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MtnGoat
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PostTue Aug 13, 2002 10:29 am 
On another board in a discussion about the Rowell's plane crash someone made the comment that their continual flying made it more likely they would die in a plane crash. Another person held that the odds of dying in a plane crash remain the same regardless of the number of flights. I do not believe this to be the case, the chances of dying in any *single* plane flight are fixed, but the cumulative odds over multiple events must change, right? My post: "I don't think that's quite right. The odds of any *particular* game or flight are indepent of how often they are engaged in, but the *cumulative* odds change the more of either you engage in. For example, with Russian Roulette each time you spin the cylinder at random and then pull the trigger, you face 1 in 6 odds. And each time after that, for each spin, it's 1 in 6. But the cumulative risk of a Colt lobotomy from one of those spins is increasing with each spin. " The response: from "axe" "Chris, your example violates the random rule, unless the revolving part is respun with each turn. Probability calculations (odds) are based upon the principle of randomness and, if those folks betting against a Colt lobotomy reshuffle the revolver, it's always 1 chance in 6. It is only cumulative if they don't spin it each time, and it is not cumulative at all if you are the lucky chap to follow mr/mrs brainsonthewall. I tool stats AND slept in a Holiday Inn Express. -Axe'" I don't find it hard to believe he "tooled" stats....anyway, it's been so long since I took probability I can't remember how the math works out exactly. I know with the brains on this board someone will remember how to show how this works with the Russian Roulette example. I know for each spin, there will be a 1:6 chance of shooting yourself, what is the equation for the *cumulative* odds after spinning say, 6 times? The number of events does play a role in the cumulative odds, does it not?

Diplomacy is the art of saying 'Nice doggie' until you can find a rock. - Will Rogers
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PostTue Aug 13, 2002 10:48 am 
Whoa! It's been a long time, but it's nice to put college courses to use. I took probabilities and statistics. The first theory taught to us was the "Monkey Theory". You take an event such as an infinite number of monkeys. You give each monkey a typewriter. In a certain amount of time, one of the monkeys will write "War & Peace". In other words, the longer time passes, repetition of the same activity, probability increases that more of a variety of events can and will happen. Every scenario that can be speculated on, either negative or positive results, and possibly not the results that were intended. Makes you think, doesn't it? confused.gif BW dizzy.gif

"Wait by the river long enough and the bodies of your enemies will float by"...Sun Tsu
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PostTue Aug 13, 2002 10:57 am 
Fair.russian.roulette How can you play Russian Roulette with each person having a 50/50 chance of losing? Assume you only have one bullet and you may not take the bullet out of the gun. The gun has six chambers. Also assume that the players spin the cylinder before each turn. The problem is to find a sequence of turns which makes the game fair. -------------------------------------------------------------------------------- Solution: Calculate the accumulated probability of losing for each player and whichever has the lower number takes the next turn. If you ever end up with a position where both players have the same accumulated probability, then flip a coin (you have to do this on the first turn anyway, it's not clear that you'll ever have to do it again). Just to make it clear, the probability of losing which you're calculating is the a priori probability that you will have lost at or before that point in the game. It is NOT conditioned on the information that both players have survived up to that point. You're trying to make a game where, at the start, both players have a 1/2 chance of winning. Specifically, write down a geometric sequence starting with 1/6 and with each entry 5/6 of the previous entry. Both players start with a total of zero. Player A goes first, and adds 1/6 to his/her total, so that A now has 1/6 while B has 0. On the next turn, since B's total is less than A's, it's B's turn. B now has 5/36. Next turn is B's again, B now has 5/36 + 25/216, which is greater than 1/6, so it's now A's turn again . . . Continue until one player is dead. The opening sequence is: ABBABAABBAABBAABBAABBABAABB . . . Clearly, if the game lasts indefinitely, the sum of A's and B's numbers will approach 1. The only way this algorithm can fail is if one of the two numbers ever exceeds 1/2 (else both numbers will approach 1/2 from below). This problem can only occur if the sum of the lesser of A's and B's numbers and the next number on the list exceeds 1/2, which would mean that more than half of the remaining probability is in the next number on the list. But each number on the list is equal to only 1/6 of the sum of all the numbers that follow it, so this can never happen. In fact, there are a bunch of sequences that will work. You can relax the rule so that, as long as there's no possibility of one player's total exceeding 1/2, you flip a coin. This actually appears to be the general solution. All solutions could be generated by this method. In fact, you'll flip a coin on the majority of the turns (around 2/3). You can end up with a sequence where A goes once, B goes 5 times, A goes 29 times, . . . from google search

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MtnGoat
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PostTue Aug 13, 2002 11:21 am 
I appreciate the help with the last example, but my honorable opponent is never, ever going to understand it! I need something along the lines of: overall odds = 1:6 x occurences, something relatively clear, after all this guy tooled statistics.

Diplomacy is the art of saying 'Nice doggie' until you can find a rock. - Will Rogers
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PostTue Aug 13, 2002 12:27 pm 
The odds of surviving a particular spin are always 5/6 the cumulative odds increase the more spins with an infinite number death is inevitable. For two spins it is 5/6 x 5/6 = 25/36 three 125/212 four 525/1272 ... As you can see the numbers get smaller the more spins.

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PostTue Aug 13, 2002 1:00 pm 
isn't 5*5*5*5 = 625, and 6*6*6*6 = 1296? Not trying to be a butthead, just was working through the math in your example and the second spin numbers are consistent but the 4th spin numbers aren't. Am I missing something? thanks for the effort, I'm grateful for the help.

Diplomacy is the art of saying 'Nice doggie' until you can find a rock. - Will Rogers
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PostTue Aug 13, 2002 2:00 pm 
Earning my living as an actuary I think I'm competent to answer this one. All else being equal, it makes no difference how many times you've previously flown on an airplane, your odds of dying in a crash are the same the next time you fly. GIVEN you have survived up to a certain point, the only relevant statistic is how many times you plan to fly again, not how many times you have flown. For example, let's say the odds are 10% of dying in a crash on any given flight. Ignoring other causes of death, the following probabilities apply: # times you plan to fly again => probability of dying in a future plane crash 0 => 0% 1 => 10.0% 2 => 19.0% 3 => 27.1% 4 => 34.4% 5 => 41.0% 10 => 65.1% 15 => 79.4% 20 => 87.8% 25 => 92.8% 50 => 99.5% n => (odds for n-1) + (1 - 0.10)^(n-1) x 0.10 Now why do I sense a GPS navigation related joke on the horizon tongue.gif.

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PostTue Aug 13, 2002 2:25 pm 
Well you guys left me in the dust! huh.gif I guess I'll console myself by joining the monkeys! I'll let you know when our first book is published. tongue.gif BW

"Wait by the river long enough and the bodies of your enemies will float by"...Sun Tsu
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PostTue Aug 13, 2002 2:32 pm 
no GPS jokes here, your answer, combined with the Russian Roulette example is all the info I need! Thanks guys.

Diplomacy is the art of saying 'Nice doggie' until you can find a rock. - Will Rogers
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PostTue Aug 13, 2002 3:23 pm 
Apparently this is not good enough. Now another guy is going on about how for probabilty, you add events as in 5/6+5/6= 10/12 odds! Tom, if you want to read this mania for grins, here's the site..the additive fellow needs a little book larnin', and I think you may the be the guy. http://pub58.ezboard.com/fmtheadsfrm1 rowell's thread

Diplomacy is the art of saying 'Nice doggie' until you can find a rock. - Will Rogers
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PostTue Aug 13, 2002 3:52 pm 
My employer blocks that site so I guess I'll have to save the grins for later. That sure is an interesting way of adding fractions, not to mention 5/6 = 10/12. lol.gif Basically this is how it works in laymans terms: If the odds are 1/10 you die on the first flight then you have a 9/10 chance of surviving to the next flight. You cannot simply add another 1/10 for the next flight - you have to haircut it 9/10 since you will only get there 9/10 of the time. Thus, the cumulative probability for surviving two future flights is: 1/10 + (9/10 x 1/10) = 19/100 = 19% For 3 flights the odds are: 1/10 + (9/10 x 1/10) + (9/10 x 9/10 x 1/10) = 271/1000 = 27.1% and so on...

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PostTue Aug 13, 2002 3:54 pm 
5/6 + 5/6 = 10/6 = 5/3 = 1 2/3

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PostTue Aug 13, 2002 4:10 pm 
From a post over there, for Tom bait! "Actuaries are oddsmakers with a white shirt and tie. "

Diplomacy is the art of saying 'Nice doggie' until you can find a rock. - Will Rogers
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PostTue Aug 13, 2002 7:08 pm 
First of all, I have to take issue with an infinite number of monkeys being able to confabulate War and Peace. Hasn't the internet disproved this? We've got almost an ininite number of people typing gibberish each day and nothing remotely like War and Peace has occurred biggrin.gif biggrin.gif biggrin.gif ok, nwhikers.net is pretty close wink.gif While were on the statistics subject, here is the famous Ask Marilyn Let's Make a Deal puzzle with all sorts of statistics lingo that might translate into War and Peace for all I know. http://stat-www.berkeley.edu/users/stark/SticiGui/Text/ch10.htm

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Bushwacker
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PostTue Aug 13, 2002 7:24 pm 
lol.gif lol.gif lol.gif 1+1=2 1+2=3 1+3=5 huh.gif Wait a minute. confused.gif Let me check with a monkey. I'll get back to you dizzy.gif BW

"Wait by the river long enough and the bodies of your enemies will float by"...Sun Tsu
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