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#1




ASM Manual. Exercise 51.1.
Hi,
I don't know whether it is forbidden or not to write out the question here so I will just give a general description. This problem asks for Buhlman probability. There are two dice, each with different numbers. You are told one dice is selected from the pair and rolled 4 times. You are told the result of its first three rolls and asked for the expected value of the 4th. The solution mechanically works through the Buhlman steps. EPV and VHM using both dices etc. However the dices have different numbers. So you know that the selected dice has to be dice A. It seems like a trick question to me, but it's actually not. Or am I misunderstanding the question? 
#2




First the die is the singular and dice is the plural. Don't know the specifics of the problem but it could be illustrating the difference between the Bayes and Buhlman estimates.

#3




I'll paraphrase
Two fair dice from which you pick one. First is numbered 1,2,3,4,5,6 Second is numbered 6,7,8,9,10,11 The chosen dice is rolled four times times and you are told that the first three rolls get 1,2,3 Using Buhlman, What is expected value for fourth roll? (I use dice rather than die. I also used "dices" above which was incorrect but I did that just to prevent confusion. Die is used more in US English. ) The answer is 2.538 which in itself seems suspicious...........you basically are almost assuming that rolling the dice is a nonMarkovian process. I mean you know for certain which of them it is. And you are saying that the expected value of the next roll is less because the first three rolls were low...I guess maybe what you are updating is the assumption that it is fair i.e. each one is uniformly distributed...... Last edited by tcd; 02092018 at 10:21 AM.. 
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