I@chompers:
I dunno, without the Japs having some real teeth to their starting army (perhaps not in the form of 5 billion planes) they won’t be able to keep pace with the US economy and will fall far short of Moscow in the same time frame that the Allies are crushing Europe.
With the allies adequately powered-down, it is probably possible to give the japanese the time to cause enough damage to give them a chance.
@San:
Agreed. :-) For a more exciting game, play with tokens or reduce the cost of a researcher to 4 or even 3. Or, keep it at 5 and don’t use tokens, but allow multiple breakthroughs per turn (ala Classic).
Thanks for the intelligent discussion, San Chillos.
Yes, we do play with tokens costing 6 IPC each. Always fun to have to throw a die or two every turn. :-D By the way, you’re welcome. :wink:
My question was when should one expect to hit their first tech, on average, when rolling 1 die per round. In other words, at one point does one get to a 50/50 chance of getting a tech?
In more technical terms, the “average” is the “population mean”. In our case, each “individual” of the population is a single rolling sequence until you get a tech and its value is the number of rolls needed to get it. If you try until you get a tech an infinite or at least an extremely high number of times (let’s say a million times or even more), you will find that the average number of rolls done to get a tech is 6. It’s that simple. Yes, you can get it on your first roll; however, you could need 20 rolls to get it. On average however, you will need 6 rolls i.e. you will get it with your 6th roll.
You have about a 50/50 chance (48.2%) to miss your first 4 rolls so the fifth roll may be the “point” you’re looking for…
In fact, I hit 4 straight techs with the Axis in a game I’m playing from round 2 into round 3. So my point is, you have a 100% of getting zero tech if you never buy a research die and if you never buy a research die you are not projecting the threat of hitting an instant tech to your opponent, who can play less defensively in accordance with the lack of threat. :-)
I agree with you. However, the effective (average) cost for acquiring techs could be much lower and it would not hurt the game at all. I know Larry doesn’t like techs and it shows more than ever. IMO, the AA50 system is much more interesting (and fun).
You have 42.1% chances of getting at least one technology by the 3rd roll. (The probability of missing on your 3 first rolls is (5/6)3 = 0.579 or 57.9% so you will have at least one tech 42.1% of the time.)
By the way Gamerman, what was exactly your question?
I know that 6 is the average number of turns that you would get a given result (in this case, a “6”). But if you take a set of 6 rolls and assume that one is a hit (a “6”) wouldn’t that hit, on average, be in the middle of the 6 rolls? Wouldn’t the average (expected, if you will) first hit of a 6 (a breakthrough) come between attempt #3 and attempt #4?
Average simply means that you should expect to need 6 turns to get a tech if you buy one die every turn i.e, you will get it on your 6th turn with absolute average luck. If it was to hit in the middle of the 6 rolls, the average would be 3.5. However, it’s interesting to note that the probability to hit with the first die (1/6 = 16.7%) is higher than with any other die simply because the probability to need a specific die decreases with the number of dice rolled before it. You only have a probability of (1/6)*(5/6)5 = 6.7% to hit on your 6th turn in fact. However, you have more chances to need at least 10 rolls (you need to miss (5 chances out of 6) on your 9 first rolls (exponent 9) -> (5/6)9 = 19.4% probability) than to hit on your first roll. So, yes, 6 is the average but you should not rely too much on “average luck” simply because the probability to get a very different result (better or worse) is relatively high. (In probability theory, this is related to the variance if you’re interested…)
The number of turns you need on average with x die each turn is given by 1/(probability to get a tech each turn). So, it is n_turnsaverage = 1/(1-(5/6)x).
So, with 1 die every turn, on average, you will get a tech every 6 turns.
Ok… I have more to do than read every post on these forums but as far as I’ve been, allweneed is totally correct. The average cost to get a technology if you buy a number x of die every turn is given by the relation (cost by turn)/(probability to get a tech) where the probability is expressed by a number between 0 and 1. In mathematical terms, we have costaverage = 5*x/(1-(5/6)x).
-Buying 1 die a turn gives an average of 5*1/(1-(5/6)1) = 30 IPCs.
-Buying 2 dice a turn gives an average of 5*2/(1-(5/6)2) = 32.7 IPCs.
-Buying 6 dice a turn gives an average of 5*6/(1-(5/6)6) = 45.1 IPCs.
assuming you only roll 1 die at a cost of 5ipc per die roll then the average number of rolls to acquire a tech is 6. 5ipc * 6rolls = 30ipc = average cost to acquire tech.
This is not how averages are calculated. The question is largely meaningless. If you really wanted to calculate the average cost then you’d have to record every game of Axis and Allied played and find out how many IPC’s were spent to aquire tech by the multitude of players. Then you’d have to add those numbers all together and divide by the total number of players.
You’re just making up some arbitrary fomula that sort of seems that it should fit and calling it the average cost. Then your claiming that your average cost cannot be disputed and building an entire thread around it.
No, allweneed is right. Holden, your method doesn’t work in theory unless you try it an infinite number of times. Good luck… :wink:
In conclusion, yes, techs really suck in this game… :-(
I@chompers:
I dunno, without the Japs having some real teeth to their starting army (perhaps not in the form of 5 billion planes) they won’t be able to keep pace with the US economy and will fall far short of Moscow in the same time frame that the Allies are crushing Europe.
With the allies adequately powered-down, it is probably possible to give the japanese the time to cause enough damage to give them a chance.
The problem is not Italy. It’s the awfully bad setup of Pacific that has about 3 times the number of aircrafts that it should have. I doubt Larry will ever propose a brand new setup so it looks like it’s gonna be “don’t play global” or “do it yourself”. Personally, I prefer the second option by far…
IMO, the global game will never show it’s full potential as long as the Pacific setup won’t be fixed. By fixed I mean scaling this setup at the same level as that of Europe. At least 2/3 of the aircrafts have to be removed along with a few land troops here and there (and a lot of troops from the ugly stack in eastern Russia). Only then, the game will have some chances of being balanced (and fun).
I can’t believe Larry never saw this coming when he approved the official setup in the Pacific game. Its current attempts at trying to balance the Pacific game are like putting a plaster on a cancer.
At least now I’m conviced to try my modified setup at my first global game.
@Imperious:
Cruisers don’t do this in the real world.
Its not realistic. They don’t do anything to enemy destroyers in that regard. The solution must be somewhat realistic.
or drop to 11 and add in ASW and one round AA fire
Yeah, I see your point. But I see the problem on the opposite angle. The purpose of my rule is not to increase the strength of cruisers but rather to decrease that of destroyers against larger and better armed vessels. As far as I know, large ships have been removed from fleets after WW2 because they were getting too vulnerable to aircrafts. I don’t think that strengtening cruisers would be a very good idea because of balance vs aircrafts that is already fine.
There are other options for destroyers though. Maybe preventing a hit from a roll of 2 to be allocated to a cruiser, a battleship or an aircraft (or something like this)…