Help with statistical question!
On February 20th I was at a game of Global war 1936. This is the third day long session we’ve spent trying to finish the same game. It was a 16 hour day lol; and the game still isn’t done….
There was a HUGE debate, over a simple statistical question, (That occurred TWICE in the game) that lead to a number of nasty comments -
#1. that my mind is too stupid to comprehend basic statistics, and
#2. that because I mentioned I attended church on occasion, sometime 5+ years ago, that I must not understand ‘science’ because I’m religious.
Please appreciate we were about 15 hours deep the 2nd time it happened
Humorous elements aside lol… I want to post the question to the community of what happened so here goes.
- This occurred on two occassions
- The game is played with D12’s
- Britain attacks a German position with 3 infantry 1 artillery 1 marine or other unit…
- Infantry attack at 2 on a D12
- Artillery attack at 3 on a D12
- The Artillery is designated to support the marine/other unit.
Britain rolls it’s attack and makes an error, instead of rolling 3@2 and 1@3 separately, Britain has accidently rolled all four dice together.
The Result of Britain’s roll is as follows: 8,9,6,3
BOTH parties agree a full reroll is not required,
BOTH parties agree that a simple statistical roll can determine whether or not the “3 result” would associate with the artillery and be a HIT, or associate with an infantry piece and be a MISS.
Roll a D12. on a 1,2 or 3, the artillery would be considered a hit; and the odds IMO are exactly the same as just rerolling the artillery attack. The chance being 1 in 4.
I envisioned this as reaching in to a bag and removing a random die. You would have a 1/4 chance of pulling the “3” result.
My Opponents interpretation
Please bare in mind that I do NOT understand my opponents position; so it’s hard to do it justice. (perhaps someone can elaborate?)
But in plain words it was his interpretation that on the reroll dice, that any roll from 4-12 would be a hit, a 3/4 chance.
I believe this came from his inverse look at the question, that because there were so many infantry, its a fact they will get 2 misses+; and therefore its more likely that an infantry will get the undesirable dice; and that somehow from that look at the question - it is more likely the artillery would get the “3” result dice.
My opponent rolled a 12, and demanded that the hit stand despite my expressed intent that only a 1,2 or 3, should be a hit.
To move the game forward I begrudgingly let the hit stand on both occasions this occurred. My opponent also apologized profusely for his over the top commentary. An Apology I accept.
We probably spent upwards of 40 minutes to an hour disputing this stupid and mostly irrelevant debate; I said I would post this question on the community boards for comment, and my opponent insisted that he has a degree in statistics, and that unless a PROFESSOR told him he was drunk, that he didn’t care what anyone else had to say.
So before I pursue a professor’s comment, I’m opening this up to the community for review. -in the event my interpretation is somehow wrong!?
Please let me know what you guys think!
Well - I don’t have a degree in pure statistics, but in a subject which involved a fair degree of it. Then again it was long ago and I have forgotten far more than I remember!
As I understand it your position is that there were 4 dice, so a one in 4 chance that the 3 was the artillery. Makes sense to me. Whether that’s roll of 1,2 or 3 or any other selection of three numbers of 12 is irrelevant, so long as those numbers are agreed beforehand.
Your opponent’s position makes no sense to me, unless:
1. You were not agreed on the 3 numbers that would apply? Sorry - I am not suggesting such an obvious cause, but it would account for the problem. Or
2. I don’t know Global War 1936, but there may be some omissions in your description? You seem to list 5 units but 4 dice. You don’t say what the marine/other unit has to roll. You don’t say how the artillery supports? These gaps may explain your opponent’s position.
If you are lucky someone who knows this game will save you having to explain any of this.
I haven’t studied in detail the situation you described, so my analysis may be faulty, but my impression is as follows. I think the root cause of the problem is that you and your opponent confused two different concepts.
Your opponent was approaching the problem from the point of view of figuring the percentage chances of an infantry hit and the percentage chances of an artillery hit, on the basis of the combat conditions that existed on the battle board. You were approaching the problem from the point of view of figuring out whether the “3”-rolled die (out of four pre-rolled dice) should be applied as an artillery hit, on the basis of the argument that “one of those dice needs to be treated differently from the other ones because it should have been rolled separately in the first place.” In other words, you wanted to achieve this fourth-die isolation on a purely random basis (a 25% chance) whereas he wanted to achieve this fourth-die isolation on a basis that reflected the probabilities that governed the original dice roll; in effect, what he wanted to do was to use that single supplementary roll as a kind of re-roll (in miniature) of the original 4-dice roll.
A cleaner solution would have been to simply re-do the entire original roll. For some reason, both parties agreed that a full reroll was not required – which in and of itself wasn’t a bad decision. The real error was in cobbling together on-the-spot a resolution mechanism without reaching a full and clear prior agreement on exactly how it would work. And my feeling is that if working out such a mechanism proves to be excessively complex or contentious, then this shows that the solution is worse than the problem it’s trying to solve and that it would be better to just re-roll the original faulty roll.
Hey Panic, I will answer.
#1. Â It was not a matter of dice “agreement”. Â I went in detail to explain that I don’t think it’s reasonable that on his “roll to see if he hit” that he should have a 75% chance of success. Â He was adamant that 4-12 would be a hit for him.
#2. The Marine/other unit had yet to roll (as it was rolled separately after). Â SO essentially it doesn’t matter for the question.
Fair points Mark.
I was opposed to the reroll on the basis I didn’t want my opponent getting another chance at 3 hits
But you are 100% right, that would have been the better way to resolve it. Mistaken roll is garbage - no matter the result.
If you don’t agree on a re roll, which I do agree with Gar ( plus his 4 -12 thats BS ) then I would have given him 1 hit for the 3 and moved on.
We have done it both ways. To be honest never had a blow up about it.
OOHH wait I rolled 4 dice on mistake. 1,2,2,4. Yes I agree no re roll. :evil: :evil: :evil:
After 14 hours I would have dropped a baby nuke on the game. Allies win !
what he wanted to do was to use that single supplementary roll as a kind of re-roll (in miniature) of the original 4-dice roll.
Mmm! The original dice roll can be aggregated into a single probability event:
(3 inf x 1/6) + (1 art x 1/4) does equal a 3/4 chance of 1 hit
I cannot see any merit in this approach. It ignores the original roll entirely. If that is the intention it would be best represented by a re-roll of all the dice.
The only other possibility I can see, Garg, is that your opponent is aggregating the remaining rolls - the probability of the art having hit + the supported marine/other unit/whatever (I am still not clear) into that single dice. If the supported marine/other unit/whatever have a 50% chance of a hit then he would get to his 3/4. That would make more sense, but in your earlier reply you say that the remaining unit was rolled separately afterwards.
Might either of these be the case? :?
The most likely explanation is probably (statistics again!) an excess of alcohol.
P.S. I see that your opponent - or someone who agrees with him - has voted. Perhaps he can explain?
PP we all know that vote was imperious leader. For certain.
Another poor choice to add to his record:)
Thanks for comments guys!
Clyde85 last edited by
Your opponents logic is extremely flawed. In a throw of the dice his Artillery is just as likely to have rolled the 8 as it is the 3. If you were to allow a roll of 4 dice, but singled out the artillery die with a separate color, you would see that regardless of its to hit value it can just as easily roll something undesirable, like an 8 or 9, as it could roll a hit.
The best solution would have been to consider the roll of 4 dice all misses for his infantry, and then have him roll for his artillery again.
Kreuzfeld last edited by
This might be late, but I feel like I could contribute.
If you don’t want to reroll the question is simple. The question is: “Which result was rolled by which unit?”. There are several ways of doing this I will first describe the simple and then the longer (Which will give the same result).
Simple: Since most of the result where misses, they are not interesting, the only question is where to assing the 3. There is 1 in 4 units that could have been the unit rolling a 3, terefore it is 1 in 4 chance that unit was the artillery.
In basic uniform probability, we first start with the assumption that every result is equally likely. Every string of lottery numbers have the same odds. all we have to do then is to count the number of outcomes we like, and compare that to the number of possible outcomes when assigning the results. I will call your units I1, I2, I3 and A. There will be 24 ways of assinging the dies, and 6 of them will assign the 3 to the art. The ways of assigning them are:
A: 3 I1: 6 I2:8 I3:9 = 1 Hit
A: 3 I1: 6 I2:9 I3:8 = 1 Hit
A: 3 I1: 8 I2:6 I3:9 = 1 Hit
A: 3 I1: 8 I2:9 I3:6 = 1 Hit
A: 3 I1: 9 I2:6 I3:8 = 1 Hit
A: 3 I1: 9 I2:8 I3:6 = 1 Hit
A: 6 I1: 3 I2:8 I3:9 = 0 Hit
A: 6 I1: 3 I2:9 I3:8 = 0 Hit
A: 6 I1: 8 I2:3 I3:9 = 0 Hit
A: 6 I1: 8 I2:9 I3:3 = 0 Hit
A: 6 I1: 9 I2:3 I3:8 = 0 Hit
A: 6 I1: 9 I2:8 I3:3 = 0 Hit
A: 8 I1: 3 I2:6 I3:9 = 0 Hit
A: 8 I1: 3 I2:9 I3:6 = 0 Hit
A: 8 I1: 6 I2:3 I3:9 = 0 Hit
A: 8 I1: 6 I2:9 I3:3 = 0 Hit
A: 8 I1: 9 I2:3 I3:6 = 0 Hit
A: 8 I1: 9 I2:6 I3:3 = 0 Hit
A: 9 I1: 3 I2:6 I3:8 = 0 Hit
A: 9 I1: 3 I2:8 I3:6 = 0 Hit
A: 9 I1: 6 I2:8 I3:9 = 0 Hit
A: 9 I1: 6 I2:9 I3:8 = 0 Hit
A: 9 I1: 8 I2:6 I3:3 = 0 Hit
A: 9 I1: 8 I2:3 I3:6 = 0 Hit
In total. That is 6 possible hits, of 24 possible scenarios. Which is 1 in 4.