ok as far as I can see the who statistics thing starts here: "Let's say you play poker at a casino. It's a low probability bet, but there's a chance you might win. So you play and you lose $1000. Do you stay at the table? Because your odds go down the longer you stay."
The odds of a win in poker are different than for a coin flip because the cards that remain to be dealt can affect the outcome, as can the opponent's hand and the bluffing ability each of you may or may not have. But it is knowable. So the odds of winning for any given hand depend on complex but knowable factors, and let's call that x. You are correct in saying that the odds of winning six hands in a row are x*x*x*x*x*x. However your odds of winning the next hand are still x, whatever the value of x is for that hand, no matter how many hands you have played. The soldier's odds of survival likewise depend on factors particular to the situation *that very second.* If nobody is trying to kill him they are pretty damn good. If someone is it may boil down to who is the better shot. But they don't get lower because he has been in a war zone for a certain length of time. I would have to think about whether you can apply this model to a situation as complex as Iraq, where there are hundreds of thousands of factors and most likely not all of them are known. Granted that the Drake equation also deals with very complex situations, it is a model that knows it's a model and an oversimplification. Its main point is proving that in an infinite universe even the infinitely likely *can* occur. But probably won't ;) But even though your analogy doesn't prove it, being a bad analogy, I agree with your main point, which is that Iraq is a question of throwing good money after bad, and it isn't a situation that persistence is likely to help. On 11/12/06, Dana <[EMAIL PROTECTED]> wrote: > the thing that is outstanding here is that you are still > condescendingly explaining where he is wrong ;) Crack a statistics > text, dude. > > If the coin toss is a reference model, it's yours, and it sucks. Let > me scroll up and see what your exact argument is, and see if I can't > straighten this out. I think you may need somehting like the drake > equation. > > On 11/12/06, Gruss Gott <[EMAIL PROTECTED]> wrote: > > > RoMunn wrote: > > > No, no, no. if 1 in 167 people is killed each year, > > > > Here's where you're stuck: There's a difference between the odds of a > > favorable outcome in one iteration and that favorable outcome > > **without losing** **multiple times in a row**. So the odds of death > > do indeed double the longer you stay because you're extending the odds > > of a favorable outcome *without dying* across 2 years, not 1. > > > > If one could die in year 1 and then return from the dead to fight > > again in year 2, you'd be right. But that's not possible so you're > > wrong. I'll show you the math again: > > > > Every time a soldier goes into combat they have **TWO** probabilities > > that are important: > > > > (1.) The probability that they will die **on any given day**. This is > > where you and Dana are stuck. > > > > (2.) The probability of NOT dying **multiple days in a row**. If you > > can't make it multiple days in a row WITHOUT DYING, you can't live and > > return to combat. > > > > Therefore when you calculate the probability you MUST calculate the > > probability of the favorable outcome multiple days in a row. Here are > > those calculations: > > > > ..999983^365=.993814 This is the calculation that shows the probability > > of living 365 days in a row. It says that, based on current casualty > > rates, that probability is 99.3814% > > > > 0.999983^730 = 0.987667 This is calculation that shows the > > probability of living 2x365=730 days in a row. That probability is > > 98.7667% > > > > 98.7667% < 99.3814% which means that the longer you stay in combat the > > more likely you are to die. > > > > Now let's address your concern: > > -------------------------------------------------------- > > The probability of living is 98.7667 so 1 - 0.987667 = 0.012333 or > > 1.2% which is probability of dying. > > > > That means 12.33 / 1000 will die based on current rates or > > (12.33/12.33) / (1000/12.33) = 1/81.103 > > > > or 1 in 81 **of those in combat 2 years in a row** will die. > > > > Here's the key: [listen closely now] > > > > Of those that are in combat for 730 days in a row, 1 in 81 will die. > > Or, put another way, those that stay in combat 730 days in a row are > > more likely to die than those that are in combat only 365 days in a > > row. > > > > ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~| Introducing the Fusion Authority Quarterly Update. 80 pages of hard-hitting, up-to-date ColdFusion information by your peers, delivered to your door four times a year. http://www.fusionauthority.com/quarterly Archive: http://www.houseoffusion.com/groups/CF-Community/message.cfm/messageid:220460 Subscription: http://www.houseoffusion.com/groups/CF-Community/subscribe.cfm Unsubscribe: http://www.houseoffusion.com/cf_lists/unsubscribe.cfm?user=11502.10531.5
