to us … Bret staying loyal to his guys …

Second quote … Can confirm … Lot of guys already ALL IN … Someone turn on the DMX … 😎😎😎

Third quote … Couple of those former guys are dying to come back here … They left and realized how good they had it here … Willing to come back for $0 NIL … We run a professional program here and focus on development … Other programs … Not so much …
wDq6qVv.gif
 
#123      
You've got the right idea, but let's tweak it slightly. If Event X has a 95% probability (0.95), and the combined probability of both Event X and Event Gabe happening is between 65% and 70%, then we're looking to find the probability of Event Gabe occurring given that Event X has already occurred.

The formula for the probability of two independent events happening together is:

P(X \cap Gabe) = P(X) \times P(Gabe)

Given:
P(X) = 0.95
P(X ∩ Gabe is) between 0.65 and 0.70

We can find P(Gabe) by rearranging the formula:

P(Gabe) = P(X ∩ cap Gabe)
____________
P(X)


Substitute the values:

P(Gabe) = 0.65
____________ ≈ approx 0.684
0.95

P(Gabe) = 0.70
____________ ≈ approx 0.737
0.95


So, the probability of Event Gabe occurring given that Event X has occurred is approximately between **68.4% and 73.7%**.

Keep in mind, probabilities are inherently uncertain, but this should give you a solid range to work with!
revenge of the nerds GIF
 
#125      
A general question for those on this thread: when's the most recent time you were this satisfied about the state and and optimistic about the trajectory of the football program?

For me: 1990, and it was unfounded. I was 24 and had ~ 1% of the information I have access to now through all of you here (and particularly our beloved Insiders) so that '90 optimism was a stab in the dark. Today, I feel pretty good. This seems stable. Waited ~ 35 subsequent years to get here.

Josh, BB, donors: 🙏
 
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