You sit down with a well mixed deck containing A cards marked "+" and B cards marked "—". You may draw cards from this deck as long as you want, i.e., you can stop playing at any point. Each time you draw a + card you are given $1 and each time you draw a — card you have to pay $1. Cards are
not replaced after having been drawn.
What would be a fair amount to pay for the right to play (i.e., what is the expected payoff) and under what circumstance should a player cease drawing?
(In reply to
re: Incorrect assumption by ed bottemiller)
Leming and I have been going on the assumption that the "fair amount" is paid once at the beginning of the game, rather than at each draw of a card.
Based on the table (extended farther to the right than I've posted), the following are the largest numbers of -'s that would be allowable to continue playing with a given number of +'s:
# +'s/max -'s to keep trying
1 1
2 3
3 4
4 6
5 7
6 8
7 9
8 11
9 12
10 13
11 14
12 15
As a sampling, the following program simulates starting with 8 +'s and 10 -'s.
DEFDBL A-Z
DIM hitOn(12)
DATA 1,3,4,6,7,8,9,11,12,13,14,15
FOR i = 1 TO 12: READ hitOn(i)
PRINT i; hitOn(i)
: NEXT
RANDOMIZE TIMER
totWinnings = 0
DO
winnings = 0
plus = 8: minus = 10
DO
r = INT(RND(1) * (plus + minus) + 1)
IF r <= plus THEN
winnings = winnings + 1
plus = plus - 1
ELSE
winnings = winnings - 1
minus = minus - 1
END IF
LOOP UNTIL minus > hitOn(plus) OR plus = 0
totWinnings = totWinnings + winnings
gameCt = gameCt + 1
PRINT totWinnings, gameCt, totWinnings / gameCt
LOOP
I halted it at 109,459 trials of this game using the above strategy for stopping and the accumulated winnings were 39,124, for an average winnings per full game of 0.35743. The table previously presented shows, to the limit of accuracy of that table, 0.36.
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Posted by Charlie
on 2008-03-12 15:20:55 |
re(2): Incorrect assumption
