is it double standards for the ai?

hawkyh1 · July 2017

I have noticed recently I'm unable to dodge leaving
the ai an obvious match 5. anyone else get this?
(I used to never leave the ai a match 5 that can be
avoided)

also my continuous multiple draws of the same card.
if the ai got this, would the ai not have full charged
hands of demolish more often? (I play against the ai
the same number of times as it has played against
me. so the odds should be the same)

(just feels like it's a loaded dice every time it happens)

HH

TheDragonHermit · July 2017

I haven't had the obvious five match issue too often, though when it drops from the ceiling it is irritating enough to note. However I have been having odd streaks of drawing the same card as my entire opening hand and the my second round draw. Not hardly every game, but enough to note the improbability of seeing it so often. Wouldn't say it is favoring the AI, but maybe a new randomizer string is in order for the deck shuffler.

p2dr · July 2017

kinda feel the same about the unavoidable swap profiting the AI, also like, if you have almost won, the AI gets a last hurrah and gets like 4 swaps and great draws

5 or 6 of the same cards drawn happens sometimes too, yeah

Tilwin90 · August 2017

p2dr said:

kinda feel the same about the unavoidable swap profiting the AI, also like, if you have almost won, the AI gets a last hurrah and gets like 4 swaps and great draws

5 or 6 of the same cards drawn happens sometimes too, yeah

Omg so annoying when you are running a support deck and they go Demolish - Demolish - Demolish - Demolish. But I guess that just happens.

Slypenslyde · August 2017

Odds are finicky things. There's no Guardian Angel of RNG that enforces proper balance between random events. But you CAN sort of figure out your odds for a 4-identical-card draw without having to do the math.

This is a hypergeometric distribution calculator. Put simply, it does the math for selecting a "sample" (your hand) from a "population" (your library of 40) with a "number of successes) (the number of interesting cards left in the library) and determining the odds of getting some number of "successes" (you drew the card.) I lived on this page when I played MtG.

So let's get a feel for it. We start with 40 cards in our library, and we know there are 4 of each of our cards in it. So we know up-front there's a 10% chance of drawing the card we want on the first draw. So plug that into the calculator. "Population size" is 40. "Number of successes in population" is 4. "Sample size" is 1. "Number of successes in sample" is 1. Here's how you read the results.

"P(X = 1)" is the probability we get exactly the number we asked for. It comes out as 0.1, which is 10%!

"P(X < 1)" is the probability we get fewer than the number we want (which would be 0 here.) It comes out as 0.9, which is 90%. We expected this!

"P(X <= 1)" is the probability we get 1 or 0 in this case, that's kind of silly so let's ignore it.

"P(X > 1)" is the probability we get more than 1. Since we drew only 1 card that's 0. Another silly case for this test.

"P(X >= 1)" is the probabilty we get 1 or more. Since we only drew 1, it's 0.1, the same as our probability for 1.

We start with 2 cards in our hand. What are the odds that both are the same card? We can plug in the numbers again. Try it. Same numbers, only change "Sample size" to 2 since we're drawing 2. Change "number of successes in sample" to 2, since we want to pull 2 of our "interesting" card. What are our odds?

It turns out this should happen about 0.7% of the time [P(X = 2) = 0.0076].

An unrelated, but fun question. We start with 2 cards in hand, what if I just care how often one specific card is there? The only change we make is to change "number of successes in sample" to "1". This is where another interesting way to read the numbers comes out.

Our opening hand should have exactly 1 specific card like Sacred Cat in it 18% of the time. [P(X=1) = 0.184]. But having 2 still sort of counts as having 1, doesn't it? So the stat I care about is [P(X>=1)], which is roughly 19.2%. Neat.

Now let's talk what really happens in a couple of scenarios. I can be brief since now you should have an idea how to read the calculator.

You start with a 2-card hand, then you draw a third card. So the odds you start with:

3 identical cards: 0.04%

2 identical cards: 2.19%

1 of a certain card: 25.5%

But when we really get ticked is when the computer pulls out 4 Demolish in a row on some critical mid-game turn. Let's say that usually happens around turn 8. We've drawn 3 cards on turn 1, then drawn 7 more cards by turn 8. So we can model this by starting with a 40 card deck, drawing 10, and asking questions like, "How likely am I to find 4 Demolish by this time?" The odds for finding:

0 -> 29%

1 -> 44%

2 -> 21.4%

3 -> 3.9%

4 -> 0.2%

So the lucky 4-chain of Demolish is pretty rare, but not impossible. I frequently grind out RPG equipment with 1/255 drops, and that's 0.3%. 2 is kind of spooky frequent. And the odds of getting "1 or more" are actually a scary 70%.

Now let's add card draw to the mix to show how powerful cycling can be. Let's say you know you can draw and extra 4 by turn 8. Now you're effectively drawing 14. The odds for finding Demolish are:

0 -> 16%

1 -> 39%

2 -> 32%

3 -> 10%

4 -> 1%

Yikes. That's pretty consistent for 2, and a very high rate for "1 or more". Definitely draw cards if you can. The odds of finding 1 or more are 83%! Just a handful of extra draws has made it far more likely we can expect to see 2 cards, and almost certain we'll see 1!

Play around with the calculator and you can often find things are way more likely than you thought!

is it double standards for the ai?

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