When you say “that is unbelievable control”, you seem to be assuming the exact outcome with trick dice would be exactly and entirely predetermined. But there’s no reason to think that. The trick dice would only have to make winning much more likely to pull your “impossible” odds down into the realm of the possible. What you describe as a die that “occasionally rolled every number except a 1” is what you would expect to see if the “1″ side were weighted a bit—it would often roll a 6, sometimes roll a number adjacent to 6, and never roll 1. Contrariwise, it’s possible that the three dice facing it could have been rigged to do poorly. If a die with the “1” side weighted faced three dice with the 6 side weighted, that could do the trick.
Some amount of dice rigging could make your loss expected or reasonably likely but not guaranteed. And yes, it’s unlikely your friend would (a) weight your dice, (b) waste this ability on a meaningless game of risk, and (c) keep up the act all these years, but it’s not 1-in-100-billion unlikely. People playing little tricks or experimenting on their friends is something that does happen in the world as we know it, therefore it could have happened to you.
″...it would often roll a 6, sometimes roll a number adjacent to 6...”
Assuming standard probability applied to my three dice, the odds of my rolling at least one 6 are 1 - (5/6)^3 or approximately 0.4. Assume that the “trick” die rolls a 6 half the time. (Remember I was watching as my opponent also rolled 5′s, 4′s, and 3′s.) Then the probability that I would win a battle is at least 0.4 x 0.5 = 0.2. The attacker odds are actually higher since the attacker would usually win if the defender rolls anything but a 6. My estimate is that with the trick die, the defender would win with frequency around 0.6. So the probability that the defender would win 24 battles is around 0.6^24 or about 1-in-100,000.
“And yes, it’s unlikely your friend would (a) weight your dice, (b) waste this ability on a meaningless game of risk, and (c) keep up the act all these years, but it’s not 1-in-100-billion unlikely.”
There is also (d), even with a “trick” die the event would only be expected to happen with frequency 1-in-100,000. Now combine that low probability with the low probabilities of (a), (b), and (c) also being true. I agree that it is more likely that (a), (b), (c), and (d) are all true is more likely than that a 1-in-100-billion event happened. However, I’m not claiming a 1-in-100-billion event happened. I’m claiming that it is more likely that something unknown occurred, i.e., I have no scientific explanation for the event.
Yes, you do: all four dice were weighted. You did your math assuming only one of them was weighted, but if they all were then the event you saw wasn’t unlikely at all. Assume that a weighted die rolls the side that it favors with probability p, each of the sides adjacent to it with probability (1-p)/4, and never rolls the side opposite the favored side. How strongly weighted do the dice have to be (that is, what should p be) for 26 consecutive victories for the defender are assured?
The defender automatically wins on a 5 or 6, which come up with probability p + (1-p)/4. If the defender rolls a 2, then for the defender to win, each of the attacker’s dice must either be a 1 (which it is with probability p) or a 2 (with probability (1-p)/4), so the defender wins in this case with probability (p+(1-p)/4)^3. The cases where the defender rolls a 3 or 4 are similar. Summing all the cases, we get that the defender wins with probability
To win 26 times in a row with 50% probability, the defender would have to win each battle with probability 0.974. To win 26 times in a row with 95% probability, the defender would have to win each battle with probability 0.998.
(1/64)(-9p^4-6p^3+54p+25) > .974
--> p > .841
(1/64)(-9p^4-6p^3+54p+25) > .998
--> p > .958
In other words, to produce the event you saw with 50% reliability would require weighted dice that worked 84% of the time. To produce the event you saw with 95% reliability would require weighted dice that worked 96% of the time. I’m unable to find any good statistics on the reliability of weighted dice, but 84% sounds about right.
I used three reddish, semi-transparent plastic dice with white dots (as I always did). My opponent used standard opaque, plastic ivory dice with black dots. I noticed nothing unusual about the dice and by the end of the run I was examining dice, cups, methods of rolling closely.
“Assume that a weighted die rolls the side that it favors with probability p, each of the sides adjacent to it with probability (1-p)/4, and never rolls the side opposite the favored side.”
This assumption does not match my recollection of the dice rolls. As I stated previously, I rolled 6′s, 5′s, 4′s, 3′s, 2′s, and 1′s. I also never rolled a 1,1,1 which should happen frequently if my dice were heavily weighed to roll 1′s. Nor do I remember rolling large numbers of 1′s.
Your probability model for a trick die also fails to match my observations of my opponents die rolls. E.g., in your model my opponent would be expected to roll similar numbers of 5′s, 4′s, 3′, and 2′s. However, he only rolled a 2 once and he rolled far more 5′s than 3′s.
Besides with your probability model for trick dice, I would have easily noticed if my opponent rolled a 6 84% of the time and I never rolled a 6 at all.
PS You used 26 in the above calculation. I had 26 armies and in Risk the attacker must have at least 4 armies to roll three attack dice. So the 3vs1 dice scenario only happened 23 times.
When you say “that is unbelievable control”, you seem to be assuming the exact outcome with trick dice would be exactly and entirely predetermined. But there’s no reason to think that. The trick dice would only have to make winning much more likely to pull your “impossible” odds down into the realm of the possible. What you describe as a die that “occasionally rolled every number except a 1” is what you would expect to see if the “1″ side were weighted a bit—it would often roll a 6, sometimes roll a number adjacent to 6, and never roll 1. Contrariwise, it’s possible that the three dice facing it could have been rigged to do poorly. If a die with the “1” side weighted faced three dice with the 6 side weighted, that could do the trick.
Some amount of dice rigging could make your loss expected or reasonably likely but not guaranteed. And yes, it’s unlikely your friend would (a) weight your dice, (b) waste this ability on a meaningless game of risk, and (c) keep up the act all these years, but it’s not 1-in-100-billion unlikely. People playing little tricks or experimenting on their friends is something that does happen in the world as we know it, therefore it could have happened to you.
Though I like Jack’s explanation too.
″...it would often roll a 6, sometimes roll a number adjacent to 6...”
Assuming standard probability applied to my three dice, the odds of my rolling at least one 6 are 1 - (5/6)^3 or approximately 0.4. Assume that the “trick” die rolls a 6 half the time. (Remember I was watching as my opponent also rolled 5′s, 4′s, and 3′s.) Then the probability that I would win a battle is at least 0.4 x 0.5 = 0.2. The attacker odds are actually higher since the attacker would usually win if the defender rolls anything but a 6. My estimate is that with the trick die, the defender would win with frequency around 0.6. So the probability that the defender would win 24 battles is around 0.6^24 or about 1-in-100,000.
“And yes, it’s unlikely your friend would (a) weight your dice, (b) waste this ability on a meaningless game of risk, and (c) keep up the act all these years, but it’s not 1-in-100-billion unlikely.”
There is also (d), even with a “trick” die the event would only be expected to happen with frequency 1-in-100,000. Now combine that low probability with the low probabilities of (a), (b), and (c) also being true. I agree that it is more likely that (a), (b), (c), and (d) are all true is more likely than that a 1-in-100-billion event happened. However, I’m not claiming a 1-in-100-billion event happened. I’m claiming that it is more likely that something unknown occurred, i.e., I have no scientific explanation for the event.
Yes, you do: all four dice were weighted. You did your math assuming only one of them was weighted, but if they all were then the event you saw wasn’t unlikely at all. Assume that a weighted die rolls the side that it favors with probability p, each of the sides adjacent to it with probability (1-p)/4, and never rolls the side opposite the favored side. How strongly weighted do the dice have to be (that is, what should p be) for 26 consecutive victories for the defender are assured?
The defender automatically wins on a 5 or 6, which come up with probability p + (1-p)/4. If the defender rolls a 2, then for the defender to win, each of the attacker’s dice must either be a 1 (which it is with probability p) or a 2 (with probability (1-p)/4), so the defender wins in this case with probability (p+(1-p)/4)^3. The cases where the defender rolls a 3 or 4 are similar. Summing all the cases, we get that the defender wins with probability
p + (1-p)/4 + (1-p)/4 * ((p+(1-p)(3/4))^3 + (p+(1-p)(2/4))^3 + (p+(1-p)(1/4))^3)
Which simplifies to
(1/64)(-9p^4-6p^3+54p+25)
To win 26 times in a row with 50% probability, the defender would have to win each battle with probability 0.974. To win 26 times in a row with 95% probability, the defender would have to win each battle with probability 0.998.
(1/64)(-9p^4-6p^3+54p+25) > .974 --> p > .841
(1/64)(-9p^4-6p^3+54p+25) > .998 --> p > .958
In other words, to produce the event you saw with 50% reliability would require weighted dice that worked 84% of the time. To produce the event you saw with 95% reliability would require weighted dice that worked 96% of the time. I’m unable to find any good statistics on the reliability of weighted dice, but 84% sounds about right.
I’m unable to find any good statistics on the reliability of weighted dice, but 84% sounds about right.
here is a set of loaded dice for sale that are advertised to roll a seven (6 on one, 1 on the other, I think) 80% of the time.
“all four dice were weighted”
I used three reddish, semi-transparent plastic dice with white dots (as I always did). My opponent used standard opaque, plastic ivory dice with black dots. I noticed nothing unusual about the dice and by the end of the run I was examining dice, cups, methods of rolling closely.
“Assume that a weighted die rolls the side that it favors with probability p, each of the sides adjacent to it with probability (1-p)/4, and never rolls the side opposite the favored side.”
This assumption does not match my recollection of the dice rolls. As I stated previously, I rolled 6′s, 5′s, 4′s, 3′s, 2′s, and 1′s. I also never rolled a 1,1,1 which should happen frequently if my dice were heavily weighed to roll 1′s. Nor do I remember rolling large numbers of 1′s.
Your probability model for a trick die also fails to match my observations of my opponents die rolls. E.g., in your model my opponent would be expected to roll similar numbers of 5′s, 4′s, 3′, and 2′s. However, he only rolled a 2 once and he rolled far more 5′s than 3′s.
Besides with your probability model for trick dice, I would have easily noticed if my opponent rolled a 6 84% of the time and I never rolled a 6 at all.
PS You used 26 in the above calculation. I had 26 armies and in Risk the attacker must have at least 4 armies to roll three attack dice. So the 3vs1 dice scenario only happened 23 times.