roll bias within a game

[gobruinss]

I have a question. I realize that we all remember the bad rolls and forget about the good rolls, but I was wondering if anyone had ever seen a situation where in one particular game they got consistently bad rolls. A few times I had a dominant position in one particular game (2556424) and by having bad rolls, let the person back in the game. Yet I have had games where I kept getting good rolls and noticeably fewer bad rolls over the course of the game. I was wondering if anyone else had noticed this phenomenon and whether there could be some programming bias? It just seems to be more than randomness.

Thanks!

[Timminz]

It happens. Random, means, anything can happen.

[nukie]

You also have to take into fact that other people are rolling dice a the exact same time. All together will be random, what you get, may not appear as random.

[detlef]

I have a question. I realize that we all remember the bad rolls and forget about the good rolls, but I was wondering if anyone had ever seen a situation where in one particular game they got consistently bad rolls. A few times I had a dominant position in one particular game (2556424) and by having bad rolls, let the person back in the game. Yet I have had games where I kept getting good rolls and noticeably fewer bad rolls over the course of the game. I was wondering if anyone else had noticed this phenomenon and whether there could be some programming bias? It just seems to be more than randomness.

Thanks!

To answer your first question, absolutely. I recall off the top of my head a few games where I merely needed marginal dice to put the thing away and failed to even get that. Mind you, I was in a game recently where I was about to get pounded and my opponent lost like 13 taking 2 in a 14v3 attack and that was enough to give me an edge that I never lost the rest of the game.

So, for the 1,000,000th time, suck it up. The dice are not rigged or biased and they can't be made "more random".

[MeDeFe]

My games are usually too long for bad rolls to happen all the time, but I remember a particular game where I got bad rolls at exactly the worst moment possible, I had 35 armies against 20 on 12 territories or thereabout, should have been able to take out one player, get the cards and then cripple the other, but I lost my first 1 or 2 rolls against every country I attacked and the other team won.

But I've also auto-attacked 600vs600 and ended up with 150 left.

[detlef]

But I've also auto-attacked 600vs600 and ended up with 150 left.

At first glance that didn't seem so extraordinary as most people tend to forget how much the attacker is favored in large battles. I dropped it into the odds calculator but it can't handle battles that large so I broke it down into 100 v 100 battles looking at what the odds would be of ending up with an extra 25 guys per batch of 100. Well, long story short, that's some nice freaking dice! Though I'm sure there's some reason why working it out this way isn't 100% accurate, it would have to be way off. After all, I came up with something like .02% chance of that happening!

[Thezzaruz]

Though I'm sure there's some reason why working it out this way isn't 100% accurate, it would have to be way off. After all, I came up with something like .02% chance of that happening!

Yea you are missing something. A 100 v 100 battle has about 26 % chance of leaving 25 or more attackers and as those odds will just get even better with bigger armies I can't see it being that outrageous tbf. But still WD.

[e_i_pi]

Though I'm sure there's some reason why working it out this way isn't 100% accurate, it would have to be way off. After all, I came up with something like .02% chance of that happening!

Yea you are missing something. A 100 v 100 battle has about 26 % chance of leaving 25 or more attackers and as those odds will just get even better with bigger armies I can't see it being that outrageous tbf. But still WD.

Average armies lost vs 600 is about 480, assuming you are attacking with 3 the entire time.

Kj

[MeDeFe]

Though I'm sure there's some reason why working it out this way isn't 100% accurate, it would have to be way off. After all, I came up with something like .02% chance of that happening!

Yea you are missing something. A 100 v 100 battle has about 26 % chance of leaving 25 or more attackers and as those odds will just get even better with bigger armies I can't see it being that outrageous tbf. But still WD.

Average armies lost vs 600 is about 480, assuming you are attacking with 3 the entire time.

Kj

So 25% above the expected value.

[FabledIntegral]

I've had games where I've had the advantage and lost it due to craptacular dice, but also moments where I should have had it np, had shit dice, and barely just gotten it.

I still remember one time I had 15 v 3 for the win (eliminating another player and getting a sweep), and another player Maniacmath was coming from Greenland (speed freestyle) to get to Japan to try to beat me (I was on Kamchatka, so he could have autoed me). I auto 15 v 3... end result is 3v2 left. But I won the 3v2, thank god! Things like that just make you hold your breathe...

Another time I autoed 45 v 22 and came with 3v1 left... oh man I would have been pissed if I lost that

I've lost 9v1 before 3x though. I think it's a particular strand out of the 500,000 recycled numbers used (afaik, that's how it works), that I just manage to continue to stumble upon unluckily (correct me if I'm wrong, but it doesn't matter how many other people are rolling at the same time if you use auto attack?).

[detlef]

Though I'm sure there's some reason why working it out this way isn't 100% accurate, it would have to be way off. After all, I came up with something like .02% chance of that happening!

Yea you are missing something. A 100 v 100 battle has about 26 % chance of leaving 25 or more attackers and as those odds will just get even better with bigger armies I can't see it being that outrageous tbf. But still WD.

Why wouldn't it work this way:

For each 100v100, you have 25% chance of having 25 left over. So, you put those 25 in the bank and roll another 100v100, again, you have a 25% chance of having 25 left over, and so on.

Why wouldn't the odds of ending up with 150 left over be .25*.25*.25*.25*.25*.25?

[MeDeFe]

Though I'm sure there's some reason why working it out this way isn't 100% accurate, it would have to be way off. After all, I came up with something like .02% chance of that happening!

Yea you are missing something. A 100 v 100 battle has about 26 % chance of leaving 25 or more attackers and as those odds will just get even better with bigger armies I can't see it being that outrageous tbf. But still WD.

Why wouldn't it work this way:

For each 100v100, you have 25% chance of having 25 left over. So, you put those 25 in the bank and roll another 100v100, again, you have a 25% chance of having 25 left over, and so on.

Why wouldn't the odds of ending up with 150 left over be .25*.25*.25*.25*.25*.25?

You're counting exactly 25 left, you should really be calculating with 25 or more.

[Thezzaruz]

Average armies lost vs 600 is about 480, assuming you are attacking with 3 the entire time.

So 25% above the expected value.

Not really. 5 % less armies lost is a better way to see it, i.e nothing spectacular at all.

Why wouldn't it work this way:

For each 100v100, you have 25% chance of having 25 left over. So, you put those 25 in the bank and roll another 100v100, again, you have a 25% chance of having 25 left over, and so on.

Why wouldn't the odds of ending up with 150 left over be .25*.25*.25*.25*.25*.25?

You're counting exactly 25 left, you should really be calculating with 25 or more.

And even with that he'll still be off. Can't just cut it up and consider it to be 6 different events.

[detlef]

Though I'm sure there's some reason why working it out this way isn't 100% accurate, it would have to be way off. After all, I came up with something like .02% chance of that happening!

Yea you are missing something. A 100 v 100 battle has about 26 % chance of leaving 25 or more attackers and as those odds will just get even better with bigger armies I can't see it being that outrageous tbf. But still WD.

Why wouldn't it work this way:

For each 100v100, you have 25% chance of having 25 left over. So, you put those 25 in the bank and roll another 100v100, again, you have a 25% chance of having 25 left over, and so on.

Why wouldn't the odds of ending up with 150 left over be .25*.25*.25*.25*.25*.25?

You're counting exactly 25 left, you should really be calculating with 25 or more.

No I'm counting with at least 25. Which, of course, adds it's own set of complications to the equation.

[detlef]

Average armies lost vs 600 is about 480, assuming you are attacking with 3 the entire time.

So 25% above the expected value.

Not really. 5 % less armies lost is a better way to see it, i.e nothing spectacular at all.

Why wouldn't it work this way:

For each 100v100, you have 25% chance of having 25 left over. So, you put those 25 in the bank and roll another 100v100, again, you have a 25% chance of having 25 left over, and so on.

Why wouldn't the odds of ending up with 150 left over be .25*.25*.25*.25*.25*.25?

You're counting exactly 25 left, you should really be calculating with 25 or more.

And even with that he'll still be off. Can't just cut it up and consider it to be 6 different events.

Actually, you sort of can considering that you're not calculating the chance of pulling off the attack, but doing so with a bunch left over. The only reason why you can't chop it up is because doing so brings in the chance of wearing your attackers down to 3 or 2 where you don't get to roll 3 dice. If anything, it favors the attacker doing it that way because calculating it this way means the defender rolls 1 die 6 times as opposed to once. Of course, that's a pretty small wrinkle.

Now, I said right off the bat that this wasn't completely exact, but I do think you are overestimating the effect of breaking it up. Again, you need to average having 25 of each 100 left over after each attack, so for every 100 of his guys you take, you need to only lose 75. I don't really think that breaking it up affects the numbers dramatically.

Would you care to explain why?

[Thezzaruz]

Now, I said right off the bat that this wasn't completely exact, but I do think you are overestimating the effect of breaking it up. Again, you need to average having 25 of each 100 left over after each attack, so for every 100 of his guys you take, you need to only lose 75. I don't really think that breaking it up affects the numbers dramatically.

Would you care to explain why?

Because you can't cut it up the way you did. That assumes it to be 6 different battles instead of 1 big one and that screws massively with the odds.

A simple example. 100v100 the odds of the attacker being left with at least 1 army is 82.5%. The odds of a 600v600 leaving at least 6 would then, if doing it your way, be .825 x/y 6 that comes out at about 31%. Or just below 70% chance that the attacker is left with 5 or less armies. You don't have to know/understand much about the 3v2 advantage to realize that ain't realistic...

(btw, 70% chance of 5 left is reached at about 15v15 battles, anything bigger and it will be more probable )

[Thezzaruz]

(btw, 70% chance of 5 left is reached at about 15v15 battles, anything bigger and it will be more probable )

Hmmm, that was poorly thought and worded late last night. The rest still stands though.

[detlef]

Now, I said right off the bat that this wasn't completely exact, but I do think you are overestimating the effect of breaking it up. Again, you need to average having 25 of each 100 left over after each attack, so for every 100 of his guys you take, you need to only lose 75. I don't really think that breaking it up affects the numbers dramatically.

Would you care to explain why?

Because you can't cut it up the way you did. That assumes it to be 6 different battles instead of 1 big one and that screws massively with the odds.

A simple example. 100v100 the odds of the attacker being left with at least 1 army is 82.5%. The odds of a 600v600 leaving at least 6 would then, if doing it your way, be .825 x/y 6 that comes out at about 31%. Or just below 70% chance that the attacker is left with 5 or less armies. You don't have to know/understand much about the 3v2 advantage to realize that ain't realistic...

(btw, 70% chance of 5 left is reached at about 15v15 battles, anything bigger and it will be more probable )

Well, just because it obviously doesn't make sense at the extreme you've used to illustrate the folly doesn't mean it's not a reasonable approximation at the other. The higher the odds of having 'at least' x amount of guys left, the higher the odds of significantly more than x guys left. So, in your 82.5% chance deal, it is very likely that you'll have more than those extra 5 in the bank by the end of your first 100 and won't need to do any better than trade evenly the rest of the way to get there. After all, there's a 50/50 chance that you'll be 16 up after any given batch of 100 and you've got 6 chances for that 50/50 prop to come through. So, you've got all kinds of reasons why you'll out-pace the rate you need to take armies and only an 18% chance of ever losing any ground at all to the pace over any stretch of 100.

In the case being discussed, however, considering that there's pretty slim odds that you'll end up with an extra 25 per 100 (which is the pace you need to keep them to end up with 150 at the end of the battle) means that it's very, very slim odds that you'll actually get ahead of the curve on any batch of 100. I mean, they go down quickly. To get 10 ahead of the curve on any batch of 100 is a 1 in 12 shot and all that buys you in the big picture is that you've got a 58% chance of having enough left over on one other of the 5 remaining batches to keep pace. So despite managing to pull off a very unlikely advantage, it's barely better than a coin flip that you'll even be able to stay on pace despite the head start. Even if you do, you're still back to square one and still have 4 more batches remaining where you need to succeed at a pace that requires pulling out a 1 in 4 chance.

So, that's why, in terms of an approximation, breaking it up the way I did can make sense even if the same formula doesn't make any at all for the example you used to discredit it.

Now, again, I understand that this equation is not going to deliver an entirely accurate number. However, I don't think it's as far off as you claim.

[Thezzaruz]

In the case being discussed, however, considering that there's pretty slim odds that you'll end up with an extra 25 per 100 (which is the pace you need to keep them to end up with 150 at the end of the battle) means that it's very, very slim odds that you'll actually get ahead of the curve on any batch of 100. I mean, they go down quickly. To get 10 ahead of the curve on any batch of 100 is a 1 in 12 shot and all that buys you in the big picture is that you've got a 58% chance of having enough left over on one other of the 5 remaining batches to keep pace. So despite managing to pull off a very unlikely advantage, it's barely better than a coin flip that you'll even be able to stay on pace despite the head start. Even if you do, you're still back to square one and still have 4 more batches remaining where you need to succeed at a pace that requires pulling out a 1 in 4 chance.

Slim odds??? There is a 26% chance of being left with 25 armies in a 100v100 battle, that's anything but slim tbf. And calling the 1 in 12 for "very, very slim" or "very unlikely" is not at all true.

So, that's why, in terms of an approximation, breaking it up the way I did can make sense even if the same formula doesn't make any at all for the example you used to discredit it.

Now, again, I understand that this equation is not going to deliver an entirely accurate number. However, I don't think it's as far off as you claim.

No, it isn't anywhere close to being a good estimate. It is mathematically completely wrong and hence makes no sense at all.

[detlef]

In the case being discussed, however, considering that there's pretty slim odds that you'll end up with an extra 25 per 100 (which is the pace you need to keep them to end up with 150 at the end of the battle) means that it's very, very slim odds that you'll actually get ahead of the curve on any batch of 100. I mean, they go down quickly. To get 10 ahead of the curve on any batch of 100 is a 1 in 12 shot and all that buys you in the big picture is that you've got a 58% chance of having enough left over on one other of the 5 remaining batches to keep pace. So despite managing to pull off a very unlikely advantage, it's barely better than a coin flip that you'll even be able to stay on pace despite the head start. Even if you do, you're still back to square one and still have 4 more batches remaining where you need to succeed at a pace that requires pulling out a 1 in 4 chance.

Slim odds??? There is a 26% chance of being left with 25 armies in a 100v100 battle, that's anything but slim tbf. And calling the 1 in 12 for "very, very slim" or "very unlikely" is not at all true.

So, that's why, in terms of an approximation, breaking it up the way I did can make sense even if the same formula doesn't make any at all for the example you used to discredit it.

Now, again, I understand that this equation is not going to deliver an entirely accurate number. However, I don't think it's as far off as you claim.

No, it isn't anywhere close to being a good estimate. It is mathematically completely wrong and hence makes no sense at all.

If you don't see a 1 in 4 chance as slim or a 1 in 12 as very unlikely, than I wish I was using your dice. Certainly neither are impossible but slim enough, especially in light of the fact that, after pulling off either of the chances listed, you're still nowhere near out of the woods, the overall odds are quite slim indeed.

Forgive me, but if you can't get on board that something that something only happening 1 out of 12 times is, "very unlikely", than I'm not sure this conversation is worth continuing. And, to be frank, makes me inclined to take your argument with a grain of salt.

[Thezzaruz]

Forgive me, but if you can't get on board that something that something only happening 1 out of 12 times is, "very unlikely", than I'm not sure this conversation is worth continuing. And, to be frank, makes me inclined to take your argument with a grain of salt.

Oh I wouldn't put my hope on a 1 in 12 event in a risk game if I can help it there we agree. But calling it "very very slim" just isn't true in a wider perspective. Unless you're using a very loose definition of "slim".

[detlef]

Forgive me, but if you can't get on board that something that something only happening 1 out of 12 times is, "very unlikely", than I'm not sure this conversation is worth continuing. And, to be frank, makes me inclined to take your argument with a grain of salt.

Oh I wouldn't put my hope on a 1 in 12 event in a risk game if I can help it there we agree. But calling it "very very slim" just isn't true in a wider perspective. Unless you're using a very loose definition of "slim".

I'm sorry, I missed the day in class when "very, very slim" was specifically defined. I fail to see what sort of argument you're trying to make if the best you can do is question whether a proposition that fails over 90% of the time is not "very, very slim".

[Thezzaruz]

I'm sorry, I missed the day in class when "very, very slim" was specifically defined. I fail to see what sort of argument you're trying to make if the best you can do is question whether a proposition that fails over 90% of the time is not "very, very slim".

I agree, discussing semantics won't do us any good. But as you still haven't realized that the way you tried to calculate the original question is wrong I'll give it one last try.

Take a 100v100 battle and ask "what is the odds of being left with at least 25% of the armies?". We'll start by doing it your way (splitting it into several small battles, lets say 5 this time) and that gives us a 45.15% chance of ending with at least 5 armies. That gives us 0.4515 x/y 5 = 0.0187, i.e almost a 2% chance. And the odds when doing a straight calculation is (as we know from earlier in this thread) 25.97%.


So we have:
(5 left in 20v20) * 5 = 2%
25 left in 100v100 = 26%

That huge difference means that your way of doing it isn't a good or even a reasonable estimate but rather incredibly wrong both intuitively and mathematically.

[detlef]

I'm sorry, I missed the day in class when "very, very slim" was specifically defined. I fail to see what sort of argument you're trying to make if the best you can do is question whether a proposition that fails over 90% of the time is not "very, very slim".

I agree, discussing semantics won't do us any good. But as you still haven't realized that the way you tried to calculate the original question is wrong I'll give it one last try.

Take a 100v100 battle and ask "what is the odds of being left with at least 25% of the armies?". We'll start by doing it your way (splitting it into several small battles, lets say 5 this time) and that gives us a 45.15% chance of ending with at least 5 armies. That gives us 0.4515 x/y 5 = 0.0187, i.e almost a 2% chance. And the odds when doing a straight calculation is (as we know from earlier in this thread) 25.97%.


So we have:
(5 left in 20v20) * 5 = 2%
25 left in 100v100 = 26%

That huge difference means that your way of doing it isn't a good or even a reasonable estimate but rather incredibly wrong both intuitively and mathematically.

Well then, I'll concede here. Though I don't think Pi's assertion that losing 120 is "average" is correct either. According to another odds calculator, attacking 3v2, the attacker loses .85 armies for every 1 army the defender loses. Thus, "average" losses on such attack would be 505 or 95 armies left over. So, by that, MeDeFe ended up with more than 50% more guys than he "should " have. So, I tried to think about what could be wrong with this and the first thing that came to mind is the linear nature of the equation vs the fact that the attacker has an ever increasing armies advantage over the defender. However, regardless of how many more guys he has, he's still rolling 3v2, so assuming that the .85 number is true, he's never going to out pace that ratio until he's rolling 3v1 which won't happen until the very last roll or couple. Ironically, the longer he's given the advantage of the increased odds of 3v1, the worse his winnings will be.

I thought I'd try another stab at approximating the chances. I took a smaller battle 150 v 150 and broke that up into 6 pieces like I tried with the first and looked at how many left over after a 25v25 would have the same odds of having 25 after 100v100. At 25v25, there's a 27% of having 9 left over. So, I ran a 150v150 to see what the odds of having 54 left over on that attack. It came to 3%. Which is obviously several orders of magnitude more likely than the .02% that I originally thought.

There seem to be two factors at play here. One being that there's a rather unavoidable trend for the attacker to amass extra armies vs the defender. However, there's also the fact that the longer you roll dice, the more likely you are to trend back to neutral odds. So, any proposition requiring favorable odds is less and less likely to work out over the long haul.

Doing something that has a 1 in 4 chance of working out once is no big deal. However, defying those same odds continually over a long period of time is.

The first factor obliterates my theory, but the second seems to do the same for Pi's

Of course, as has been rather exposed by Thezzaruz, my math is incredibly rusty.

[Thezzaruz]

However, there's also the fact that the longer you roll dice, the more likely you are to trend back to neutral odds. So, any proposition requiring favorable odds is less and less likely to work out over the long haul.

Doing something that has a 1 in 4 chance of working out once is no big deal. However, defying those same odds continually over a long period of time is.

Yea I agree, and my initial statement of it not needing that much luck might be as far off as your 0.02%.