Re: poker

In article <1139282339.260273.161690@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
<matt271829-news@xxxxxxxxxxx> wrote:
Robert Israel wrote:
In article <1139180329.641246.96860@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
<matt271829-news@xxxxxxxxxxx> wrote:
PARASHAR wrote:
The two players each ante 1 unit to a pot. Then each player receives a
random number uniformly distributed between 0 and 1. Each player knows
the value of his number but not the value of his opponent's number. The
first player is then given an opportunity to bet one additional unit. If
the first player doesn't bet there is a showdown and the player with the
highest number collects the antes. If the first player bets the second
player may call by matching the bet or drop out (giving the antes to the
first player). If the second player calls there is again a showdown and
the player with the highest number collects the pot (consisting of 4
units, the bets and the antes). If both players follow their optimal
strategy what is the value of the game? In other words if they play
(optimally) a large number of games how much is the first player
expected to win (or lose if the value is negative) per game?

Following the suggestion in some of the other replies, I have worked
through the whole thing again under the assumption that A's optimal
strategy is to bet if and only if his number is less than some
predetermined number c or greater than some other predetermined number
d. Rather astonishingly, since my calculations cover about six pages, I
arrived at the same answer as others have given - namely that the best
choice is c = 0.1, d = 0.7, when the value of the game for A is 0.1.

If A follows this strategy, I make it that in order to peg his losses
to the best case of -0.1, B should call if his number is greater than
0.7, should fold if his number is less than 0.1, and between these
values can do either.

I did the calculations under the assumption that A and B both know each
other's optimal strategies and behave accordingly. (This must be the
case since if we can work it out then so can they.)

At the risk of repeating what has already been said, I am still very
unclear about several points:

1. How do we know that A's optimal strategy is to bet if his number is
less than some number c or greater than some number d? The "most
obvious" strategy (my first attempt, assume that A will bet if his
number is greater than some predetermined cutoff) is not optimal, so
what confidence do we have that this new one - the "second most
obvious" - has any better chance of being optimal?

2. How do we know that A cannot profit by deviating from this supposed
optimal strategy under the assumption that B will continue playing as
if A was following it? The more I think about this the more confused I
get... it's a kind of double-triple-quadruple bluff thing spiralling on
forever.

3. How do we know that A's optimal strategy will be to follow the same
rule for every game (if a series of games is to be played)?

The optimal strategies for A and B should satisfy the basic optimality
conditions:
1) if A follows his optimal strategy, the expected payoff to A is always
at least v, no matter what B does.
2) if B follows his optimal strategy, the expected payoff to A is always
at most v, no matter what A does.

Unless I've made an error, these conditions are both satisfied for the
claimed optimal strategies, with v = 1/10.

Thus if A follows his claimed optimal strategy and B bets with probability
g(t) when A bets and he is dealt t, the expected payoff is

int_0^{1/10} ds (int_0^s dt (1 + g(t)) + int_s^1 dt (1 - 3 g(t)))
+ int_{1/10}^{7/10} ds (int_0^s dt - int_s^1 dt)
+ int_{7/10}^1 ds (int_0^s dt (1 + g(t)) + int_s^1 dt (1 - 3 g(t)))
= 7/25 + int_0^{1/10} dt (2/5 - 4 t) g(t)
+ int_{7/10}^{1} dt (14/5 - 4 t) g(t)
Since 2/5 - 4 t > 0 for 0 < t < 1/10 and 14/5 - 4 t < 0 for
7/10 < t < 1, this is minimized if g(t) = 0 for t < 1/10 and g(t) = 1
for t > 7/10, which makes the expected payoff 1/10.

Next, suppose B follows his claimed optimal strategry and A bets with
probability f(s) when he is dealt s. The expected payoff is

int_0^{2/5} ds (int_0^s dt +int_s^{2/5} dt (-1 + 2 f(s)) +
int_{2/5}^1 dt (-1 - f(s)))
+ int_{2/5}^1 ds (int_0^{2/5} dt + int_{2/5}^s dt (1 + f(s))
+ int_s^1 dt (-1 - f(s))
= int_0^{2/5} ds (1/5 - 2 s) f(s) + int_{2/5}^1 ds (-7/5 + 2 s) f(s)

This is maximized when f(s) = 1 for 0 < s < 1/10, 0 for 1/10 < s < 7/10,
and 1 for 7/10 < s < 1, which makes the expected payoff 1/10.


Thanks for your reply Robert. I confess at the moment I don't really
follow what you are doing here. I have tried the same calculation -
assume that A's strategy is to bet with probability p(a) when his
number is a. The idea is then to find the function p such that A's
expectation is maximised.

No, that's not the idea. The idea is to find the function p(a) that
maximizes the minimum, over all possible strategies for B, of the
expected payoff. Since that's not so easy to find directly, I started
by making some assumptions about the form of the optimal strategies,
which luckily turned out to be correct. The clincher is to show, as
I did above, that

1) if A follows the strategy I found for him (bet with < 1/10 or > 7/10,
otherwise pass), then no matter what B does, the expected payoff is at
least 1/10. This is done by minimizing the expected payoff when A follows
this strategy and B uses probability g(t) to bet when he is dealt t.

2) if B follows the strategy I found for him (bet with > 4/10, otherwise
fold), then no matter what A does, the expected payoff is at most 1/10.
This is done by maximizing the expected payoff when B follows this
strategy and A uses probability f(s) to bet when he is dealt s.

Robert Israel israel@xxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.

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