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A player throws an ordinary die. If the player gets a six, he can throw the die once more. Then his score is the sum of the two throws. Otherwise his score will be the number of his first throw. Assume the player throws the die no more than two times. Let X be the random variable denoting the player’s score. Find the probability distribution of X.

Can anyone please help?

Henry Tsui
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1 Answers1

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It is easy to generate a PDF for this distribution, given there are only $12$ ($11$ if you don't count 6) possible states: $$f_X(1) = f_X(2) = f_X(3) = f_X(4) = f_X(5) = \frac16$$ $$f_X(6) = 0$$ $$f_X(7) = f_X(8) = \cdots =f_X(12)= \frac16\cdot\frac16$$

Obviously there is a $\frac16$ chance of getting $1-5$ on the first roll. $f_X(6) = 0$ because the minimum sum on two dice is $7$. The numbers $7-12$ have a $\frac16$ possibility each from the second roll, which has a $\frac16$ chance of occuring. Therefore, their probabilities are all $\frac1{36}$.

Given that the player has a $0\le p\le1$ chance of rolling the second die, the probability distribution changes to:

$$f_X(1) = f_X(2) = f_X(3) = f_X(4) = f_X(5) = \frac16$$ $$f_X(6) = \frac{1-p}6$$ $$f_X(7) = f_X(8) = \cdots =f_X(12)= \frac16\cdot\frac p6$$

Austin Weaver
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  • But in the question, it said "he can throw the die once more", so it isn't mean the player can decide he throw or not throw the die once again? – Henry Tsui Oct 06 '17 at 15:30
  • Does he choose to throw twice exactly half of the time he has the opportunity? – Austin Weaver Oct 06 '17 at 15:31
  • When i ask my teacher, he reply "Note that the player can throw 1 time or two times.". – Henry Tsui Oct 06 '17 at 15:32
  • The question didn't provide such information... – Henry Tsui Oct 06 '17 at 15:33
  • Why would the player not choose to throw a second time, since the score is the sum of the two throws, so can only go up. Thus, your bolded can is silly. I would assume a second throw is automatic if the first throw is a $6$. – quasi Oct 06 '17 at 15:37
  • @quasi Perhaps they are trying to get the closest sum to 8 or something similar, so it is a gamble to get a better score. – Austin Weaver Oct 06 '17 at 15:39
  • There's no reason to assume that. By default, unless specified otherwise, I would assume the player is trying to maximize his or her score. No need to invent scenarios that aren't there. A problem can't specify everything, so normal common sense needs to be applied as to what the default assumptions are. – quasi Oct 06 '17 at 15:42