A gambler choice
A gambler has $2. He is allowed to play a game of chance four times,
and
his goal is to maximize his probability of ending up with a least
$6. If
the gambler bets $ b dollars on a play of the game, then with
probability 0.4, he wins the game and increase his capital position
by
$ b dollars with probability 0.6, he loses the game and decreases
his
capital position by $ b dollars. On any play of the game, the
gambler
may not bet more money than he has available. Determine a betting
strategy
that will maximize the gambler's probability of attaining a wealth of
at least
$ 6 dollars by the end of the fourth game. We assume that bets of
zero dollars
(that is, not betting) are permissible.
Define ft(d) to be the probability that by the end of game 4, the
gambler will have
at least $6, given that she acts optimally and has d dollars
immediately before
the game is played for the tth time.
If we give the gambler a reward of 1 when his ending wealth is at
least $6 and a reward of 0
if it is less than, ft(d) will equal the maximum expected award
that can
be earned during games t, t+1, ... 4. if the gambler has d
dollars immediately
before the tth play of the game.
If the gambler is playing the game for the fourth and final time, his
optimal
strategy is clear: if he has $6 or more, don't bet anything, but if
he has less
than $6, bet enough money to ensure (if possible) that he will have
$6 if he wins
the last game. Note that if he begins game 4 with $0, $1, or $2,
there is no way
to win (no way to earn a reward of 1). This reasoning yield
the following results:
|
f4(0) = f4(1) = f4(2) = 0 |
|
|
f4(3) = f4(4) = f4(5) = 0.4 |
|
For t £ 3, we can find a recursion for ft(d) by noting that if
the gambler
has d dollars, is about to about to play the game for the tth
game, and bets
b dollars, then the following summarizes what can occur:
- with probability 0.4 win the game t, expected reward
ft+1(d+b);
- with probability 0.6 lose the game t, expected reward
ft+1(d-b).
Thus if the gambler has d dollars at the beginning of game t and
bets b dollars,
the expected reward will be
0.4 ft+1(d+b) + 0.6 ft+1(d-b). This leads to the following
recursion:
|
ft(d) = |
max
0 £ b £ d
|
|
max
| { 0.4 ft+1(d+b) + 0.6ft+1(d-b) }. |
|
Compute f1(2) ?
|
