Just want to say i used to watch your content relating to MTG 5 years ago and stopped because of how MTG ran. I am really glad to now able to watch your data driven content on pokemon tcg
Hayley's guest appearance at the end was the sweetest! My heart melted at the "hi, chat!" Every time I see Jeff interacting with his kids I get one step closer to wanting my own! hahahahha
glad to rediscover you playing this. I watched your mtg stuff and the start of marvel snap when it came out but if im honest snap melted my brain a bit and i didnt want to think that hard playing a card game on my phone.
After opening more than 120 packs, Omastar is the only card I am missing to have the Kanto pokedex completed. After seeing the notification for this video, I am tempted to craft two copies.
I would probably have used the omastar over the jolteon at 9:00. I also always forget to count the other successful outcomes in my quick odds calculations. I was thinking it was 50%, but like you said a bit later: it is closer to 75%. We are happy if it is three or four heads as well.
Jeff, I finally built a deck that feels as good as pickachu EX thanks to you! Who knew adding two more basics (one jynx, one kangaskhan) to mewto gardevoir would make it waaaaay more consistent. 😅
If you pick Omastar over Jolteon i think you lose? Turn 1 he swaps Mewtwos and Psychic Spheres you for 50, turn 2 he swaps in Jynx and hits you for lethal. If you Sabrina he just swaps in Gardevoir then immediatly swaps back to his health Mewtwo which only delayes one turn.
Care to showcase the Melmetal/Butterfree deck? I've been doing very well with it. No other basics, all metal energy, either starting Pokemon does something productive with their attack. Melmetal never dies once it hits the board successfully.
This is just me, but a Giovanni instead of an X-speed might be better. It would allow you to guarantee kills in several situations, such as the first game with the two Mewtwos.
When you decided to go with the Jolteon pin missile line to knock out the Mewtwo, you said it was 75% to win there. I calculated 85%~ with a hypogeometric calculator but i think I did it wrong. I set population size as 20 (0 Heads + 4 tails, 1 Heads + 3 tails... 5 heads + 0 tails), number of successes in population = 12 (the 3 sets of results that have at least 2 heads), sample size = 4, and number of successes in sample = 2. What did I do wrong? the omastar line seemed pretty safe - Hard to say if it's better than a 75% shot to win but Im struggling to think of a combination of cards that gets them out of that spot.
You used it wrong, but that's understandable. First off, a Hypergeometric Calculator is not necessary for coinflips, and will actually give a wrong result regardless of if you input all the information in a way that appears correct. Hypergeometric Calculators are for populations that decrease as each sample is pulled. A deck of cards is the prime example; each card pulled cannot be pulled again. For coinflips we don't need to do any of that. The results are independent of each other, so we can just multiply odds to find specific results, and add the odds of successful results. So for four fair coin flips, two heads is 6/16, three heads is 4/16, and four heads is 1/16, making the total odds of flipping at least two heads 11/16 or about 69% (nice). As for how to actually use a Hypergeometric Calculator, the population size is not all possible outcomes, it is how many objects there are to be randomly decided between. Let's try an example. We'll see how often a dedicated water aggro deck can draw Misty on turn 1. First, a Pokemon Pocket deck would have a population size of 20. But, we know one of those cards is guaranteed to be a basic in our starting hand, so we only put 19. The number of successes in the population would be how many cards we are trying to find in that population, so let's say 2 for copies of Misty. The sample size would be how many cards are drawn, so 6 would be your starting hand plus the card you drew on your first turn, but we already know one is the guaranteed basic, so we only put 5 for the sample size. Number of successes in the sample would be how many in the sample size you want to be a success, so 1 for a copy of Misty. After plugging those numbers in, we can see that getting exactly one Misty is about 41%, and getting at least one Misty is about 47%.
@@connorhamilton5707 Thank you! That was super helpful and you did a great job explaining! I forgot those key points since the last time I took stats lol. I referenced an example for how to pull from a deck of cards to set mine up and forgot the key difference that the coins are not "without replacement" like cards are. Btw, is there a quick way to mentally math out what the odds are for successful results for coin flips? for example, the odds of getting 2 heads is 6/16 - I can see that when I map out all of the combinations and count them up, but is there a shorter way of determining that? going back to teh video, I guess Jeff had just rounded up from 69 to 75%.
@@donsmith2588 hmm, looks like my response was not allowed for some reason. Maybe too long, or too many spaces in a row made it look like spam. Okay, much simpler reply, you are looking for (n choose k). If you look that up, you can get a more detailed explanation. And the 75% value seems like a misuse of the Hypergeometric Calculator. Population size 8, total successes 4, sample size 4, sample successes 2. Edit: clarified successes.
I love Jolteon but it puts a bad taste in my mouth because I once got 12 tails in a row over the span of 3 consecutive turns (seriously no joke) and that’s not something you recover from lol
Just want to say i used to watch your content relating to MTG 5 years ago and stopped because of how MTG ran. I am really glad to now able to watch your data driven content on pokemon tcg
Hayley's guest appearance at the end was the sweetest! My heart melted at the "hi, chat!"
Every time I see Jeff interacting with his kids I get one step closer to wanting my own! hahahahha
Really awesome showing and explanation of the decklist.
I put this deck in my editor after watching it on the stream, I have it named Lightning Helix, OMG 😂
Damn, and here I was thinking that I was smart for naming it that myself xD Great minds think alike!
Does it deal 3 damage and heal 3 health? (Mtg)
@@eleclerbarreto7512 If you squint, Vaporeon does!
glad to rediscover you playing this. I watched your mtg stuff and the start of marvel snap when it came out but if im honest snap melted my brain a bit and i didnt want to think that hard playing a card game on my phone.
I agree with Hayley's card evaluation at the end
As always, great vid Jeff! Thanks for making excellent content
I really like omastar. Feels like an underrated pick in dual energy decks. I've been loving it with weezing lately.
Great highlight! I’ve been jamming a TON of games with a similar shell with Greninja over Omastar and been having an absolute ball
After opening more than 120 packs, Omastar is the only card I am missing to have the Kanto pokedex completed. After seeing the notification for this video, I am tempted to craft two copies.
The editor is killing it with the intro deck animations!
I would probably have used the omastar over the jolteon at 9:00. I also always forget to count the other successful outcomes in my quick odds calculations. I was thinking it was 50%, but like you said a bit later: it is closer to 75%. We are happy if it is three or four heads as well.
It's actually about 69% or 11/16 to get 2 or more heads. Nice odds, but not quite as nice as Jeff says.
As always, Jeff's children are adorable!!
Great games, love the slower approach compared to aggro decks.
Nice, don't have either of these cards, so it's cool to see them in action
That's all fine and good Jeff. I didn't hear a single thing you said the first 40s of the video. Your daughter is absolutely adorable!
Jeff, I finally built a deck that feels as good as pickachu EX thanks to you! Who knew adding two more basics (one jynx, one kangaskhan) to mewto gardevoir would make it waaaaay more consistent. 😅
Most of the top finishing Mewtwo decks in events are on 5-6 basics now as well. Feels good to have my deck building theories validated.
Jolteon is fire.
this team is a banger just want the trade feature to come sooner… I’m down bad for a Kang
If you pick Omastar over Jolteon i think you lose? Turn 1 he swaps Mewtwos and Psychic Spheres you for 50, turn 2 he swaps in Jynx and hits you for lethal. If you Sabrina he just swaps in Gardevoir then immediatly swaps back to his health Mewtwo which only delayes one turn.
I enjoy watching you i like pokemon tcg its very f2p friendly
Care to showcase the Melmetal/Butterfree deck? I've been doing very well with it. No other basics, all metal energy, either starting Pokemon does something productive with their attack. Melmetal never dies once it hits the board successfully.
Mewtwo EX has entered the chat.
@HooglandiaPocket But of course. The games have played against it I ramped faster than Sir Two.
I would say try the exeggutor butterfree it has 160 hp and 80 dmg turn 1 so stronger vs pika & mewtwo
This is just me, but a Giovanni instead of an X-speed might be better. It would allow you to guarantee kills in several situations, such as the first game with the two Mewtwos.
Nice deck Jeff. Strange nobody trying the legendary trainer Ash deck.
List?
Nice :)
maaaan ur so epic
When you decided to go with the Jolteon pin missile line to knock out the Mewtwo, you said it was 75% to win there. I calculated 85%~ with a hypogeometric calculator but i think I did it wrong. I set population size as 20 (0 Heads + 4 tails, 1 Heads + 3 tails... 5 heads + 0 tails), number of successes in population = 12 (the 3 sets of results that have at least 2 heads), sample size = 4, and number of successes in sample = 2. What did I do wrong?
the omastar line seemed pretty safe - Hard to say if it's better than a 75% shot to win but Im struggling to think of a combination of cards that gets them out of that spot.
You used it wrong, but that's understandable.
First off, a Hypergeometric Calculator is not necessary for coinflips, and will actually give a wrong result regardless of if you input all the information in a way that appears correct. Hypergeometric Calculators are for populations that decrease as each sample is pulled. A deck of cards is the prime example; each card pulled cannot be pulled again.
For coinflips we don't need to do any of that. The results are independent of each other, so we can just multiply odds to find specific results, and add the odds of successful results. So for four fair coin flips, two heads is 6/16, three heads is 4/16, and four heads is 1/16, making the total odds of flipping at least two heads 11/16 or about 69% (nice).
As for how to actually use a Hypergeometric Calculator, the population size is not all possible outcomes, it is how many objects there are to be randomly decided between.
Let's try an example. We'll see how often a dedicated water aggro deck can draw Misty on turn 1.
First, a Pokemon Pocket deck would have a population size of 20. But, we know one of those cards is guaranteed to be a basic in our starting hand, so we only put 19.
The number of successes in the population would be how many cards we are trying to find in that population, so let's say 2 for copies of Misty.
The sample size would be how many cards are drawn, so 6 would be your starting hand plus the card you drew on your first turn, but we already know one is the guaranteed basic, so we only put 5 for the sample size.
Number of successes in the sample would be how many in the sample size you want to be a success, so 1 for a copy of Misty.
After plugging those numbers in, we can see that getting exactly one Misty is about 41%, and getting at least one Misty is about 47%.
@@connorhamilton5707 Thank you! That was super helpful and you did a great job explaining! I forgot those key points since the last time I took stats lol. I referenced an example for how to pull from a deck of cards to set mine up and forgot the key difference that the coins are not "without replacement" like cards are.
Btw, is there a quick way to mentally math out what the odds are for successful results for coin flips? for example, the odds of getting 2 heads is 6/16 - I can see that when I map out all of the combinations and count them up, but is there a shorter way of determining that?
going back to teh video, I guess Jeff had just rounded up from 69 to 75%.
@@donsmith2588 hmm, looks like my response was not allowed for some reason. Maybe too long, or too many spaces in a row made it look like spam.
Okay, much simpler reply, you are looking for (n choose k). If you look that up, you can get a more detailed explanation.
And the 75% value seems like a misuse of the Hypergeometric Calculator. Population size 8, total successes 4, sample size 4, sample successes 2.
Edit: clarified successes.
Kangaskhan is extremely hard to obtian for me, any replacement suggestions for this elusive mama?
You can play Farfetch'd, but your Pikachu match up will be harder.
omar what a star
Please play marowak ex 😁
I've got a Machamp-Maro & Sandslash-Maro in the queue to get up this week hopefully!
@@HooglandiaPocket looking forward to it !
I love Jolteon but it puts a bad taste in my mouth because I once got 12 tails in a row over the span of 3 consecutive turns (seriously no joke) and that’s not something you recover from lol
I'm a simple man: I see 4 tails on my first turn, I reset my app.