Operation Hollywood : Flying the Japanese Air Force off the coast of Gilbraltar

It might be better if done as a 1-2 punch.

On Italy -1 …The one Italian TR unloads a Tank+Inf on Malta.  ( The UK CRU being eliminated by Italian  Sub+Air)… Mayyybe  also Use Italian SBR  on the Land attack… (hoping for the AAA to miss)
  The Tank hit it likely taken by the UK on the AAA

ON Germany -2…  Hit with 4 SBRs

Very very risky… but doable… more so on live games… than in Low-luck Triple A

4 Ftr’s  and 3 Bmbrs.

I don’t have setup charts in front of me for Global 40. In my games we have British Commandos, Marines, Germany Air borne and others. You could have it were you buy paratrooper troop just for air transport like you suggested. Give it a few. Other guys will be posting too with there ideas.

If we can get a live game on triple A, that would be sweet.  I want to see this in action.

@wittmann:

Agreed. That NO could even replace the Strategic outer defensive perimeter(Guam, Midway etc.).
Conquering China is difficult and much effort is required; this would be the reward.

Yeah, and it is at least possible too. That 5 island NO is next to impossible. I have had games where Japan won on the Pacific board and still didn’t get that NO. The only time I have gotten it was when the Allies spent most of their resources in Europe and Japan was allowed to go hog wild.

the best Sydney strike I have seen is when Japan takes West Aus, lands planes on it next round, and sacks Sydney the round after.

If you only have one type of troop, it’s a binomial distribution:

Prob(k) = (n choose k) * p^k * (1-p)^(n-k),

where k is the number of hits, n is the total number of units, and p is the probability of a hit.

If you have different units, you need a product of binomial distribution for each permutation of possible events.

So, for the example of two tanks and two infantry, what is the probability of three hits? I’ll in words, then write it out:  You can have either 2 infantry hit, and one tank, or two tanks and one infantry.  Those are the only two ways that you can get two hits:

Prob(3) = (2 choose 2) * (1/6)^(2) * (5/6)^(2-2) * (2 choose 1) * (3/6)^(1) * (3/6)^(2-1) + (2 choose 1) * (1/6)^(1) * (5/6)^(2-1) * (2 choose 2) * (3/6)^(2) * (3/6)^(2-2)

Prob(3) = 1 * 1/36 * 1 * 2 * 3/6 * 3/6 + 2 * 1/6 * 5/6 * 1 * 9/36 * 1

Prob(3) = 18/1296 + 90/1298

Prob(3) = 108/1298

Now, if you have 3 INF, 2 TANKS, and 1 FGT, what is the probability of 2 hits?

Well, how many ways are there to make two hits?  I’ll write them out:

2 INF, 0 Tanks, 0 Fighters
1 I, 1 T, 0 F
1 I, 0 T, 1 F
0 I, 1 T, 1 F
0 I, 2 T, 0 F

An example of how one of these terms expands:

1 I, 1 T, 0 F:

(2 choose 1) * (1/6)^(1) * (5/6)^(2-1) * (2 choose 1) * (3/6)^(1) * (3/6)^(2-1) * (1 choose 0) * (3/6)^0 * (3/6)^(2-2)

If you want Prob(2), then you need to take the sum of the probabilities of all those possible ways to make 2 hits.

Clear as mud?

@jeffdestroyer:

if AB and NB stop the blitz, then there will be no blitzing in W. Europe.  All of the TT have something.

Except Holland