Two Plus Two Older Archives - Updating Probability Estimates

When we gain information about a particular player's cards
(due to their betting action, for example), I don't know the
mathematically correct way to update the probabilities for
the other players and the deck.

For concreteness, I give a toy problem here.
Suppose we have two players and a five card deck, with
cards A, B, C, D, and E. Each player now has two cards,
with one card remaining in the deck. We don't know
either players' cards -- we are just watching the game.

Somehow we have estimated the probability that each
player has each two card hand. Below, the two columns
give the probability of each hand for Player 1 and
Player 2. The matrix below shows the probability that
they have each card, and the conditional probability
of each second card, given the first. (The columns
and the matrix are just two representations of the same
information.)

Player 1 Prior
--------------
AB .20
AC .05
AD .05
AE .05
BC .20
BD .10
BE .10
CD .10
CE .10
DE .05

___ _ A__ B__ C__ D__ E__
.35 A ___ 4/7 1/7 1/7 1/7
.60 B 1/3 ___ 1/3 1/6 1/6
.45 C 1/9 4/9 ___ 2/9 2/9
.30 D 1/6 1/3 1/3 ___ 1/6
.30 E 1/6 1/3 1/3 1/6 ___

Player 2 Prior
--------------
AB .05
AC .05
AD .15
AE .10
BC .05
BD .10
BE .10
CD .10
CE .15
DE .15

___ _ A__ B__ C__ D__ E__
.35 A ___ 1/7 1/7 3/7 2/7
.30 B 1/6 ___ 1/6 1/3 1/3
.35 C 1/7 1/7 ___ 2/7 3/7
.50 D .30 .20 .20 ___ .30
.50 E .20 .20 .30 .30 ___

And from the above we can infer a probability distribution
on the deck, given below.

Deck Prior
----------
A .30
B .10
C .20
D .20
E .20

Now, we also somehow have an estimate of the probability
distribution of the actions Player 1 will take, depending
on which hand he has, shown below on the left.

Player 1 Action Probablities
----------------------------
___ Fold Call Raise
AB .00 .10 .90
AC .05 .25 .70
AD .05 .30 .65
AE .10 .60 .30
BC .20 .05 .80
BD .40 .10 .50
BE .50 .10 .40
CD .70 .25 .05
CE .80 .20 .00
DE .75 .25 .00

Suppose that Player 1 calls. Then we can update our
estimates of the likelihood that Player 1 has each hand.
This results in the distribution below.

Player 1 After Calling
----------------------
AB .121
AC .076
AD .091
AE .182
BC .061
BD .061
BE .061
CD .152
CE .121
DE .076

____ _ A___ B__ C___ D__ E___
.470 A ____ .257 .162 .194 .387
.304 B .398 ____ .201 .201 .201
.410 C .185 .149 ____ .371 .295
.380 D .239 .161 .400 ____ .200
.440 E .414 .139 .275 .173 ____

Now, Player 1's action should also lead us to update our
estimates for the deck and Player 2. A naive first try
is to adjust the deck's and Player 2's single card
probabilities so that each card's location probabilities
sum to 1. For example, Player 1's A-probability went from
.35 to .47, so we multiply Player 2's and Deck's A-probability
by .53/.65. Similarly, we multiply the probabilities for
B, C, D, and E by .696/.40, .59/.55, .62/.70, .56/.70,
respectively.

Naive Deck Estimate After Player 1's Call
-----------------------------------------
A .245
B .174
C .215
D .177
E .160
-----
0.971 = sum

Naive Player 2 Estimate After Player 1's Call
---------------------------------------------
A .285
B .522
C .375
D .443
E .400
-----
2.025 = sum

But these don't even sum to 1 and 2, as they must. So
this approach is too simple-minded.

Another approach would be to solve a system of linear
equations. We want

Deck(A) + Player2(A) = .530
Deck(B) + Player2(B) = .696
Deck(C) + Player2(C) = .590
Deck(D) + Player2(D) = .620
Deck(E) + Player2(E) = .560

Deck(A) + Deck(B) + Deck(C) + Deck(D) + Deck(E) = 1.0
Player2(A) + Player2(B) + Player2(C) + Player2(D) + Player2(E) = 1.0

But these equations in no way account for the prior distributions
estimated for the deck and Player 2.

So, what is the mathematically correct way to update our
estimated distributions for the deck and Player 2 in this
situation? The answer should satisfy all of the equations
above, but also in some way reflect the prior estimated
distributions.

Thanks in advance!