You would still have the BOM available for naval combat, that would remain unchanged. You just get rid of the 600 pound gorilla of how to make Heavy Bombers not unbalance the naval campaigns while still allowing the Allies to use a CBO offensive against Germany.
Heavy bombers and SBR in LHTR

it is well known that the expected value for damage on an SBR is 3.5 IPCs (given the bomber survives AA fire and given the territory is 6 IPCs or more). there are a few posts taking a stab at the distribution for heavy bombers in LHTR, but no one has formulated it correctly. the work below shows the average SBR damage is 5.472 IPCs per heavy bomber.
in LHTR, heavy bombers roll 2 dice, take the max, add one, limit to the territory value, and that’s the IPC damage inflicted.
let A and B be independent random variables corresponding to the two dice rolled. they are distributed as:
P(A=a) ={ 1/6 a=1,2,3,4,5,6
{ 0 otherwiseP(B=b) ={ 1/6 b=1,2,3,4,5,6
{ 0 otherwiselet C be the derived random variable defined as:
C = max(A,B) + 1Theorem: C is distributed as:
P(C=c) ={ 1/36 c=2
{ 3/36 c=3
{ 5/36 c=4
{ 7/36 c=5
{ 9/36 c=6
{ 11/36 c=7
{ 0 otherwiseProof:
note: <= is the lessthanorequalto operator
note: E is the summation operator
P( C<=c ) = P( max(A,B)+1 <= c )6
= E P( max(A,b)+1 <= c  B=b ) P(B=b)
b=16
= E P( max(A,b) <= c1 ) P(B=b)
b=1c1
= 1/6 * E P( A <= c1 ) c<=7
b=1c1 c1
= 1/6 * E E P(A=a) c<=7
b=1 a=1c1 c1
= 1/36 * E E 1 c<=7
b=1 a=1= 1/36 * (c1)^2 c<=7
and
P(C=c) = P(C<=c)  P(C<=c1)
={ 1/36 c=2
{ 3/36 c=3
{ 5/36 c=4
{ 7/36 c=5
{ 9/36 c=6
{ 11/36 c=7
{ 0 otherwiseAnalysis:
note: E is the expectation operator
so, if one were heavy SBR’ing a 7 (or more) territory, the expected damage is:
E = 21/36 + 33/36 + 45/36 + 57/36 + 69/36 + 711/36 = 197/36 = 5.472 IPCso is the extra ~2 IPC per bomber worth the investment? i’d like to know what other analysts think now that i’ve quantified the difference.