Probabilistic and Statistical choices for cool people

Use math to prove why the Minotaur will win more than 50% of the time. For you brainlets out there, simply finding the average won't help you.

Partial answers will receive partial insults.

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The expected damage is 1200 by the time dragonboy will die.

lack of data, attack speed isn't present

seriously doubt a dagger has the same attack speed as a war hammer

Attack speed is identical, each get one attack per turn, turns are simultaneous. Defense is simple subtraction from damage.

Goblin wins.

why do you think the goblin has the advantage, idiot?

>Hint: You are incorrect.

Expected value of damage done by goblin is 0.5*30+0.5*10= 20. Expected value of damage done by minotaur is 0.95*0+0.05*90=4.5
200/4.5 = 44.444
1200/20=60
So Minotaur is more likely to win

i think Bayes' Theorem can solve this one sec

1/20 chance to hit goblin, 3 hits needed to kill
1/1 chance to hit minotaur, 40 hits needed to kill

(1/20 * 1/3) / (1/20 * 1/3 + 1/2 * 1/40)

(0.0167) / (0.0167 + 0.0125)

0.572 = 57.2% chance for minotaur to kill goblin

(1/2 * 1/40) / (1/2 * 1/40 + 1/20 * 1/3 )

(0.0125) / (0.0125 + 0.0167)

0.428 = 42.8% chance for goblin to kill minotaur

Incorrect because the Minotaur must deal 90*3=270 damage to win, a partial infliction of 20 damage is not possible.

At what level of hitpoints is the sword on the far right no longer optimal? Obviously you can tell at 20hp the far right sword is optimal.

Minotaur wins: 57.39%
Draw: 1.14%
Goblin wins: 41.47%

The sword of evil surpasses the sowrd of thousend deaths at hp>20.

>i think Bayes' Theorem can solve this one sec
nvm idk

lizard will win in between 60 to 120 turns.
minotaur will have to miss 60 times to 58 times to lose indefinitely to lizard's best case. this occurs at a rate of:

0.95^58 + 0.95^59 + 0.95^60 gives 14.56 %

this means that 14.56% of the time, minotaur will miss enough for the little lizard to achieve best case win.

it means that 85.44% of the time, the Minotaur will win given the condition that the little lizard is at its best case

i misread as 50% of defense
lizard wins within turn 40 - 120
0.95^40 0.95^39*0.05 + 0.95^38 * 0.05 ^2
gives 13.56%. this means that Minotaur will hit under 3 times during 40 turns 13.56% of the time only if lizard is at its best. his winrate is > 86.44%

I realize minotaur has a greater chance to win, but can someone explain the logical reason?

It's a Markov chain problem. The state space is

{1200, 1190, ... , -20} x {200, 110, 20, -70}

The transistion probability from the state
>(i,j) to (i-10, j) is 19/40
>(i,j) to (i-30, j) is 19/40
>(i,j) to (i-10, j-90) is 1/40
>(i,j) to (i-30, j-90) is 1/40
if i and j are both positive. Otherwise It's a stationary state.

The minotaur wins if it ends in one of those states: {1200, 1190, ... ,10} x {-70}
The manlet wins if it ends in one of those states: {0, -10, -20} x {200, 110 , 20}
There is a draw if it ends in one of those states: {0, -10, -20} x {-70}

What does atk and def do?

HP lost per turn = Attack - Defense except when the kobold ignores the minotaur's defense.

This thread was on /v/ a couple of days ago. What I did was calculate the expected value of the random variables "how many hits it takes the kobold to kill the centaur" and "how many hits it takes the minotaur to kill the kobold" and see which was smaller:

The expected value of "how many hits it takes the minotaur to kill the kobold" is exactly 60.

The expected value of "how many hits it takes the kobold to kill the centaur" is exactly 60.250.

Thus, the centaur is most likely to win, but just barely.

where did the centaur come from?

I got
~57.39% for Minotaur wins.
~41.47% for Goblin wins.
~1.14% for draw
using this MATLAB code
pastebin.com/qLuRd3wC

Previous attempt gave the minotaur 10 HP too few but it should be correct now.

I don't understand your code, but I got similar results

This is a good description, I'm going to try to work with this

Incidentally I monte-carlo'd it like some other anons and got
57.44% minotaur wins
41.45% goblin wins
1.11% draw

Goodjob user. I would have understood this 3 years ago. The last time I used MATLAB was when I learned it.

It doesn't work that way

The odds are always 50/50, either the minotaur wins or it doesn't

I got the same results as , using this R code
pastebin.com/fcxRzMAL

Specifically
57.393403% Minotaur wins
41.467130% Goblin wins
1.139467% draw

Equations:

Average damage per turn for minotaur = 0.05(100 - 10) = 4.5

0.05 is the dodge mutliplier, 100 is the atk, 10 is the goblin's def.

This means that the minotaur will kill the goblin with an average of 200/4.5 = 44.4 hits

Average damage per turn for goblin = (30 - (0.5)20) = 20

0.5 is the ignoring of defense mutliplier, 30 is the atk, 20 is the minotaur's def.

This means that the goblin will kill the minotaur with an average of 1200/20 = 60 hits

As the minotaur needs less hits on average, it will win most of the time.

Can I roll to seduce the minotaur?

>win more than 50% of the time
Impossible. He'll either win or he'll lose. That's a 50/50 chance right there.

Actually it is possible for them to kill each other simultaneously. How do you account for when both die?

that's still 50/50, they win or the lose. If they both lose or both win, that doesn't change anything.

what's the chance they become friends?

50/50, same as everything else.

Katana Way of the Samuri looks good at 52 base damage with 9% crit at 75 speed.

Broadsword Swordmaster With Dual Wield might be the best with 75*1.7= 127 base damage, 3% crit and 65% accuracy.

for arguments sake speed 100 might as well be equal to one attack per second.

Because it would take 100ish turns for the goblin to win and in that time the minotaur should hit ~6ish times and it only needs 3 to connect