How to get a uniform 1 out of 7 chance, using only a coin and a 6-sided die [closed]

Question

There are seven days in a week. Let's say you want to pick one at random, such that each day has an equal chance of being picked as each other day. In other words, the goal is to obtain a uniform random 1/7 chance of picking any of the seven days.

But all you have is a coin and a six-sided die (not dice, you don't have 2 dice, you have only 1 die). What procedure of coin-flipping and/or die-rolling should you use to get a 1/7 chance?

(EDIT: I originally intended a coin that can only give a 1/2 result. However, there was a decent out-of-the-box answer about rollable coins that I found pretty useful.)

I know of two ways to do this, but they're slightly out-of-the-box solutions. I consider them valid solutions but want to see what you guys think. Of course, I'll let people try to answer first before revealing my solutions.

Hint: don't try anything like trying to pick a random number in your head. The only things you can use are the coin and die.

EDIT: I'm going to post the four answers I know of in the most compact and simple way (in my humble opinion). Everyone who answered so far has given the equivalent of one of these answers:

Method 1.

Roll the die and flip the coin. This gives 12 possibilities: 1T, 2T, 3T, 4T, 5T, 6T, 1H, 2H, 3H, 4H, 5H, 6H. Assign seven days of the week to seven of those values. If the result is one of the remaining five, just repeat the process until you get a valid value. This has a 5/12 (41.67%) chance of repeating.

Method 2.

Flip the coin three times (and be sure to record it in order of course). This gives 8 possibilities: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Assign seven days of the week to the first seven of those values. If your result is the eighth value, repeat the process until you get a valid value. This has a 1/8 (12.5%) chance of repeating.

Method 3.

Roll the die twice (and be sure to record them in order because a 2 and 5 is not the same thing as a 5 and 2 in the sample space!). This gives 36 possibilities: 1&1, 1&2, 1&3, 1&4, 1&5, 1&6, 2&1, 2&2, 2&3, 2&4, 2&5, 2&6, 3&1, 3&2, 3&3, 3&4, 3&5, 3&6, 4&1, 4&2, 4&3, 4&4, 4&5, 4&6, 5&1, 5&2, 5&3, 5&4, 5&5, 5&6, 6&1, 6&2, 6&3, 6&4, 6&5, 6&6. Assigned 7 days of the week, but do that 5 times. This uses up 35 values. If your two rolls results in the 36th value, just repeat the process until you get a valid value. This has a 1/36 (2.78%) chance of repeating, which is much smaller than the previous two methods.

Method 4:

Roll the coin. If it's a 7-sided rollable coin, then just see which side ends up closest to some arbitrary direction, like north, or arbitrary marker, such as a stick in the ground. If it's a circular coin, you could theoretically mark the coin to divide it into 7 sectors and then see which sector points towards your arbitrary thing. There was also an idea of constructing a Roulette Wheel with 7 sectors. I know I said you only have the coin and die, but then you could just label some sectors on the ground and use that as your roullette wheel, and presumably find some way to confine the coin there or just roll again if it escapes. So I think this idea too is valuable.

Some concluding thoughts...

Now that I think about it, you could extend the multiple rolls/tosses. If you flipped the coin 6 times, that's 64 possibilities, and 63 is evenly divisible by 7 so you can assign days of the week to that, and just repeat if u get the 64th value (1 out of 64 chance = 1.56%!). You could roll the die 4 times for 1,296 possibilites. 1,295 is divisible by 7, so assign days of the week evenly to all those values, and just repeat if u get the 1,296th value (1 out of 1,296 chance = 0.08%!!). You can probably go even further with them, but it's a trade off of doing more rolls/tosses, which takes a longer time and might take longer to compute the result, than just taking the 1/36 chance of having to repeat via the two-roll method, for example.

Answer 1

There is a way that doesn't use the dice (3+ coin-flips):

• Flip the coin three times and record 1 or 0 for heads or tails.
• If all three numbers are zero start again.
• Otherwise read the 1s and 0s as a binary number. Its value will be from 1 to 7 inclusive with each number having an equal probability of occurring.

Here's a summary of Mike's answer (2+ dice-rolls):

• Roll the dice twice call the results 𝑎 and 𝑏. If both results are 1 roll again.
• Otherwise calculate (𝑎 - 1) × 6 + (𝑏 - 1) Divide that number by 7 to find the remainder. The remainder is in the range 0 to 6 inclusive, add 1 if you want 1 to 7.

Answer 2

Roll the die twice, to generate a random number between 0 and 35. Specifically, subtract one from each roll, and have the first roll be the "tens" digit of a number in base 6, and the second roll be the ones digit.

If the result nonzero, return the remainder of the result (mod $7$). If the result is zero, then repeat until it is nonzero.

This gives one of the numbers $0$-$6$ with equal probability, which correspond to the seven days of the week.

This process will take $\frac2{1-\frac1{36}}\approx 2.06$ rolls on average.

Answer 3

Here's an answer that may take more steps to complete but is a lot more intuitive to think about:

Roll the die and flip the coin simultaneously. If the coin lands on heads, read the die value. If the coin lands on tails and the die lands on 1, get the value 7. Else, repeat the process.

Answer 4

This is a slightly lateral-thinking answer, and is definitely against the spirit of the question, if not its letter. But there's a lot to recommend this approach, enough that I felt it was worth providing.

Now, I need to begin by admitting that this approach is only convenient if you happen to live in Westeros, from Game of Thrones. (And if you do, my condolences).

However, to make up for that requirement, it's worth pointing out that this approach requires no maths or probability calculations at all. Additionally, it only ever requires a single roll. No rerolls, no use of base-6 numbers, etc. All of which is probably a bit of a relief to you if you do happen to live in Westeros, as statistically you've probably had no mathematical schooling at all. (again, my condolences)

But those of us living in the real world can use this method too; it just requires us to first purchase a very particular (and slightly expensive) licensed coin.

In the mythos of the show, one common coin is the Copper Star of Robert Baratheon

Notably, the "tails" side of the coin depicts a seven-pointed star.

So all one needs to do in order to generate an evenly distributed 1/7 chance with a die and a coin is:

1. Number the points of that star from 1-7.
2. Roll the die and the coin together into the rolling area.
3. If the coin lands face-up, flip it over to be face-down (without looking!).
4. Imagine a line drawn from the center of the coin to the center of the die. See which of the seven points of the star on the coin is closest to that line, and that's your result.

See? Probability can be easy, if you're not wedded to solving it with maths!

Answer 5

In general, you can roll any real probability with a coin. Write out the decimal expansion of the probability, say its .1010101 etc. Have the first coin flip determine the first digit after the decimal, then the second flip determines the second digit, etc. If at any point the number determined by the coin flips can be proven to be larger or smaller than the probability then we are done and we choose based on larger or smaller.

Answer 6

Throw the die into the corner of a box, giving you 8 possibilities. Add together the scores of the three exposed faces around the exposed corner:

     +---+                   +---+
   6 | 3 | 9              14 | 3 | 11
 +---+---+---+           +---+---+---+
 | 2 | 1 | 5 |           | 5 | 6 | 2 |
 +---+---+---+           +---+---+---+
   7 | 4 | 10             15 | 4 | 12
     +---+                   +---+

Now divide the score by 7 and take the remainder. Every number 0-6 occurs once, except 0, which occurs twice. Roll again on 14.

7 (or 14)-->0    (Roll again on 14.)   
       15-->1 
        9-->2  
       10-->3
       11-->4  
       12-->5 
        6-->6       

Answer 7

Generalizable suboptimal incremental-roll approach

[Gotta present this after being inspired by 2012rcampion's answer to a different question and by FarazMasroor's unacknowledged partial answer here, which would have benefited from more (and more accurate) detail. This also turned out to be eerily similar to an answer at Mathematics SE.]

Here's a close-to-optimal approach that adapts readily to any desired outcome distribution with any-sided dice. A random fraction between 0 and 1 is generated to fairly pick a point on an arbitrary inverse cumulative distribution function.

Start with fractions for $\,\textsf{Monday}=0/7\,$ through $\,\textsf{Sunday}=6/7\,$ in their base 6 expansions.

$$\begin{array}{rcr} \textsf{ Monday}= 0/7=.000000\textrm{...}\!\:_6 &~~& \textsf{ Friday}= 4/7=.323232\textrm{...}\!\:_6 \\ \textsf{ Tuesday}= 1/7=.050505\textrm{...}\!\:_6 && \textsf{ Saturday}= 5/7=.414141\textrm{...}\!\:_6 \\ \textsf{Wednesday}= 2/7=.141414\textrm{...}\!\:_6 && \textsf{ Sunday}= 6/7=.505050\textrm{...}\!\:_6 \\ \textsf{ Thursday}= 3/7=.232323\textrm{...}\!\:_6 \end{array}$$

Now begin rolling repeatedly, appending each result as the next digit of a lengthening base-6 fraction, treating each roll of ‘$6$’ as the digit $0$.

Keep rolling as long as the cumulative digit sequence matches one of the days' base-6 fractions listed above. When a digit fails to match, select the day with the largest fraction that is below what resulted from the rolls. The exception to these rules is when the first two rolls amount to $.00\!\:_6$, in which case call it a $\textsf{Monday}$ and be done.

This means that we are done after 2 rolls unless those rolls amount to some day's fraction's first 2 digits:

$$ .05\!\:_6 \qquad .14\!\:_6 \qquad .23\!\:_6 \qquad .32\!\:_6 \qquad .41\!\:_6 \qquad .50\!\:_6 \qquad $$

This gives a 6/36 = 1/6 chance of needing a 3rd roll. Each subsequent roll also leaves a compounded 1/6 chance of needing yet another roll. So:

$$\begin{array}{c} \textsf{average number of rolls} ~=~ 2 + \frac16 \Bigl( 1 + \frac16 \bigl( 1 + \frac16 ( 1 + \dots ) \bigr) \Bigr) ~=~ 2.2 \end{array}$$

Example

If the first 2 rolls were ‘$1$’ and ‘$5$’, the randomly-selected day would be $\textsf{Wednesday}$ because $.15\!\:_6$ does not begin any day's fraction and:

$$\textsf{Wednesday}=.141414\textrm{...}\!\:_6~<~.15\!\:_6~<~.232323\textrm{...}\!\:_6=\textsf{Thursday}$$

If those rolls were ‘$1$’ and ‘$4$’, however, we would roll again because $\textsf{Wednesday}$'s fraction begins with $.14\!\:_6$. If the next roll comes up ‘$0$’, which produces $.140\!\:_6$, the randomly-selected day would be $\textsf{Tuesday}$:

$$\textsf{Tuesday}=.050505\textrm{...}\!\:_6~<~.140\!\:_6~<~.141414\textrm{...}\!\:_6=\textsf{Wednesday}$$

Variation with alternating die rolls and coin flips

This is in the full spirit of the puzzle's title. Use a hybrid of bases 6 and 2, where the first fractional digit is in base 6, the next in base 2, then base 6, then 2 again, and so on. This is effectively base 12 with pairs of numerals.

$$\begin{array}{r} \textsf{ Monday}=0/7= ~.00~00~00~00~00~00~ ~~00~00~00~00~00~00~...\!\,_{6\backslash2} \\ \textsf{ Tuesday}=1/7= ~.01~40~30~50~11~21~ ~~01~40~30~50~11~21~...\!\,_{6\backslash2} \\ \textsf{Wednesday}=2/7= ~.11~21~01~40~30~50~ ~~11~21~01~40~30~50~...\!\,_{6\backslash2} \\ \textsf{ Thursday}=3/7= ~.21~01~40~30~50~11~ ~~21~01~40~30~50~11~...\!\,_{6\backslash2} \\ \textsf{ Friday}=4/7= ~.30~50~11~21~01~40~ ~~30~50~11~21~01~40~...\!\,_{6\backslash2} \\ \textsf{ Saturday}=5/7= ~.40~30~50~11~21~01~ ~~40~30~50~11~21~01~...\!\,_{6\backslash2} \\ \textsf{ Sunday}=6/7= ~.50~11~21~01~40~30~ ~~50~11~21~01~40~30~...\!\,_{6\backslash2} \end{array}$$

For a die roll, again treat ‘$6$’ as $0$. For a coin flip, call heads $0$ and call tails $1$.

Similar to before, repeatedly accumulate digits, alternately rolling and flipping, as long as the sequence of digits matches some day's (other than $\textsf{Monday}$'s) fraction:

$$\begin{array}{r} \textsf{average number of alternating rolls and flips} ~=~ \phantom{2.636363...} \\ 2 + \frac6{12} \Bigl( 1 + \frac16 \bigl( 1 + \frac12 \bigl( 1 + \frac16 \bigl( 1 + \frac12 ( 1 + \dots ) \bigr) \bigr) \bigr) \Bigr) ~=~ 2.636363... \end{array}$$

Comparison to an optimal(?) system with alternating die rolls and coin flips

A less-inspired (but optimal?) system uses an average of approximately 2.479 alternating rolls and flips. Without coin flips, the same approach matches the average of 2.057 die rolls posted in other solutions.

This table outlines a repeated 12-action pattern of whether or not a day will be selected by each successive die roll or coin flip.

\begin{array}{lcccccc} &\textrm{ } &\textrm{ } &\textrm{ } &\!\!\textrm{Potential} &\!\!\textrm{Left- } &\textrm{Proba- }\! \\[-.5ex] &\!\textrm{Probability} &\!\textrm{Continued }\! &\textrm{Combined } &\!\textrm{outcomes }\! &\!\!\textrm{over } &\!\textrm{bility } \\[-.5ex] &\!\textrm{of reaching} &\textrm{potential } &\!\textrm{potential } &\textrm{assigned } &\!\!\textrm{potential} &\textrm{of con- }\! \\[-.5ex] \!\!\!\textrm{Action} &\textrm{the action } &\textrm{outcomes } &\!\textrm{outcomes } &\textrm{to days } &\textrm{outcomes } &\textrm{tinuing } \\[1ex] \textrm{roll}_1 & 1 ~~ \!& & 6 & 0 & 6 & 6/6~ \\ \rlap{\textrm{(cycle begins)}} \\ \bf\textbf{flip}_2 & 1 ~~ \!&\bf 6 &\bf 12 &\bf 7 &\bf 5 &\bf 5/12 \\ \textrm{roll}_3 & 5/12 \!& 5 & 30 & 28 & 2 & 2/30 \\ \textrm{flip}_4 & 1/36 \!& 2 & 4 & 0 & 4 & 4/4~~\\ \textrm{roll}_5 & 1/36 \!& 4 & 24 & 21 & 3 & 3/24 \\ \textrm{flip}_6 & 1/288 \!& 3 & 6 & 0 & 6 & 6/6~~\\ \textrm{roll}_7 & 1/288 \!& 6 & 36 & 35 & 1 & 1/36 \\ \textrm{flip}_8 & 1/10368 \!& 1 & 2 & 0 & 2 & 2/2~~\\ \textrm{roll}_9 & 1/10368 \!& 2 & 12 & 7 & 5 & 5/12 \\ \textrm{flip}_{10}& 5/124416 \!& 5 & 10 & 7 & 3 & 3/10 \\ \textrm{roll}_{11}& 1/82944 ~\!& 3 & 18 & 14 & 4 & 4/18 \\ \textrm{flip}_{12}& 1/373248 \!& 4 & 8 & 7 & 1 & 1/8~~\\ \textrm{roll}_{13}& 1/2985984\!& 1 & 6 & 0 &\bf 6 & 6/6~~\\ \rlap{\textrm{(cycle repeats)}} \\ \bf\textbf{flip}_{14}&\bf1/2985984\!&\bf 6 &\bf 12 &\bf 7 &\bf 5 &\bf 5/12 \\ \textrm{roll}_{15}& 5/35831808\!&\bf 5 & 30 & 28 & 2 & 2/30 \\[-1ex] \textrm{flip}_{16}& \vdots &\vdots&\,\vdots&\vdots&\vdots&\vdots\; \\[-1ex] ~~~\vdots \end{array}

The row for $\bf\textbf{flip}_{14}$ , for example, shows that:
• $\bf 1/2985984$ is the probability that this coin flip will be required.
• $\bf 6$ equally-likely continuing outcomes, die rolls {1,2,3,4,5,6 }, are carried forward from the previous action, $\textrm{roll}_{13}$ .
• $\bf 12$ potential combined outcomes, {1,2,3,4,5,6 } × { H,T }, are now considered for this coin flip's 2 possible results, {Heads,Tails}.
• $\bf 7$ of the potential ordered outcomes are arbitrarily assigned to select a day:
    {1_^&H } = Mon, { 2_^&H } =Tues, ..., { 6_^&H } = Sat and {1_^&T } = Sun.
• $\bf 5$ unassigned potential ordered outcomes, { 2_^&T, 3_^&T, ..., 6_^&T }, remain left over to combine with the next action if the present action doesn't complete the process by selecting an assigned day.
• $\bf 5/12\:$ is the corresponding probability of proceeding to the next action, $\textrm{roll}_{15}$.

If the next action is necessary, the row for $\textrm{roll}_{15}$ shows that the $5$ unassigned potential ordered outcomes from $\textrm{flip}_{14}$ are combined with the possible results of a new die roll to produce $30$ new potential ordered outcomes:
    { 2_^&T_^&1,   3_^&T_^&1,  ...,  6_^&T_^&1,
      2_^&T_^&2 , 3_^&T_^&2 , ...,  6_^&T_^&2 ,
       $^\vdots$
      2_^&T_^&6 , 3_^&T_^&6 , ...,  6_^&T_^&6 }
Of those combinations, $28$ would select days while $2$ will be left over, leaving a $2/30$ chance of requiring yet another action, $\textrm{flip}_{16}$.

Compounding the table's rightmost column of continuation probabilities gives:

\begin{array}{l} \textsf{average number of alternating rolls or flips} = \\ \qquad 1 + \frac6{6} \Bigl( 1 + \frac5{12} \bigl( 1 + \frac2{30} \bigl( 1 + \frac4{4} \bigl( 1 + \frac3{24} \bigl( 1 + \frac6{6} \bigl( 1 + \frac1{36} \bigl( 1 + \frac2{2} ( 1 + \ldots ) \bigr) \bigr) \bigr) \bigr) \bigr) \bigr) \Bigr) \\ \phantom{\textsf{average number of alternating rolls or flips}} = 2.4794153... \end{array}

Incidentally, the same calculation with no coin flips is:

\begin{array}{l} \textsf{average rolls for die only} = 1 + \frac66 \Bigl( 1 + \frac1{36} \bigl( 1 + \frac66 \bigl( 1 + \frac1{36} ( 1 + \ldots ) \bigr) \bigr) \Bigr) \\ \phantom{\textsf{average rolls for die only}} = 2.0\overline{571428} \end{array}

Answer 8

I'm posting this solution since it is really different from what has been posted so far. Not because it is efficient. It isn't.

We are working in rounds. Initially each day is in round 0. Now we toss a coin for each day. If the coin lands on heads the day moves on to the next round. We repeat this for every new round, until we end up with a round having only one day left. This day is the answer. If accidentally we end up with a round having no days left at all, we repeat the previous round.

Why does this yield a uniform distribution? We have treated each day the same way.

Answer 9

I'm two years too late, but I feel like this is easier to explain to people playing a board-game with low probability of repeating:

Roll the dice twice (or two dice):

• If both are 6, roll both again and start over.
• Else If the first number is 6 -> Sunday
• Else look at the second number:
  • 1 => Monday
  • 2 => Tuesday
  • ...
  • 6 => Saturday

Each option (excluding the $1/36$ chance of a re-roll) has the chance of $5/35 = 1/7$.

While this is the same in spirit to some of the other answers, pragmatically this method is easier than the tables proposed.

Answer 10

Roll two dice (or roll the die twice) and flip the coin. Sum the dice. Use this table to figure out the result.

Dice Sum	Coin	Result
5 or 7	H	1
5 or 7	T	2
6	H or T	3
8	H or T	4
9, 10, 11, or 12	H	5
9, 10, 11, or 12	T	6
3 or 4	H or T	7
2	H or T	Redo

There are 36 possible dice results, which doubles to 72 when you account for the coin result. Here is the table that shows each result has an equal probability (N is the number of ways each combination can happen):

Sum	Coin	N	Assigned To
7 5	H H	6 4	1
7 5	T T	6 4	2
6 6	H T	5 5	3
8 8	H T	5 5	4
9 10 11 12	H H H H	4 3 2 1	5
9 10 11 12	T T T T	4 3 2 1	6
3 4	H or T H or T	4 6	7
2	X	2	Try Again

This method has two main advantages:

• a) The dice values can be summed, without regard to order (in other words, a 1-3 is the same as a 3-1 in this method)
• b) There is a low percentage chance of a redo (1 in 36, or 2.778%)

The main disadvantage is probably that, without the table in front of you, it might be hard to remember that, say, Double 5's and Tails means "6", while a 5 and a 2 (i.e., a 7) plus Heads is a "1".

Answer 11

Out-of-the-box solutions using just the coin, mostly coin-rolling and not coin-flipping:

1. You never said the coin was round. It could be a heptagon or Reuleaux (constant-width, hence easily rollable) heptagon, just like the British (or pre-Euro Irish) 50p coin or British 20p coin. (If it isn't, you could clip it into that, although that seems contrived.) Label the sides 1..7 clockwise starting from some recognizable point on the head or tail. Roll it.
   • NOTES: a Reuleaux polygon coin is both perfectly unbiased and rollable, unlike other approaches. Naively clipping the edges off a round coin would create something that was both biased and unrollable. And this actually cites the construction algorithm, it doesn't just conjecture that such a thing might be feasible.
   • you can read the result off a (labelled) polygonal coin near-instantly, unlike a round coin with a polygonal motif.
2. Roll the coin around a circular track, like a roulette marble, which again has been predivided into seven (or 14, 28 etc.) equal sectors (or you could flip it in the air just to see where it lands and disregard the face it lands on, but the flipping movement is unlikely to be unbiased wrt where it lands).
3. Even more far-out-of-the-box interpretations:

• a) only works one-time (your question implied you only want to generate a random number once): use the year-of-issue of the coin, if you know the distribution of years-of-issue of that type of coin, and only if it happens to divide nicely into sevenths. (This is similar to generating random numbers from the 11-digit serial number on a US dollar bill, e.g. by summing and taking them modulo 7, although 11-digit serial nos are far better than coin year-of-issue).
• b) or, to (repeatedly, unbiasedly) generate arbitrary random numbers (a la "Liar's Poker" with dollar-bill serial numbers), even more "out-of-the-box" interpretation is, we could say the coin has an n-digit RFID serial no.

British 50p coin

Old Irish 50p coin (pre-Euro, last used in 2001)

UPDATE: some lists of polygon coins of the world 1) BezalelCoins 2) CoinBrothers

Answer 12

Similar to Mike Earnest, I suggest rolling again on a double six, to give you 35 possibilities instead of 36. The question then becomes how to assign the 35 possibilities to the 7 scores you require. The following formula is reasonably easy to calculate, and does not involve any base conversions:

(7+Die1-Die2) mod 7

       Die 1
        1 2 3 4 5 6
        ===========
Die   1|0 1 2 3 4 5
2     2|6 0 1 2 3 4
      3|5 6 0 1 2 3
      4|4 5 6 0 1 2
      5|3 4 5 6 0 1
      6|2 3 4 5 6 X       X=roll again

As can be seen, each score from 0 to 6 appears in the table exactly 5 times.

In plain english, this becomes: "Roll the die twice. Subtract the second roll from the first (if this is not possible because it would give a negative number, add 7 to the value of the first roll before subtracting.) If you roll a double six, discard the score and roll again."

Answer 13

Use this method:

1. Assign a number on the die to each day of the week i.e. Day 1 = 1, Day 2 = 2 and so on.
2. For the last two days (days six and seven) in the week, give them both the number 6.
3. Roll the dice. If you land on the number 6, flip the coin and if its heads, pick day 6, if its tails, you pick day 7. Since the coin has a probability of 1/2, there is a probability of 1/12 of picking either day 6 or day 7 from the start.
4. If it lands on any other number, flip the coin and if it lands on heads accept the number, if it lands on tails then repeat the whole process. Now every other day has a probability of 1/12 of being chosen as well.

Answer 14

I would use the following method

1. roll the dice, then flip the coin
2. if the coin is head, add 6 to the number on the dice.
   This gives you an outcome between 1 and 12 with uniform probability. now if the number is greater than 7 try again.

The way I came up with this solution is trying a simpler problem: getting a number between 1 and 5 with uniform probability with one dice. The solution is to roll the dice, if it is 6 roll again. Basically doing a "modulo" operation.

If you feel comfortable that the one-dice problem works, then the proposed solution working from something that gives you a 1-12 uniform distribution and converting it to a 1-7 uniform distribution also works.

This answer is the same as 'ace'.

A more interesting problem is to prove that there is a solution with a guaranteed finite number of trials. My answer would be no, given than 7, and 2 and 3 are co-prime (their greatest common divisor is 1)

Answer 15

What we consider "optimal" depends on what we consider the "costs" of various actions. If flipping the coin and rolling the die together has the same "cost" as doing one of them, then we might as well do both. This means that we're dealing with base 12. Each round, we can take the set of possibilities, split it into seven equal subsets, with some possibilities "left over". Each round, if we get one of the "left over" possibilities, we can go another round, and take the Cartesian product of that result and the "left over" possibilities from the previous round.

For the first round, we have five "left over" results. If we get one of those, there are 5*12=60 different combinations for our second round, with four "left over" possibilities. So on the third round, there are 48 combinations, with six "left over". And so on.