# When God doesn't like a movie...

I don't remember anyone on Last Temptation of Christ being hit by lightning and Gibson's film was hit TWICE!

Too damn funny...

I don't remember anyone on Last Temptation of Christ being hit by lightning and Gibson's film was hit TWICE!

Too damn funny...

## Comments

2,407memberi read that somewhere.....a vonnegut book maybe.

8,760memberKind of like how people figure that if a coin comes up heads 10 times in a row it's 10 times more likely to come up tails the next time. Nope. Still 50/50.

Heck, one could argue that if you get hit by lightning it means you're more likely to be in an area where you are more likely to get hit, since you were *once*...

2,664memberOriginally posted by KickahaNope. Truly random events have no bearing on later probabilities.

Kind of like how people figure that if a coin comes up heads 10 times in a row it's 10 times more likely to come up tails the next time. Nope. Still 50/50.

Heck, one could argue that if you get hit by lightning it means you're more likely to be in an area where you are more likely to get hit, since you were *once*...

that reminds me of a very interesting probability question i heard in a comp sci class once, called the "Monty Hall problem." Named after the host of an old game show. The problem was like this:

Monty presents you with 3 doors. One has a prize behind it, and two have 'zonks' (non-prizes; often farm animals for comic effect). You, at random, select one door. He then messes with your head by removing one of the other doors from the list, guaranteeing that he did NOT remove the prize door, and asks if you want to change your mind or stick with your initial door.

Prudence, and basic math skills, would suggest that initially, you have a 1/3 chance (33.3333%) of getting that prize. And, depending on how you think about it, when he removes a bad door, your chances are then 50-50 (some people believe at this point your odds have not changed from 1-in-3; others would think that the new odds would be som function of 33% and 50%, often the average). None of these probabilities is very strong (I wouldn't put down any money on a fitty-fitty or 33% chance).

But, and this is the kicker, if you do this problem a huge amount of times (like automatically in a computer simulation), you can win 66% of the time, if you change your door after Monty removes a zonk. It's pretty interesting.

.

back to the topic at hand: lightning isn't truly random (nothing really is). from my understanding, it is caused by a sudden surge of electric charge at the position of the actual bolt (tho i dont know the cause of such a buildup). And that surge passes from the ground to the clouds/atmosphere. So, after having discharged, that immediate area would, in theory, be relatively safe. I of course, wouldn't hang out there, but in theory it'd be safe. Of course, "in theory, communism works.

In theory"1,120memberOriginally posted by thuh FreakBut, and this is the kicker, if you do this problem a huge amount of times (like automatically in a computer simulation), you can win 66% of the time, if you change your door after Monty removes a zonk. It's pretty interesting.I'm trying to work this out in my head and its not happening. Why did you have to make me think on a Friday?

440memberOriginally posted by kneelbeforezodI'm trying to work this out in my head and its not happening. Why did you have to make me think on a Friday?ok. If you choose one of the doors, your chance of getting the right door are 33%. This means that the chance that one of the other two doors being correct is 66%. I guess the easiest way to look at it, is that removing that invalid door would be equivalent to giving you both of the other doors... which would give you the 66% chance (if you choose it).

1,120memberOr is it that it?s like getting two choices at a 33.3% probability?

Or maybe I just didn't pay enough attention in statistics?

1,120member6,523memberHere is another way of looking at it: Only if you hold on to the chosen door the chances are as they were originally (33%). When you are given the chance to change you choice there is a 50/50 chance between the doors. So if you chose the opposite of not doing anything (which gave you 33%) you have a 2/3 chance of chosing the right door.

Here is another one: I have a lotto coupon with one row of random chosen numbers. It has a 1/100.000 mathematical chance to win. Numbers are drawn and the nice lady in the telly tells that 4 out of 200.000 rows have won.

Now is the chance of your row being the right one 1/100.000 or 1/50.000? I have arguments against both positions

1) 1/50.000: Your row was chosen in random but not all are. More rows have numbers like 3, 5, 7, 10, 13 than 14, 6, and numbers above 12 and 31. So if the first group of numbers are drawn more rows win. But your was chosen in random so it would still be the mathematically chance of 1/100.000

2) 1/100.000: What if the nice lady say noone won? Given they doesn´t have a first generation pentium to register the rows we can assume that the row cannot still have its mathematically chance of winning and that must(?) also mean something for the situation where it can possible be a winning row.

So what is right?

7,431member3,325membersame odds as winning the lottery.

anybody who's been hit twice ought to be buying extra tickets

the golfer, Lee Trevino, has been hit twice while playing golf.

his reply to other golfers about what to do during a storm on the links?

"hold up a one iron... even god can't hit a one iron properly"

6,523memberOriginally posted by ScottAll of these "trick" stats problems come out of the fact that the available information changes half way though. It's no longer the same problem.That doesn´t have anything to do with, at least not with my example. I´m asking you from a position where you have two sets of information at you hand. The question is which set of information you can judge your chances from.

4,695memberSuppose that of doors A, B, and C, A is the prize, B and C are zonks. There are six possible moves you can make: Three initial choices, multiplied by the two choices of stick or switch:

A) Pick A

B or C is taken away -- arbitrary, doesn't affect outcome.

A-stick) Stick with A: win

A-switch) Switch to C: lose

Pick B

C is taken away

B-stick) Stick with B: lose

B-switch) Switch to A: win

C) Pick C

B is taken away

C-stick) Stick with C: lose

C-switch) Switch to A: win

Sticking wins in 1 out of 3 "stick" outcomes. Switching wins in 2 out of 3 "switch" outcomes. You could further enumerate the solution space for B being the prize, then C being the prize, but the pattern is the same with the same stick/switch win ratios.

I don't think, however, that we can use a mere 2/3 odds of winning as the basis for a new Switch campaign.

1,120memberPick one of three doors. There is a 33.3% probability that you made the correct choice and a 66.6% probability that you didn't. After one door is eliminated you can stick with your original choice, and the 33.3% probability that this is the correct choice remains...so switching to the remaining door will give you a 66.6% probability of winning.

Suddenly it all makes sense

2,664member6,523memberOriginally posted by kneelbeforezodThe way it finally clicked with me was this:

Pick one of three doors. There is a 33.3% probability that you made the correct choice and a 66.6% probability that you didn't. After one door is eliminated you can stick with your original choice, and the 33.3% probability that this is the correct choice remains...so switching to the remaining door will give you a 66.6% probability of winning.

Suddenly it all makes sense

Wasn´t that what Anders said above?

17,995memberOriginally posted by KickahaNope. Truly random events have no bearing on later probabilities.

Kind of like how people figure that if a coin comes up heads 10 times in a row it's 10 times more likely to come up tails the next time. Nope. Still 50/50.

Heck, one could argue that if you get hit by lightning it means you're more likely to be in an area where you are more likely to get hit, since you were *once*...

Mathematically, that's true. But I'd argue that in actuality the coin is more likely to come up tails with each toss. Put money on the next toss and tell me which way you'd bet. I'd also argue that if one gets struck at random by lightening, the possibility of that rare event happening again is unbelievably low.

1,120memberOriginally posted by AndersWasn´t that what Anders said above?For some reason it didn't click until I thought of it in terms of 100% - 33.3% = 66.6%...must have been the fractions

5,486memberif you bet it will come up tails 11 times in a row, the odds are 0.09765%

however, if it's already been tails 10 times in a row, and you're only betting on the 11th, the odds are 50%

1,558member9,812memberOriginally posted by alcimedesthe thing with the coin is this.

if you bet it will come up tails 11 times in a row, the odds are 0.09765%

however, if it's already been tails 10 times in a row, and you're only betting on the 11th, the odds are 50%

And what's confusing is that if you flip a coin 11 times, you ARE more likely to end up with 1 head and 10 tails than 11 tails. I think people intuitively know that, and they're just over-applying that rule.

Here's another weird one, just when you think you have the gambler's fallacy and the binomial distribution figured out (goes to copy and paste) :

You meet a woman, ask how many children she has, and she replies "two." You ask if she has any boys, and she replies "yes." Now you know that at least one of her children is a boy. What are the odds that her other child is a boy?

To put it another way:

You flip two coins until at least one of them is a head. Now, what are the odds that the other is also a head?

The Monty Hall one is kinda funny, because when he actually played that game on Let's Make a Deal, he would only offer the switch to you if you had it right to begin with.