tag:blogger.com,1999:blog-6899095487697524238.post2764384376378533134..comments2014-05-05T11:35:25.551-07:00Comments on GTORangeBuilder Blog: GTO Brainteaser #2 -- Red or Blackasheivenoreply@blogger.comBlogger6125tag:blogger.com,1999:blog-6899095487697524238.post-3184202904625550062014-04-13T14:12:19.128-07:002014-04-13T14:12:19.128-07:00Actually, no, nevermind, I think I screwed up the ...Actually, no, nevermind, I think I screwed up the math on that (wrong probabilities for waiting on [2R,1B]). The Anon is correct.Bryanhttps://biophysicalchemistry.wordpress.com/noreply@blogger.comtag:blogger.com,1999:blog-6899095487697524238.post-78697457985443816612014-04-13T14:04:25.020-07:002014-04-13T14:04:25.020-07:00I don't think this is quite right. "For ...I don't think this is quite right. "For [2R,1B], betting gives EV = +100/3, but waiting gives and EV = +100/2 (0.5 chance or [2R,0B] which has EV = +100 and 0.5 chance of [1R,1B] which has EV = 0)." You have a 2/3rds chance of going to [1R,1B] and a 1/3rd of going to [2R,0B] GTO RangeBuilderhttp://www.blogger.com/profile/01506290885754080155noreply@blogger.comtag:blogger.com,1999:blog-6899095487697524238.post-13524067069736866952014-04-13T13:58:07.761-07:002014-04-13T13:58:07.761-07:00For the four card game, I actually find that the E...For the four card game, I actually find that the EV is positive. If you bet with four cards, the EV is obviously 0. If you draw, then you either end up in with [2R,1B] or [1R,2B]. For [2R,1B], betting gives EV = +100/3, but waiting gives and EV = +100/2 (0.5 chance or [2R,0B] which has EV = +100 and 0.5 chance of [1R,1B] which has EV = 0). Thus for [2R,1B] you should draw with EV = +50. For [1B,2R], the EVs are reversed with an EV of -100/3 for betting and -50 for drawing. Thus for [1B,2R] you should bet overall with EV -100/3. Overall, the EV for drawing in the 4-card game is: 0.5*EV(draw,[2R,1B]) + 0.5*EV(bet,[1R,2B]) = +25/3.<br /><br />So the strategy for the 4-card game is to first draw a card. If the card is red, then bet on the next card being red. If the card is black, keep drawing until you draw the last black card from the deck or are forced to bet on the last card.Bryanhttps://biophysicalchemistry.wordpress.com/noreply@blogger.comtag:blogger.com,1999:blog-6899095487697524238.post-23244948911761534302014-04-12T07:10:19.927-07:002014-04-12T07:10:19.927-07:00It also seems true for eight and ten, so I'm a...It also seems true for eight and ten, so I'm a little skeptical that it shifts at a critical point. Unlike a game that does shift (say the pirate game once you extend beyond 2*g) there is no numerical constraint that would alter the play. However, I'm incredibly rusty at game theory, such that I don't remember anything beyond basic backwards induction, so I could be missing something obvious. I look forward to seeing the more advanced techniques and the solution.<br /><br />As an aside, please keep posting these. They're a nice way to practice game theory.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6899095487697524238.post-53475667299352554192014-04-12T05:42:24.815-07:002014-04-12T05:42:24.815-07:00This is pretty good logic and you are right in the...This is pretty good logic and you are right in the 4 and the 6 case. There are techniques for backwards inducting through all 52 cards which I will show in the solution and there are cases where things appear true for small numbers like 4 and 6 and then shift at some critical point. I won't give away the solution by saying whether this is one of these cases or not :)GTO RangeBuilderhttp://www.blogger.com/profile/01506290885754080155noreply@blogger.comtag:blogger.com,1999:blog-6899095487697524238.post-89167571583519505272014-04-11T21:25:53.295-07:002014-04-11T21:25:53.295-07:00Here are my thoughts, which may be wrong due to ca...Here are my thoughts, which may be wrong due to calculation error but seem intuitive:<br /><br />Backwards induction for both a four and six card game suggests there is no difference between betting and not betting at the first card. The stipulation that one must bet on the final card removes the slight advantage that would be otherwise present from trying for a more favorable ratio of colors, because it counteracts every potential positive outcome with a forced loss. It is unreasonable and impractical to conduct backwards induction for the 52 card game, but the analysis can be extrapolated from that on the games of manageable size; it becomes clear that every game starting with an even ratio is composed of even subgames (with expected value zero) and offsetting uneven subgames. Therefore the optimal strategy is simply to bet on the first card every time, since you are not indifferent to the time it would take to progress through a game. Under this logic, of course, the real optimal strategy is not to play.Anonymousnoreply@blogger.com