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Subject: Re: Probability problem  Date: 9/9/2020 12:18 PM  
Author: MrPlunger  Number: 6002 of 6013  
First consider how likely it is that each of the three type of dice were the one first picked. Getting a 4 is 1/6 of a chance for the 6 sided die, 1/12 chance for the 12 sided die, 1/20 for the 20 sided. So the chance that a 6 sided die was picked is 1/6 / (1/6+1/12+1/20) = 0.556. Similar for the others Chance it was 6 sided .556 chance it was 12 sided .278 Chance it was 20 sided .167. Next given each of these one at a time let's work out what are the chances of getting a 3 or less. There is a 50% chance of picking each of the other 2 dice. From the start of the 6 sider, there is a 3/12 chance of getting a 1, 2 or 3 if the 12 sider is the second die, 3/20 if the 20 sider. Each of these multiplied by their 50% chance and added together gets a probability of 1/4, if it was the 6 sider picked first. Multiply that by the likelihood of the 6 sider being picked (0.556 chance) and we get 0.111 chance that the 6 sider was picked and of a 3 or less on the second die. Similarly the probability of the 12 sider being picked first and a 3 or less off the second die is 0.09. And with the 20 sider first and 3 or less second time 0.063. Total probability 0.264. 

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