Forum
skrum's conundrum on combinations
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skrumgaer wrote
at 3:28 PM, Sunday May 11, 2008 EDT
Suppose I have two 8 stacks adjoining an opponent's one 8 stack, and I have no reserve dice. When should I risk using both my 8's against the opponent's one stack?
The probablility of losing on a single 8 v 8 is 56%, but the probability of failing to take the opponent's stack with two 8's versus one is only 31%. With a big enough payoff, such as gaining a place, or preserving a place, a 31% chance of losing two 8's would be worth it. But the conundrum is this: suppose I lose the first 8. I have the opportunity to re-think about whether to risk the second 8. The dice have no memory. Losing an 8 v 8 does not increase my probabiliity of winning the next 8 v 8. So I should not risk the second 8 unless the payoff is sufficiently large, knowing that I lost the first 8. But if I know that I will make a second assessment before risking the second 8, I should not risk a two-8 combination unless I knew I would risk the second 8 after having lost the first one. But if I would risk an 8 after just having lost one, I would risk the first 8 anyway and still have one left. I would not have to assess the two-8 combination in the first place. Suppose that I "committed" myself to using both 8's, no matter what. I can think of all sorts of commitments that have negative value, for example, committing oneself to stay in a poker hand no matter what the outcome. Is it possible for a commitment to have positive value? Perhaps there are some professional gamblers or mathematicians out there who can comment on this conundrum. |
Replies 1 - 10 of 10
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JDizzle787 wrote
at 4:39 PM, Sunday May 11, 2008 EDT What you're talking about all has to do with psychology. However, your assessment may come true if it is based on an assumption through many games of play and understanding of a possible pattern.
Now, you math folks would say "Nay JDizzle&87! Pattern does not exist! This shiznit is random!" Au contraire, my friend. From my own experience, it has proved me well sometimes to rely on certain things to happen that have happened in the past to happen again, be it that they are not extraordinary circumstances. Take for example, restacks. While it is claimed that they are random in all sense, it can be noted, in many games, that they will replace themselves in a desirable manner, i.e., say you're neighboring a seven stack. I have witnessed on many occasions a restack that will counter that said 7 of the neighbor, either equal or just one under, like a 6. |
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skrumgaer wrote
at 4:46 PM, Sunday May 11, 2008 EDT Wishbone has made the same claim.
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Shevar wrote
at 4:26 PM, Monday May 12, 2008 EDT and the claim is based on personal observation.
The problem with this personal observation is that you only recall those events proving your theory. If you would keep a record of all the restacks (which is insane of course) any stacking theory will fail. Except for this one: The restacks are randomly distributed among your lands. on topic: since im not a professional mathematician but just a semi-professional physicist, i suggest you go around with one 8stack do as much damage as possible and keep the other where it is. |
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Vanilla Dice wrote
at 4:58 PM, Monday May 12, 2008 EDT Your conundrum is pretty similar to the Monty Hall problem in probability theory.... (http://en.wikipedia.org/wiki/Monty_Hall_problem)
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MadHat_Sam wrote
at 5:02 PM, Monday May 12, 2008 EDT If all you have is 2 8 stacks and the neighbor only has 1 8 stack, th situation sounds like the end of a game and you are jockying for the last 2 places. Roll the dice and flag to not risk losing any more dom. End game dom will get worse faster then the gain is place will help.
If it is midgame and there are lands you can take, use one 8 to take them and the other 8 to block any advance or chase the other player. Doing this of course will always depend on who are playing and who you may or maynot have allied with. |
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skrumgaer wrote
at 6:12 PM, Monday May 12, 2008 EDT Thanks, Vanilla, for the reference to the Monty Hall problem. In the kdice setting, who would play the role of the host?
Further observations: In the 8 v 8 problem, the expected outcome is +0.44 lands and -3.48 dice. In the committed 8,8 v 8 problem, the expected outcome is +0.686 lands and -5.429 dice. In either case the ratio of dice to lands exchanged is 7.914. If you think that taking the land is worth 7.914 dice, you would be indifferent in either case in regard to trying to take the land. |
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JDizzle787 wrote
at 7:34 PM, Monday May 12, 2008 EDT lol, well, observation hasn't failed me yet!
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§ilverfox wrote
at 1:08 PM, Tuesday May 13, 2008 EDT In backgammon, you roll two dice. You can then move a man the total of the dice, but you must do so only in the increments that the dice show. For example, if you roll a 4 and a 3, then you may move your man 7 spaces. However, if the spots three and four spaces away are both blocked, then you MAY NOT move that man seven spaces (thereby jumping over the blockade).
I believe there is an analogous relationship between this idea and your conundrum, Skrum. A person can argue that an attacking 8v8 rolled twice will win somewhere in the neighborhood of 70% of the time. Mathematically that is as correct as saying that the man in the backgammon analogy moves seven spaces (the mathematical total of the dice). And that is not wrong. However, that is not the complete picture. The reality is you are rolling two chances of success at 44%, not one chance at 69%. There is a difference there that mathematics can only take into account in the long run (say a bell curve of 200,000 trials). All that being said, I think it boils down to this: If the odds are worth rolling an 8v8, then go all out and pre-commit to rolling both of them. The “opportunity” to re-think the situation is really only an “opportunity” to short change your long term odds of winning the attack. On a slight side note, here is another conundrum. If the bell curve doesn’t lie (and it doesn’t), yet the dice have no memory (and they don’t), then how is it we cannot predict a streak of winning because of observing a streak of losing (and we can’t). :) Little things like that are going to make me a religious man in the end, I swear… lol |
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§ilverfox wrote
at 1:13 PM, Tuesday May 13, 2008 EDT And on another side note....
What the hell is it with this cheesy ass forum that it can't recognize a double-quote or a single quote etc....?! Ryan, Please. I absolutely guarantee that there are members here who would be willing to admin a real forum board (vBulletin or something). These packages are FREE, man FREE! Can we please work it out? Do I need to donate $20 next month or something? Just tell us what the issue is and we as a community will take care of it for you. |
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mr Kreuzfeld wrote
at 5:11 AM, Wednesday May 14, 2008 EDT i would agree with silverfox.
I say always do both attacks in the long run you will gain from it. and the a different argument says. k say you lost the 1st attack. still do the 2nd attack because if you wait you will most likely loose dom. this will make the expected values of the 2nd attack most likely positive. |