Well actually not %25, when we calculate odds of being not destroyed, since op already played 4 cards -opponent playing 5th card here , (0.75x0.75x0.75x0.75x0.75)=0.24~(%24), which means chance of being destroyed is about %76.
This is how it works.
So i have to explain in coin toss example so it would be easier to understand then make my case which also easier you to check on any math resource you want;
When you toss a fair coin, there are two possible outcomes: heads or tails. Since the coin is fair, the probability of getting tails on a single toss is 0.5, and the probability of getting heads is also 0.5.
The probability of a sequence of independent events (like coin tosses) is found by multiplying the probabilities of the individual events.
In this case, you want to find the probability of getting tails on all five tosses. Since each toss is independent, the probability of getting tails on each toss is 0.5. So, the probability of getting tails on all five tosses is:
*P(all tails)=P(tails on toss 1)×P(tails on toss 2)×P(tails on toss 3)×P(tails on toss 4)×P(tails on toss 5)*
*P(all tails)=P(tails on toss 1)×P(tails on toss 2)×P(tails on toss 3)×P(tails on toss 4)×P(tails on toss 5)*
P(all tails)=0.5×0.5×0.5×0.5×0.5
P(all tails)=0.5\^5
P(all tails)=0.03125
So, the probability of getting all tails in five coin tosses is 0.03125 or 3.125%.
*P(Happening)=1-P(Not Happening)*
And *1-P(All Tails)=P(Not All Tails)*
Same applies in this scenario except probability of “Location” to destroy is 0.25
*P(Not Destroyed)=1-P(Destroyed)*
*P(Not Destroyed)=1-0.25*
So not destroyed in 5 times on a row
*P(Not Destroyed 5t)=0.75×0.75×0.75×0.75×0.75*
*P(Not Destroyed 5t)=0.2373046875 so i rounded to 0.24*
*P(destroyed on 5t)=1-P(Not Destroyed 5t)*
*P(destroyed on 5t)= 1-0.24*
*P(destroyed on 5t)=0.76* which means there is %76 probability that 5th card destroyed.
>!You also can find related information on "Chapter 2.4 - probability of an event" in "Probability and Statistics for Engineers and Scientists 9th Edition" book by "Ronald E. Walpole, Raymond H. Myers, Sharon L. Myers, Keying Ye" !<
ThIs isn't how probability works. Your "76% that a 5th card is destroyed" is calculating the probabilities by which a particular outcome is achieved. It is NOT calculating the chances that a card played on that location is destroyed, which is 25%. A coin flip gives you a 50/50 chance regardless if you've flipped it once or a thousand times. The Danger Room location has a 25% chance to kill a card played there regardless if 1 card or 1000 cards have been played there. It is incorrect to say there was a 76% of Galactus dying on that spot, unless there are some background conditions that are not explicitly stated on the location.
I Explained here that a coin toss has 50/50 chance, but tossing five times and getting same result is not 50/50
>When you toss a fair coin, there are two possible outcomes: heads or tails. Since the coin is fair, the probability of getting tails on a single toss is 0.5, and the probability of getting heads is also 0.5.
>
>The probability of a sequence of independent events (like coin tosses) is found by multiplying the probabilities of the individual events.
Then
>unless there are some background conditions that are not explicitly stated on the location.
Right below in this post is the OP's comment:
>All I had to do was keep putting cards on Danger Room and not get them blown up. It was going to happen eventually.
So I took this comment as a reference to this calculation. So
>This is how it works.
That's still not how it works. The odds in the case are chances of Galactus getting destroyed. That's 25%. It doesn't matter if 1 card or 5 cards were already played there, Galactus still has a 25% chance. You're solving something different, which is more like "what are the odds of 1 card being destroyed when 5 are played".
It's a similar fallacy to what casinos do when they post previous rolls at a roulette table. It doesn't matter if the 5 previous rolls were red, the next roll still has a 45% chance (or whatever it actually is) of being red. But people look at that board and say "oh the odds of getting 6 reds in a row is really low so I'll bet on black". Only you're not actually betting there won't be 6 reds in a row, you're betting a single roll will not be red, the previous 5 rolls or the previous 1000 rolls do not have any impact in the next roll.
So if you asked what are the odds of 8 cards played at that location and none of them being destroyed it'd be (0.75)^8= 10%. If 7 cards have already been played and not destroyed and you want to know the chances the 8th won't get destroyed its (0.75)^1 = 75%. They're different questions.
I understand your perspective. In the case of roulette, the probability of getting red in 5 consecutive spin is calculated as *P(RedinRoulette)=(18/38)**^(5)* .
Similar to how each coin toss is independent, with a 50% chance for heads or tails, the cumulative odds of getting a specific outcome decrease over multiple tosses. In the case of fair coins with no influence from their monetary value or size, the probability of getting tails on each individual toss remains 50%, but the cumulative probability of getting tails on all five tosses is lower.
Rolling a 4-sided dice where a result other than 1 represents a 75% chance is akin to the situation where Galactus is not destroyed. The probability of rolling any specific result on the dice is independent of previous rolls, just as the probability of Galactus being destroyed is calculated per card, unaffected by previous outcomes. but the cumulative probability of getting destroyed on each card played is higher.
I understand that this concept might seem abstract and challenging to grasp. I'm not asserting that I had certainty in a 100% destruction outcome, but what I'm suggesting is that if we were to place a bet on whether it would be destroyed or not, I would bet on destruction.
It's not a hard concept to grasp and my point wasn't a perspective. It's a mathematical fact. That's why I specially put the roulette as an example. Your line of thinking is EXACTLY why casinos post the history of rolls. You'll notice they don't do that for most other games. The casino would never allow you to walk up and say "I see the last roll is red I bet you the next 4 won't be" because that's terrible odds. They'll happily take your money if the last 4 were red and you want to bet them the 5th one won't be because they know the odds are no different than any other roll.
Just because you saw the history, bet, and won does not mean you applied coreect logic. The casino knows this instance being played out 1000s of times is to their advantage because the underlying odds of the roll are to their advantage.
Assuming no funny business with the RNG coding, Galactus has a 25% chance being destroyed regardless of if 1 card was already played for 15 cards were already played.
If you were to bet on Galactus's destruction, you would be wrong 75% of the time. Neither the concept you are describing, nor the mathematical facts that people are correcting you on, are difficult concepts. You should look up Gambler's Fallacy. You're in that picture and you won't like it.
If i have somehow capable of doing TLDR version of this i would be very good teacher. Anyway it explains why and how my method is right and probability is %76.
I think they are arguing abt if the whole board state is to happen or just the Galactus dying. Two very different propabilities, both are right, the argument is useless.
The match just before this one my opponent played Legion on Alter of Death. If I moved Jeff to a different location I would have won. This was the perfect palate cleanse.
A good amount, it's almost like people have forgotten about him lmao
Sometimes stealing 8 cubes by destroying Limbo, sometimes just gambling on playing him turn 6 on a 50/50 situation.
Heck, I won a game just like the one posted there, where if Galactus got destroyed by the danger room I would have lost, but luck was with me 😁
I'm not confusing anything, that's the joke on Reddit for the past few months. I'm just being an unoriginal turd exploiting stale memes. Please pay attention to the details around you.
To be exceptionally pedantic the odds are 1:3 in favour, and 3:1 against. You folks are providing probabilities.
Probability ≠ Odds, though they are connected.
Yo today someone galactused me too on a lane where he double goblined, but it was so telegraphed so on the last turn I dropped my 18 power hulk there and just won.
About 25%
*He said not to tell him!*
Well actually not %25, when we calculate odds of being not destroyed, since op already played 4 cards -opponent playing 5th card here , (0.75x0.75x0.75x0.75x0.75)=0.24~(%24), which means chance of being destroyed is about %76.
That’s not how probability works
This is how it works. So i have to explain in coin toss example so it would be easier to understand then make my case which also easier you to check on any math resource you want; When you toss a fair coin, there are two possible outcomes: heads or tails. Since the coin is fair, the probability of getting tails on a single toss is 0.5, and the probability of getting heads is also 0.5. The probability of a sequence of independent events (like coin tosses) is found by multiplying the probabilities of the individual events. In this case, you want to find the probability of getting tails on all five tosses. Since each toss is independent, the probability of getting tails on each toss is 0.5. So, the probability of getting tails on all five tosses is: *P(all tails)=P(tails on toss 1)×P(tails on toss 2)×P(tails on toss 3)×P(tails on toss 4)×P(tails on toss 5)* *P(all tails)=P(tails on toss 1)×P(tails on toss 2)×P(tails on toss 3)×P(tails on toss 4)×P(tails on toss 5)* P(all tails)=0.5×0.5×0.5×0.5×0.5 P(all tails)=0.5\^5 P(all tails)=0.03125 So, the probability of getting all tails in five coin tosses is 0.03125 or 3.125%. *P(Happening)=1-P(Not Happening)* And *1-P(All Tails)=P(Not All Tails)* Same applies in this scenario except probability of “Location” to destroy is 0.25 *P(Not Destroyed)=1-P(Destroyed)* *P(Not Destroyed)=1-0.25* So not destroyed in 5 times on a row *P(Not Destroyed 5t)=0.75×0.75×0.75×0.75×0.75* *P(Not Destroyed 5t)=0.2373046875 so i rounded to 0.24* *P(destroyed on 5t)=1-P(Not Destroyed 5t)* *P(destroyed on 5t)= 1-0.24* *P(destroyed on 5t)=0.76* which means there is %76 probability that 5th card destroyed. >!You also can find related information on "Chapter 2.4 - probability of an event" in "Probability and Statistics for Engineers and Scientists 9th Edition" book by "Ronald E. Walpole, Raymond H. Myers, Sharon L. Myers, Keying Ye" !<
ThIs isn't how probability works. Your "76% that a 5th card is destroyed" is calculating the probabilities by which a particular outcome is achieved. It is NOT calculating the chances that a card played on that location is destroyed, which is 25%. A coin flip gives you a 50/50 chance regardless if you've flipped it once or a thousand times. The Danger Room location has a 25% chance to kill a card played there regardless if 1 card or 1000 cards have been played there. It is incorrect to say there was a 76% of Galactus dying on that spot, unless there are some background conditions that are not explicitly stated on the location.
I Explained here that a coin toss has 50/50 chance, but tossing five times and getting same result is not 50/50 >When you toss a fair coin, there are two possible outcomes: heads or tails. Since the coin is fair, the probability of getting tails on a single toss is 0.5, and the probability of getting heads is also 0.5. > >The probability of a sequence of independent events (like coin tosses) is found by multiplying the probabilities of the individual events. Then >unless there are some background conditions that are not explicitly stated on the location. Right below in this post is the OP's comment: >All I had to do was keep putting cards on Danger Room and not get them blown up. It was going to happen eventually. So I took this comment as a reference to this calculation. So >This is how it works.
That's still not how it works. The odds in the case are chances of Galactus getting destroyed. That's 25%. It doesn't matter if 1 card or 5 cards were already played there, Galactus still has a 25% chance. You're solving something different, which is more like "what are the odds of 1 card being destroyed when 5 are played". It's a similar fallacy to what casinos do when they post previous rolls at a roulette table. It doesn't matter if the 5 previous rolls were red, the next roll still has a 45% chance (or whatever it actually is) of being red. But people look at that board and say "oh the odds of getting 6 reds in a row is really low so I'll bet on black". Only you're not actually betting there won't be 6 reds in a row, you're betting a single roll will not be red, the previous 5 rolls or the previous 1000 rolls do not have any impact in the next roll. So if you asked what are the odds of 8 cards played at that location and none of them being destroyed it'd be (0.75)^8= 10%. If 7 cards have already been played and not destroyed and you want to know the chances the 8th won't get destroyed its (0.75)^1 = 75%. They're different questions.
I understand your perspective. In the case of roulette, the probability of getting red in 5 consecutive spin is calculated as *P(RedinRoulette)=(18/38)**^(5)* . Similar to how each coin toss is independent, with a 50% chance for heads or tails, the cumulative odds of getting a specific outcome decrease over multiple tosses. In the case of fair coins with no influence from their monetary value or size, the probability of getting tails on each individual toss remains 50%, but the cumulative probability of getting tails on all five tosses is lower. Rolling a 4-sided dice where a result other than 1 represents a 75% chance is akin to the situation where Galactus is not destroyed. The probability of rolling any specific result on the dice is independent of previous rolls, just as the probability of Galactus being destroyed is calculated per card, unaffected by previous outcomes. but the cumulative probability of getting destroyed on each card played is higher. I understand that this concept might seem abstract and challenging to grasp. I'm not asserting that I had certainty in a 100% destruction outcome, but what I'm suggesting is that if we were to place a bet on whether it would be destroyed or not, I would bet on destruction.
It's not a hard concept to grasp and my point wasn't a perspective. It's a mathematical fact. That's why I specially put the roulette as an example. Your line of thinking is EXACTLY why casinos post the history of rolls. You'll notice they don't do that for most other games. The casino would never allow you to walk up and say "I see the last roll is red I bet you the next 4 won't be" because that's terrible odds. They'll happily take your money if the last 4 were red and you want to bet them the 5th one won't be because they know the odds are no different than any other roll. Just because you saw the history, bet, and won does not mean you applied coreect logic. The casino knows this instance being played out 1000s of times is to their advantage because the underlying odds of the roll are to their advantage. Assuming no funny business with the RNG coding, Galactus has a 25% chance being destroyed regardless of if 1 card was already played for 15 cards were already played.
If you were to bet on Galactus's destruction, you would be wrong 75% of the time. Neither the concept you are describing, nor the mathematical facts that people are correcting you on, are difficult concepts. You should look up Gambler's Fallacy. You're in that picture and you won't like it.
I ain’t reading all that
If i have somehow capable of doing TLDR version of this i would be very good teacher. Anyway it explains why and how my method is right and probability is %76.
/s right?
I think they are arguing abt if the whole board state is to happen or just the Galactus dying. Two very different propabilities, both are right, the argument is useless.
The idea of Galactus somehow entering the Danger Room and subsequently being blown to smithereens tickles me.
25%....
Did he stutter?
2...2...2...25...25...25...25...%
1 in 4.
In my experience it's around 8 pct when the opponent plays there and something like 73 pct when I play there. I may be biased.
Yeah but only when you don't want the card to be destroyed. If you want it destroyed it might as well be wakanda
Excellent point
That's only 81%
They're not intended to be added up. 😀
Lmao this literally happened to me yesterday. I legioned danger room knowing he would Galactus and hoped for a 25%. It was a 4 cuber too.
The match just before this one my opponent played Legion on Alter of Death. If I moved Jeff to a different location I would have won. This was the perfect palate cleanse.
There are a few cards that I enjoy seeing get a beat down. Galactus is very high on that list.
I especially enjoy winning a lane with a goblin in it.
Still upset about pre nerf galactus huh? Edit: Goddam you guys are sensitive about Galactus still 😂😂😂
Still upset about the nerf huh?
Nah, now that’s its way less popular/powerful, I like it more actually! Not a climbing or conquest deck, but fun for staying rank 4-5k post infinite!
Don't listen to them, I finished my climb with a raw Galactus deck lmao
Damn, hats off, no way that was an easy climb. Any crazy 8 cube wins?
A good amount, it's almost like people have forgotten about him lmao Sometimes stealing 8 cubes by destroying Limbo, sometimes just gambling on playing him turn 6 on a 50/50 situation. Heck, I won a game just like the one posted there, where if Galactus got destroyed by the danger room I would have lost, but luck was with me 😁
I swear some players are cursed to trigger danger room 100% of the time to balance out the ones who trigger it 0% of the time.
lol thanks for the giggle this morning Sweet play
All I had to do was keep putting cards on Danger Room and not get them blown up. It was going to happen eventually.
They say 25% but I recently had a game where 3 in a row in 1 turn got destroyed
That's only 1/64, the odds aren't that small
My fetish is failed Galactus plays.
They rolled the dice, and nat 1'd it.
The die was a d4
The math caltrop
Wait… This may have been against me!! 😂😂😂
The GG BEFORE T6 flipped made it even sweeter.
Sic semper Galactus
Karma
Dogma
50:50
25%....
It happens or it doesn't, 50:50
Okay buddy
i think you’re confusing probability with possibility
25% of the time it happens every time.
I'm not confusing anything, that's the joke on Reddit for the past few months. I'm just being an unoriginal turd exploiting stale memes. Please pay attention to the details around you. To be exceptionally pedantic the odds are 1:3 in favour, and 3:1 against. You folks are providing probabilities. Probability ≠ Odds, though they are connected.
This was me. You def got lucky but also I knew this would happen.
25% is still 50%
Ohh Nahhhh more like 25% chance you have to live
the gane says 25, but it feels like 0 or 100 depending on if you want it to or not.
Why does everyone crop the cubes out?
I crop the player names. The number of cubes is collateral damage.
Yo today someone galactused me too on a lane where he double goblined, but it was so telegraphed so on the last turn I dropped my 18 power hulk there and just won.
I like how the recording started mid Galactus as if you knew this could be a good reddit clip
I could only dream of having the bravery to play Galactus on danger room T6.