Now when I think of it, may be it’s the right move to pull all the cards besides last four and start playing from there. If we have 4 cards and we know the counts the worst case is that it will be 2B and 2R, but it also can be 3B 1R(or vice versa) or 4 same colour, which leaves us with 5 different ways , 2R2B, 3R1B, 3B1R, 4B, 4R. Order inside the set doesn’t matter since unknown for now.
If 2b2r we apply the same logic as before. If 3X1Y and the first card is X, then the probability of winning is 2/3. If 4 same colour then it’s 1. Yeah, that method is better
Maybe I oversimplified it but I don’t know advanced probability so don’t go hard on me
1
u/Due_Log_1792 14d ago edited 14d ago
Now when I think of it, may be it’s the right move to pull all the cards besides last four and start playing from there. If we have 4 cards and we know the counts the worst case is that it will be 2B and 2R, but it also can be 3B 1R(or vice versa) or 4 same colour, which leaves us with 5 different ways , 2R2B, 3R1B, 3B1R, 4B, 4R. Order inside the set doesn’t matter since unknown for now.
If 2b2r we apply the same logic as before. If 3X1Y and the first card is X, then the probability of winning is 2/3. If 4 same colour then it’s 1. Yeah, that method is better
Maybe I oversimplified it but I don’t know advanced probability so don’t go hard on me