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BGonline.org Forums
A discrete random walk
Posted By: Bob Koca In Response To: A discrete random walk (Maik Stiebler)
Date: Friday, 24 July 2009, at 4:03 a.m.
"1. What is the optimal strategy for both players?"
Player A doubles at +25 at which point B has an optional take/pass. That is because B is then has teh requirede 1/4 win chance. A does not double before then as ther is no market loss
"2. What is the theoretical(=assuming optimal play from both sides) equity of the starting position?"
A has + 1/3 equity. Going from -50 to +25 is 75 and A starts 2/3 of the way there. 2/3 - 1/3 = 1/3.
"3. Assume Player A deviates from optimal strategy by doubling if and only if the current score is +30. What is the practical equity of the starting position then?"
A wins 50/80 of the games for an equity of +1/4
"4. How much does Player A's deviation from perfect play cost him on average per game?"
1/3 - 1/4 = 1/12
Now put yourself in the position of a bot that knows the optimal strategy and observes the game, not knowing Player A's complete strategy, but noting the wrong plays that follow from the strategy.
"5. a) At which points in the game does Player A, following the non-optimal strategy, blunder away theoretical equity by making a wrong play? b) How much equity does each of these wrong plays lose?"
Only when the game is at exactly 25 does A lose equity. Values below +25 is a no- double and values above +25 are an optional double. The cash is not lost in the next sequence.
At +25 B's has an optional take/pass. Let's suppose he would pass. Then the equity lost by not doubling at + 25 is 1/2 the equity lost by the game reaching 24 (if it reaches +26 it is a cash anyways). At +24 A's theoretical equity is 74/75 - 1/75 = 73/75, a loss of 2/75. So failing to double at +25 costs 1/75 equity.
"6. How often will the opportunity for Player A to make a wrong (=equity losing) play arise in a game? Compute both an average value (a) and a distribution (b). "
It happens at least once in 2/3rds of the games. Suppose that the game is exactly +25. We need to find the probability of that state occurring again. This is harder than the previous questions. When you are at +25 you go to +26 half the time. From there you go to 25 before 30 with probability 4/5. The other half the time you go to 24 and then the chance of returning to 25 before -50 is 74/75. (1/2)(4/5)+ (1/2)(74/75) = 67/75 chance of a repeat visit to 25 if start from 25. The expected number of times needed to not repeat has a geometric dist with p = 8/75 and has an expected value of 1/(8/75) = 75/8. The expected total number of visits is thus (2/3)(75/8) = 25/4
The distribution is as follows:
Exactly 0 visits occurs with probability 1/3
Exactly 1 visit occurs with probability (2/3)(8/75)
Exactly 2 visits occurs with probability (2/3)(67/75)(8/75)
Exactly 3 visits occurs with probability (2/3)(67/75)**2(8/75)
...
Exactly n visits occurs with probability (2/3)(67/75)**(n-1)(8/75)
(This was jointly done with Chris Yep).
7. Verify that the average number of blunder opportunities (6a) times the cost of a blunder (7b) equals the total cost of Player A's misguided strategy (4).
(25/4)(1/75) = 1/12.
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