but I don't think the odds of winning are fifty-fifty. every card dealt changes the odds that you'll win based upon how many later cards could have you win, as well as the rest of the table/dealer winning first. Say you get a face-value card first. you odds of winning are pretty good, as you can get really close to 21 without busting with whatever your next card is, there are 11 cards that could make you a lock on victory the next round, so the odds of winning in two rounds are 12/52 times 11/50. But, someone else could also get those two cards theoretically, and their odds are 11/51 times 10/50. but in actuality, your odds change when they get these cards, so i you both get face-value cards on the first deal the odds you' both get them the second deal are 12/52 time 10/50 and 11/51 times 9/49 respectively. Then either you bust first or get an ace. The odds of getting an ace are 4/48 at this point. meanwhile, the odds of busting are 44/48. so all in all, the odds of winning a perfect game would be 12/52 times 11/50 times 4/48, which equals a 0.423(rounded) percent chance. That is just one way to win, obviously, but my point is that the game is not 50/50, but a complex interaction of different probabilities.
No? I think you're not very good at maths. The chances are 50%, because you'll either win or lose
so true bestie
but I don't think the odds of winning are fifty-fifty. every card dealt changes the odds that you'll win based upon how many later cards could have you win, as well as the rest of the table/dealer winning first. Say you get a face-value card first. you odds of winning are pretty good, as you can get really close to 21 without busting with whatever your next card is, there are 11 cards that could make you a lock on victory the next round, so the odds of winning in two rounds are 12/52 times 11/50. But, someone else could also get those two cards theoretically, and their odds are 11/51 times 10/50. but in actuality, your odds change when they get these cards, so i you both get face-value cards on the first deal the odds you' both get them the second deal are 12/52 time 10/50 and 11/51 times 9/49 respectively. Then either you bust first or get an ace. The odds of getting an ace are 4/48 at this point. meanwhile, the odds of busting are 44/48. so all in all, the odds of winning a perfect game would be 12/52 times 11/50 times 4/48, which equals a 0.423(rounded) percent chance. That is just one way to win, obviously, but my point is that the game is not 50/50, but a complex interaction of different probabilities.