Board Game Arena

Posted: **15 July 2025, 12:22**

euklid314 wrote: ↑04 July 2025, 13:45 Your game had 66 moves, i.e. 43 (overlapping) sequences of 24 consecutive rolls. One of these 43 sequences did contain neither 6 and 8.

The expected value of this to happen is
43*(26/36)^24=0.0137=1:73

Thus you are expected to have one such event per 73 games.

Your formula greatly overstates the odds of this happening. There are two reasons why.

The first and most significant relates to multiplying by 43. In order for multiplying by 43 to be valid, all 43 events must be independent. That is not the case here. For example, of all the possible ways that the first 24 rolls did contain a 6 or 8, there is only one very unlikely result that will still allow the 2nd through 25th rolls to not have either a 6 or 8 and that is the following: The first roll is a 6 or 8, and the following 23 are never a 6 or 8.

The second reason (and I can't begin to estimate the impact of this) is that many of the scenarios may never be realized in an actual game. That is there are scenarios where some player has already won the game, thus the game never gets to the point of having those 24 straight rolls without a 6 or 8.

Posted: **15 July 2025, 13:36**

LaMyrsian wrote: ↑15 July 2025, 12:22
euklid314 wrote: ↑04 July 2025, 13:45 Your game had 66 moves, i.e. 43 (overlapping) sequences of 24 consecutive rolls. One of these 43 sequences did contain neither 6 and 8.

The expected value of this to happen is
43*(26/36)^24=0.0137=1:73

Thus you are expected to have one such event per 73 games.
Your formula greatly overstates the odds of this happening. There are two reasons why.

The first and most significant relates to multiplying by 43. In order for multiplying by 43 to be valid, all 43 events must be independent. That is not the case here. For example, of all the possible ways that the first 24 rolls did contain a 6 or 8, there is only one very unlikely result that will still allow the 2nd through 25th rolls to not have either a 6 or 8 and that is the following: The first roll is a 6 or 8, and the following 23 are never a 6 or 8.

The second reason (and I can't begin to estimate the impact of this) is that many of the scenarios may never be realized in an actual game. That is there are scenarios where some player has already won the game, thus the game never gets to the point of having those 24 straight rolls without a 6 or 8.

You have to give a little bit of grace when doing probability calculations, especially including concepts like "someone has already won" is basically impossible when thinking about the reasonable effort for someone responding to a board game forum post is considered.

For what it's worth, based on simulations, you're right euklid314 has missed a factor somewhere and the likelihood of getting 24 (or more) rolls in a row that are neither 6 or 8 is around 0.5% or 1 in 200 games. Which is I should point out, still not unusual. It's basically an ordinary everyday event that will happen pretty much every day on the site.

Just for funs, the longest run of rolls without a six or an eight out of a million games simulated with 66 rolls per game was 52 rolls in a row.

Posted: **15 July 2025, 14:39**

LaMyrsian wrote: ↑15 July 2025, 12:22
euklid314 wrote: ↑04 July 2025, 13:45 Your game had 66 moves, i.e. 43 (overlapping) sequences of 24 consecutive rolls. One of these 43 sequences did contain neither 6 and 8.

The expected value of this to happen is
43*(26/36)^24=0.0137=1:73

Thus you are expected to have one such event per 73 games.
Your formula greatly overstates the odds of this happening. There are two reasons why.

The first and most significant relates to multiplying by 43. In order for multiplying by 43 to be valid, all 43 events must be independent. That is not the case here. For example, of all the possible ways that the first 24 rolls did contain a 6 or 8, there is only one very unlikely result that will still allow the 2nd through 25th rolls to not have either a 6 or 8 and that is the following: The first roll is a 6 or 8, and the following 23 are never a 6 or 8.

The second reason (and I can't begin to estimate the impact of this) is that many of the scenarios may never be realized in an actual game. That is there are scenarios where some player has already won the game, thus the game never gets to the point of having those 24 straight rolls without a 6 or 8.

Yes, I was not stating it mathematically incorrect. In order to interpret my calculated number of 0.0137 I should have phrased it in the following:

If you take a random game of 66 moves and you look for exactly 24-long sequences that do not contain either 6 or 8 the expected value is 0.0137.

Thus, if you take 1000 random games of length 66 moves those are expected to contain 13.7 such sequences.

But my interpretation of 1:73 was wrong of course (partly because I wanted to simplify, partly because I thought the effect would not be huge). I assumed that 13 games would contain exactly 1 such sequence and 987 games would contain 0 such sequences - resulting in an expected value of 0.013 occurrences per game.

But the reality is of course that if you find one such sequence in a game you will quite likely find more of them. Because if there are, e.g., 27 consecutive non-6/8-numbers, in a game this counts as 4 sequences of 24-long non-6/8-numbers.

The simulation of Jontia seems to suggest that something like the following is to be expected if you take 1000 games of length 66 each:

995 games contain zero 24-long sequences
2 games contain one 24-long sequence each
1 game contains a 25-long sequence (i.e. two 24-long sequences)
1 game contains a 27-long sequence (i.e. four 24-long sequences)
1 game contains a 28-long sequence (i.e. five 24-long sequences)

In such a case we have 5:1000=1:200 ratio to find such a sequence in a game but a 13:1000=0.013 expected value of total such sequences per game.

@Jontia: I would be happy if you will verify my number with your simulation. This time I know I did my maths correctly.

If you take your million games of length 66 then you will find approx. 13700 sequences. Please note that your exceptional game of 52 consecutive non-6/8-numbers (btw: wow!) will alone contribute 52-24+1=29 sequences to this total...

Posted: **15 July 2025, 16:05**

It gets boring to wait for 1million games, so to make sure I don't forget I stuck to 100,000.
Sticking to 1,000 with a situation that comes up 1/200 times isn't a very stable sample.

Streak length 24 in 155 games
Streak length 25 in 102 games
Streak length 26 in 86 games
Streak length 27 in 53 games
Streak length 28 in 42 games
Streak length 29 in 42 games
Streak length 30 in 19 games
Streak length 31 in 9 games
Streak length 32 in 6 games
Streak length 33 in 6 games
Streak length 34 in 6 games
Streak length 35 in 3 games
Streak length 36 in 1 games
Streak length 37 in 1 games
Streak length 38 in 2 games
Streak length 39 in 1 games
Streak length 40 in 2 games

537 Games with streaks 24 rolls or longer of no 6|8
So the number of 24 length sequences are 1*155 + 2*102 + 3*86... = 1823

That's a little higher than your calculation, but a single long streak in that 100,000 games will pull it up. While 24 length streaks are reasonably common the frequency of very long runs (35 in a row is a 1 in 10,000 chance) is variable in relatively small samples.

Posted: **15 July 2025, 22:59**

Thanks for your simulation. Since your numbers were unexpectedly far off my calculation, I retyped and found

43*(26/36)^24 = 0.1744

So my original formula 43*(26/36)^24 was always correct (as I was sure it was), but I must have made a typo on my calculator and ended up with a wrong decimal numer of 0.137 instead of 0.1744.

Thus 1744 sequences are expected within 100.000 games and your simulated 1823 fit well enough...

Posted: **16 July 2025, 19:33**

LaMyrsian wrote: ↑15 July 2025, 12:22 In order for multiplying by 43 to be valid, all 43 events must be independent.

No: The condition is that they must be mutually exclusive. (though this too is not the case here)
The probability of getting at least one 1 from 7 rolls of a single d6, is not 7*1/6 , even though the rolls are independent.

euklid314 wrote: ↑15 July 2025, 22:59 Thanks for your simulation. Since your numbers were unexpectedly far off my calculation, I retyped and found

43*(26/36)^24 = 0.1744

So my original formula 43*(26/36)^24 was always correct (as I was sure it was), but I must have made a typo on my calculator and ended up with a wrong decimal numer of 0.137 instead of 0.1744.

Thus 1744 sequences are expected within 100.000 games and your simulated 1823 fit well enough...

This time, you presumably made a typo in transcribing your calculator's output

: 43*(26/36)^24 is approximately 0.01744 , not approximately 0.1744 .
Also, remember that this being correct is only for the average number of times it happens, not for the probability that it happens at least once.

I get that the probability of it happening at least once is strictly between 1 / 195 and 1 / 194 .

Code: Select all

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julia> "I use the last entry for
       there was at least one streak of lenth at least 24
       .

       Otherwise, the array index corresponds to the _current_ streak-length,
       and the possible values for this streak-length are  0,1,2,...,22,23 .";

julia> typeof(big"0"//big"1")
Rational{BigInt}

julia> after_even_num_of_rolls,after_odd_num_of_rolls = zeros(Rational{BigInt},25),zeros(Rational{BigInt},25);

julia> after_even_num_of_rolls[1+0] = big"1"//big"1"    #  Julia uses 1-based indexing.
1//1

julia> for _ in 1:div(66,2)
               after_odd_num_of_rolls[1+0] = big"10"//big"36"
               for newstreaklen in 1:23
                       after_odd_num_of_rolls[1+newstreaklen] = (big"26"//big"36")*(after_even_num_of_rolls[1+newstreaklen-1])
               end
               after_odd_num_of_rolls[1+24] = ((big"26"//big"36")*(after_even_num_of_rolls[1+24-1]))+(after_even_num_of_rolls[1+24])
               after_even_num_of_rolls[1+0] = big"10"//big"36"
               for newstreaklen in 1:23
                       after_even_num_of_rolls[1+newstreaklen] = (big"26"//big"36")*(after_odd_num_of_rolls[1+newstreaklen-1])
               end
               after_even_num_of_rolls[1+24] = ((big"26"//big"36")*(after_odd_num_of_rolls[1+24-1]))+(after_odd_num_of_rolls[1+24])
       end

julia> big"1"//big"195" < after_even_num_of_rolls[1+24] < big"1"//big"194"
true

julia>

Board Game Arena

For your comedic amusement- 24 straight rolls without a 6 OR 8.

Re: For your comedic amusement- 24 straight rolls without a 6 OR 8.

Re: For your comedic amusement- 24 straight rolls without a 6 OR 8.

Re: For your comedic amusement- 24 straight rolls without a 6 OR 8.

Re: For your comedic amusement- 24 straight rolls without a 6 OR 8.

Re: For your comedic amusement- 24 straight rolls without a 6 OR 8.

Re: For your comedic amusement- 24 straight rolls without a 6 OR 8.