Hi, I am currently a first year math major and while playing the game i came across an interesting situation that i couldn't seem to math out if i had lethal, if i have a 2/4 axeflinger, a 5/7 axe flinger, and a 5/4 tharssian... my opponent is at 16 health and i play bouncing blades, what are the odds i have lethal... i found myself calculating this for hours to no solution
Assuming that you can deal 12 damage, you only need 4 more...
Assuming the following - (1) you don't have any damage dealing card at hand, (2) - your opponent does not fatigue for 2 next turn, (3) - your axe flinger is not silenced... you just take the probability that it will hit axe flinger twice.
P(axe flinger) = 1/3. It is an AND condition (It must hit axe flinger AND axe flinger) so we follow multiplication rule:
P(axe flinger) * P(axe flinger) = 1/9.
EDIT: There are TWO axe flingers on board (didn't read too well).
P(axe flinger) = 2/3. Since it is an AND condition then
This is wrong. This equation works if bouncing blades reads "deal two random damage two random minions".
the equation has to be 100% subtract the odds that Thaurrisan will get hit four times before the axe flingers get hit twice. in other words, the odds of failing are if Thaurrisan gets hit 4 times out of five tries (since any more tries is a win), so 1/3 to the power of 4/5.
you lose ONLY if Thaurrisan gets hit 4 out of five tries.
odds of Thaurrisan getting hit 4 times in a row= 1/3x1/3x1/3x1/3= 1/81. Not sure how to implement the "out of five attempts" but I assume you just multiple this by 5. So 5/81, which gives you about a 6.2 %chance of losing...
you only miss lethal if you hit thaurissan 4 times(1) or hit thaurissan 4 times and an axe flinger 1 time(2)
the chance of (1) is 1/3^4 = 1/81 the chance of (2) is 4*2/3*1/3^4 = 8/243
so you miss lethal which a chance of 1/81+8/243 = 3/243 + 8/243 = 11/243. which means you have a chance of 232/243 (95.47%) to kill him
edit: the people before me who calculated ~98% ignored the fact that there are 4 different ways to hit thaurissan 4 times and an axe flinger once: atttt, tattt, ttatt, tttat
You have 2 axe flingers with 2 or more HP, so since you only need 2 hits on any of them to win, the calculation is actually pretty simple.
Assuming there are no other minions on board and you already attacked, you need 4 more damage, and Axe Flinger deals 2 damage every time it takes a hit.
Then, the only time you don't have lethal with Bouncing Blade is when the blade kills the [card/Emperor Thaurissan[/card] after dealing at most 1 hit to any of the Axe Flingers.
You have 2/3 of change to hit an Axe Flinger each time, and 1/3 to hit [card/Emperor Thaurissan[/card] each time.
So the possible outcomes are:
1- 2 hits on AF and win 2- 2 hits on AF, 1 on T and win 3- 2 hits on AF, 2 on T and win 4- 2 hits on AF, 3 on T and win 5- 1 hit on AF, 4 on T and lose (lethal) 6- 0 hits on AF, 4 on T and lose (lethal)
In case 1 we have the possible combinations:
Taking T for Thaurissan, A for AF1 and B for AF2:
AA, AB, BA and BB. 4 combinations.
However, case 2 is only the same cases in 1, but with the addition of 1 T for each, keeping in mind that there are multiple orders to that so:
Case 2: AAT, ATA, TAA ABT, ATB, TAB BAT, BTA, TBA BBT, BTB, TBB,
From that example, we can also conclude that the number of combinations for each of the other cases is the cases in Case 1, multiplied by the number of different ways of ordering the T hits.
If we were to do the mathematical calculation for Case 1, we take XY as the variables for case 1, and on case 2, we have to calculate the number of different combinations of that XY with one T hit, but keeping in mind that X and Y are already determined, so we only need to calculate the number of different positions for the T hit to be in.
This can be done by looking at the ordering as _X_Y_ with _ being blank spaces. So for 1 T (case 2) we have 3 possibilities, being calculated as 3!/2, since we have 3 unknowns (X, Y and T) but we can only count half the cases, which are when X appears before Y. Which we multiply by the combinations of XY (4) to give us 12 combinations for case 2.
As of now: Case 1: 4 combinations (XY) Case 2: 12 combinations (4 from XY times 3 from ordering)
Case 3: TTXY 4!/2!2 = 6 combinations of ordering, times 4 from XY combinations = 24 combinations.
Case 4: TTTXY 5!/3!2 = 10 combinations of ordering, times 4 from XY combinations = 40 combinations.
But on cases 5 and 6 you can't use XY anymore, since only 1 AF is hit. We have the following combinations: Taking AF1 as A and AF2 as B:
Case 5: Since there's only 1 hit, it happens either on A or B, so combinations are of the type: ATTTT or BTTTT, each of which have 5 combinations 5!/4!, so total 10 combinations.
Case 6 has only one combination TTTT.
So: Case 1: 4 comb. Case 2: 12 comb. Case 3: 24 comb. Case 4: 40 comb. Case 5: 10 comb. Case 6: 1 comb.
So, counting the win chances we have 4 + 12 + 24 + 40 combinations where we win in a total of 4 + 12 + 24 + 40 + 10 + 1 which gives
80/91, so you had 80 out of 91 chance of lethal.
Whew this took long. I'm not the best teacher, but I hope you can understand. I deleted and retyped a lot so might be badly written a bit, please ignore that.
you only miss lethal if you hit thaurissan 4 times(1) or hit thaurissan 3 times and an axe flinger 1 time(2)
the chance of (1) is 1/3^4 = 1/81 the chance of (2) is 4*2/3*1/3^3 = 8/81
so you miss lethal which a chance of 1/81+8/81 = 9/81 = 1/9. which means you have a chance of 8/9 (88.888%) to kill him
edit: the people before me who calculated ~98% ignored the fact that there are 4 different ways to hit thaurissan 3 times and an axe flinger once: attt, tatt, ttat, ttta
P.S. I have a B.Sc in computer science
Although having a degree in playing hearthstone doesn't impress me much, the calculation is correct, short and simple. Edit: I was wrong, the explanation below (next post) is correct. Here we forgot in the "chance of (2)" to hit the emperor 4 times to kill him and make bouncing blade stop.
Eh... did your opponent have any minions on the board? If not, it should be 1 - P(Thaurissan is hit 4 times in a row) - 4P(Thaurissan is hit 3 times and only a single Axe Flinger is hit once, then the Thaurissan is hit again). Which is 1 - (1/3)^4 - 4(2/3)(1/3)^4 = 232/243. 95.5% chance for lethal, basically.
Copy pasted my reply from the thread in General Discussions.
you only miss lethal if you hit thaurissan 4 times(1) or hit thaurissan 3 times and an axe flinger 1 time(2)
the chance of (1) is 1/3^4 = 1/81 the chance of (2) is 4*2/3*1/3^3 = 8/81
so you miss lethal which a chance of 1/81+8/81 = 9/81 = 1/9. which means you have a chance of 8/9 (88.888%) to kill him
edit: the people before me who calculated ~98% ignored the fact that there are 4 different ways to hit thaurissan 3 times and an axe flinger once: attt, tatt, ttat, ttta
P.S. I have a B.Sc in computer science
Although having a degree in playing hearthstone doesn't impress me much, the calculation is correct, short and simple. Edit: I was wrong, the explanation below is correct. Here we forgot in the "chance of (2)" to hit the emperor 4 times to kill him.
I agree, stating your degree in these cases always sounds like a argumentum ad verecundiam to me. The initial post was incorrect, I do believe that the now updated version is correct. The odds of winning are 95.47%
To wrap this up I wrote a simple Matlab script to simulate 10.000.000 cases of this problem and determined a winrate of 95.4721%. below the source code:
ngames = 10000000; wongame = zeros(1,ngames); axehit = 0; for i = 1:ngames
a=4; b=4; c=4; axehit=0; while a>0^b>0^c>0
hit = ceil(rand(1)*3+0); if hit == 1 axehit = axehit+1; a=a-1; elseif hit == 2 axehit = axehit+1; b=b-1; elseif hit == 3 c=c-1; end
you only miss lethal if you hit thaurissan 4 times(1) or hit thaurissan 3 times and an axe flinger 1 time(2)
the chance of (1) is 1/3^4 = 1/81 the chance of (2) is 4*2/3*1/3^3 = 8/81
so you miss lethal which a chance of 1/81+8/81 = 9/81 = 1/9. which means you have a chance of 8/9 (88.888%) to kill him
edit: the people before me who calculated ~98% ignored the fact that there are 4 different ways to hit thaurissan 3 times and an axe flinger once: attt, tatt, ttat, ttta
P.S. I have a B.Sc in computer science
Although having a degree in playing hearthstone doesn't impress me much, the calculation is correct, short and simple. Edit: I was wrong, the explanation below (next post) is correct. Here we forgot in the "chance of (2)" to hit the emperor 4 times to kill him and make bouncing blade stop.
I already edited my post with the correction before you posted this. (look at the time stamps)
you only miss lethal if you hit thaurissan 4 times(1) or hit thaurissan 3 times and an axe flinger 1 time(2)
the chance of (1) is 1/3^4 = 1/81 the chance of (2) is 4*2/3*1/3^3 = 8/81
so you miss lethal which a chance of 1/81+8/81 = 9/81 = 1/9. which means you have a chance of 8/9 (88.888%) to kill him
edit: the people before me who calculated ~98% ignored the fact that there are 4 different ways to hit thaurissan 3 times and an axe flinger once: attt, tatt, ttat, ttta
P.S. I have a B.Sc in computer science
Although having a degree in playing hearthstone doesn't impress me much, the calculation is correct, short and simple. Edit: I was wrong, the explanation below (next post) is correct. Here we forgot in the "chance of (2)" to hit the emperor 4 times to kill him and make bouncing blade stop.
I already edited my post with the correction before you posted this. (look at the time stamps)
It doesn't matter anyway, we were both wrong in the first place.
you only miss lethal if you hit thaurissan 4 times(1) or hit thaurissan 3 times and an axe flinger 1 time(2)
the chance of (1) is 1/3^4 = 1/81 the chance of (2) is 4*2/3*1/3^3 = 8/81
so you miss lethal which a chance of 1/81+8/81 = 9/81 = 1/9. which means you have a chance of 8/9 (88.888%) to kill him
edit: the people before me who calculated ~98% ignored the fact that there are 4 different ways to hit thaurissan 3 times and an axe flinger once: attt, tatt, ttat, ttta
P.S. I have a B.Sc in computer science
Although having a degree in playing hearthstone doesn't impress me much, the calculation is correct, short and simple. Edit: I was wrong, the explanation below is correct. Here we forgot in the "chance of (2)" to hit the emperor 4 times to kill him.
I agree, stating your degree in these cases always sounds like a argumentum ad verecundiam to me. The initial post was incorrect, I do believe that the now updated version is correct. The odds of winning are 95.47%
To wrap this up I wrote a simple Matlab script to simulate 10.000.000 cases of this problem and determined a winrate of 95.4721%. below the source code:
ngames = 10000000; wongame = zeros(1,ngames); axehit = 0; for i = 1:ngames
a=4; b=4; c=4; axehit=0; while a>0^b>0^c>0
hit = ceil(rand(1)*3+0); if hit == 1 axehit = axehit+1; a=a-1; elseif hit == 2 axehit = axehit+1; b=b-1; elseif hit == 3 c=c-1; end
end if axehit > 1 wongame(i)=1; end
end winpercentage = sum(wongame)/ngames*100
stating my degree was to show that I learned how to calculate probabilities, unlike some of the posters here. unfortunately, it doesn't help with remembering how whirling blade works. anyway, I corrected my post even before your post. (look at the time stamps)
Yes I noticed this, with 'I do believe that the now updated version is correct.' I was referring to your correction. Although I must admit that I can imagine that this was not completely clear from my post. In any case your initial post had the correct reasoning, which is more important than a small error.
I understand that you stated your degree to give your post more authority. In general, it is my experience that people are not impressed by these kind of statements so I usually refrain from stating my degree. I'd rather let the evidence speak for itself, it is a much more powerful tool to convince people.
Many of you would benefit from being better at explaining the solution. The first part of the solution formula is clear to everyone who attended high school maths classes, please elaborate on the rest.
Also this was a fun exercise. Let's make it harder:
Same setup as before but the opponent has a 1/6 Armorsmith (to unable easy trading) and a 5/3 Sylvanas on the board. Tackle that one!
Ha, good one. I yield.
But you could have started easier with decreasing health of one of the axe flingers. So one 2/1 axe Flinger, one5/7 axe flinger and one 5/4 Thaurissan. That's solvable but not as trivial as the OP.
Many of you would benefit from being better at explaining the solution. The first part of the solution formula is clear to everyone who attended high school maths classes, please elaborate on the rest.
Also this was a fun exercise. Let's make it harder:
Same setup as before but the opponent has a 1/6 Armorsmith (to unable easy trading) and a 5/3 Sylvanas on the board. Tackle that one!
Play blade and pray to the RNG gods.
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Hi, I am currently a first year math major and while playing the game i came across an interesting situation that i couldn't seem to math out if i had lethal, if i have a 2/4 axeflinger, a 5/7 axe flinger, and a 5/4 tharssian... my opponent is at 16 health and i play bouncing blades, what are the odds i have lethal... i found myself calculating this for hours to no solution
Assuming that you can deal 12 damage, you only need 4 more...
Assuming the following - (1) you don't have any damage dealing card at hand, (2) - your opponent does not fatigue for 2 next turn, (3) - your axe flinger is not silenced... you just take the probability that it will hit axe flinger twice.
P(axe flinger) = 1/3. It is an AND condition (It must hit axe flinger AND axe flinger) so we follow multiplication rule:P(axe flinger) * P(axe flinger) = 1/9.EDIT: There are TWO axe flingers on board (didn't read too well).
P(axe flinger) = 2/3. Since it is an AND condition then
P(axe flinger) * P(axe flinger) = 4/9
Very very good odds
If you haven't attacked, and there are no other minions, you need two hits on flingers before 4 hits on Thaurrisan. (12 attack+ 2x2procs= 16.
your odds of failing are rolling 1/3 four out of five times. So 1/3^4/5 =.415 so your success rate should be about 58.5%
This is wrong. This equation works if bouncing blades reads "deal two random damage two random minions".
the equation has to be 100% subtract the odds that Thaurrisan will get hit four times before the axe flingers get hit twice. in other words, the odds of failing are if Thaurrisan gets hit 4 times out of five tries (since any more tries is a win), so 1/3 to the power of 4/5.
Edit: the math is this:
you lose ONLY if Thaurrisan gets hit 4 out of five tries.
odds of Thaurrisan getting hit 4 times in a row= 1/3x1/3x1/3x1/3= 1/81. Not sure how to implement the "out of five attempts" but I assume you just multiple this by 5. So 5/81, which gives you about a 6.2 %chance of losing...
so you will probably win!
you only miss lethal if you hit thaurissan 4 times(1) or hit thaurissan 4 times and an axe flinger 1 time(2)
the chance of (1) is 1/3^4 = 1/81
the chance of (2) is 4*2/3*1/3^4 = 8/243
so you miss lethal which a chance of 1/81+8/243 = 3/243 + 8/243 = 11/243. which means you have a chance of 232/243 (95.47%) to kill him
edit: the people before me who calculated ~98% ignored the fact that there are 4 different ways to hit thaurissan 4 times and an axe flinger once: atttt, tattt, ttatt, tttat
P.S. I have a B.Sc in computer science
You have 2 axe flingers with 2 or more HP, so since you only need 2 hits on any of them to win, the calculation is actually pretty simple.
Assuming there are no other minions on board and you already attacked, you need 4 more damage, and Axe Flinger deals 2 damage every time it takes a hit.
Then, the only time you don't have lethal with Bouncing Blade is when the blade kills the [card/Emperor Thaurissan[/card] after dealing at most 1 hit to any of the Axe Flingers.
You have 2/3 of change to hit an Axe Flinger each time, and 1/3 to hit [card/Emperor Thaurissan[/card] each time.
So the possible outcomes are:
1- 2 hits on AF and win
2- 2 hits on AF, 1 on T and win
3- 2 hits on AF, 2 on T and win
4- 2 hits on AF, 3 on T and win
5- 1 hit on AF, 4 on T and lose (lethal)
6- 0 hits on AF, 4 on T and lose (lethal)
In case 1 we have the possible combinations:
Taking T for Thaurissan, A for AF1 and B for AF2:
AA, AB, BA and BB. 4 combinations.
However, case 2 is only the same cases in 1, but with the addition of 1 T for each, keeping in mind that there are multiple orders to that so:
Case 2:
AAT, ATA, TAA
ABT, ATB, TAB
BAT, BTA, TBA
BBT, BTB, TBB,
From that example, we can also conclude that the number of combinations for each of the other cases is the cases in Case 1, multiplied by the number of different ways of ordering the T hits.
If we were to do the mathematical calculation for Case 1, we take XY as the variables for case 1, and on case 2, we have to calculate the number of different combinations of that XY with one T hit, but keeping in mind that X and Y are already determined, so we only need to calculate the number of different positions for the T hit to be in.
This can be done by looking at the ordering as _X_Y_ with _ being blank spaces. So for 1 T (case 2) we have 3 possibilities, being calculated as 3!/2, since we have 3 unknowns (X, Y and T) but we can only count half the cases, which are when X appears before Y. Which we multiply by the combinations of XY (4) to give us 12 combinations for case 2.
As of now:
Case 1: 4 combinations (XY)
Case 2: 12 combinations (4 from XY times 3 from ordering)
Case 3: TTXY 4!/2!2 = 6 combinations of ordering, times 4 from XY combinations = 24 combinations.
Case 4: TTTXY 5!/3!2 = 10 combinations of ordering, times 4 from XY combinations = 40 combinations.
But on cases 5 and 6 you can't use XY anymore, since only 1 AF is hit. We have the following combinations:
Taking AF1 as A and AF2 as B:
Case 5: Since there's only 1 hit, it happens either on A or B, so combinations are of the type: ATTTT or BTTTT, each of which have 5 combinations 5!/4!, so total 10 combinations.
Case 6 has only one combination TTTT.
So:
Case 1: 4 comb.
Case 2: 12 comb.
Case 3: 24 comb.
Case 4: 40 comb.
Case 5: 10 comb.
Case 6: 1 comb.
So, counting the win chances we have 4 + 12 + 24 + 40 combinations where we win in a total of 4 + 12 + 24 + 40 + 10 + 1 which gives
80/91, so you had 80 out of 91 chance of lethal.
Whew this took long. I'm not the best teacher, but I hope you can understand. I deleted and retyped a lot so might be badly written a bit, please ignore that.
correct, short and simple. Edit: I was wrong, the explanation below (next post) is correct. Here we forgot in the "chance of (2)" to hit the emperor 4 times to kill him and make bouncing blade stop.Eh... did your opponent have any minions on the board? If not, it should be 1 - P(Thaurissan is hit 4 times in a row) - 4P(Thaurissan is hit 3 times and only a single Axe Flinger is hit once, then the Thaurissan is hit again). Which is 1 - (1/3)^4 - 4(2/3)(1/3)^4 = 232/243. 95.5% chance for lethal, basically.
Copy pasted my reply from the thread in General Discussions.
ngames = 10000000;
wongame = zeros(1,ngames);
axehit = 0;
for i = 1:ngames
a=4;
b=4;
c=4;
axehit=0;
while a>0^b>0^c>0
hit = ceil(rand(1)*3+0);
if hit == 1
axehit = axehit+1;
a=a-1;
elseif hit == 2
axehit = axehit+1;
b=b-1;
elseif hit == 3
c=c-1;
end
end
if axehit > 1
wongame(i)=1;
end
end
winpercentage = sum(wongame)/ngames*100
anyway, I corrected my post even before your post. (look at the time stamps)
Yes I noticed this, with 'I do believe that the now updated version is correct.' I was referring to your correction. Although I must admit that I can imagine that this was not completely clear from my post. In any case your initial post had the correct reasoning, which is more important than a small error.
I understand that you stated your degree to give your post more authority. In general, it is my experience that people are not impressed by these kind of statements so I usually refrain from stating my degree. I'd rather let the evidence speak for itself, it is a much more powerful tool to convince people.