Does the time of day affect dice roll results?

  • '18 '17 '16 '11 Moderator

    Believe it or not, I actually have a set of dice with sharp corners (cut myself on the corner once.)  As a DM, I used to force players to use my dice instead, cause I know how to throw a D6 and get the result I want!


  • Cmdr Jennifer!

    I can believe that!  Ah brings back memories, I had a friend who made his own loaded dice, you could tell it was loaded because it was cut and glued back on (didn’t do a great job; it would have been one thing if it looked professional).  Also had loaded dice that rolled 6’s but it was very easy to tell it was loaded (it was heavy!!).

    The things we used to do!


  • Back to the random dice generators.

    They can be attacked if you know the algorithm and the seed, sort of like how you would go about cracking a password.

    In theory, a program could be written that uses the algorithm and the seed that would allow you to influence the die results as it could “look” into the future or wait for a favorable instance then pull the trigger on the roll request to get the result.  You would never be guaranteed of getting a specific die result, but you could choose to get a certain percentage of "1"s or a percentage of results < 3.


  • @gnasape:

    whenever I make a move late at night usually around or close to midnight, I have noticed that the dice rolls turn out poorly.

    Example:
    2 inf, 1 art, 1 arm, 2 ftr, 1 bmb, 2 bb shot vs. 3 inf
    Result: held, 1 ftr, 1 bmb retreats.

    gnasape, be careful here that you are not falling victim to a statistical/psychological phenomenon known as selection bias.  here, the emotional pain you (and us all) feel when you get hosed by the dice makes the event so memorable that it appears to occur more often than it actually does.

    in your example, you’ve cited all of the units from a battle and the exact outcome.  i bet you couldn’t do that with a battle that was very favorable for YOU during that same game.  could you?  the painful battles are just more memorable.

    also in your example, you aren’t giving the actual dice results which could be analyzed for statistical significance, just that you got hosed.  and big time.  yeah, sure, your rolls in the example were all misses, but the opponents were all hits.  so the pseudo-random number generator was probably producing samples in it’s normal distribution, but just not sorted in an order that you liked.  it happens.

    i imagine what happened is that once you got hosed, you looked at the clock for some reason and noticed it was around midnight.  probably the next night it happened again at midnight, and your brain decided these samples were statistically significant (selection bias) and created an completely illusory artifact that “i get hosed at midnight”.

    if you like, record a bunch of samples (~100) and post them here.  i’d be happy to statistical significance test for you.


  • I suspect what you are noticing happens between say 1 or 2 and 3 or 4 Eastern US time.

    And Rechecking your post I see that is the time frame you are mentioning. I have noticed it myself as I post all kinds of hours.

    If one of you statistical gurus feels inclined you can search through my games and note the time stamps and the results. I don’t need to I have noticed the dice. And it is not just getting hosed all sides are rolling crap.


  • @bigchris:

    @gnasape:

    whenever I make a move late at night usually around or close to midnight, I have noticed that the dice rolls turn out poorly.

    Example:
    2 inf, 1 art, 1 arm, 2 ftr, 1 bmb, 2 bb shot vs. 3 inf
    Result: held, 1 ftr, 1 bmb retreats.

    gnasape, be careful here that you are not falling victim to a statistical/psychological phenomenon known as selection bias.  here, the emotional pain you (and us all) feel when you get hosed by the dice makes the event so memorable that it appears to occur more often than it actually does.

    in your example, you’ve cited all of the units from a battle and the exact outcome.  i bet you couldn’t do that with a battle that was very favorable for YOU during that same game.  could you?  the painful battles are just more memorable.

    also in your example, you aren’t giving the actual dice results which could be analyzed for statistical significance, just that you got hosed.  and big time.  yeah, sure, your rolls in the example were all misses, but the opponents were all hits.  so the pseudo-random number generator was probably producing samples in it’s normal distribution, but just not sorted in an order that you liked.  it happens.

    i imagine what happened is that once you got hosed, you looked at the clock for some reason and noticed it was around midnight.  probably the next night it happened again at midnight, and your brain decided these samples were statistically significant (selection bias) and created an completely illusory artifact that “i get hosed at midnight”.

    if you like, record a bunch of samples (~100) and post them here.  i’d be happy to statistical significance test for you.

    It’s not me, that’s what my friend got against me.  I’m ok with the dice but did see that it has the certain quirks.  In the scenario above, I’m the defender!!  Yay!!


  • Exact dice rolls:

    2 bb shot, 2 inf, 1 art, 1 arm, 2 ftr, 1 bmb vs. 3 inf

    Round 1:
    Attack: 6,5,6,5,4,6,5,6,6 (no hits)
    Defense: 2,2,1 (3 hits)

    Round 2:
    Attack: 4,3,5,4 (1 hit)
    Defense: 3,4,2 (1 hit)

    Round 3:
    Attack: 6,3,6 (no hits)
    Defense: 1 (1 hit)

    You can calculate the odds.  There are other battles for my friend that went bad.  I’m ok with it as dice is part of the game.  Due to that we’ve tried LL.  I’m not advocating anything.  I was curious if others who post at certain times noticed a pattern in the dice.


  • @gnasape:

    Exact dice rolls:

    Round 1:
    Attack: 6,5,6,5,4,6,5,6,6 (no hits)
    Defense: 2,2,1 (3 hits)

    Round 2:
    Attack: 4,3,5,4 (1 hit)
    Defense: 3,4,2 (1 hit)

    Round 3:
    Attack: 6,3,6 (no hits)
    Defense: 1 (1 hit)

    histogram for entire sequence (23 rolls total)
    6s: 6666666
    5s: 5555
    4s: 4444
    3s: 333
    2s: 222
    1s: 11

    Ho: the dice are uniformly distributed
    Ha: the dice are not uniformly distributed

    pearson chi-square test:
    with 23 rolls, we expect 23/6=3.83 rolls in each bin according to Ho
    (7 - 3.83)^2 / 3.83 +
    (4 - 3.83)^2 / 3.83 +
    (4 - 3.83)^2 / 3.83 +
    (3 - 3.83)^2 / 3.83 +
    (3 - 3.83)^2 / 3.83 +
    (2 - 3.83)^2 / 3.83 = 3.870

    chi-square value = 3.870
    degrees-of-freedom = 5
    Probability (One-Tailed): 0.5684

    you would need a chi-square value of 11 before you would adopt Ha (at the 5% level), and you are only at 3.870.

    that is, if the dice are in fact uniformly distributed (Ho), the probability of getting results as or more extreme than these is 0.5684.  this is insufficient evidence to reject Ho.  therefore, we do not reject Ho.

    now, the a priori probability of your friend winning that battle was likely greater than 99.5%, but that’s another issue.


  • @bigchris:

    @gnasape:

    Exact dice rolls:

    Round 1:
    Attack: 6,5,6,5,4,6,5,6,6 (no hits)
    Defense: 2,2,1 (3 hits)

    Round 2:
    Attack: 4,3,5,4 (1 hit)
    Defense: 3,4,2 (1 hit)

    Round 3:
    Attack: 6,3,6 (no hits)
    Defense: 1 (1 hit)

    histogram for entire sequence (23 rolls total)
    6s: 6666666
    5s: 5555
    4s: 4444
    3s: 333
    2s: 222
    1s: 11

    Ho: the dice are uniformly distributed
    Ha: the dice are not uniformly distributed

    pearson chi-square test:
    with 23 rolls, we expect 23/6=3.83 rolls in each bin according to Ho
    (7 - 3.83)^2 / 3.83 +
    (4 - 3.83)^2 / 3.83 +
    (4 - 3.83)^2 / 3.83 +
    (3 - 3.83)^2 / 3.83 +
    (3 - 3.83)^2 / 3.83 +
    (2 - 3.83)^2 / 3.83 = 3.870

    chi-square value = 3.870
    degrees-of-freedom = 5
    Probability (One-Tailed): 0.5684

    you would need a chi-square value of 11 before you would adopt Ha (at the 5% level), and you are only at 3.870.

    that is, if the dice are in fact uniformly distributed (Ho), the probability of getting results as or more extreme than these is 0.5684.  this is insufficient evidence to reject Ho.  therefore, we do not reject Ho.

    now, the a priori probability of your friend winning that battle was likely greater than 99.5%, but that’s another issue.

    Thanks bigchris, what if we assume that the dice rolls of the attacker and defenders are two separate events?

    How does that change your test?

    16 rolls for attacker, 7 rolls for defender.

    Round 1 was rolled on 06/25/09 06:24:32PM
    Round 2 was rolled on 06/29/09 06:05:27PM
    Round 3 was rolled on 06/29/09 06:06:35PM

    Same time but different day, similar results.


  • @gnasape:

    Thanks bigchris, what if we assume that the dice rolls of the attacker and defenders are two separate events?

    16 rolls for attacker, 7 rolls for defender.

    sure, that’s fine.  as long as you are separating the rolls based on some arbitrary labeling and not based on their values, you can test any group/sub-group you like.

    Ho: the dice are uniformly distributed
    Ha: the dice are not uniformly distributed

    Attacker: 6,5,6,5,4,6,5,6,6,4,3,5,4,6,3,6
    histogram for entire sequence (16 rolls total)
    6s: 6666666
    5s: 5555
    4s: 444
    3s: 33
    2s:
    1s:

    pearson chi-square test:
    with 16 rolls, we expect 16/6=2.67 rolls in each bin according to Ho.
    (7 - 2.67)^2 / 2.67 +
    (4 - 2.67)^2 / 2.67 +
    (3 - 2.67)^2 / 2.67 +
    (2 - 2.67)^2 / 2.67 +
    (0 - 2.67)^2 / 2.67 +
    (0 - 2.67)^2 / 2.67 = 13.25

    chi-square value = 13.25
    degrees-of-freedom = 5
    Probability (One-Tailed): 0.0211

    that is, for the attacker, if Ho is in fact true, the probability of getting results as or more extreme than these is 0.0211.  These data are sufficient evidence (at the 5% level) to reject Ho and adopt Ha.

    Defender: 2,2,1,3,4,2,1
    histogram for entire sequence (7 rolls total)
    6s:
    5s:
    4s: 4
    3s: 3
    2s: 222
    1s: 11

    pearson chi-square test:
    with 7 rolls, we expect 7/6=1.17 rolls in each bin according to Ho.
    (0 - 1.17)^2 / 1.17 +
    (0 - 1.17)^2 / 1.17 +
    (1 - 1.17)^2 / 1.17 +
    (1 - 1.17)^2 / 1.17 +
    (3 - 1.17)^2 / 1.17 +
    (2 - 1.17)^2 / 1.17 = 5.857

    chi-square value = 5.857
    degrees-of-freedom = 5
    Probability (One-Tailed): 0.3204

    that is, for the defender, if Ho is in fact true, the probability of getting results as or more extreme than these is 0.3204.  this is insufficient evidence to reject Ho.  therefore, we do not reject Ho.


  • @bigchris:

    @gnasape:

    Thanks bigchris, what if we assume that the dice rolls of the attacker and defenders are two separate events?

    16 rolls for attacker, 7 rolls for defender.

    sure, that’s fine.  as long as you are separating the rolls based on some arbitrary labeling and not based on their values, you can test any group/sub-group you like.

    Ho: the dice are uniformly distributed
    Ha: the dice are not uniformly distributed

    Attacker: 6,5,6,5,4,6,5,6,6,4,3,5,4,6,3,6
    histogram for entire sequence (16 rolls total)
    6s: 6666666
    5s: 5555
    4s: 444
    3s: 33
    2s:
    1s:

    pearson chi-square test:
    with 16 rolls, we expect 16/6=2.67 rolls in each bin according to Ho.
    (7 - 2.67)^2 / 2.67 +
    (4 - 2.67)^2 / 2.67 +
    (3 - 2.67)^2 / 2.67 +
    (2 - 2.67)^2 / 2.67 +
    (0 - 2.67)^2 / 2.67 +
    (0 - 2.67)^2 / 2.67 = 13.25

    chi-square value = 13.25
    degrees-of-freedom = 5
    Probability (One-Tailed): 0.0211

    that is, for the attacker, if Ho is in fact true, the probability of getting results as or more extreme than these is 0.0211.  These data are sufficient evidence (at the 5% level) to reject Ho and adopt Ha.

    Defender: 2,2,1,3,4,2,1
    histogram for entire sequence (7 rolls total)
    6s:
    5s:
    4s: 4
    3s: 3
    2s: 222
    1s: 11

    pearson chi-square test:
    with 7 rolls, we expect 7/6=1.17 rolls in each bin according to Ho.
    (0 - 1.17)^2 / 1.17 +
    (0 - 1.17)^2 / 1.17 +
    (1 - 1.17)^2 / 1.17 +
    (1 - 1.17)^2 / 1.17 +
    (3 - 1.17)^2 / 1.17 +
    (2 - 1.17)^2 / 1.17 = 5.857

    chi-square value = 5.857
    degrees-of-freedom = 5
    Probability (One-Tailed): 0.3204

    that is, for the defender, if Ho is in fact true, the probability of getting results as or more extreme than these is 0.3204.  this is insufficient evidence to reject Ho.  therefore, we do not reject Ho.

    Thanks, I am basing it as usually the attacker’s dice roll is submitted first than defender.

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