dlsm @ 27.06.2013, 14:28
What AI weighs solutions from the point of view of probability is correct. But what he accepts chooses solutions for strict comparisons, is wrong. By kind, it would be necessary to choose one of the existing actions with the same probability with which it will work, i.e. With the ratio of the gain probability at two versions, as 90/10, you need to choose the first option with a 90% probability, and with a 10% probability - the second one. In this case, when the ratio of the probability of the winnings as 49.999 / 50.001 AI will be equally poked into one of the options, and will become more close to the real player.
In fact it's not even true. We must consider several factors:
1. The probabilities of each catch hole under the condition that the second is not caught.
2. The probability that caught both holes.
3. The probabilities of locomotives of different lengths for each of the holes in p. 2.
4. The presence of interceptions postoronkah.
5. Strategy interceptions mizeryaschego ...
And so forth.
I tried to solve it in a general way, but too much scribbling - and in my handwriting even I could hardly understand sometimes. Hence - the constant slips, mistakes and lack of results.
I decided only in the particular case of the simplest:
In the hands of alignment with the holes in the two suits. Only 1 bribe can take in each leaky suit (not including the locomotive in otsushennyh suits).
After any demolition is only one hole in one of those suits.
Interception impossible or unprofitable unacceptable on religious grounds. Length of steam locomotives in both suits are the same and constant. Those. if possible locomotive is only a single length of any of the holes.
In this simplest case, the solution takes only a few lines. However, with the use of partial derivatives: D ...
denote:
P0 - The probability that none of the holes is not caught.
P1 - probability that caught the first hole and the second is not caught.
P2 - The probability that the second hole is caught and not caught first.
P12 - The probability that the two holes are caught. Then the fishers will have to guess.
So, if | P1 - P2 |>P12, the strategy mizeryaschego degenerates into pure - always have to leave a hole with the least probability.
Otherwise, it is necessary to use a mixed strategy of {0.5, 0.5}. Yes, yes ... it is equally probable, even if the holes are caught with different probabilities.
But the fishermen thus should use a more complex mixed strategy, which depends on the probability of catching, but the frequencies are not correlated, as the probability of catching these holes. And how exactly is calculated - this is not to say.
It is caused by the fact that such policies are sustainable (non-operated). Those. change its strategy by either party unilaterally will increase the MoD win (to exploit others' strategy). A pure strategy applied in marriage, which in this case are exploited. As it has been shown in this example with pictures.
Interesting conclusions follow from this:
{7 8 9 10} {7 8 9 10} {8} T {K} - always endure worms and catch always tambourine.
{7 8 9 10} {7 8 9 10 9} {} {} T K - demolish brood or worm + tambourine with equal frequencies. Catch the tambourine with the probability seems to 3/4 (or 2/3 - I do not remember), and worms with the rest.
On Gamblere it discussed not too long ago:
https://www.gambler.ru/forum/index.php?showtopic=503067https://www.gambler.ru/forum/index.php?showtopic=503126Prada, the water there is also complete. Just as the creator of "algorithm" catching schitet all demolitions are equally probable. A second vehicle, in contrast, believes it is necessary to bear always fishing hole more likely.
Post has been editedPochemuk - 27.06.13, 14:16