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I encountered a tricky homework question

We toss a dice repeatedly until the first 6. Given that all the results are even, what is the probability that the total number of tosses is 1?

There are two equally convincing yet incompatible answers:

  1. Since all the results are even, the possible outcomes of the first toss are $\{2, 4, 6\}$. For the total number of tosses to be 1, the outcome must be 6, so the probability is $\frac{1}{3}$.

  2. By Bayes' theorm,

$$\begin{equation}\begin{aligned} &P(\text{the total number of tosses is 1}|\text{all the results are even}) \\ =& \frac{P(\text{all the results are even | the total number of tosses is 1}) \times P(\text{the total number of tosses is 1})}{P(\text{all the results are even})} \\ =& \frac{1 \times \frac{1}{6}}{\frac{1}{6} + \frac{1}{3}\cdot\frac{1}{6} + \frac{1}{3}^2\cdot\frac{1}{6} + \cdots} \\ =& \frac{2}{3}. \end{aligned}\end{equation} $$

To elaborate, in the expression for $P(\text{all the results are even})$, $\frac{1}{3}$ is the probability of getting an even number which doesn't end the tossing procedure (i.e. 2 or 4), whereas $\frac{1}{6}$ is the probability of getting a 6.

Which one is correct?

nalzok
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    The second is the correct answer. While true that the possible outcomes for the first toss are originally $1,2,3,4,5,6$ and these are equally likely when not looking at a conditional probability... and while true that the possible outcomes become $2,4,6$ when looking at the conditional probability that it was all evens rolled... it is not true that those three outcomes are still equally likely for the first roll. Your analysis in the second attempt proves this. – JMoravitz Sep 17 '21 at 16:08
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    For intuition, a hand-wavy explanation might be that since we know the experiment ended with a 6, and it ended "before something bad happened" and we rolled an odd number, we are more inclined to believe that the 6 came early so as to not give us as much of a chance to let an odd be rolled. – JMoravitz Sep 17 '21 at 16:10
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    On another note, I found a similar type question you may be interested in: https://math.stackexchange.com/questions/2779681/expected-number-of-die-rolls-to-get-6-given-that-all-rolls-are-even – Igor Sep 17 '21 at 16:13
  • @JMoravitz Then, consider the alternative problem. A die is rolled repeatedly until a 6 shows. All odd rolls are ignored (i.e. not counted). Then the probability that the number of counted rolls until the 1st 6 equals $(1)$ would be $(1/3)$. If you agree, that implies that this alternative problem is different from the offered problem. Interesting. – user2661923 Sep 17 '21 at 17:14

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