Let X be a random variables representing number of fair dice thrown with mass:

$P(X=x)=\frac{1}{2^x}$ for $x=1,2,....$

Let also Y be the sum of numbers appearing on the faces of dice. Then, find $P(X=2|Y=3)$

My approach

Although I have tried to solve the problem but I am a little doubtful as my answer didn't match with manual.

This is Bayes Problem. The sum of 3 can only occur if die is thrown one or two or three times. If we throw more dice, then sum would exceed. Now, $P(X=1)=\frac{1}{2}$ $P(X=2)=\frac{1}{4}$ $P(X=3)=\frac{1}{8}$

Also, $P(S=3|X=1)=\frac{1}{6}$ $P(S=3|X=2)=\frac{2}{36}$ $P(S=3|X=3)=\frac{1}{216}$

Now, using the Bayes formula, I am getting the required probability as $P(X=2|Y=3)=\frac{24}{169}$

What am I doing wrong here? The answer does not match with the manual.

Any help?

  • $\begingroup$ What is the manual's answer? Does it give any indication of how it got that? $\endgroup$ Apr 27, 2018 at 4:09
  • $\begingroup$ @GrahamKemp I don't how it got $\frac{13}{45}$. $\endgroup$
    – userNoOne
    Apr 27, 2018 at 4:12

1 Answer 1


Yes indeed, if the sum of pips is $3$ it is impossible to have thrown more than $3$ die, so we do seek:

$$\mathsf P(X=2\mid Y=3)= \dfrac{\mathsf P(X=2, Y=3)}{\mathsf P(X=1, Y=3)+\mathsf P(X=2, Y=3)+\mathsf P(X=3, Y=3)}$$

And indeed you have evaluated these correctly:

  • $\mathsf P(X=1, Y=3)=\tfrac 12\cdot \tfrac 16\qquad$ throw a 3 on one dice.
  • $\mathsf P(X=2, Y=3)=\tfrac 14\cdot \tfrac {2}{36}\quad~~$ throw a 1 and 2 on two die.
  • $\mathsf P(X=3, Y=3)=\tfrac 18\cdot \tfrac {1}{216}\quad~$ throw 1 on each from three die.
  • $\mathsf P(X=4, Y=3)=\tfrac 1{16}\cdot \tfrac {0}{~}\quad~$ cannot do it for anything beyond three die.

Finally that does give us: $$\mathsf P(X=2\mid Y=3)= \dfrac{2\cdot 2\cdot 6}{4\cdot 36+2\cdot 2\cdot 6+1} = \dfrac{24}{169}$$

You are vindicated!   The manual is clearly wrong.   Bad manual, bad.


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