i was just about to comment the same thing

You comment this on every video give it a rest

@@drawnrex8573 they have 1 comment in total on this channel..?

@@drawnrex8573 literally has commented once on the channel

What if I told you it was itani dodging me?

@@AJ3 we'd all believe it

Whats an everton legend? Bar rooney

@@banton9368 baines Cahill Dixie dean neville southall

Rooney> His card is so bad

Your getting mad over a squad builder showdown …

@@banton9368 football existed before 1992

Probably first and last hungarian icon ever😅 but we are proud of him

@@Doncika99 that rb leipzig guy looks good

lol

Hi oliver lol

Oliver

Hi

Hi Oliver

but surely it doesn't change the comparative chance of getting a Ronaldo compared to Daka for example, ie although the actually chance of getting either would go up if Eduoard wasn't in it the the percentage difference of getting Daka instead of Ronaldo wouldn't therefore it only effects those who actually got Eduoard as an option as it removed a choice. I do economics at uni so have some statistics modules but dont have the best understanding but that's what I can make sense of based on conditional probabilities, but please correct me if I'm wrong.

How far would you regress with that logic though? You could regress all the way to the actual player pick, the odds of me getting x player given no other players (bar two randoms) is 100%. I think andy is right in this case, I don’t think conditional probability plays a part because Edouard WAS available, the fact I didn’t get him doesn’t allow me to exclude him from the probability pool. In this original case there’s nothing special about Edouard, he is just a card among hundreds, he only becomes relevant after he is removed, then the odds change.

It does actually apply and let me try to explain why: Lets simplify this by applying it to the die example discussed in the video. Let's say you're looking for a result between 1 and 4 but accidentally get handed a 5 sided die. Now still every result between 1 an 4 is equally likely but in case you roll a 5, the mistake is noticed and they hand you a new correct 4 sided die to get the proper result. Conditional probability applies, because the only legit scenario is the one where you don't get a faulty result. In a sense you could also think it the way that you keep rolling the die until you get a valid result and for those valid results the odds would still remain the same as they would be in a die with no invalid results. Now obviously this still does not make it entirely fair for everyone. Those lucky enough to get the invalid result get an unfair advantage because in addition to getting to roll again, they are able to keep the result of the invalid roll (or a player pick). This could be used as an argument that everyone should get a new pick item but it does not change the odds of valid results and in case of player pick where you get offered two other players in addition to the invalid result, it actually increases the odds of getting desired result, even if very marginally.

@@kallekalliomaki4141 ur so bothered

@@kallekalliomaki4141 I was about to comment esentially exacty what you said. I will however, give another expample: Imagine you have a wheel with the numbers 1-8, and another wheel with the numbers 1-8 and a spin again (basically Edouard). The odds of getting, say, number 2 is the same on either wheel (after spinning again if landed on). The fairest way to do it would be to give people who got palyer x, player y and Edouard another pick which gives them x and y guaranteed, and another random player (who isn't x, y or Edouard). Also they don't get to keep their original pick unless they pick said player again. That way it's fair for everyone.

As things stand he is Spanish mate on the paper 😂 (just kidding)

@@suthir_nadar loool😂

Legend

They're joking around, his name is Jack.

Get well soon mate

Basically you got three players from the proper pool if you didn't get Edouard, so you weren't affected

If you had edouard it was like a extra chance but it doesnt change the probably i mean AJ is a bit lost if he really if he really thinks a 1/4 times to chance is 50% by that logic 4 tryes would guarantee u a think what is impossible without taking the other options away therfore a 25% is a 25% even with 50000 tryes

@@ehrlichgesagt863 I mean a single player pick is three draws from a pool without replacement, so in this situation where the pool is 4 cards and you get 4 picks without replacement it is a 100% chance

Rolling the dice twice will give 16 possible outcomes with 7 of them including the spesific number, so therefore the chance of getting a specific number is just under 50%. Lets say we chose the number 2. the possible diceroles we could Get with 2 in it is: 1,2-2,2-2,3-2,4,-2,1-3,2-4,2 The diceroles Are independent of eachother, but the chance of getting the specifikk number increases with the number of diceroles

@@sondrerpnorway6290Unfortunately statistics doesn't work like that. You take each dice roll independently. The only time that logic works is if you're rolling 2 dice simultaneously. Or else it's 25% for the first roll and another 25% for the second roll

@@Mo_C_ rolling the die one by one and simultaniously will have the same possible outcomes. By your Logic rolling the die 100 times will only give me a 25% chance of getting the specific number. The chance of getting the specific number on each roll is 25% and the rolles Are not dependant on each other, but there Are still 16 possible outcomes with 7 of them being a 2 rolled

@@sondrerpnorway6290 That's exactly what I'm saying. If you roll a 4 sided die 100 times, you are expected to get a specific number 25 times. Because you completely ignore the first roll so if you didn't get the number you want then it's still 25% chance of getting it the second time.

We’re talking about Fifa picks though If you roll a 1, 1 gets added to the player pick and can’t be chosen again However you then have a 1/3 chance of landing on the other 3 and then being added So I think technically we are both wrong but I have just got up and cba to properly do the maths (it’s 0.75x0.66 I think as they’re the chances of not getting it on both rolls)