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Guest Message by DevFuse

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A shady game?!

Best Answer Prime, 25 October 2013 - 07:42 PM


To calculate the payoff for the best dynamic strategy, we need a computer program, which traverses all variations choosing optimal path.

In this case such program is relatively simple, while variations aren’t that many. Thus the payoff for the best strategy is found. To produce a formulation of the strategy complicates matters a little bit. However, in this case with the maximum of only 4 card draws, the best strategy can be expressed in simple terms as well.


Spoiler for Optimal game


Was this result calculated?

The first two cards didn't seem to have that much effect in simulations.


Spoiler for My simulation



The code calculates the exact probability for the average payoff with the best strategy (unless I messed up.)

Both simulation and code must take into account the cards leaving the deck. Also, must discard invalid (in terms of strategy) variations. E.g., a draw like (13, 13, 8).

The criteria for the staying card on the third turn can be established analytically (see the spoiler.)


Spoiler for programming code

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#21 Prime


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Posted 11 November 2013 - 11:14 PM



It would be interesting to see your code's four payoff values for 10 9 7 and 10 9 8 when the first two cards are above and below 14. The code would have to be modified from "search" mode to where 10 9 are given, and results sorted into two categories with respect to 14. I think I see how to calculate it directly, at least the EV for 10 9. The EV contribution for the third draw, in the two cases is a little complicated, and I didn't finish it last night.


It suddenly downed on me, what you meant. The (10, 9, 8) and (10, 9, 7) are not the actual cards drawn, but rather representations of two different strategies. So, (10, 9, 8) means: stay on 10 and up on the first card, on 9 and up – on the second, and on 8 - on the third.

Spoiler for In that case

Edited by Prime, 11 November 2013 - 11:18 PM.

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Past prime, actually.

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