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The odds, assuming a ROF of 3, are around 1 in 262144. While that's impressive, it isn't going to win you the powerball.How about 18 ROFs?

JR

Kieth Spurlock did that to me. Only stopped because he had nothing left to shoot at. -- jimHow about 18 ROFs?

Gravity sucks. (Slow reverse camera pan revealing I said this while looking at myself in the mirror)Sorry, have to disagree. The force in this case is gravity and mass. If a die is not weighted correctly it will effect how it rolls. Now that doesn't mean you need "precision" dice but getting "precision" dice takes that one fear off the table. It's because I use those dice that it never pops into my head that maybe the dice are weighted incorrectly. - Also why are "precision" dice not "normal" dice? Is it not the desired definition of normality for a dice to be well-weighted?

As far as your decision making portion - Iwantto agree but I've seen too many players make all the right choices/decisions but have those decisions nullified by bad dice rolls vs good dice rollsat the right time or juncturein a game. To me there are 3 pillars to an ASL game and you need two to win:

1. Skill - this is obvious. If your skill is better than the other guys or you're playing up to your level, and the other guy isn't.

2. Situation - mostly game balance, but sometimes this can be when you're a good player but you don't do as well with Japanese as some but you're great playing Russians.

3. Fate - I don't mean dice rolls, I mean the right dice rolls at the right time. You can roll the same number of snakes as the other guy but if he's rolling snakes on shots when he really needs them to happen and you're rolling snakes on things like PTC's - that's fate, imo. Fate is a part of the game although I see tons of people refusing to admit this.

If your opponent is a better player, then you better have great fate and perhaps a scenario slightly favoring your side.

If you're playing a game that is weighted against you, you better be a better player and a little better fate wouldn't hurt.

If you're playing a guy who is going through a long streak of luck, then you better be a better player....and so on.

Doubtless I will get a lot of responses to this arguing that everything about ASL is skill, but deep down we know that's simply not true, just as it's not exactly true in actual combat. Fate and situation play a huge role. Same with sporting events too. That's actually one of the reasons I love ASL - it's what makes it more "combat-like" or "sport-like". If I want skill only and a perfectly balanced scenario with no fate, I'll play chess.

How about 18 ROFs?

I was there, Jack went on a tear. I believe he finally ran out of targets.The odds, assuming a ROF of 3, are around 1 in 262144. While that's impressive, it isn't going to win you the powerball.

JR

If you haven't Chi-square tested your precision dice (or 'normal' dice) for fairness, don't be so sure. And even if they do pass a Chi-square test for fairness, if you are banging them around in a glass cup, you may need to re-test them after awhile. How's that for popping something into your headSorry, have to disagree. The force in this case is gravity and mass. If a die is not weighted correctly it will effect how it rolls. Now that doesn't mean you need "precision" dice but getting "precision" dice takes that one fear off the table. It's because I use those dice that it never pops into my head that maybe the dice are weighted incorrectly. ...

A definite YES!The less I care the better I roll.

Was it a precision glass cup?If you haven't Chi-square tested your precision dice (or 'normal' dice) for fairness, don't be so sure. And even if they do pass a Chi-square test for fairness, if you are banging them around in a glass cup, you may need to re-test them after awhile. How's that for popping something into your head

I played a guy who rolled 9s, 10s, 11s, and 12s on all his MCs and half way through the scenario he did throw his dice across the room. The new set did improve his checks! I know, it was the averaging of the dice.I don't care if people do this as long as it doesn't result in dice being hurled across the room

Somehow it's your fault for having enough targets in sight for 18 shotsKieth Spurlock did that to me. Only stopped because he had nothing left to shoot at. -- jim

I've never seen that long a streak; the longest I've experienced (on the good side of it) was 7 or 8, and I did stop because there were no targets in sight. Fewer than 10 is still better than one chance in a thousand, so it's bound to happen over the years if you play enough games.

My units where in +1 and +2 TEM. He was shooting a .50 cal / 9-2 combo. The worst possible DR is a 9 (1MC final). He also managed to TH/TK a lightly armored AFV. When he was done, he still had rate, and I had a single broken HS on the map. He said he didn't have the heart to kill him and asked if I wanted to get a beer instead. -- jimSomehow it's your fault for having enough targets in sight for 18 shots

I've never seen that long a streak; the longest I've experienced (on the good side of it) was 7 or 8, and I did stop because there were no targets in sight. Fewer than 10 is still better than one chance in a thousand, so it's bound to happen over the years if you play enough games.

10 consecutive [3] ROF shots is (1/2)^10 = 1/1024, 10 [2] shots is (1/3)^10 = 1/59049, 10 [1] shots is (1/6)^10 = 1/60466176.

Probability of not getting 10 [3] in a game thus is 1023/1024 (0.9990234375). Let us take a high estimate of 50 games in a year. So the probability of getting no 10+ [3] runs = (1023/1024)^50 (0.952322), leaving a probability of at least one 10 [3] streaks at 1-(1023/1024)^50 = 0.0476779 or 4.76%. Which is not too far off 50/1024 that a quick and crude estimate would give.

To break even (50/50) you need a little more complex maths. You need the probability of something (P) or not something (Q) = 0.5 and I will use P(N)/Q(N) to denote the probability of something/not something after N attempts. So Q(N) = (Q(1))^N = 0.5. Q(1) = 1023/1024. (1023/1024)^N = 0.5 which if we take the log of each side, log((1023/1024)^N) = log(0.5), N*log(1023/1024)=log(0.5), N=log(0.5)/log(1023/1024). Whether you use 10 or e based logarithms doesn't matter, but that works out at N=709.436. So you need 710 games to have a 50/50 chance of getting a streak of 10.

For a streak of 9, P(1) = 1/512, Q(1) = 511/512 (0.998046875) , Q(50) = (511/512)^50 thus P(50) = 1 - (511/512)^50 = 0.0931259 or 9.31%. Again let's see how many games for a 50/50 chance. N=log(0.5)/log(511/512) = 354.545, so you need 355 games to break just even.

Approaching from another direction what would the probability of a single streak need to be to hit 50/50 in 50 games. We saw N=log(0.5)/log(Q(1)), N=50 so we get 50=log(0.5)/log(Q(1)), 50*log(Q(1))=log(0.5), log(Q(1))=log(0.5)/50, Q(1)=10^(log(0.5)/50) = 50th root of 0.5 = 0.986233. Thus P(1) = 1-Q(1) = 0.0137673 or 1/72.636. That's between a streak of 6 (1/64) and 7 (1/128). So only a streak of 6 or less is required to have a 50/50 or better in 50 games.

Of course the assumption of only 1 qualifying streak in a game might be fine for streaks of 8 or 9 or greater, but is much shakier for below that, but the above estimates for streaks of 9 and 10 will be very close with 355 and 710 games required just to get to 50/50 respectively. To get to 75/25 in favour just double 355 or 710 for the number of required games (710, 1420).

The whole point of the above is that gut feelings in such matters is too often wrong. While I strongly doubted a year would do it, I suspected that a few years would also be an underestimation and wanted to find out. For 50 games/year, which I would regard as a good number, it would take 7 and 14 years for 9 and 10 ROF streaks to just have a 50% chance, never mind become quite likely (75%+, ~ 14 and 28 years).

You are right about < 10 shots (9 [3] = 1/512, 8 [3] = 1/256, etc), but let's see what the probability of getting at least one 10 [3] in a year. I will make an assumption that more than a single 10 streak is highly unlikely within a single game, mainly as others have pointed out, you tend to run out of targets.Fewer than 10 is still better than one chance in a thousand, so it's bound to happen over the years if you play enough games.

Probability of not getting 10 [3] in a game thus is 1023/1024 (0.9990234375). Let us take a high estimate of 50 games in a year. So the probability of getting no 10+ [3] runs = (1023/1024)^50 (0.952322), leaving a probability of at least one 10 [3] streaks at 1-(1023/1024)^50 = 0.0476779 or 4.76%. Which is not too far off 50/1024 that a quick and crude estimate would give.

To break even (50/50) you need a little more complex maths. You need the probability of something (P) or not something (Q) = 0.5 and I will use P(N)/Q(N) to denote the probability of something/not something after N attempts. So Q(N) = (Q(1))^N = 0.5. Q(1) = 1023/1024. (1023/1024)^N = 0.5 which if we take the log of each side, log((1023/1024)^N) = log(0.5), N*log(1023/1024)=log(0.5), N=log(0.5)/log(1023/1024). Whether you use 10 or e based logarithms doesn't matter, but that works out at N=709.436. So you need 710 games to have a 50/50 chance of getting a streak of 10.

For a streak of 9, P(1) = 1/512, Q(1) = 511/512 (0.998046875) , Q(50) = (511/512)^50 thus P(50) = 1 - (511/512)^50 = 0.0931259 or 9.31%. Again let's see how many games for a 50/50 chance. N=log(0.5)/log(511/512) = 354.545, so you need 355 games to break just even.

Approaching from another direction what would the probability of a single streak need to be to hit 50/50 in 50 games. We saw N=log(0.5)/log(Q(1)), N=50 so we get 50=log(0.5)/log(Q(1)), 50*log(Q(1))=log(0.5), log(Q(1))=log(0.5)/50, Q(1)=10^(log(0.5)/50) = 50th root of 0.5 = 0.986233. Thus P(1) = 1-Q(1) = 0.0137673 or 1/72.636. That's between a streak of 6 (1/64) and 7 (1/128). So only a streak of 6 or less is required to have a 50/50 or better in 50 games.

Of course the assumption of only 1 qualifying streak in a game might be fine for streaks of 8 or 9 or greater, but is much shakier for below that, but the above estimates for streaks of 9 and 10 will be very close with 355 and 710 games required just to get to 50/50 respectively. To get to 75/25 in favour just double 355 or 710 for the number of required games (710, 1420).

The whole point of the above is that gut feelings in such matters is too often wrong. While I strongly doubted a year would do it, I suspected that a few years would also be an underestimation and wanted to find out. For 50 games/year, which I would regard as a good number, it would take 7 and 14 years for 9 and 10 ROF streaks to just have a 50% chance, never mind become quite likely (75%+, ~ 14 and 28 years).

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10 consecutive [3] ROF shots is (1/2)^10 = 1/1024, 10 [2] shots is 1/59049, 10 [1] shots is 1/60466176.

You are right about < 10 shots ([9] = 1/512, [8] = 1/256, etc), but let's see what the probability of getting at least one 10 [3] in a year. I will make an assumption that more than a single 10 streak is highly unlikely within a single game, mainly as others have pointed out, you tend to run out of targets.

Probability of not getting 10 [3] in a game thus is 1023/1024 (0.9990234375). Let us take a high estimate of 50 games in a year. So the probability of getting no 10+ [3] runs = (1023/1024)^50 (0.952322), leaving a probability of at least one 10 [3] streaks at 1-(1023/1024)^50 = 0.0476779 or 4.76%. Which is not too far off 50/1024 that a quick and crude estimate would give.

To break even (50/50) you need a little more complex maths. You need the probability of something (P) or not something (Q) = 0.5 and I will use P(N)/Q(N) to denote the probability of something/not something after N attempts. So Q(N) = (Q(1))^N = 0.5. Q(1) = 1023/1024. (1023/1025)^N = 0.5 which if we take the log of each side, log((1023/1024)^N) = log(0.5), N*log(1023/1024)=log(0.5), N=log(0.5)/log(1023/1024). Whether you use 10 or e based logarithms doesn't matter, but that works out at N=709.436. So you need 710 games to have a 50/50 chance of getting a streak of 10.

For a streak of 9, P(1) = 1/512, Q(1) = 511/512 (0.998046875) , Q(50) = (511/512)^50 thus P(50) = 1 - (511/512)^50 = 0.0931259 or 9.31%. Again let's see how many games for a 50/50 chance. N=log(0.5)/log(511/512) = 354.545, so you need 355 games to break just even.

Approaching from another direction what would the probability of a single streak need to be to hit 50/50 in 50 games. We saw N=log(0.5)/log(Q(1)), N=50 so we get 50=log(0.5)/log(Q(1)), 50*log(Q(1))=log(0.5), log(Q(1))=log(0.5)/50, Q(1)=10^(log(0.5)/50) = 50th root of 0.5 = 0.986233. Thus P(1) = 1-Q(1) = 0.0137673 or 1/72.636. That's between a streak of 6 (1/64) and 7 (1/128). So only a streak of 6 or less is required to have a 50/50 or better in 50 games.

Of course the assumption of only 1 qualifying streak in a game might be fine for streaks of 8 or 9 or greater, but is much shakier for below that, but the above estimates for streaks of 9 and 10 will be very close with 355 and 710 games required just to get to 50/50 respectively. To get to 75/25 in favour just double 355 or 710 for the number of required games (710, 1420).

The whole point of the above is that gut feelings in such matters is too often wrong. While I strongly doubted a year would do it, I suspected that a few years would also be an underestimation and wanted to find out. For 50 games/year, which I would regard as a good number, it would take 7 and 14 years for 9 and 10 ROF streaks to just have a 50% chance, never mind become quite likely (75%+, ~ 14 and 28 years).

OK. Now my brain broke.

You are missing an important thing here - each time your HMG (or other ROF 3 weapon) starts firing, it has a one in 1024 chance to get 10 consecutive shots (significantly less than that if it's a MG without leader direction because of cowering). But in a typical game, the same weapon will tend to start firing more than once - once per player turn at the maximum. Let's say it's about once per game turn, so in a 6 turn game, your chances for such a long streak are closer to 0.6%. At 50 games a year, you're close to 30% in a year (I know probabilities don't add this way, but it's close enough for estimates). If something "rare" has a 30% chance to happen to you in a given year, then as I said, play long enough and it's highly likely that it will happen.You are right about < 10 shots ([9] = 1/512, [8] = 1/256, etc), but let's see what the probability of getting at least one 10 [3] in a year. I will make an assumption that more than a single 10 streak is highly unlikely within a single game, mainly as others have pointed out, you tend to run out of targets.

Probability of not getting 10 [3] in a game thus is 1023/1024 (0.9990234375). Let us take a high estimate of 50 games in a year. So the probability of getting no 10+ [3] runs = (1023/1024)^50 (0.952322), leaving a probability of at least one 10 [3] streaks at 1-(1023/1024)^50 = 0.0476779 or 4.76%. Which is not too far off 50/1024 that a quick and crude estimate would give.

The mythical "dice bot" effect.

Hugga Bugga.

Hugga Bugga.

You have a point, that's assuming there are enough targets or targets that survive long enough to receive 10 shots. Breakdown is NA in such a rate tear (though a captured B11, now B9, could seem like a B#/ROF combination possibility, by being captured weapon its ROF is reduced by 1, so 8 now is the highest ROF capable DR).You are missing an important thing here - each time your HMG (or other ROF 3 weapon) starts firing, it has a one in 1024 chance to get 10 consecutive shots (significantly less than that if it's a MG without leader direction because of cowering). But in a typical game, the same weapon will tend to start firing more than once - once per player turn at the maximum. Let's say it's about once per game turn, so in a 6 turn game, your chances for such a long streak are closer to 0.6%. At 50 games a year, you're close to 30% in a year (I know probabilities don't add this way, but it's close enough for estimates). If something "rare" has a 30% chance to happen to you in a given year, then as I said, play long enough and it's highly likely that it will happen.

You are quite right to point out that each time a MG initially fires (twice per GT), it could have a chance of such a tear, 12 times in a game. However running out of targets is likely with long tears. As well, when you have a sufficient multitude of targets, it's likely that you fail ROF after a small number or no extra shots. Simple ROF factors (IE losing ROF) are not the only terminating condition, target elimination and/or player risk analysis ("I'm just as likely to break my MG as break the enemy") are external and independent of the ROF mechanism. You need the combination of both ROF

You did raise an important and correct point, but given the other variables involved, the real result is going to be something in between our rough calculations.

(And yes it happened not later than five days ago, i lost 35% of my OB in one fire phase and conceded shortly after).