tag:blogger.com,1999:blog-9170413323750351865.comments2024-01-27T13:08:16.378-08:00Quantifying StrategyUnknownnoreply@blogger.comBlogger7125tag:blogger.com,1999:blog-9170413323750351865.post-71295477963762127422024-01-27T13:08:16.378-08:002024-01-27T13:08:16.378-08:00Wow!Wow!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-9170413323750351865.post-19283935910793512072022-06-07T12:00:46.919-07:002022-06-07T12:00:46.919-07:00This is a good point. I stopped the plots at 5 mo...This is a good point. I stopped the plots at 5 moves, but perhaps extending it to 6 would make this more clear, as rare as that would be to happen.Timhttps://www.quantifyingstrategy.com/noreply@blogger.comtag:blogger.com,1999:blog-9170413323750351865.post-77529709501178909542022-06-05T07:22:37.935-07:002022-06-05T07:22:37.935-07:00Your calculation is correct but in the FAQ you wil...Your calculation is correct but in the FAQ you will see that 1 is considered always a miss for everything. Thus Frodo can move against 5 eyes with 5 moves before that and still have a slim chance of being unseen even with 5 rerolls :)<br />Andrei Munteanu PrincePhocsnoreply@blogger.comtag:blogger.com,1999:blog-9170413323750351865.post-3929509707355990992020-06-23T20:30:17.585-07:002020-06-23T20:30:17.585-07:00Upon further review, the equation above is incompl...Upon further review, the equation above is incomplete. Specifically, $\binom{4}{k}$ does not specify the number of ways to choose unpaired asymmetric cards when $k > 1$. This is because we must not only select the pair, but which card inside the pair. When $k=1$ this cancels out because you're looking at the unique cases. Consider $k=2$, though. If we consider each pair to have a left and right version, then we could pick left left, right right, left right, or right left. The first and second two sets are isomorphs of each other, but left left is not an isomorph of left right. Thus we have an additional factor of $2^k$ to select each card inside the pairs, and $1/2$ to remove isomorphs. In my errata, now linked above, I use this approach integrated into my own, and check it against $n_1$ to make sure I'm not over or under counting.Timhttps://www.blogger.com/profile/11170445239976241795noreply@blogger.comtag:blogger.com,1999:blog-9170413323750351865.post-86849772353646088082020-06-09T07:05:15.682-07:002020-06-09T07:05:15.682-07:00Bob, I see what you mean about over counting and e...Bob, I see what you mean about over counting and especially like your $n_3 > n_1$ check.<br /><br />I think I follow your equation but need time to digest and update my post.<br /><br />Thank you!Timhttps://www.blogger.com/profile/11170445239976241795noreply@blogger.comtag:blogger.com,1999:blog-9170413323750351865.post-47511210922185305022020-06-09T02:33:27.465-07:002020-06-09T02:33:27.465-07:00As to where my summation comes from: I'm count...As to where my summation comes from: I'm counting sets with unpaired asymmetric cards from $k$ pairs, (as we're counting modulo isomorphism, it doesn't matter which cards) and $p$ pairs of asymmetric cards. There are $\binom{4}{k}$ ways of choosing the pairs for the unpaired asymmetric cards, $\binom{4-k}{p}$ ways of choosing the pairs of asymmetric cards, and $\binom{8}{5-k-2p}$ ways of choosing the remaining symmetric cards.Bob Zwetsloothttps://www.blogger.com/profile/16637233225000423493noreply@blogger.comtag:blogger.com,1999:blog-9170413323750351865.post-89484462147170807232020-06-09T02:28:50.303-07:002020-06-09T02:28:50.303-07:00While I like your idea, there is an error in the c...While I like your idea, there is an error in the calculations.<br />Clearly, as n_3 is the number of sets of 5 cards satisfying a certain condition while n_1 is the number of sets of 5 cards, we should have that n_3 is at most n_1, yet according to your calculations n_3>n_1. <br /><br />The problem is that you're counting sets with multiple unpaired asymmetric cards multiple times: once for each unpaired asymmetric card, as any of them can be the initial card you pick.<br /><br />In fact, I'd count the entire thing differently. The initial positions of any set of cards are exactly the set of isomorphs of the initial positions of any other set of cards that can be obtained by just flipping unpaired asymmetric cards to the other card in the pair. That means that it literally doesn't matter which exact unpaired asymmetric cards are in a set; you just need to know which pair it comes from.<br /><br />Hence the number of starting positions modulo isomorphisms is actually $\frac{5!}{2\cdot 2}\cdot \sum_{k=0}^4\sum_{p=0}^{\lfloor \frac{5-k}{2}\rfloor}\binom{4}{k}\binom{4-k}{p}\binom{8}{5-k-2p} = 45360$.Bob Zwetsloothttps://www.blogger.com/profile/16637233225000423493noreply@blogger.com