0
$\begingroup$

What is the probability you roll a $4$ before rolling $2$ odd numbers?

I'm not too sure on how to solve this question, here's what I did. There are two cases, one where you first get the four, or you get an odd then a four. The probability of just rolling a 4 first is $\frac{1}{6}$. If you roll one odd number before the $4$, that has a chance of $\frac{1}{2}*\frac{1}{6}$ Now, you can roll $2$ or $6$ infinitely many times and it won't matter. So, it is a infinite geometric series.

So, the sum of $\frac{1}{6}+\frac{1}{6}*\frac{1}{3}+\frac{1}{6}*(\frac{1}{3})^2+...$ and $\frac{1}{2}*\frac{1}{6}+\frac{1}{2}*\frac{1}{6}*\frac{1}{3}+\frac{1}{2}*\frac{1}{6}*(\frac{1}{3})^2+...$ The answer is $\frac{1}{4}+\frac{1}{8}=\frac{3}{8}$

However, this is wrong. Could someone please point out where I made my mistake? Thanks!

  • 0
    Which sum? I'm not sure what you mean. – 2017-01-31
  • 0
    My comment was in error, deleted. – 2017-01-31

2 Answers 2

2

This problem is probably best solving recursive techniques. Let me define a state $x$ to be the number of odd numbers you have already rolled in previous rolls.

If $x=1$, with probability $\frac{1}{6}$ you roll $4$ and transfer to state with value of $1$ (ending with $4$), with probability $\frac{1}{2}$ you roll odd number and transfer to state with value of $0$ (ending with $2$ odd) and with probability $\frac{2}{6}$ transfer to state $x=1$ with value $v_{1}$. Hence $v_{1}=\frac{1}{6}\cdot1+\frac{1}{2}\cdot0+\frac{2}{6}\cdot v_{1}$, or, equivalently, $v_{1}=\frac{1}{4}$.

If $x=0$, similar argument gives probabilities $\frac{1}{6}$, $\frac{1}{2}$ and $\frac{2}{6}$ of transferring to $4$, $v_{1}$ and $v_{0}$ respectively. Hence $v_{0}=\frac{1}{6}\cdot1+\frac{1}{2}v_{1}+\frac{2}{6}\cdot v_{0}$. Substituting and solving gives me $v_{0}=\frac{7}{16}$.

  • 0
    wow! Great explanation! However, is it possible to solve it in the way I was trying to solve it? – 2017-01-31
  • 0
    @user406996 I don't know. But besides the answer, I was hoping my explanation would lead you to appreciate the recursive methods and that these can be useful. – 2017-01-31
  • 0
    It was a good solution, I didn't see the recursive way! :) – 2017-01-31
3

There are six equally likely outcomes for rolling a fair die: $1,2,3,4,5,6.$ But rolling $2$ or $6$ is irrelevant for this question, so it's best to ignore them and consider only the four equally likely outcomes $1,3,4,5.$ In other words, a "trial" consists of rolling the die repeatedly until it comes up something different from $2$ or $6.$

Restating the question: if each trial results in one of the equally likely outcomes $1,3,5,$ or $4,$ what's the probability that $4$ comes up on one of the first two trials? The answer is clearly $$1-\left(\frac34\right)^2=\boxed{\frac7{16}}.$$

The trouble with your attempt is that you don't consider the possibility of getting $2$ or $6$ on the first roll. Here is a corrected version of your solution:

The probability of rolling $m$ twos-or-sixes, then a four, is $$\left(\frac13\right)^m\cdot\frac16.$$ The probability of rolling $m$ twos-or-sixes, then an odd then $n$ two-or-sixes, and then a four, is $$\left(\frac13\right)^m\cdot\frac12\cdot\left(\frac13\right)^n\cdot\frac16.$$ The probability or rolling a four before rolling two odds is $$\frac16\sum_{m=0}^\infty\left(\frac13\right)^m+\frac1{12}\sum_{m=0}^\infty\sum_{n=0}^\infty\left(\frac13\right)^{m+n}=\frac14+\frac3{16}=\boxed{\frac7{16}}.$$

  • 0
    Answer updated to include a corrected version of your solution. – 2017-01-31