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LOL - Math Edition |
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Darur
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Topic: LOL - Math EditionPosted: 08 December 2007 at 1:22am |
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So tonight my family wanted to watch "Deal or No Deal", for those not familiar its another mindless game show where everything is almost always pure luck. Tonight's show was interesting though with a classic Monty Hall problem.
For those that didn't see it, the contestant went through all the cases down to hers and the last on the board. The two possible values were $200,000 and $1,000. The host offered her a chance to switch cases which she declined. Had she accepted, she would have had a 96% chance of selecting the $200,000 case with her original odds being 1/26 of picking the $200,000 case first and 25/26 odds with it being in the last case opened. Proof of how valuable math can be ![]() |
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carl_the_sniper
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Posted: 08 December 2007 at 1:37am |
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So how do you figure 96%?
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pntbl freak
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Posted: 08 December 2007 at 1:40am |
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Since you have two cases left, wouldnt it be a 50% chance?
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Darur
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Posted: 08 December 2007 at 1:47am |
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Nope. Initially, the odds of selecting the $200,000 case is 1/26. That means there is a 25/26 chance of the $200,000 case being in the cases to be opened. As the player eliminates cases, the odds remain 1/26 and 25/26 for the player and the unopened cases respectively regardless of how many cases are open. In the end the case the player selected still has a 1/26 chance of holding $200,000; meanwhile, the last unopened case has a 25/26 chance of holding $200,000 because it keeps the initial odds of all the unopened cases. 25/26 = 96.15% 1/26 = 3.85% Wikipedia for more info |
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carl_the_sniper
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Posted: 08 December 2007 at 1:52am |
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Aww crap I knew that sounded familiar.
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RoboCop
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Posted: 08 December 2007 at 6:42am |
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So did she not choose correctly or did she take the deal?
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Skillet42565
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Posted: 08 December 2007 at 9:44am |
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Thats hilarious.
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.357 Magnum
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Posted: 08 December 2007 at 9:49am |
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It is pure luck. |
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Darur
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Posted: 08 December 2007 at 6:47pm |
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RoboCop, she decided to go with what was in her case, which was $1,000, hoping for the 4% chance she had the money. A bad move.
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oreomann33
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Posted: 08 December 2007 at 6:50pm |
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No it's not, math comes into play like Darur said. DUH |
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pntbl freak
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Posted: 08 December 2007 at 7:17pm |
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Maybe Im still missing something.
She selected the $1,000 dollar case to begin with. Once she got down to the final two cases she had a chance to switch giving her a 96% chance that the case contains 200,000. However she doesnt know what she picked. Lets say she picked the 200,000 case to begin with. She gets down to the final two cases again and the 1,000 case is still out there. She would then have a 96% that the case contained 1,000. She would have no clue if its a good move to switch or not. Since there are two cases left I still think she would have a 50% chance. |
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FROG MAN
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Posted: 08 December 2007 at 7:30pm |
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its how you look at it. if I roll a dice there is a 1/6 chance i will get 1, if I roll it now 10 times, and 1 has yet to come up, the odds of rolling it have increased. tho if you ignore those previous 10 rolls, it is still 1/6 chance. |
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pntbl freak
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Posted: 08 December 2007 at 7:33pm |
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Maybe I just feel like arguing right now and Im just looking for an error, but I still dont see how this changes anything. Just because a 1 has yet to be rolled doesnt change the probability that you will roll at 1 on your next roll. The dice doesnt care what you rolled previously and you will always have a 1/6 chance. Edited by pntbl freak - 08 December 2007 at 7:34pm |
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FROG MAN
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Posted: 08 December 2007 at 8:27pm |
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the probability of throwing a dice 6 times and not getting a 1 is .335, the probability of getting at least one 1, is .665 with a mutiple independent out comes, you can still find a single probability that is more likley. Though if you treat each throw as an independent, it stays at 1/6 chance to get a 1. ps. im agreeing with you. |
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Darur
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Posted: 08 December 2007 at 9:46pm |
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Your trying to look at the odds from all angles, which doesn't work. We're looking for the specific odds of the contestant having picked $200,000. In the second to last round, there was $10, $1000 and $200,000 in play. Because she was shooting for $200,000, we consider the odds of her selecting that case, which is 1/26. We lump together all the other cases as n/26 (n for the number of cases below $200,000, I don't remember how many this is). The odds of her selecting one of these lower cases however is much much higher. Regardless, when we look at her situation at the last round, the odds of her having the $200,000 case remain at 1/26, and the odds the last case to open is $200,000 remains 25/26. |
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Real Men play Tuba
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Bolt3
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Posted: 08 December 2007 at 10:45pm |
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It's called conditional probability and its a real thing.
It's looking at the situation from a mathematical standpoint. |
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Dye Playa
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Posted: 09 December 2007 at 12:13am |
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My brain hurts.
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RoboCop
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Posted: 09 December 2007 at 12:17am |
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It's an easy concept of the math invovled. I'm surprised teachers haven't taught you these things before.
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Dye Playa
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Posted: 09 December 2007 at 12:22am |
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They probably did (no pun intended), but whether or not I was paying attention at the time is a different story. |
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carl_the_sniper
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Posted: 09 December 2007 at 12:24am |
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I understand it but I could never explain it.
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