How many tries to roll a 6?

Question

Suppose you roll a (fair, 6-sided, perfectly ordinary) die repeatedly until you roll a 6. As is well known, the expected (i.e., long-term average over many trials) number of rolls required is 6.

Now suppose we ask this question conditional on never having rolled any odd numbers. That is, suppose you learn that someone has followed the procedure above, and that they rolled only even numbers up to the point where they rolled that 6. What is the expected number of rolls they took?

(That is: What is the expected number of rolls taken to roll a 6, conditional on having got no odd numbers in the process of rolling that 6? The question is not anything like "what's the expected number of rolls next time, given that you didn't get any odd numbers last time?". I've added this paragraph and slightly reworded the question above because it seems from comments that some readers interpreted the question in an unintended way.)

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This question is interesting because there is an extremely tempting wrong answer.

Some history: The question seems to have first been asked by Elchanan Mossel while teaching an undergraduate probability course at the University of Pennsylvania in 2014-15. Gil Kalai posted it on his blog as part of his Test Your Intuition series, and then posted a followup. There is interesting discussion in the comments of both. And here is a nice little article about the question, in PDF form.

WARNING: All the links in the paragraph above may take you to places that give away some or all of the solution. If you want to solve the puzzle for yourself, don't follow those links until you've done it.

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[See this Meta question for some relevant PSE-specific context, which in particular explains why this is community wiki. An older question has been merged into this one, and it's possible that some details of answers or comments may look odd as a result.]

Answer 1

The answer is indeed...

         1¹⁄₂ rolls

            ...because the question is equivalent to...

What is 1 more than the average number of consecutive rolls of only 2s and/or 4s before any other roll terminates the series?

One way to see this equivalence is to note that only 2/4 streaks terminated by 6 are considered. (A first-roll 6 terminates a 2/4 streak of length 0.) Yet the average length of those streaks is the same as of streaks terminated by 1, 3 or 5. Collect all such streaks, including those terminated by 6, and the result is all streaks of 2/4, each with the same expected length.

  Calculations:

    E = expected number of consecutive even rolls up to and including the first 6
     = (expected number of consecutive 2s and/or 4s) + (one more roll of 6)
     = T + 1
where
    T = expected number of consecutive 2s and/or 4s before some other roll
     = (probability of 2 or 4)   ×   (1 for this roll + expected subsequent 2s and/or 4s)
     = ¹⁄₃ (1 + T)
   so
    T = ¹⁄₂
   thus
    E = T + 1
     = 1¹⁄₂

Answer 2

This surprisingly beguiling puzzle may also be solved with a surprisingly unsophisticated approach. Symmetry, by itself, predicts the average length of evens-only sequences ending with 6 to be...

                  ...1 ¹⁄₂  throws.

Start with T  many random throws: 2153664315121226553111444142566363625461525 . . 3644464461

Sift them into 4 groups that, due to symmetric criteria, are each expected to include ^T⁄₄  throws.

Every 1 and any preceding 2s/4s:   21......1.121......1114441.............1... . . .........1
Every 3 and any preceding 2s/4s:   ...3..43..........3............3.3......... . . 3.........
Every 5 and any preceding 2s/4s:   ..5......5......55........425......25...525 . . ..........
Every 6 and any preceding 2s/4s:   ....66.......226.............66.6.6..46.... . . .64446446.

The last group, of 6s and their preceding 2s/4s, precisely contains every evens-ending-with-6 sequence considered by the puzzle.   The number of such sequences is just the original number of 6s, as every 6 is in exactly one such sequence.   From 6’s symmetry with the other 5 equally-likely throws —1, 2, 3, 4 and 5 — that number is expected to be ^T⁄₆  sequences.

Thus...

(Average length)   =   (number of throws in these sequences)   /   (number of sequences)

         =   ^T⁄₄ throws   /   ^T⁄ ₆ sequences

         =   1 ¹⁄₂  throws /sequence.

Notes

The expected number of rolls seems suspiciously small but, as seen among the example throws above, it reflects the likelihood that most evens-ending-with-6 sequences will be just 1 roll long, a lone 6. This does make sense after all by noting that each sequence-ending 6 has ⁴⁄₆  chance of following a 1, 3, 5 or 6(!) that allows no prior rolls to count in that 6’s sequence.

The condition “all throws gave even numbers” is a potent red herring that suggests 6s have equal status to 2s and 4s.   Their status is not equal, however, after retaining all 6s originally rolled but discarding the ³⁄₄  of 2s/4s that did not lead to 6s.

This solution’s approach does have a loose end, literally, as commented by Artur Kirkoryan. Any stretch of T  throws could end with a streak of 2s/4s that might extend to become a qualifying sequence of any length if allowed to continue. That comment and Deusovi’s reminder to take density into account lead to the realization that possible 2s/4s loose ends would need an infinite expected length to make a difference because any bounded expectation would have a vanishing effect as T  increases without limit. This further leads to acknowledgment that the evens-ending-with-6 sequences in question are only assumed, not proven, to have a finite expected length. Presumption of their nonetheless having a bounded expectation forces any 2s/4s loose ends to also have a bounded expectation and thus not undermine this solution’s approach.

This solution ensued from surprise at the match between the 4 kinds of non-2s/4s throws and the ¹⁄₄  fraction of undiscarded throws exhibited by a Lisp routine’s simulating a million throws at a time.

 ( defun RollEm (many)
   ( let ( (i many) (sixes 0) (streakSum 0) (streak 0) )
     ( while  (>= (setq i (1- i)) 0)
              ( pcase  (1+ (random 6))
                       (   2  ( setq streak   (1+ streak)             ) )
                       (   4  ( setq streak   (1+ streak)             ) )
                       (   6  ( setq sixes    (1+ sixes)              )
                              ( setq streakSum (+ streakSum streak 1) )
                              ( setq streak     0                     ) )
                       ( else ( setq streak     0                     ) )
     )        )
     ( insert ( format "\n%d (%d%%) undiscarded throws / %d streaks  =  %.3f\n"
                       streakSum
                       ( / ( + (* 200 streakSum) many) (* 2 many) )
                       sixes
                       ( / (float streakSum) sixes )
 ) ) )        )

 (RollEm 1000000)     249372 (25%) undiscarded throws / 166527 streaks  =  1.497
        ""            250234 (25%) undiscarded throws / 166972 streaks  =  1.499
        ""            249503 (25%) undiscarded throws / 166256 streaks  =  1.501
        ""            250777 (25%) undiscarded throws / 166953 streaks  =  1.502
        ""            249947 (25%) undiscarded throws / 166791 streaks  =  1.499

Answer 3

Let $X_n$ be the event that the dice takes $n$ rolls to get the first 6, given all the rolls are even. Let $A_n$ be the event that it takes $n$ rolls to get the first 6, and let $B$ be the event that all rolls up to the first 6 are even.

$P(X_n)=P(A_n|B)=\dfrac{P(B|A_n)P(A_n)}{P(B)}$ (using Bayes' theorem)

Now:

$P(A_n)=\dfrac{1}{6}\cdot\left(\dfrac{5}{6}\right)^{n-1}$ (since the first $n-1$ rolls are not 6 and the last roll is)

$P(B)=\displaystyle\sum^\infty_{i=0}\dfrac{1}{6}\cdot\left(\dfrac{1}{3}\right)^{i}$ $=\dfrac{1}{6}\cdot\displaystyle\sum^\infty_{i=0}\left(\dfrac{1}{3}\right)^{i}$ $=\dfrac{1}{6}\cdot\dfrac{3}{2}=\dfrac{1}{4}$ (since at each stage there is a 1 in 3 chance that the roll is 2 or 4 and a 1 in 6 chance of getting a 6 at each roll)

$P(B|A_n)=\left(\dfrac{2}{5}\right)^{n-1}$ (since the last roll is even but for all the other rolls there are 2 even outcomes and 3 odd outcomes)

So $P(A_n|B)$ $=\dfrac{\dfrac{1}{6}\cdot\left(\dfrac{2}{5}\right)^{n-1}\cdot\left(\dfrac{5}{6}\right)^{n-1}}{\dfrac{1}{4}}$ $=\dfrac{2}{3}\cdot\dfrac{2^{n-1}\cdot5^{n-1}}{5^{n-1}\cdot6^{n-1}}$ $=\dfrac{2}{3}\cdot\left(\dfrac{1}{3}\right)^{n-1}$

Now, we wish to find

\begin{align} \sum^{\infty}_{i=1}i\cdot P(X_i) &=\sum^{\infty}_{i=1}i\cdot\dfrac{2}{3}\cdot\left(\dfrac{1}{3}\right)^{i-1} \\ &=2\cdot\sum^{\infty}_{i=1}i\left(\dfrac{1}{3}\right)^{i} \\ &=2\cdot\sum^{\infty}_{i=0}\left(\left(\dfrac{1}{3}\right)^{i}\cdot\sum^{\infty}_{j=1}\left(\dfrac{1}{3}\right)^{j}\right) \\ &=2\cdot\sum^{\infty}_{i=1}\left(\dfrac{1}{3}\right)^{i}\cdot\dfrac{3}{2} \\ &=3\cdot\sum^{\infty}_{i=1}\left(\dfrac{1}{3}\right)^{i} \\ &=\dfrac{3}{2} \end{align}

So it takes on average 1.5 rolls for the dice to get a 6, given that all the rolls are even.

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Gamow wants me to keep my previous solution: if you want to see it, here it is. It is wrong because the definition of $B_n$ is faulty - for example, a sequence such as 26143 isn't counted in $B_5$ when it should be, and a sequence such as 24(16) is counted in $B_2$ when it shouldn't be, since we are given that all the rolls up to the first six are even, not that the rolls up to a certain point are even.

Answer 4

I believe the answer is

$\frac{3}{2}$

This is computational calculation, so it is not statistical answer.

Here is the simple code

It is for the people who try to find it probabilistically.

Here is the probabilistic solution:

First of all, we know that the probability of getting 6 on the first roll is $\frac{1}{6}$, then getting 6 after an even roll is $\frac{2}{6}\frac{1}{6}$, and so on as the equation below:

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

so if we calculate all possible solutions and sum them we will get the probability set among all other possibilities (such as getting odd after getting some even):

$\sum_{1}^{\infty }\left (\frac{1}{6} \right )\left (\frac{2}{6} \right )^{n-1}=0.25$

So if we multiply this by $4$ we will normalize the probability and find the actual expected value of getting $6$ given that all the rolls are even as below such as;

1. Getting $6$ in the first roll expected value:

$E_1=4\left (\frac{1}{6} \right )\left (\frac{2}{6} \right )^{1-1}=\frac{2}{3}$

2. Getting $6$ in the second roll (requires not getting $6$ in the first roll)

$E_2=\left [1-4\left (\frac{1}{6} \right )\left (\frac{2}{6} \right )^{1-1} \right ]+4\left (\frac{1}{6} \right )\left (\frac{2}{6} \right )^{2-1}=\frac{5}{9}$

In general, we can generalize the formula as below,

$E=\sum_{1}^{n}(1-A_{n-1})+B_n$

$A_{n}$ represents the sum of previous normalized probabilities and $B_{n}$ is the normalized probability of getting 6 at that roll.

firstly:

$A_n=\sum_{1}^{n}(4\times\frac{1}{6}\frac{2}{6}^{n-1})=1-3^{-n}$

secondly,

$B_n=4\times\frac{1}{6}\frac{2}{6}^{n-1})=2\times 3^{-n}$

so

$E_{2,\inf}=\sum_{2}^{\inf}(1-1-3^{1-n})+2\times 3^{-n}=\frac{5}{6} $

lastly, we are going to add the first result since the general sum will work after 1:

$ E=E_1+E_{2,\inf}=\frac{2}{3}+\frac{5}{6}=\frac{3}{2}$

So if we sum all possible outcome expected values, we will get the actual expected value of getting $6$ given that all the rolls are even:

$E =\sum_{1}^{\infty}E_n=\frac{3}{2}$

Answer 5

Somewhat related to Oray's solution.

The answer is

$\frac{3}{2}$

Why?

The average chance to get a 6 can be modelled using an infinite sum:
$$\sum_{n=0}^\infty \frac{1}{3} \times \frac{2}{3} ^ r$$
Where $\frac{1}{3}$ is the chance to get a 6, $\frac{2}{3}$ is the chance to not get a 6 and $r$ is the number of fails.
This is equivalent to a geometric series with $a_0 = \frac{1}{3}$ and $r=\frac{2}{3}$
$$\begin{align}S_\infty &= \frac{ar}{1-r}\\ &= \frac{\frac{1}{3} \times \frac{2}{3}}{1-\frac{2}{3}}\\ &= \frac{\frac{1}{3} \times \frac{2}{3}}{\frac{1}{3}}\\ &= \frac{2}{3}\end{align}$$
The average number of rolls needed is just $\frac{1}{S_\infty} = \frac{1}{\frac{2}{3}} = \frac{3}{2}$

Answer 6

Going by definition of what is the expected number we get

$$ \sum_{n=1}^{\infty} n \cdot 1/3 \cdot (2/3)^{n-1} = 3 $$
where 1/3 part is the probability to get "6" after reaching the n'th roll and $ (2/3)^{n-1} $ - probability to reach that step.

Edit: well, it looks like the "reroll" [only the steps that produce odd number] and "reset" [the whole sequence] interpretations are not equal and the later is the intended meaning of the question. Under the later the probability to roll "6" on each step is in fact

not equal to 1/3 as in the first case.
Let $p_1$ be the conditional probability to roll "6" (equal for all steps because after roll of "2" or "4" all outcomes are still conditionally as probable as before the roll, just with the chain of rolls being longer by one). Conditional probability to roll "2" or "4" would be of course $1-p_1$.
Let "A" be the event that the first roll is "6", "B" be the event that the sequence ended on "6" and did not contain any odd numbers.
$$ \begin{align} p_1 & = P(A|B) \\ & = P(A \& B)/P(B) \\ & = (1/6)/(P(B)) \end{align}$$ $$ P(B)= \sum_{n=1}^{\infty} 1/6 \cdot (2/6)^{n-1}=1/4 $$ $$ p_1=(1/6)/(1/4)=2/3 $$ which is indeed rather counterintuitive.

The final answer in this approach:

$$ \sum_{n=1}^{\infty} n \cdot p_1 \cdot (1-p_1)^{n-1}=\sum_{n=1}^{\infty} n \cdot 2/3 \cdot (1/3)^{n-1}=3/2$$

Answer 7

I was struggling to get my head around this so I decided to test it programatically to see if I had misunderstood the question. Here's the code:

#!/usr/bin/perl

use strict;
use warnings;

my $die_size = 6;
my $number_of_runs = 100000;
my $total_count = 0;

sub count_rolls {
    my $number_of_rolls = 0;
    my $number_rolled = 0;
    while ($number_rolled != $die_size) {
        $number_of_rolls++;
        $number_rolled = 1 + int rand($die_size);
        if ($number_rolled % 2 == 1) {
            return count_rolls();
        }
    }
    return $number_of_rolls;
}

for (1..$number_of_runs) {
    my $roll_count = count_rolls();
    $total_count += $roll_count;
}

my $mean_count = $total_count / $number_of_runs;
print "$mean_count\n";

Try it online!

The result this gives is:

1.5

Hopefully this is helpful to anyone else who is a bit confused here.