Coin flip percentage?

dWarriors88 said:

Is there an easy way to find out what the start of game coin toss w/l ratio is for Alabama under CNS?

And then to take it further what about other coaches? Lets see who the luckiest coach in the top ten is.

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I believe we defer if we win the coin toss, so if we win we kick off. You can probably find that in the stats section on Rolltide.com - football - stats


Quiz:
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

TiderGreg said:

I believe we defer if we win the coin toss, so if we win we kick off. You can probably find that in the stats section on Rolltide.com - football - stats


Quiz:
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

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Although I disagree with the logic behind it, I know the accepted answer to this is yes, switch to Door #2. Increases the chance from 33.3% to 66.6%.

Bear Disciple said:

Although I disagree with the logic behind it, I know the accepted answer to this is yes, switch to Door #2. Increases the chance from 33.3% to 50%.

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The essence of the improved probability is in the host's prior knowledge of where the car is. If he opened the door randomly, your probability would not change.

TiderGreg said:

I believe we defer if we win the coin toss, so if we win we kick off. You can probably find that in the stats section on Rolltide.com - football - stats


Quiz:
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

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If the host knows, it is statistically to your advantage to switch.

50/50

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Bear Disciple said:

Although I disagree with the logic behind it, I know the accepted answer to this is yes, switch to Door #2. Increases the chance from 33.3% to 50%.

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Let's see if I can explain the logic another way.

Instead of 3 doors, let's say there are 100. One door has a prize. You randomly pick door 12. The host (who knows where the prize is) then opens all doors except 12 and 37, and the prize is still not revealed. Therefore, it has to be either 12 or 37. Would you switch now?

Is it the off season already?

Quiz:
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

cbi1972 said:

The essence of the improved probability is in the host's prior knowledge of where the car is. If he opened the door randomly, your probability would not change.

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This always puzzled me too, because your odds are better picking the other door, but if someone just came on the scene, his are lower, so you have inside information.
The players who swap have a 2/3 chance of winning the car and players who stick have a 1/3 chance of winning the car, is based on the premise that the host knows which door hides the car and intentionally reveals a goat. If the player selected the door hiding the car (1/3), then both remaining doors hide goats and the host may choose either door at random, and switching doors loses. On the other hand, if the player initially selected a door that hides a goat (a 2-in-3 chance), then the host's choice is no longer at random, as he has no options but to show the other goat, the second goat, and switching doors wins for sure.

Someone that disagrees

""You look away, and I put a pea under one of three shells. Then I ask you to put your finger on a shell. The odds that your choice contains a pea are 1/3, agreed? Then I simply lift up an empty shell from the remaining other two. As I can (and will) do this regardless of what you’ve chosen, we’ve learned nothing to allow us to revise the odds on the shell under your finger."



Back to the thread, don't want to hijack.

mikes12 said:

Let's see if I can explain the logic another way.

Instead of 3 doors, let's say there are 100. One door has a prize. You randomly pick door 12. The host (who knows where the prize is) then opens all doors except 12 and 37, and the prize is still not revealed. Therefore, it has to be either 12 or 37. Would you switch now?

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I have had it explained to me that way too, which I think is a better example but I still don't buy it completely. To me, in both scenarios, new information has been presented. I view it that in the beginning door 12 and 37 both have a 1% chance, I don't see why door 37 gets a 98% bump while 12 stays at 1%. If there's a 100% chance it's behind one of them, then I figure it's 50/50.

mikes12 said:

Let's see if I can explain the logic another way.

Instead of 3 doors, let's say there are 100. One door has a prize. You randomly pick door 12. The host (who knows where the prize is) then opens all doors except 12 and 37, and the prize is still not revealed. Therefore, it has to be either 12 or 37. Would you switch now?

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Or, you have a 1/3 chance of picking the right door the first time. So, there is a 2/3 chance it is another door. Since you know it is not the door the host revealed, there is a 2/3 chance it is the other available door.

I don't get the analysis. The improved probability occurs when the wrong door is revealed, regardless of whether the host knows the prize's location (and the fact that he revealed the wrong door doesn't mean that he knows the true location, only that it's behind one of two doors). The probability is 50/50 for either door remaining.

Does this even matter?


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The odds change when the goat door is revealed, and the odds improve. But they improve from 33% to 50%, regardless of whether or not you switch your pick and regardless of whether or not the host knew about it.

As for the original question, the luckiest coach has to be the mad hatter. How many times have we seen him make the most asinine coaching decisions and have it miraculously work out in his favor?

Edit: After researching the problem, I now realize that: I was wrong. If the host knows, it's better to switch doors.

The wiki entry explains it better than I could in the limited time that I have.

Contestants who switch have a 2/3 chance of winning the car, while contestants who stick have only a 1/3 chance. One way to see this is to notice that there is a 2/3 chance that the initial choice of the player is a door hiding a goat. When that is the case, the host is forced to open the other goat door, and the remaining closed door hides the car. "Switching" only fails to give the car when the player had initially picked the door hiding the car, which only happens one third of the time.

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tidefan23 said:

I don't get the analysis. The improved probability occurs when the wrong door is revealed, regardless of whether the host knows the prize's location (and the fact that he revealed the wrong door doesn't mean that he knows the true location, only that it's behind one of two doors). The probability is 50/50 for either door remaining.

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The probability that you selected the correct door first is 1/3. That doesn't change. Basically, the bet is "do you think you chose the correct door first?"

I think Mythbusters did this and it was pretty revealing. It was better to switch on the three door example.

In school while studying the bell curve, the coin flip scenario was discussed and the percentage was that tails comes up more often, 52%. They said nothing about goats.

In statistics class at C&BA they teach every first year to get over the Monte Hall problem. More importantly, if you make at least a C in strategy you know that if you want to receive and the opponent wants to kick, you're pretty much happy as long as the thing doesn't stick in the turf on its side!

BamaFlum said:

I think Mythbusters did this and it was pretty revealing. It was better to switch on the three door example.

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Someone actually wrote a program a few years ago that did this thousands of times and you were something like 17% more likely to win by switching than by standing pat. Think about it like this - what if Monty asked you if you would prefer to pick two doors, or only one? You would pick two doors every time, right? Well, this is an example of that, but you are only being given a chance to pick two doors after you are shown what you already knew - that one of those two doors was going to be wrong.

Stasticians want to believe that it is a 33%/67% ratio, but the computer model proved that the difference was smaller - 50%/67%. Why? Because you really have no worse than a 50% chance at that point if you stand pat, but your odds go up to 67% if you switch.