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Posted: 3/21/2002 11:12:45 AM EDT
Since you fellows are pretty smart and good at math, I want to pose
the following question? Who is right? Here is the scenario: You are forced into playing a game of Russian Roulette. You are presented with two options. Option #1 allows you to load 1 round into the wheel of a six-shooter. You must spin the wheel in regular RR fashion, and pull the trigger against your own head. You must repeat this operation six times in order to win. Option #2 allows you to load 5 rounds into the wheel, but you only have to spin the wheel and pull the trigger once to win. Which option would provide you the greater chance of Winning? Our Marketing teacher told us that most people would choose Option #1 because it "seems" safer. He says that people are more likely to choose the option with seemingly less risk. This is true, but he said however, that choosing Option #2 would give you the best chance of survival because of the actual probabilities. Here is where I disagreed. I maintain that you have a greater chance of survival by choosing Option #1. His reasoning is that on Option #1 you have a 1 in 6 chance of being killed. This seems great, but if you take this chance 6 times, you have a 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 6/6 chance, or a virtual certainty of being killed. You are better off by taking a 1 time 5/6 chance of being killed than a 6 times 1/6 chance. His chance of living to see another day is 16.7% My argument is that you have a decent chance of living through all six trigger pulls. Therfore the odds cannot be anywhere close to 6/6 or 100% certain. This means that the professor has incorrectly figured his odds on Option #1. My arguement is that you are dependent upon a string of outcomes in order to survive. Each event has an equal probability of occuring each time, but the chances of stringing together desirable outcomes shrinks with each trigger pull. For illustration purposes I will use a coin. Heads, you lose, Tails, you win. Game: You have to flip 3 tails in a row to win. Question: What are your odds of winning? Well, the first flip yields you a 1/2 chance of remaining in the game. Remember that if you ever flip a heads, you will immediately lose. The second flip, while still giving you a 1/2 chance of staying in the game, gives you a 50% chance on a 50% chance. The third flip adds another 50% chance to the equation. We can now calculate our odds of winning three tails in a row to be 1/2 * 1/2 * 1/2 = 1/8. We have a 12.5% chance of winning this game. This is the same formula that can be applied to Option #1 in Russian Roulette. Each time you pull the trigger you have a 5/6 chance of advancing to the next trigger pull. If we must perform this operation 6 times, we have 5/6 * 5/6 * 5/6 * 5/6 * 5/6 * 5/6 = 15625/46656, or about 33.5% chance of surviving the game. Each time you pull the trigger in this scenario, you decrease your chances of moving on to the next round. After many many trigger pulls, say 50, your survivability rate is .011%. What I am asking you to do is give me your input on who is correct: The (Marketing) Professor: Option #1 = 6/6 chance vs. Option #2 = 5/6 chance of being killed. The (Mediocre) Student: Option #1 = 33.5% chance vs. Option #2 = 16.7% chance of survival. Thanks for your input. |
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The (mediocre) student is correct. This must be a public school teacher. I am no mathematician, but it is obvious the effect of six 1/6 chances following one another is NOT cumulative. You have a 1/6 chance of death EACH TIME. Subsequent trigger pulls stand alone each time. Please tell him I said he is a dumbass.
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I will tell him thanks!
He is a Professor at Bradley University. You are probably familiar with it (being from IL). I am taking Post-Graduate classes there because my employer requires continuing education. They pay, so I don't complain. |
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I just picked up on the whole key to the problem: He's a [b][i]marketing[/i][/b] professor. I've never yet met anyone in marketing with the slightest link to reality.
BTW, you're precisely correct in your calculations. He hasn't a clue. |
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Being a non-pro mathematician, you always have a 1/6 chance with the one bullet.
Your proffesor is an idiot. |
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Quoted: Since you fellows are pretty smart and good at math, I want to pose the following question? Who is right? Here is the scenario: You are forced into playing a game of Russian Roulette. View Quote Well, a six-shooter has a 1 in 6 chance of ruining your day, so I'd ask for one of them semi-auto Tec-9 junk guns, which has a 1 in 3 chance of jamming or misfiring. [;D] [BD] Okay, since nobody likes a wise arse, I'll go with Option 1. Option 2 has a 5 times out of 6 likelihood that your brains will land in the next county. I like one chance in 6, no matter HOW many times I gotta do it. The "one in 6" for six times can't be added the way you did it becasue theres the possibility if the round is in chamber 1, you could land on chamber 2 all six times, or chamber 3 all six times. (Or chamber 4 or 5 or 6 all six times.) Or if you can add up probabilities that way, Look at it another way, each time you do option 1 you have a 5 in 6 chance of NOT being killed. Do that six times, and you end up with 30 chances out of 36 that your blue suade shoes will walk away with no grey matter on them. I like 30 chances in 36 of being okay better than I like 5 chances outta six of seeing Kurt Kobain. (But then, I didn;t care for him when he was alive, either) [}:D] |
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Option 1 poses less risk because of the physics and properly done the math does reflect this. The type of simplified probability analysis that your teacher is applying is not correct (not withstanding the fact that his math is wrong) .
I would go with option 1. Fis_Prod edited to say: You math is correct by the way. however remember that the physics makes it such that the probability on any given chamber has to be weighted. |
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Hmmm... Could a lack of analytical skills be why your professor wound up marketing? [:D]
The probability of shooting yourself in option #1 is 6/6 [b]only[/b] if the cylinder isn't spun after each pull of the trigger. Otherwise, as you calculated, the pulls are independent events and the probability of survival is indeed (5/6)^6. If Professor Badmath is too stubborn to admit his mistake, ask him to put his money where his mouth is and play a little game of dice with you: He gets one (cubic) die and you get six. In each round of the game, each player antes $5 and then both players roll their dice simultaneously. If he rolls a six (analogous to dropping the hammer on the one unloaded chamber in Russian Roulette), he "survives". If you roll anything but a one on all six of your dice (analogous to [b]not[/b] dropping the hammer on the one loaded chamber in six rounds of Russian Roulette), you "survive". If both players "survive", the money stays in the pot and you play another round. If neither player "survives", the money stays in the pot and you play another round. If one player "survives" and the other doesn't, the "survivor" gets all the money in the pot. Tell him that you want to play for ten rounds, with any money left in the pot after the tenth round to be split evenly between the players. Of course, if he insists at that point that you're only winning because you're on a lucky streak, by all means, indulge the poor man and play another ten rounds. [;)] |
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Quoted: He is a Professor at Bradley University. You are probably familiar with it (being from IL). View Quote Why, oh why, does this not surprise me? Yes, I have a passing familiarity with BU. He is probably among the best and brightest there. Fis_Prod, you are damned eloquent and diplomatic in the way you called the idiot an idiot. LOL [:D] |
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Quoted: Well, a six-shooter has a 1 in 6 chance of ruining your day, so I'd ask for one of them semi-auto Tec-9 junk guns, which has a 1 in 3 chance of jamming or misfiring. [;D] [BD] View Quote Ha ha! Wasn't there a Darwin award about someone who tried to play Russian Roulette with a Semi-auto? -snip- I like one chance in 6, no matter HOW many times I gotta do it. View Quote By my calculations, you would drop below a 16.7% survivability rate on the 10th attempt, where your odds would be 16.2% The "one in 6" for six times can't be added the way you did it becasue theres the possibility if the round is in chamber 1, you could land on chamber 2 all six times, or chamber 3 all six times. (Or chamber 4 or 5 or 6 all six times.) View Quote Exactly, there is a 33.5% chance that all six pulls will land on chambers 2-6 assuming that chamber 1 is hot. Or if you can add up probabilities that way, Look at it another way, each time you do option 1 you have a 5 in 6 chance of NOT being killed. Do that six times, and you end up with 30 chances out of 36 that your blue suade shoes will walk away with no grey matter on them. I like 30 chances in 36 of being okay better than I like 5 chances outta six of seeing Kurt Kobain. (But then, I didn;t care for him when he was alive, either) [}:D] View Quote Oops, don't forget to multiply across the numerators and denominators, and not add. [:)] Anyway, I think we are all in agreement! |
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Your teacher is an idiot. You are absolutely right. Tell that idiot to take a ststistics course before he says something stupid like that.
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Quoted: Fis_Prod, you are damned eloquent and diplomatic in the way you called the idiot an idiot. LOL [:D] View Quote Beekeeper, Thank you...I have been required to take a course in interpersonal skills (everyone at work has..new politically correct world [rolleyes] ) and I am suppose to be nice to people....even those who show obvious signs of deficiency.. Take Care Fis_Prod |
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One small point that I may have missed...
In the one hot game, is the cylinder spun BETWEEN each pull of the trigger, or are the six clicks one after another directly? Click, Click, BOOM! If you spin after each trial, I will take option #1. If you have to pull six times in a row, I will take #2. 16.666...% chance is better than 0%... FFZ |
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Quoted: Hmmm... Could a lack of analytical skills be why your professor wound up marketing? [:D] The probability of shooting yourself in option #1 is 6/6 [b]only[/b] if the cylinder isn't spun after each pull of the trigger. Otherwise, as you calculated, the pulls are independent events and the probability of survival is indeed (5/6)^6. If Professor Badmath is too stubborn to admit his mistake, ask him to put his money where his mouth is and play a little game of dice with you: He gets one (cubic) die and you get six. In each round of the game, each player antes $5 and then both players roll their dice simultaneously. If he rolls a six (analogous to dropping the hammer on the one unloaded chamber in Russian Roulette), he "survives". If you roll anything but a one on all six of your dice (analogous to [b]not[/b] dropping the hammer on the one loaded chamber in six rounds of Russian Roulette), you "survive". If both players "survive", the money stays in the pot and you play another round. If neither player "survives", the money stays in the pot and you play another round. If one player "survives" and the other doesn't, the "survivor" gets all the money in the pot. Tell him that you want to play for ten rounds, with any money left in the pot after the tenth round to be split evenly between the players. Of course, if he insists at that point that you're only winning because you're on a lucky streak, by all means, indulge the poor man and play another ten rounds. [;)] View Quote I like this idea! Your professor is a moron! If the whole point was to illustrate that people make a certain kind of decission with given circumstances and it turns out to be the wrong choice, then he is still a moron for picking such a dumb example and for explaining it so poorly to his students. Go tell your teacher to re-read his stats book. Better yet-read a stats book yourself. That stuff is actually pretty useful, and next time you can really shut him down. |
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Hey, here's a math problem for you: Rosie O'Donuts weighs 475 pounds and her lesbian lover weighs 355 pounds. Since their mattress can only support 700 pounds of blubbery, cellulite fat at any one time, they do their strap-on slogging of each other's fat lesbo asses out on the living room floor every night. When Rosie is on top she generates between 1200-1600psi of pressure while ramming her squealing pigfriend below. Now, given the above information, what is the best caliber bullet to use to ensure an 99% probability of hitting both of those pigs with a single lethal shot? |
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Quoted: Ha ha! Wasn't there a Darwin award about someone who tried to play Russian Roulette with a Semi-auto? View Quote I don't know whether he was listed as a Darwin award winner or not, but the Houston Chronicle had a news item about a guy killing himself by playing RR with a semi-auto. |
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Quoted: Hey, here's a math problem for you: Rosie O'Donuts weighs 475 pounds and her lesbian lover weighs 355 pounds. Since their mattress can only support 700 pounds of blubbery, cellulite fat at any one time, they do their strap-on slogging of each other's fat lesbo asses out on the living room floor every night. When Rosie is on top she generates between 1200-1600psi of pressure while ramming her squealing pigfriend below. Now, given the above information, what is the best caliber bullet to use to ensure an 99% probability of hitting both of those pigs with a single lethal shot? View Quote Oh God! 30-06 and Hurry!!!! |
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Ask the professor to prove it to you by using the gun with the 5 full cylinders. No big loss if he looses.
BTW, being a marketing guy AND a professor is two strikes. Marketing guys have only the slightest idea of reality & professors have none. |
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To calculate the correct chance of losing you must add the probability of losing each round together. The probability of each round is calculated by multiplying the chance of losing this round by the chance of making it this far.
The correct formula for chance of losing: 1*1/6 + 5/6*1/6 + (5/6)^1*1/6 + (5/6)^2*1/6 + (5/6)^3*1/6 + (5/6)^4*1/6 + (5/6)^5*1/6 = 0.665 Adding my result to your previous result equals 1, confirming the formula is correct: 0.665 + 0.335 = 1 Try asking one of the math professors there to have a "talk" with your marketing professor. |
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Quoted: Hey, here's a math problem for you: Rosie O'Donuts weighs 475 pounds and her lesbian lover weighs 355 pounds. Since their mattress can only support 700 pounds of blubbery, cellulite fat at any one time, they do their strap-on slogging of each other's fat lesbo asses out on the living room floor every night. When Rosie is on top she generates between 1200-1600psi of pressure while ramming her squealing pigfriend below. Now, given the above information, what is the best caliber bullet to use to ensure an 99% probability of hitting both of those pigs with a single lethal shot? View Quote LMAO!!! I dunno. Do nuclear bombs come in range of calibers???? [;D] The_Macallam - Dude, you are in rare form these days. But, PLEASE, man, can ya cut back on the imagery???? After your verbal fusillade of that five-for-a-dime (bag) ho Pammy and now Rosie and her lesbo fat slapper, I'm gonna be scarred for life. [}:D] |
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Quoted: Quoted: Hey, here's a math problem for you: Rosie O'Donuts weighs 475 pounds and her lesbian lover weighs 355 pounds. Since their mattress can only support 700 pounds of blubbery, cellulite fat at any one time, they do their strap-on slogging of each other's fat lesbo asses out on the living room floor every night. When Rosie is on top she generates between 1200-1600psi of pressure while ramming her squealing pigfriend below. Now, given the above information, what is the best caliber bullet to use to ensure an 99% probability of hitting both of those pigs with a single lethal shot? View Quote Oh God! 30-06 and Hurry!!!! View Quote do you really think thats enough? for a clean sure kill i would use a 30mm twin cannon from a us fighter plane. strafeing run. saw them in half. |
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Quoted: do you really think thats enough? for a clean sure kill i would use a 30mm twin cannon from a us fighter plane. strafeing run. saw them in half. View Quote Hey, be careful now. Does anyone remember that beached whale in Oregon they tried to get rid of with dynamite a few years ago? I think it's still raining bits of blubber over there. |
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Sorry sir! That was all that was available at the time! It wouldn't stop..... Ahhhhhhh!
[scarred for life] |
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" There are lies, damn lies, and statistics."
Proving yet again that any idiot can make a point with bad math.[:D] You sir are correct in your assumptions - I hope that professor isn't tenured! |
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First of all How old is this Professor?
Second I guess he doesn't shoot guns, or he would not use that has an example. Third if he wants to try make sure he ties a bandana around his head. It helps a little from what I remember. Things I try to forget always pop-up no matter how many years have passed. |
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Option #1 allows you to load 1 round into the wheel of a six-shooter. You must spin the wheel in regular RR fashion, and pull the trigger against your own head. [red]You must repeat this operation six times in order to win. [/red] View Quote Does this mean that you will have to load another round everytime or not?? If not, then it is a 1/6 chance everytime else the chances is 2/6 for the second time, 3/6 for the 3rd time,........6/6 on the sixth time which will surely blow your head off. On the other hand, there is no mention how many rounds can the revolver hold. It may be a 22LR revolver that can hold 9 rounds for all we know. It will not matter whether someone chooses option #1 or option #2 if you are unlucky. Probability can be calculated but as long as there is a chance that things will go south, it will go south. [:)][:)][:)][:)] Did I mention that I am a pessimist??? |
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Quoted: First of all How old is this Professor? View Quote I would guess his age at about 38 yrs. old. Second I guess he doesn't shoot guns, or he would not use that has an example. View Quote I don't know, seems to be a decent enough example. Too bad he sucks at math. He used other examples too, some positive and some negative. Third if he wants to try make sure he ties a bandana around his head. It helps a little from what I remember. View Quote Why? So you can find all the pieces easier for reassembly? |
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Quoted: Option #1 allows you to load 1 round into the wheel of a six-shooter. You must spin the wheel in regular RR fashion, and pull the trigger against your own head. [red]You must repeat this operation six times in order to win. [/red] View Quote Does this mean that you will have to load another round everytime or not?? If not, then it is a 1/6 chance everytime else the chances is 2/6 for the second time, 3/6 for the 3rd time,........6/6 on the sixth time which will surely blow your head off. On the other hand, there is no mention how many rounds can the revolver hold. It may be a 22LR revolver that can hold 9 rounds for all we know. It will not matter whether someone chooses option #1 or option #2 if you are unlucky. Probability can be calculated but as long as there is a chance that things will go south, it will go south. [:)][:)][:)][:)] Did I mention that I am a pessimist??? View Quote No the option was clearly stated to involve loading 1 round, and playing Russian Roulette with it six times. Actually, it is widely believed that a six-shooter will hold about 6 rounds. What kind of mutant six shooter do you have?! [:)] |
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Your professor is obviously wrong. In reality with a 1/6 chance of dieing - six times - you have a 6/36 chance of dieing which equals 1/6.
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Hey, Torf--
I have an idea. When this thread plays out, how about printing it and presenting it to him? You may get "extra credit!" LOL I'll bet he would stroke out over the Rosie post. [}:D] Edited: I forgot to mention, explain to him that as long as it's Thursday, the probability ratio is 6/6 to the tenth power that he MUST give you extra credit. He'll understand what that means, or at least pretend to. LMAO |
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I hate math- I would have said that the probabilities were equal.
frickin math...gotta hate it |
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Quoted: Hey, Torf-- I have an idea. When this thread plays out, how about printing it and presenting it to him? You may get "extra credit!" LOL I'll bet he would stroke out over the Rosie post. [}:D] View Quote Actually I was thinking of showing this to my boss (who was also there), until The_Macallen showed up. Now? Forget it! |
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The way to figure the first case is to determine the probability of your not dying on all of the trials. For independent trials, that's (5/6)^6 or 0.33489. The probability of your being killed is the complement of that, 1 - 0.33489, or 0.6651.
The probability of your being killed with five rounds in a six-shooter is 5/6 or 0.8333. You're better off playing six times. "Seven years of college down the drain" --Bluto Blutarsky, Faber College '63. |
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Quoted: Your professor is obviously wrong. In reality with a 1/6 chance of dieing - six times - you have a 6/36 chance of dieing which equals 1/6. View Quote |
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Here is your answer. Are you mad, know your invincible in death and life? Then you win. For when these games were played in my day and age it meant something. It had nothing to do with Statistics. We couldn't even count half the time. SHIT this was a way out from misery for some Gooks and Americans macho BS.
Anyone remember the game along with the Cox fight and the little suckle while the ref picked the bullet up said, @ men sit 1 man lives 1 man dies and neither reaches his GOD. They are both cowards. You remember that line. Who cared. You get pulled out dead stripped thrown to the dog. The winner big heroes until he bit the next round. It was nothing like the Dear Hunter, or what ever BS Movie depicts our attitude has fighting men there. GOD FOR GIVE ME FOR THE EVIL I PERFORMED AND THE LIVES I LEFT WASTED. GOD FOR GIVE ME FOR DOING NONTHING THEN AND NOT SAVING ANYONE ALIVE ON 9/11/ WHY DID this stupid game have to be -played with airplanes and towers. IT WAS A HIT AND MIS SITUATION...FORGIVES ME MY BROTHERS AND SISTERS I WILL FIND YOU ALL. WHERE EVER YOU LAY. Please find peace at this time. You are in my mind 24/7 along with the insanity we live with now GOD BLESS ALL OUR FALLEN & INJURED WORKERS |
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Quoted: Actually I was thinking of showing this to my boss (who was also there), until The_Macallen showed up. Now? Forget it! View Quote [:P] |
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Torf,
You are right, your professor is wrong. Your problem is a binomial distribution, and you are calculating the probability of getting 6 successful outcomes in a row. The formula is: P(X)= n!/[X!(n-X)!] * p^X * (1-p)^(n-X) where P(X) is the probability of X successes n is the sample size (six in this case) p is probability of success (5/6 in this case) 1-p is the probability of failure (1/6) X is the number of successes The stuff with ! after are factorials - ex. 3!=3*2*1. If you plug in all the numbers, your final equation is just (5/6)^6, which is what you calculated intuitively - I'm just backing you up with the equations. In closing, your MARKETING professor is a dumbass. Tell him to stick with building systems to sell crap to stupid people. |
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After getting through my last psychotic episode about the past and now. The answer is depends how lucky you are that day. I think play either way it's a 1*6 chance.
I have seen and I mean seen people play that game. This is how it was played. 1 bullet, spin the chamber. Depending on whom was loading he would cock it then give it to you. If I figure it out there were guys who could play for hours and survive. Some first shot out of the game. Some played for weeks. Sooner or later they lost. Your Professor should be sat down given a gun. You ask him 5 or 1 bullets. If he wins you get a (D). If he loses you get a (A). See if he will take the odds he only has to play once. Forgive me for my ranting last post. It is a 1*6 chances either way you play. |
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