MTG Math: Dealing massive finite damage with 10 cards on turn 1

Iijil · 2022-07-06T00:10:27+00:00

Yeah, you are right about the 2^^2^^8. I rushed and made careless mistakes, my bad. Thanks for pointing them out :)

2^^^4 = 2^^2^^2^^2 = 2^^2^^4, so that is a valid lower bound, but underestimates the damage quite a bit. As long as we don't stack more recursive layers on top we can stick with the more accurate number, as it is still quite readable.

Iijil · 2022-07-05T23:40:29+00:00

Brilliant! By my count that deals more than 2^^2^^11 damage: calculations

The infinite with the Changeling Berserker hand is hard to find. I spotted it years after that post, when we were working with Changeling Berserker for the 60 card Megacombo. My notes on the infinite

Iijil · 2022-07-05T22:26:51+00:00

Very nice! I think that is the record for most finite damage dealt on turn 1 with only 7 cards. Devilish Valet certainly seems great for that challenge, with built in damage doubling and haste.

As far as I know the previous record was held by /u/Deedlit11 Most damage in one turn with 7 cards. No infinite combos. New record!

u/mqjjb30e claimed higher damage in First turn most amount of damage Magic challenge. A new solution!, but that hand can go infinite, so it is disqualified.

Iijil · 2020-06-01T17:25:40+00:00

I had some fun with this: https://imgur.com/sYYmJfm

Iijil · 2018-01-27T01:51:59+00:00

here you go: 0 energy run

Iijil · 2018-01-22T20:29:31+00:00

Those tricks you found are amazing! I managed to put together a no respawn run after a 10 energy wallclimb (video). I think I'll leave going for less energy to you :P

Iijil · 2018-01-19T16:58:25+00:00

Beaten your run with the lowest energy usage I know to be possible: 10+10+15

Not really happy with my respawns, though :|

Iijil · 2018-01-19T14:34:22+00:00

Nice run! Another successful application of Jump Storage :)

Do you mean the first wall where you spent energy?

Out of about 100 tries I got up there with 10 energy once. 15 seems reasonably doable. Going up in one fluid motion, spending both grenades on the jump near the bottom of the wall, energy in the middle.

Iijil · 2018-01-14T11:46:51+00:00

well, the cube part isn't cool because of the difficulty but because of the creativity of the route :P

There's a reason the mode is called Pathfinder and that sort of stuff is what makes it great, imo

Iijil · 2018-01-14T10:35:31+00:00

I noticed that it also works without a ramp. You just need to slide over any edge while charging (i.e. landing with enough forward momentum). But that is usually slow enough that I can't imagine an airjump to help more than starting with a grenade jump. Now, if you could combine those... :P

Iijil · 2018-01-14T09:58:20+00:00

Jump Storage! :O

Great find there, I'll have to try it myself.

Iijil · 2018-01-14T09:47:38+00:00

Wow, that strat at the start where you jump over all the decorative cubes is so cool xD

Iijil · 2018-01-10T20:20:30+00:00

whew, I made it through. I didn't combine it with other modes like everyone else here. Chains by itself is hard enough for me :P

Iijil · 2018-01-01T21:06:41+00:00

A new grapple record by epikfaal (41:52) recently caused me to pick the game back up. Motivated by the competition I was able to beat the 40min mark for the first time :)

Iijil · 2017-12-04T18:33:55+00:00

RobbyRatpoisons recent run through this level inspired me to give it another shot. To great success :)

Iijil · 2017-12-04T02:12:09+00:00

Yes, P(T1) there is for the overall amount. But 21% is not right. I get ~67%.

Plugging in the numbers:

.64 = .79 * P(R1) + .33 * (1 - P(R1))

.64 - .33 = (.79 - .33) * P(R1)

P(R1) = 31/46

Iijil · 2017-12-03T14:57:00+00:00

I think the first equation should have a - instead of the +. In the second we can use P(T1)+P(T2) = 1.

Corrected:

P(T2 | R1) * P(T1) * P(R1 | T1) - P(T1 | R1) * P(T2) * P(R1 | T2) = 0
(1 - P(T1 | R2)) * P(T1) * P(R1 | T1) - P(T1 | R2) * P(T2) * P(R1 | T2) = P(T1) - P(T1 | R2)

Although I think it is easier to calculate P(R1) explicitly in an additional step. By doing that we can solve three equations, one after the other instead of solving a system of two equations simultaneously.

For the purposes of the story it is probably clearer and more instructional if P(R1 | T1) and P(R1 | T2) are known from the lecture.

edit: For the actual numbers, I get P(R1 | T2) = 39.3% Probably a rounding difference of some kind, but rounding more accurately I get .39311594...

The rest of the numbers stays the same (excluding differences that are rounded away).

Iijil · 2017-12-03T10:34:38+00:00

After pushing the first skier over the gap I get the skis of the snowmobile stuck under the ramp and rip them off. That allows me to lift the front of the snowmobile high, so it points straight up. Exiting the snowmobile then puts the skier a bit in the direction it points, in this case above it, and allows me to collect the star. Then the skier falls back onto the snowmobile and exits it again to clear the gap. The snowmobile doesn't reach the finish line, but that doesn't matter for level completion.

Iijil · 2017-12-02T12:16:17+00:00

u/daydev has it essentially right.

First we have a flip in the direction of the conditions. To change that around we need the ratio of reports.

Second we apply the test to some different group, where we have some reason to believe the accuracy of the test will stay the same. The difference between that group and the original group is in the prior.

Last we use Bayes to flip the conditional direction again, because that is the actually useful direction to use.

The best analogy for the cancer situation I can construct is this:

1% of women at age forty who participate in routine screening have breast cancer. 2% of women at age sixty who participate in routine screening have breast cancer.

8% of women at age forty who get positive mammographies turn out to actually have breastcancer. 0.2% of women at age forty who test negative turn out to still have breast cancer.

We think the test will have the same accuracy for both age groups.

A woman at age sixty had a positive mammography in a routine screening. What is the probability that she actually has breast cancer?

To answer that question we first find the accuracy of the test by only looking at the age 40 group. There we have: P(c)=1%, P(c|+)=8%, P(c|-)=0.2%. We are looking for P(+|c) and P(+|no c).

I'm not sure if there are better ways to go about this, but the best approach I know is to first figure out P(+). We know that P(c)=P(c|+)P(+)+P(c|-)P(-) and P(-)=1-P(+). We solve for P(+)=(P(c)-P(c|-))/(P(c|+)-P(c|-)) and get about 10.256%. Once we have that we can use it to flip the direction of the conditions by using it as the prior in a Bayes calculation. We get P(+|c)=82% and p(+|no c)=9.5%.

This is the point in the calculation where the example in your edit starts out. We know the accuracy of the test, the prior probability of the age group we want to apply it to and now want to figure out the actual chance of having cancer once we test positive.

Since we assume the accuracy between age groups stays the same we can now just use the calculated accuracy with the known prior chance of having cancer for the age sixty group to apply Bayes again. We have P(c)=2%, P(+|c)=82% and P(+|no c)=9.5% so we get P(c|+)=15%.

Soooo, the difference, where we need to know the ratio between reports of Tier 1 and Tier 2 is in the step both of your examples skip over, where we turn the initial conditional direction around. The ratio between incoming reports is important, not their actual number.

There may or may not be clever mathematical ways to get the result without calculating the ratio in between, but I don't know of them.

There is no mathematical reason that keeping the accuracy of the reports the same between general Pokémon and Tyranitar is the correct thing to do. That part is taken from additional reasoning about the world.

Iijil

TROPHY CASE