Rookie Running Backs and a Famous Statistics Problem

[Lotto4Life]

I'm not familiar with Monty Hall. One question I have - does the host always pick a door that has a goat behind it?

[Lotto4Life]

I thought we had moved to baseball in this topic?

[Life of Pablo]

I'm not familiar with Monty Hall. One question I have - does the host always pick a door that has a goat behind it?

Yep. Let's say you have doors A B and C. You pick door A. If A has a goat behind it (not that you know this), the host will show you the other door that has a goat behind it (either B or C). If A has the prize behind it, then the host can show you B or C, it doesn't matter which, since they both would have goats behind them.

I thought we had moved to baseball in this topic?

Idk lol

[ArrylT]

Idk if the last few comments are talking about rookie RBs or the Monty Hall problem anymore lol. But fyi as jeffster said, the Monty Hall problem is always a legit 67% chance of winning if you understand how it works. You always want to switch doors. I'd explain it better but I'll just save time by directing to the wiki instead: https://en.wikipedia.org/wiki/Monty_Hall_problem

Having said that. I get the idea you're going for jeffster, but I don't think that problem really applies well to draft success/bust rates lol. Those are all just independent cases loosely backed by sample size. In your example of Chubb, RoJo, and another RB (let's say Sony). You pick Chubb, and are shown RoJo who busts. Monty Hall would tell you to always switch to Sony (the 67%). But even if the "success rate" of these RBs is 33%, this isn't an actual Monty Hall problem. Chubb could be a stud, Sony could be a stud, both of them could be studs, or both could bust. And whatever Chubb does, it has no effect on what Sony does (much like how RoJo busting doesn't affect either of them in the first place). The only way this would be a Monty Hall problem, is if we 100% knew for a fact that EXACTLY ONE of Chubb/RoJo/Sony would be a stud, and the other two would bust. But we don't know that at all.

TL;DR - what UATahoe said earlier.

Unless jeffster has a time machine, and knows the answer already. IE in this situation he is the host. We have selected Door A - Chubb. Door B is Michel and Door C is Jones. Assuming we have Door A, and jeffster has just opened Door C - revealing that Jones is a bust/goat, the odds are that the best choice is now Michel, even if we have no new info on Chubb - jeffster has hinted Chubb is also a goat, by virtue of his certainty (via time travel) that Jones is a goat/bust and opening said Door C (ie 66% odds Michel = car, 33% odds Chubb = car).

[jeffster]

My comments had nothing to do with Rb's. LOL. Strictly on the probability theorem that was in discussion. The probabilities i reference are in direct correlation to the number of unknown variables you are dealing with. In the Monty Hall problem, that would be 3. Changing the number of variables here completely changes the principles of the theorem.

Not sure what you mean there. Changing the number of variables doesn't change the principles of the theorem - if you follow the link in my OP you can simulate it with however many you like. Well, I suppose dropping it below 3 doors would change everything, because the game wouldn't work.

that is just wrong. changing your pick to the other door doesnt increase your odds and the 66% stuff is totally wrong too. once the one door is revealed then both of the remaining doors, including your original door, have a 50% chance of being the right one (as there are now only 2 options, not 3...simple maths)

No, that's exactly how it works, sorry!

Having said that. I get the idea you're going for jeffster, but I don't think that problem really applies well to draft success/bust rates lol. Those are all just independent cases loosely backed by sample size.

I actually agree, and posed this more as an exercise for anyone who was interested. I did warn about that in the first sentence, then pose it as a question at the end. The reason the problem doesn't really apply is that we don't gain new information about the other backs just because one of them busts, whereas in the Montey Hall problem, we *do* gain information about the other doors after the reveal, because we know the host considered them when picking the one to open, but then opted not to open them.

Anyway, I just love the Montey Hall problem. At the least it should make everyone feel more humble about trusting your instincts when it comes to probability!

[UATahoe]

My point was if you change the vairables from 3 to 10, you would need 8 of the remaining variables all removed to leave only 2 choices still. Meaning you would need all of the other rb's removed leaving only 2 remaining. Then the choice becomes keep what you chose or switch to the one remaining option. And actually, your probability of choosing right goes up as the number of variables goes up assuming you are only left with 2 to choose from.

[ArrylT]

My point was if you change the vairables from 3 to 10, you would need 8 of the remaining variables all removed to leave only 2 choices still. Meaning you would need all of the other rb's removed for injury leaving only 2 remaining. Then the choice becomes keep what you chose or switch to the one remaining option. And actually, your probability of choosing right goes up as the number of variables goes up assuming you are only left with 2 to choose from.

Now we know who to blame for Guice!

[UATahoe]

Well played. LOL. Injury was the wrong usage anyways. Need to be removed based on talent disqualification. LOL

[Reljac]

I didn't read all the posts here, but I will say that I am an actuary and I can assure you this is not a Monty Hall situation. In a Monty Hall situation there is 100% certainty of a winning selection being included in the choice of options remaining, once you remove one. It is not inconceivable, albeit it is very unlikely, that ALL running backs in this class bust. It is also not inconceivable that Guice still ends up as the top back from this class.

[hockeyBjj]

Your theory goes out the window as in the game show example, there's a fixed 1/3 chance. We know there is a 100% chance one of the doors is a winner

For the first round RBs, the odds may in fact be 1/3, but it is not fixed. There very well could be 0 winners out of any 3, or 2 out of the 3

[Ruggenater]

I didn't read all the posts here, but I will say that I am an actuary and I can assure you this is not a Monty Hall situation. In a Monty Hall situation there is 100% certainty of a winning selection being included in the choice of options remaining, once you remove one. It is not inconceivable, albeit it is very unlikely, that ALL running backs in this class bust. It is also not inconceivable that Guice still ends up as the top back from this class.

Ayyy, hello fellow actuary!

[Hahaclintondix]

that is just wrong. changing your pick to the other door doesnt increase your odds and the 66% stuff is totally wrong too. once the one door is revealed then both of the remaining doors, including your original door, have a 50% chance of being the right one (as there are now only 2 options, not 3...simple maths)

sorry got lost on the 3 door contest thing haha but i dont really get what the point of this post is. nobody is calling anyone a bust before any of these dudes play at least a season or two so there would never be an opportunity to say oh this guy is a bust but this one might not be

This is wrong actually.