Note to the Teacher: This is just an introduction to both games with arbitrary parameters. the analyses are done later, seperately, either in student projects or in text pages.
The two families of games described and partially analyzed in this chapter are classics. VNM POKER has been introduced by von Neumann and Morgenstern in their monograph [von Neumann/Morgenstern]. KUHN POKER (in the variant (3,1,1,2)) was introduced by Kuhn in [Kuhn 1950].
In both games we have four paramters, S, r, n, and m. There are cards of value from 1 to S in a stack, and each value occurs r times. So overall there are S·r cards. There are two players, Ann and Beth. Each player randomly gets a card from the deck, looks at her card, but doesn't show the value to the opponent. The ante (initial bet) is m, both player put m dollars into the pot.
VNMPOKER(S,r,m,n): Now Ann moves first by either checking (playing for m) or raising (playing for n).
KUHNPOKER(S,r,m,n): These games extend VNMPOKER insofar as they add VNMPOKER with Ann and Beth playing reversed roles in case of Ann checking. Again Ann moves first by either checking or raising.
Before we analyze both games by looking at their reduced normal forms,
let us show the extensive form, since with them the descriptions above may become a little clearer.
Actually we first will not show the full extensive form, but rather "modules",
which are put into a skeleton to give the full extensive form for the games.
Note that in both games Nature moves first by giving both players a card. Ann can get any of the cards 1,2,...S, and Beth as well (provided r>1), thus Nature has S2different alternatives if r>1, and S·(S-1) alternatives if r=1. This alternatives for the random move are not all equally likely. All alternatives with equal cards for Ann and Beth have the probability
Let me give four examples of extensive forms of these games. Can you identify the modules? Interestingly the information sets span different modules.
What is a strategy for the first player Ann, what is a strategy for the second player Beth? Although there are S·S different positions after dealing one card for each player, Ann cannot distinguish those where Ann has the same card and Beth has different cards. Therefore Ann has only S information sets, depending on the card ("1" to "S") she has in her hand. In each of these situations she can decide to check or to raise. T, all of them reduced. Again we display these strategies by sequences of "R"s and "C"s--- for instance "RCCR" for S=4 means that Ann raises with the lowest and the highest card and checks with the two middle-valued cards.
What are Beth's pure strategies? In case Ann checks, Beth doesn't have any decision to make, so it suffices to consider the other case, when Ann raises. Beth has S informations sets for her move, since she doesn't know Ann's card, only her own. In each of these S information set, she will either call or fold. So she has also 2S pure strategies, which are also obviously all reduced, since Beth moves only at most once. Again the strategies are encoded by sequences of "C"s and "F"s---"FFCC" for S=4 means that she folds with the two lower-valued cards and calls with the two higher-valued cards.
In order to find the payoff for Ann if both player use a given pure strategy, we have to add the S·S products of probabilities for that outcome and payoff in that case--- m, 0, or -m in the case where Ann checks, m in case Ann raises and Beth folds, and n, 0, or -n in the case where Ann raises and Beth calls. For example, if S=2, and if Ann plays CR and Beth CC, Ann's expected payoff equals equals
Fortunately some of these pure strategies are weakly dominated. When Ann holds a highest valued card, raising weakly dominates checking. Any of Ann's strategies "X...YC" where Ann would check with a card of value S would be weakly dominated by the strategy "X...YR" that coincides with "X...YC" in all information sets except the one for the card value S, where Ann would raise. When Beth holds a highest valued card, calling dominates folding. Any of Beth's strategies "X...YF" with Beth folding with a card of value S would be weakly dominated by "X...YC" the strategy identical to "X...YF" except for the information set holding a card of value S, where Ann would raise. Thus, in the case S=4, there only remain the 8 pure strategies CCCR, CCRR, CRCR, CRRR, RCCR, RCRR, RRCR, RRRR, for Ann, and only the 8 pure strategies FFFC, FFCC, FCFC, FCCC, CFFC, CFCC, CCFC, CCCC, for Beth.
One can even go a little further:
Actually we will show that a pure strategy for Beth, let's call it "Strategy 1",
to call with card value "i" and fold with card value "i+1"
is weakly dominated by the modified "Strategy 2" where everything remains the same except that now Beth
folds with a card value of "i" and calls with card value "i+1".
Out of the S·S terms that add to Beth's payoff, all are the same for the two strategies
except those 2S for card values "i" and "i+1" for Beth.
However, the contribution where Beth has a "i" and Ann a "k" for Strategy 1 equals the contribution
where Beth has a "i+1" and Ann a "k" in Strategy 2, provided k is different from i and i+1. Therefore
all contributions to the payoff except the four where Ann and Beth both have cards of value "i" and "i+1"
can be matched in both strategies.
All what remains to be done is to compare these four parts of the payoff in the two strategies.
Thus for S=4 and n/m < (2r-1)/(r-1), there only remain the 4 pure strategies FFFC, FFCC, FCCC, and CCCC for Beth.
The first player Ann has 2·S information sets in this game: Before her first move, she can either hold a card from 1 to S. Or Ann is in the position where she has checked in this move and Beth has raised. Again she knows only her own card in this situation, thus there are also S of these information sets. Beth has also 2·S information sets, determined by her own card from 1 to S and whether Ann has raised (the first S information sets) or checked (the second S information sets).
Therefore Ann has 22·S pure strategies, indicated by S letters "R" or "C" for her choice of raising or checking depending on her card from low to high, and then S letters "F" or "C" for folding or checking when she has checked at the beginning and Beth did raise, again depending on Ann's card from low to high. For instance, for S=2 we get the pure strategies "CCFF", "CCFC", "CCCF", "CCCC", "CRFF", "CRFC", "CRCF", "CRCC", "RCFF", "RCFC", "RCCF", "RCCC", "RRFF", "RRFC", "RRCF", and "RRCC". But if Ann raises with a given card it doesn't matter if she would later check or fold with the same card. For this reason, the reduced strategy with an "R" in position x (1 ≤ x ≤S) would have a "•" in position S+x. In the case S=2 the reduced pure strategies are "CCFF", "CCFC", "CCCF", "CCCC", "CRF•", "CRC•", "RC•F", "RC•C", and "RR••".
Beth has also 22·S pure strategies. ... In the case S=2 these are "FFCC", "FFCR", "FFRC", "FFRR", "FCCC", "FCCR", "FCRC", "FCRR", "CFCC", "CFCR", "CFRC", "CFRR", "CCCC", "CCCR", "CCRC", and "CCRR". All these strategies are already reduced, for every S.
Next some of these (reduced) pure strategies can be eliminated because they are weakly dominated: If Ann holds a highest card, and checked while Beth did raise, it is always better for her to call. When she would fold, she would always lose m units, wheras with calling she cannot lose. But note that if Ann holds a highest value card, raising does not dominate checking. It all depends on Beth's pure strategy. Assume Beth would always fold when Ann raises, but does always raise when Ann checks. Then, in case Ann holds a highest valued card but beth not, raising would give Ann a payoff of m units, but checking would give her a payoff of n.
If Beth holds a highest valued card, calling weakly dominates folding, and also raising weakly dominates checking.
This is the example analyzed by Kuhn in his 1950 paper. It has a rather interesting property that Ann has a contiuunm of optimal strategies, whereas Beth has only one ..... The game is slightly in favor of Beth, since the expected payoff for Ann is -1/18 when both play optimally.
In this variant, Ann and Beth both have just one optimal strategy: If Ann has a card of value "3" she always raises. If she has a "2" she never raises. All this seems reasonable. But now the "bluffing" occurs if Ann has a "1". Then she would raise (bluff) in 1/5 of the cases and check in the remaining 4/5 of the cases. Beth's behavior is more linear monotone, depending on the card she holds. If Ann raises, then Beth would always call if she has a card of value "3", in 3/5 of the cases if she holds a "2", and never if she holds a "1". If Ann checks in the first round, then Beth would always raise if she holds a "3", but never otherwise. Now if Beth raises, then Ann would always fold with a "1" (there is no one to bluff left, since Beth will not move anymore), but will call in 4/5 of the cases with a "2", but only in 1/2 of the cases with a "3". Why the latter is the case I don't know. The game is still a little unfair to Ann, since he expected payoff for Ann when both play optimally is -1/30.
SIMULTANEOUS VNMPOKER(S,r,m,n): In the simultaneous version both players decide simultaneously whether they want to raise for m or for n. If one of them raises for n and the other for m, the one daring the higher amount (n) wins m from the other one, regardless what the cards show. If both raise the same amount, the one with the higher card wins n respectively m from the other one, again, no win for draws of identical cards.
Both player have four pure strategies: Betting low in both cases ("LL"), Betting low with a card of value "1" and high with a card of value "2" ("LH"), the counterintuitive strategy of raising with a card of value "1" and low with a card of "2" ("HL"), and raising high for both cards ("HH"). Remember that the probability of both having a card of calue "1" is (r-1)/(2(2r-1)), wheras the probability for Ann getting a "1" and Beth getting a "2" is the slightly higher r/(2(2r-1)). Then for each pair of strategies, the four cases of card distributions have to be condsidered, the payoffs noted, and the expected value as sum of probability multiplied by payoff, for all these four cases has to be computed. We get the following payoff matrix:
| LL | LH | HL | HH | |
| LL | 0 | -m(r-1)/(2(2r-1)) | -m(3r-1)/(2(2r-1)) | -m(4r-2)/(2(2r-1)) |
| LH | m(r-1)/(2(2r-1)) | 0 | r(n-m)/(2(2r-1)) | (nr-2mr+m)/(2(2r-1)) |
| HL | m(3r-1)/(2(2r-1)) | r(m-n)/(2(2r-1)) | 0 | (1-r(n+2m))/(2(2r-1)) |
| HH | m(4r-2)/(2(2r-1)) | (-nr+2mr-m)/(2(2r-1)) | (r(n+2m)-1)/(2(2r-1)) | 0 |
Choose m=1, n=3, r=4:
| LL | LH | HL | HH | |
| LL | 0 | -3/14 | -11/14 | -6/14 |
| LH | 3/14 | 0 | 8/14 | 6/14 |
| HL | 11/14 | -8/14 | 0 | (1-r(n+2m))/(2(2r-1)) |
| HH | 6/14 | -6/14 | (r(n+2m)-1)/(2(2r-1)) | 0 |