So earlier today I came across Elchanan Mossel's Dice Paradox, and I am having some trouble understanding the solution. The question is as follows:

You throw a fair six-sided die until you get 6. What is the expected number of throws (including the throw giving 6) conditioned on the event that all throws gave even numbers?

Quoted from Jimmy Jin in "Elchanan Mossel’s dice problem"

In the paper it goes on to state why a common wrong answer is $3$. Then afterwards explains that this problem has the same answer to,

"What is the expected number of times you can roll only $2$’s or $4$’s until you roll any other number?"

I don't understand why this is the case. If the original problem is asking for specifically a $6$, shouldn't that limit many of the possible sequences?

I also attempted to solve the problem using another method, but got an answer different from both $3$ and the correct answer of $1.5$.

I saw that possible sequences could have been something like:

$$\{6\}$$ $$\{2,6\}, \{4,6\}$$ $$\{2,2,6\}, \{2,4,6\}, \{4,2,6\}, \{4,4,6\}$$ $$\vdots$$

To which I set up the following summation and solved using Wolfram Alpha:

$$\text{Expected Value} =\sum_{n=1}^\infty n\left( {\frac{1}{6}} \right)^n 2^{n-1} = 0.375$$ Obviously this is different and probably incorrect, but I can't figure out where the error in the thought process is.

Any help on understanding this would be greatly appreciated.

A blog post discussing the problem can be found here.

  • $\begingroup$ really really late reply, but can someone explain this? repl.it/repls/UsedScalyTurtledove $\endgroup$ – Quintec Dec 1 '17 at 2:40
  • $\begingroup$ @thecoder16 Your code should not throw away odd rolls. Instead, if an odd roll is seen, you should declare the sequence in progress to be unsuccessful, and omit it from your running total when a 6 is eventually rolled. See the edit to my answer. $\endgroup$ – grand_chat Mar 3 '18 at 2:33
  • $\begingroup$ @grand_chat yeah, thanks, I solved this problem a while ago :) $\endgroup$ – Quintec Mar 3 '18 at 18:36
  • $\begingroup$ hey there--this is Jimmy Jin. I wanted to thank everyone for pointing out the confusing parts! I've since updated my writeup. My original goal in using the "subtraction" idea was to explain why most people get the wrong answer by oversimplifying to the space of a single die roll. But as @MichaelHardy pointed out below, this is actually the right principle except on the enriched space of a sequence of die rolls. $\endgroup$ – gogurt May 29 '18 at 23:36

When you roll a die until $6$ appears, you can represent the sample space as all possible finite sequences from the set $\{1, 2, 3, 4, 5, 6\}$ ending in $6$, with probability of any sequence of length $k$ being $(1/6)^k$. The original question is asking for

$(1)$ the expected length of a sequence conditional on all throws being even.

You've correctly enumerated all sequences from $\{2,4,6\}$ that end in $6$, and calculated the sum $$\sum_{n=1}^\infty n\left( {\frac{1}{6}} \right)^n 2^{n-1} = 0.375$$ properly, but you forgot to divide this by the probability of the event you are conditioning on, which is $$ \sum_{n=1}^\infty\left(\frac16\right)^n2^{n-1}=1/4. $$ So your approach does yield the correct answer, namely $4\times 0.375=1.5$.

The act of conditioning on all throws being even is tantamount to restricting the sample space to all possible finite sequences from the set $\{2, 4, 6\}$ that end in $6$, and rescaling the probability function (by a factor $4$) so that this new sample space has total mass $1$.

As for the Jin paper, he claims that the original question $(1)$ is equivalent to

$(2)$ the expected number of times you can roll only $2$'s or $4$'s until you roll a $6$.

I disagree with $(2)$; it is incorrect to compute an unconditional expectation, as he just explained in his previous paragraph. He still needs an expectation conditional on some event, and I would argue the original question $(1)$ is equivalent to computing

$(2')$ the expected number of times you can roll only $2$'s or $4$'s until you roll any other number, given that the other number is $6$.

The reason is that conditioning on the event "the other number is $6$" results in the same restricted sample space as before. In fact his subsequent argument that it suffices to compute the unconditional expectation

$(3)$ the expected number of times you can roll only $2$'s or $4$'s until you roll any other number.

(i.e., that $(2') = (3)$, which is what he actually proves) is relevant only if we intend $(2')$ instead of $(2)$.

EDIT: Here's a Python simulation of the experiment, based on code provided by @thecoder:

import random

times = 0 #number of times a successful (all-even) sequence was rolled
rolls = 0 #total of all number of rolls it took to get a 6, on successful sequences
curr = 0
alleven = True

for x in range(0, 100000):

  num = random.randint(1,6)
  if num % 2 != 0:
    alleven = False
    if num == 6:
      if alleven:
        times += 1
        rolls += curr + 1
      curr = 0
      alleven = True
      curr += 1

print(rolls * 1.0 / times)
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  • 1
    $\begingroup$ The division by $1/4$ is what I was missing. Thank you $\endgroup$ – WaveX Oct 9 '17 at 3:32

Part of Jimmy Jin's explanation is right, but his account of "the subtraction fallacy" is unfortunate. Conditioning does consist of discarding part of the sample space.

Suppose one throws a die repeatedly and gets this sequence: $$ 2, 5, 3, 2, 1, 4, 3, 6 $$ Next suppose one excludes the odd numbers: $$ \require{cancel} 2, \xcancel5, \xcancel3, 2, \xcancel1, 4, \xcancel3, 6 $$ One then has this sequence: $$ 2,2,4,6 $$ The average length of this sequence will be $3.$

But now suppose you get this sequence: $$ 2, 5, 3, 2, 1, 4, 3, 6 $$ Then try again and get this: $$ 3,2,5,2,1,6 $$ Then try again and get this: $$ 2,4,4,2,6 $$ Then discard the first two since one got odd numbers before getting a $6:$ $$ \xcancel{2, 5, 3, 2, 1, 4, 3, 6} $$ $$ \xcancel{3,2,5,2,1,6} $$ But keep that last one: $$ 2,4,4,2,6 $$ Among the ones you keep the average length is $3/2,$ not $3.$

You are in fact excluding that part of the sample space in which at least one odd number appears before the first $6,$ and in what's left of the sample space the average position of the first $6$ is $3/2.$ Thus Jimmy Jin is wrong about the "subtraction fallacy."

However, he is right that the average position of the first $6,$ given that only even numbers appear, is $3/2.$

In sequences pared down by excluding all the odd numbers that appear, the average position of the first $6$ is $3.$

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  • $\begingroup$ +1: Nice clarification. $\endgroup$ – copper.hat May 14 '18 at 4:10
  • 2
    $\begingroup$ Just wanted to thank you Michael for pointing this out (this is Jimmy). My intention was to highlight the incorrect deletion of the sample space treating the problem through the lens of a single die roll. I'm in complete agreement that the correct perspective is the sequential one. I've since updated the writeup to clear this up! $\endgroup$ – gogurt May 30 '18 at 0:12
  • $\begingroup$ How do you conclude that the average length in the restricted sample space is 3/2? $\endgroup$ – Roland May 28 '19 at 8:44

I think I found another reasonable explanation (would appreciate feedback on this) that clicks in better in my brain:

I also tried adding up the lengths with the probabilities, and I got this:

$$ \sum_{n=1}^{\infty}n \frac{1}{6} \left(\frac{2}{6}\right)^{n-1} = \frac{3}{8} $$

Here, $n$ is the length of the sequence itself, $1/6$ would be the probability of the last throw being a 6, and $(2/6)^{n-1}$ is the probability that every previous throw is a $2$ or a $4$. However, what I was missing is that you need not count anything where previous throws are $1$, $3$, or $5$. So, this becomes:

$$ \sum_{n=1}^{\infty}n \frac{1}{6} \left(\frac{2}{3}\right)^{n-1} = \frac{3}{2} $$

where $(2/3)^{n-1}$ is the probability that previous throws are $2$ or $4$, but not $6$. I don't know if my reasoning is correct, but I can follow this a bit better than other explanations.

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Here is a correct and intuitive way to understand it. Suppose you roll a die about a million times and record the rolls as one long string of numbers. Every $6$ in there is the end of a legal sequence of some length $L$. The probability that the number before the $6$ is $1,3,5$ or $6$ is $\frac46$. In this case the $6$ is the end of a legal sequence of length $1$, i.e., only the $6$ itself. So the probability that $L=1$ is $\frac23$.

That leaves probability $\frac13$ for a longer sequence with $L$ being $2$ or more when there is at least one $2$ or $4$ preceeding the $6$. The expected length is thus $L = \frac23 \cdot 1 + \frac13 \cdot (2 + TF)$ where $TF$ is the expected length of the run of further $2$’s and $4$’s preceeding the $2$ or $4$ we already found, including run length $0$ if there are no further $2$’s and $4$’s.

Next we determine TF, the expected length of a run of $2$’s and $4$’s only. The probability that the run length is $0$ is $\frac23$. The probability that the run length $1$ is $\frac13 \cdot \frac23$ since we first need a $2$ or $4$, and then something that is not a $2$ or $4$. That leaves probability $\frac13 \cdot \frac13$ for a run length of $2$ or more $2$’s and $4$’s. For the “or more” part we can use $TF$ itself again. This leads to $TF = (\frac23 \cdot 0 + \frac13 \cdot \frac23 \cdot 1 + \frac13 \cdot \frac13(2 + TF)$. Rewriting we get $\frac89TF = \frac29 + \frac29 = \frac49$ so that $TF = \frac12$. Substituting $TF$ in $L$ we get expected sequence length $L = \frac23 + \frac13 \cdot (2 + \frac12) = \frac23 + \frac56 = 1.5$.

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