From Newsgroup: rec.puzzles

Its been raining again...

I have six unbiased six-sided dice. The game comprises two rolls. In the
first roll I roll a single die. In the second roll, I roll as many dice as there are spots in the first roll. So, if the first roll comes up with a
one, I just roll one die in the second roll. If the first roll comes up
with a six, I roll six dice in the second roll. The outcome of the game is
the total of the spots in the second roll. We can see that an outcome of
one, (roll a 1 followed by 1) and and outcome of thirty-six, (roll a 6 followed by 6, 6, 6, 6, 6, 6) are the most unlikely outcomes, as there is
only one way of achieving either, but what is the most likely outcome of
the game?

I got my answer the long way - is there a better way?
--
David Entwistle
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From Newsgroup: rec.puzzles

On Sun, 15 Feb 2026 10:50:13 -0000 (UTC), David Entwistle wrote:

I got my answer the long way - is there a better way?

Possibly interesting to have a guess before working on the solution. My
guess was somewhat better than AI's "solution".
--
David Entwistle
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From Newsgroup: rec.puzzles

In article <10ms8d5$3uabf$1@dont-email.me>,
David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:

I have six unbiased six-sided dice.

Now suppose you have N N-sided dice. I was initially very surprised
by the result but on reflection I can see why it is so.

-- Richard
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From Newsgroup: rec.puzzles

On 15/02/2026 10:50, David Entwistle wrote:

Its been raining again...

I have six unbiased six-sided dice. The game comprises two rolls. In the first roll I roll a single die. In the second roll, I roll as many dice as there are spots in the first roll. So, if the first roll comes up with a
one, I just roll one die in the second roll. If the first roll comes up
with a six, I roll six dice in the second roll. The outcome of the game is the total of the spots in the second roll. We can see that an outcome of
one, (roll a 1 followed by 1) and and outcome of thirty-six, (roll a 6 followed by 6, 6, 6, 6, 6, 6) are the most unlikely outcomes,

On my first reading I read over this, but actually this isn't right! 36 is definitely the least
likely outcome, but it turns out each of the outcomes from 22 up to 36 are also less likely than
outcome 1. (In an earlier post I presented a calculation of the number of ways for making each
outcome 1-36.)

Just intuitively, it's much much more likely you'll role two 1s in a row compared to seven 6s!


Mike.

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From Newsgroup: rec.puzzles

On 15/02/2026 11:00, David Entwistle wrote:

On Sun, 15 Feb 2026 10:50:13 -0000 (UTC), David Entwistle wrote:

I got my answer the long way - is there a better way?

Possibly interesting to have a guess before working on the solution. My
guess was somewhat better than AI's "solution".

I couldn't see any simplification that gave me a solid reasoning for a definite answer. Clearly the
counts for lower numbers of dice are going to be scaled up relative to using higher numbers of dice.
Hmm, what I mean can be seen with how many "ways" there are to make a total of 1 vs a total of 36:
- to get total 1, we must first roll 1, then roll 1.
- to get total 36, we must roll 6 on all 7 rolls.
So at first we might think they both have one way to make them - but that ignores how likely those
"ways" are! To level things up,We can imagine we always make 7 rolls, but only total up the number
of subsequent dice dictated by the first roll. So now we have 6^7 = 279936 outcomes for the 7 dice
rolls, all equally likely, and those are the "ways" we should be counting. Revising the above
"ways" of making 1 vs 36:
- to get total 1, we must first roll 1, then roll 1, then any of 6 outcomes for rolls 3,4,5,6,7.
- to get total 36, we must roll 6 on all 7 rolls.
So in fact there are 6^5 = 7776 ways to make 1, but only 1 to make 36.

So there is a heavy weighting applied towards the number of ways of making a total using fewer dice.

But that was all I could see, so I guessed maybe the most likely total [equating with "most number
of ways of making that total"] might be around 10.

I still have no mathematical closed form solution for the problem, but a computer calculation for
the number of ways for each total (assuming I've not messed up) is:

#ways for total 1 : 7776
#ways for total 2 : 9072
#ways for total 3 : 10584
#ways for total 4 : 12348
#ways for total 5 : 14406
#ways for total 6 : 16807
#ways for total 7 : 11832
#ways for total 8 : 12507
#ways for total 9 : 13076
#ways for total 10 : 13482
#ways for total 11 : 13650
#ways for total 12 : 13482
#ways for total 13 : 12852
#ways for total 14 : 12897
#ways for total 15 : 12772
#ways for total 16 : 12453
#ways for total 17 : 11928
#ways for total 18 : 11207
#ways for total 19 : 10332
#ways for total 20 : 9387
#ways for total 21 : 8292
#ways for total 22 : 7101
#ways for total 23 : 5880
#ways for total 24 : 4697
#ways for total 25 : 3612
#ways for total 26 : 2667
#ways for total 27 : 1876
#ways for total 28 : 1251
#ways for total 29 : 786
#ways for total 30 : 462
#ways for total 31 : 252
#ways for total 32 : 126
#ways for total 33 : 56
#ways for total 34 : 21
#ways for total 35 : 6
#ways for total 36 : 1
------------------ ---------
sum of totals above : 279936

and indeed 279936 = 6x6x6x6x6x6x6, which gives me some confidence!

So the most likely total is 6, with probability 16807/279936 = 0.0600387...

A graph for the above shows a general pattern of a rise then a falling off as the target total
increases, which seems intuitive, but the falling off is not completely smooth - e.g. the peak is
for total=6 after which there's a considerable drop for total 7, but totals for 8, 9, 10 and 11 are
increasing again!


Regards,
Mike.

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From Newsgroup: rec.puzzles

On Sun, 15 Feb 2026 11:00:36 -0000 (UTC), David Entwistle wrote:

My
guess was somewhat better than AI's "solution".

Or possibly not...
--
David Entwistle
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From Newsgroup: rec.puzzles

In article <10mu4cu$iom8$1@dont-email.me>,
Mike Terry <news.dead.person.stones@darjeeling.plus.com> wrote:

A graph for the above shows a general pattern of a rise then a
falling off as the target total increases, which seems intuitive,
but the falling off is not completely smooth - e.g. the peak is for
total=6 after which there's a considerable drop for total 7, but
totals for 8, 9, 10 and 11 are increasing again!

I've put a graph at https://www.cogsci.ed.ac.uk/~richard/dice.png

The lower graphs are the probabilty of getting N with 1,2,..,6 dice.
As the number of dice increases the distribution changes from
rectangular to a smooth distribution approaching (central limit
theorem!) normal.

The upper graph is sum of these - i.e. the overall chance of getting
N.

The big drop between 6 and 7 is because the probability for one die
drops to zero.

(The use of a continuous line graph for discrete variables is
obviously wrong, but I didn't spend time trying to find an alternative
that was still clear.)

-- Richard
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From Newsgroup: rec.puzzles

On Mon, 16 Feb 2026 04:29:26 +0000, Mike Terry wrote:

On my first reading I read over this, but actually this isn't right! 36
is definitely the least likely outcome, but it turns out each of the
outcomes from 22 up to 36 are also less likely than outcome 1. (In an earlier post I presented a calculation of the number of ways for making
each outcome 1-36.)

Yes, I see I had that all wrong. It turns out that my intuition was even
worse than the AI's solution, but neither of us were close.
--
David Entwistle
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From Newsgroup: rec.puzzles

On Mon, 16 Feb 2026 09:59:03 -0000 (UTC), Richard Tobin wrote:

I've put a graph at https://www.cogsci.ed.ac.uk/~richard/dice.png

The lower graphs are the probabilty of getting N with 1,2,..,6 dice.
As the number of dice increases the distribution changes from
rectangular to a smooth distribution approaching (central limit
theorem!) normal.

Excellent. That makes things very clear.
--
David Entwistle
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From Newsgroup: rec.puzzles

On Sun, 15 Feb 2026 13:04:22 -0000 (UTC), Richard Tobin wrote:

Now suppose you have N N-sided dice. I was initially very surprised by
the result but on reflection I can see why it is so.

Interesting question. I'll work on getting the correct answer to the
initial question first, and then think about moving on.
--
David Entwistle
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From Newsgroup: rec.puzzles

On Mon, 16 Feb 2026 09:59:03 -0000 (UTC), Richard Tobin wrote:

I've put a graph at https://www.cogsci.ed.ac.uk/~richard/dice.png

The lower graphs are the probabilty of getting N with 1,2,..,6 dice. As
the number of dice increases the distribution changes from rectangular
to a smooth distribution approaching (central limit theorem!) normal.

The upper graph is sum of these - i.e. the overall chance of getting N.

The big drop between 6 and 7 is because the probability for one die
drops to zero.

(The use of a continuous line graph for discrete variables is obviously wrong, but I didn't spend time trying to find an alternative that was
still clear.)

I've run a simulation, with 10^7 plays, the result is here:

https://ibb.co/Ps906g3Z

I should have done that in the first place.
--
David Entwistle
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From Newsgroup: rec.puzzles

On 16/02/2026 12:03, David Entwistle wrote:

On Mon, 16 Feb 2026 04:29:26 +0000, Mike Terry wrote:

On my first reading I read over this, but actually this isn't right! 36
is definitely the least likely outcome, but it turns out each of the
outcomes from 22 up to 36 are also less likely than outcome 1. (In an
earlier post I presented a calculation of the number of ways for making
each outcome 1-36.)

Yes, I see I had that all wrong. It turns out that my intuition was even worse than the AI's solution, but neither of us were close.

So what did the AI say?




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From Newsgroup: rec.puzzles

On Mon, 16 Feb 2026 17:15:19 +0000, Mike Terry wrote:

So what did the AI say?

To be fair, it was closer than I was.

ChatGPT answer:

Final Step: Determining the Most Likely Outcome

With frequency counts set up, you can see which total sum has the highest likelihood based on contributions from all potential first rolls.

The most likely outcome seems to trend around 14 or 15, taking into
account the combination frequencies. You would find that:

14 and 15 likely emerge as the highest counts, both receiving contributions from several first roll scenarios.

If you need the complete breakdown or calculations for other numbers, let
me know!
--
David Entwistle
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From Newsgroup: rec.puzzles

In article <10mv162$r16i$3@dont-email.me>,
David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:

Now suppose you have N N-sided dice. I was initially very surprised by
the result but on reflection I can see why it is so.

Interesting question. I'll work on getting the correct answer to the
initial question first, and then think about moving on.

Answer
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Just as the most common result for 6-sided dice is 6, for N-sided dice
it seems to be N. As with the 6-sided case the probability increases
from 1 to N, and then there's the sudden drop when you reach the first
sum can't be obtained with a single die.

Once N gets up into the teens a surprising phenomenon occurs: the
graph is quite flat in the central part. For example, when N=20
the probability of sums from 71 to 137 are all equal (to 5dp).

See https://www.cogsci.ed.ac.uk/~richard/dice20.png

-- Richard
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From Newsgroup: rec.puzzles

On Mon, 16 Feb 2026 19:01:44 -0000 (UTC), Richard Tobin wrote:

Answer .
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Just as the most common result for 6-sided dice is 6, for N-sided dice
it seems to be N. As with the 6-sided case the probability increases
from 1 to N, and then there's the sudden drop when you reach the first
sum can't be obtained with a single die.

Once N gets up into the teens a surprising phenomenon occurs: the graph
is quite flat in the central part. For example, when N=20 the
probability of sums from 71 to 137 are all equal (to 5dp).

See https://www.cogsci.ed.ac.uk/~richard/dice20.png

-- Richard

Yes, very nice.

Simulations:

https://ibb.co/ttmMThV
--
David Entwistle
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