This card is very fun, and I enjoy playing it alot. Just thought I would provide some stats about this card.
These are the natural probabilites of getting 5 counters on turn 7 and up, barring of course Krark's Thumb or Paradox Hazes, etc.
After getting a turn two Goblin Bomb:
7: 3% 8: 1.6% 9: 5.5& 10: 3% 11: 7% 12:4.4%
These probabilities follow from a restricted Pascal's Triangle. It took me awhile but I was able to realize that the pattern is sort of relfected about the middle.
The result is this
1
1 1
2 1 1
3 3 1 1
6 4 4 1 1
10 10 5 5 1 1
Someone correct me if I am wrong, but this should be the pattern. It would be intersesting to see what the maximum probability is, if allowed to take infinite turns. Also the probability utilizing the Krark's Thumb, Paradox Haze, or Doubling Season would also be fun.
DeathDark
★★★★★ (5.0/5.0)(3 votes)
@RiverWolf3
If you play it on turn 2, it will be at least turn 8 before you can use the ability as you will play it on 2, and the ability will first activate on turn 3.
Assuming a fair coin, you have an equal chance to land on either heads or tails. Thus, at each step, you can have 2 possibilities, giving us a binary branching pattern. However, the only path that will lead us to max capacity at turn 8 is 5 successes. At that point, there are 2^5 possibilities, meaning there are 32 other things that can happen. Your probability at turn 8 of having all 5 counters is 1/32. Decent chance, but I wouldn't bet my deck on it.
Also, there's some green card that doubles the number of counters on target permanent, or at least I could have sworn I saw it somewhere. If you have that, then getting 3 at any time would be all you need (with a 1/8 chance).
Duskdale_Wurm
☆☆☆☆☆ (0.0/5.0)
If the ability goes off, *BOOM!* Oh my ass!
sarroth
☆☆☆☆☆ (0.0/5.0)
@DeathDark: The card you are referring to is the G/U hybrid Gilder Bairn. If you get to 3 counters with this, use Gilder Bairn to get to 6 and activate the ability.
@riverwolf13 The probability of winning BY a specific turn (after playing a single goblin bomb) would follow the pascals triangle, but your probabilities are incorrect you can find the probability of winning ON turn X (After playing goblin bomb) by going to the 5th column of the Xth row and dividing that number by the sum of the numbers in that row. So the probability of winning on the 7th turn (5th turn after playing goblin bomb) would be 1/16 or 6.25% To find the probability of winning BY a specific turn you can find the sum of the integers of from the 5th column and to the right of the 5th column divided by the sum of the row again
To those who do not know how to construct pascals triangle you can find it here: http://math.about.com/od/algebralessons/ss/Pascal_2.htm
If someone can come up with a elegant equation to calculate to solve for this please post it, the closest i've come needs to be extended for each new turn(whenever you add another flip)
The only obstacle to proliferate is getting it started. It doesn't use Charge Counters, so Power Conduit doesn't work... But all you have to do is get one right. Then, you use a few Thrummingbirds or Contagion Engines or something, and instawin.
Unless your opponent plays lifegain. But who plays that?
@DeathDark - with unlimited turns, wouldn't your probability be 100%? Eventually it'll go off unless it's removed.
tavaritz
★★★★★ (5.0/5.0)(1 vote)
This is a little bit trickier to calculate than pure binary distribution. The reson being that if you lose the first toss then you can still win with next five as there was no fuse counter to remove.
So to get the boom on turn seven you have 1/32 probabilty and the same is for turn eight if you lost the first toss but won the next five (which means we have 1/64 chance for turn 8 boom). For turn 9 boom you need one lost toss on turn 4, 5, 6, 7, or 8 while all others are won, so you have 5 times 1/128 which is 5/128 chance to get boom on turn 9. Cu-mulatively this means that you have 11/128 chance to inflict 20 damage before turn 10.
Totema
☆☆☆☆☆ (0.0/5.0)
GROND! GROND! GROND!
Wait, no...
BOMB! BOMB! BOMB!
Tinno
★★★★★ (5.0/5.0)(1 vote)
@djflo
Actually no, it would never be able to reach 100% probability with ulimited turns. The probability of getting it at some point during an ulimited amount of turns would be as close to 100% as possible, but not exactly 100%, that would never be possible for the probability to reach, as you can never force the coin flip to flip your way :P
For an infinity it could flip against you, and keep on doing that, you could never predict if it were ever going to flip the other way, but the probability for that goes up and up each flip, but would never be able to reach 100% :D
I don't think it's a bad strategy but if you base your deck solely around this and proliferate, there are hundreds of cards that can leave your opponent just that little bit out of your range.
Cyber_Squirrel
★★★★★ (5.0/5.0)(1 vote)
@Tinno But as the number of turns converges on infinity, the probability of success converges on 100%. For any given finite number of turns, there will be less than 100% probability, but if you somehow put the game into an infinite loop that could only be escaped by setting the bomb off, I would argue that you hit 100% probability right there. It cannot go against you for an infinity, there's just no upper limit on the finite number of times it can go against you.
I'm not sure how the official MTG rules would cope with that, though.
JJBrazman
☆☆☆☆☆ (0.0/5.0)
This is a very simple Markov chain; a quick analysis shows that the probability of eventual explosion (if not removed, and given time as required) is in fact 1.
More interesting is the average number of turns required to explode (again assuming no interference of any kind), which is 30 (or 31 if you include the turn required to play it, which I don't tend to).
If you have a single proliferate card, and no other interference exists, your best bet is to save it for when you're on 4 counters; this cuts off an estimated 10 turns, rather than using it when you only have 1 counter (you can't proliferate 0 counters) which would only save you an estimated 4 turns.
If you have a Paradox Haze this cuts all times in half (assuming you played the haze first).
Even with paradox haze and proliferating in twice, you're still only expecting to make it in 6 turns, in which time your opponent should easily be able gain 1 life, set up his own strategy and start laughing, and that's if he doesn't have enchantment removal.
If you were required to sacrifice the enchantment upon losing the toss with no counters you'd have only a 1/6 chance of it going off at all, and expected time estimates wouldn't work.
If you only added counters, and merely did nothing upon losing the coin toss you would again have probability 1 of making it (excluding interference) and it would take you an average of 10 turns.
OlvynChuru
☆☆☆☆☆ (0.0/5.0)
@djflo
Keep in mind that, if you have infinite turns, you still draw a card each turn. This means that there is a good chance that you will deck yourself before you are able to trigger the bomb.
Comments (28)
Demonic Fan is pretty fun with this card too.
And Chance Encounter.
These are the natural probabilites of getting 5 counters on turn 7 and up, barring of course Krark's Thumb or Paradox Hazes, etc.
After getting a turn two Goblin Bomb:
7: 3% 8: 1.6% 9: 5.5& 10: 3% 11: 7% 12:4.4%
These probabilities follow from a restricted Pascal's Triangle. It took me awhile but I was able to realize that the pattern is sort of relfected about the middle.
The result is this
1
1 1
2 1 1
3 3 1 1
6 4 4 1 1
10 10 5 5 1 1
Someone correct me if I am wrong, but this should be the pattern. It would be intersesting to see what the maximum probability is, if allowed to take infinite turns. Also the probability utilizing the Krark's Thumb, Paradox Haze, or Doubling Season would also be fun.
If you play it on turn 2, it will be at least turn 8 before you can use the ability as you will play it on 2, and the ability will first activate on turn 3.
Assuming a fair coin, you have an equal chance to land on either heads or tails. Thus, at each step, you can have 2 possibilities, giving us a binary branching pattern. However, the only path that will lead us to max capacity at turn 8 is 5 successes. At that point, there are 2^5 possibilities, meaning there are 32 other things that can happen. Your probability at turn 8 of having all 5 counters is 1/32. Decent chance, but I wouldn't bet my deck on it.
Also, there's some green card that doubles the number of counters on target permanent, or at least I could have sworn I saw it somewhere. If you have that, then getting 3 at any time would be all you need (with a 1/8 chance).
The probability of winning BY a specific turn (after playing a single goblin bomb) would follow the
pascals triangle, but your probabilities are incorrect
you can find the probability of winning ON turn X (After playing goblin bomb) by going to the 5th column of the Xth row and dividing that number by the sum of the numbers in that row. So the probability of winning on the 7th turn (5th turn after playing goblin bomb) would be 1/16 or 6.25%
To find the probability of winning BY a specific turn you can find the sum of the integers of from the 5th column and to the right of the 5th column divided by the sum of the row again
To those who do not know how to construct pascals triangle you can find it here: http://math.about.com/od/algebralessons/ss/Pascal_2.htm
If someone can come up with a elegant equation to calculate to solve for this please post it, the closest i've come needs to be extended for each new turn(whenever you add another flip)
Boom.
Unless your opponent plays lifegain. But who plays that?
(Shut up, Felidar Sovereign)
So to get the boom on turn seven you have 1/32 probabilty and the same is for turn eight if you lost the first toss but won the next five (which means we have 1/64 chance for turn 8 boom). For turn 9 boom you need one lost toss on turn 4, 5, 6, 7, or 8 while all others are won, so you have 5 times 1/128 which is 5/128 chance to get boom on turn 9. Cu-mulatively this means that you have 11/128 chance to inflict 20 damage before turn 10.
Wait, no...
BOMB! BOMB! BOMB!
Actually no, it would never be able to reach 100% probability with ulimited turns. The probability of getting it at some point during an ulimited amount of turns would be as close to 100% as possible, but not exactly 100%, that would never be possible for the probability to reach, as you can never force the coin flip to flip your way :P
For an infinity it could flip against you, and keep on doing that, you could never predict if it were ever going to flip the other way, but the probability for that goes up and up each flip, but would never be able to reach 100% :D
Provided you hit every flip it goes off on turn 7. 2+5=7. But with proliferate you don't even really need to worry about that anymore.
I don't think it's a bad strategy but if you base your deck solely around this and proliferate, there are hundreds of cards that can leave your opponent just that little bit out of your range.
But as the number of turns converges on infinity, the probability of success converges on 100%. For any given finite number of turns, there will be less than 100% probability, but if you somehow put the game into an infinite loop that could only be escaped by setting the bomb off, I would argue that you hit 100% probability right there. It cannot go against you for an infinity, there's just no upper limit on the finite number of times it can go against you.
I'm not sure how the official MTG rules would cope with that, though.
More interesting is the average number of turns required to explode (again assuming no interference of any kind), which is 30 (or 31 if you include the turn required to play it, which I don't tend to).
If you have a single proliferate card, and no other interference exists, your best bet is to save it for when you're on 4 counters; this cuts off an estimated 10 turns, rather than using it when you only have 1 counter (you can't proliferate 0 counters) which would only save you an estimated 4 turns.
If you have a Paradox Haze this cuts all times in half (assuming you played the haze first).
Even with paradox haze and proliferating in twice, you're still only expecting to make it in 6 turns, in which time your opponent should easily be able gain 1 life, set up his own strategy and start laughing, and that's if he doesn't have enchantment removal.
If you were required to sacrifice the enchantment upon losing the toss with no counters you'd have only a 1/6 chance of it going off at all, and expected time estimates wouldn't work.
If you only added counters, and merely did nothing upon losing the coin toss you would again have probability 1 of making it (excluding interference) and it would take you an average of 10 turns.
Keep in mind that, if you have infinite turns, you still draw a card each turn. This means that there is a good chance that you will deck yourself before you are able to trigger the bomb.