Editing Talk:1252: Increased Risk
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I think it's worth mentioning that this comic doesn't [[985|distinguish between percentages and percentage points]]. --[[User:DiEvAl|DiEvAl]] ([[User talk:DiEvAl|talk]]) 12:35, 16 August 2013 (UTC) | I think it's worth mentioning that this comic doesn't [[985|distinguish between percentages and percentage points]]. --[[User:DiEvAl|DiEvAl]] ([[User talk:DiEvAl|talk]]) 12:35, 16 August 2013 (UTC) | ||
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Is it the case that doing something three times increases risk by 50% over two times inherently? I feel like this is the case, but it's early, here. Also, I'm not sure Randall is attacked by a dog, he may be using it as a diversion. I think that he's done this before. [[User:Theo|Theo]] ([[User talk:Theo|talk]]) 12:56, 16 August 2013 (UTC) | Is it the case that doing something three times increases risk by 50% over two times inherently? I feel like this is the case, but it's early, here. Also, I'm not sure Randall is attacked by a dog, he may be using it as a diversion. I think that he's done this before. [[User:Theo|Theo]] ([[User talk:Theo|talk]]) 12:56, 16 August 2013 (UTC) | ||
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Saying that unfortunately Cueball is mistaken in his calculations because he said 50% instead of 49.99999992% is a bit of an exaggeration. [[User:Xhfz|Xhfz]] ([[User talk:Xhfz|talk]]) 20:19, 16 August 2013 (UTC) | Saying that unfortunately Cueball is mistaken in his calculations because he said 50% instead of 49.99999992% is a bit of an exaggeration. [[User:Xhfz|Xhfz]] ([[User talk:Xhfz|talk]]) 20:19, 16 August 2013 (UTC) | ||
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:As far as I understand it: doing something twice doubles your chance of getting the desired outcome. For example, you want to role a dice and get a six. If you role it twice, you have double the chance of getting at least one six. If you role it three times you have triple the chance of getting a six; in other words you increase it from two chances to three chances, which is an increase of 50%. {{unsigned ip|213.86.4.78}} | :As far as I understand it: doing something twice doubles your chance of getting the desired outcome. For example, you want to role a dice and get a six. If you role it twice, you have double the chance of getting at least one six. If you role it three times you have triple the chance of getting a six; in other words you increase it from two chances to three chances, which is an increase of 50%. {{unsigned ip|213.86.4.78}} | ||
::It doubles the likely number of sixes, but does not double the chance of getting at least one six. This is because there is a small chance of getting two sixes, and while that counts as two sixes for the number of occurrences, it still only counts as one chance of getting at least one six. The easiest way to visualize this is to look at the probability that you won't get a six in any given roll of the die, which is 5/6ths. Each time you roll, the probability you won't get a six at all goes down by 5/6ths. So the probability for two rolls is 25/36ths, and thus the probability of getting one or more sixes in two rolls is 11/36ths. This is 1/36th less than 2/6ths, and 1/36th is the probability of getting two sixes. Similar (although more complicated) logic applies to rolling it three times, for which the probability of getting at least one 6 is 91/216ths (not 108/216ths, as the naive approach would imply). As others (CFoxx) have pointed out, if you roll a die 6 times, there is still a chance you won't get any sixes. If you roll it a million times, it is still possible (albeit very, very, very unlikely) that you wouldn't get any sixes! As far as the 50% and 16.67% figures given by the original poster, I believe those were calculated for events that have a 50% probability for each event. The increase in probability from 1 to 2 events where 1/x is the probability looks like (1-(1-1/x)^2)/(1/x)-1, which is (1-(1-2/x+1/x^2))*x-1 or (2/x-1/x^2)*x-1 or (2-1/x)-1 or 1-1/x. Thus for an event like a fair coin toss, the increase in probability for two tosses over one toss is 1/2. For a 6-sided die, the increase in probability is 5/6th. For a 1/billion, the increased probability for one or more occurrence for two events compared with one event is 0.999999999. Finally, the probability of the second event being the desired event is always the same. It is unchanged by the first event. It is the probability of either (or both) of the events being desired that we are calculating here. If the first die roll is a six, the probability of the second being a six is still 1/6. If the first die roll is not a six, the probability of the second being a six is still 1/6 (assuming a fair die). But the probability of either or both being a six is the absence of any information about the two rolls is not 2/6, but rather 11/36! [[Special:Contributions/206.174.12.203|206.174.12.203]] 17:06, 21 August 2013 (UTC)Toby Ovod-Everett | ::It doubles the likely number of sixes, but does not double the chance of getting at least one six. This is because there is a small chance of getting two sixes, and while that counts as two sixes for the number of occurrences, it still only counts as one chance of getting at least one six. The easiest way to visualize this is to look at the probability that you won't get a six in any given roll of the die, which is 5/6ths. Each time you roll, the probability you won't get a six at all goes down by 5/6ths. So the probability for two rolls is 25/36ths, and thus the probability of getting one or more sixes in two rolls is 11/36ths. This is 1/36th less than 2/6ths, and 1/36th is the probability of getting two sixes. Similar (although more complicated) logic applies to rolling it three times, for which the probability of getting at least one 6 is 91/216ths (not 108/216ths, as the naive approach would imply). As others (CFoxx) have pointed out, if you roll a die 6 times, there is still a chance you won't get any sixes. If you roll it a million times, it is still possible (albeit very, very, very unlikely) that you wouldn't get any sixes! As far as the 50% and 16.67% figures given by the original poster, I believe those were calculated for events that have a 50% probability for each event. The increase in probability from 1 to 2 events where 1/x is the probability looks like (1-(1-1/x)^2)/(1/x)-1, which is (1-(1-2/x+1/x^2))*x-1 or (2/x-1/x^2)*x-1 or (2-1/x)-1 or 1-1/x. Thus for an event like a fair coin toss, the increase in probability for two tosses over one toss is 1/2. For a 6-sided die, the increase in probability is 5/6th. For a 1/billion, the increased probability for one or more occurrence for two events compared with one event is 0.999999999. Finally, the probability of the second event being the desired event is always the same. It is unchanged by the first event. It is the probability of either (or both) of the events being desired that we are calculating here. If the first die roll is a six, the probability of the second being a six is still 1/6. If the first die roll is not a six, the probability of the second being a six is still 1/6 (assuming a fair die). But the probability of either or both being a six is the absence of any information about the two rolls is not 2/6, but rather 11/36! [[Special:Contributions/206.174.12.203|206.174.12.203]] 17:06, 21 August 2013 (UTC)Toby Ovod-Everett | ||
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