- #1
bahamagreen
- 1,014
- 52
3 for 3 means guessing the first, second, and third cards of a pack without trying a fourth time... and 3 for 4 means guessing the first, second, and third correctly, but trying and missing the fourth. I think both are conditional (dependent) without replacement.
I see a form for 2 guesses that reads p(a and b) = p(a) x p(b|a) where the "and" means "and then",
I suppose for 3 guess it would be:
p(a and b and c) = p(a) x p(b|a) x p(c|b|a)
I'm not sure how to interpret the "|" symbol arithmetically... is it just a reminder that the actual calculation must be without replacement?
The story... A long time ago someone grabbed a deck of cards and asked me to guess the first one. Amazingly I guessed right, and then guessed correctly the second and the third. She wanted to pull a fourth card, but intuitively I felt that guessing three out of three was more improbable than guessing three out of four if I missed the last one, so I declined and said let's stay with the improbability of just three tries...
Recalling this event, I decided to check if my intuition was right, but am a bit confused on the calculation... and it may be that the calculation is additionally complicated by the guesser being allowed to halt the series of card guesses whenever he wants - another aspect of conditional dependent?
Any help on untangling the process to get a calculation is appreciated.
I see a form for 2 guesses that reads p(a and b) = p(a) x p(b|a) where the "and" means "and then",
I suppose for 3 guess it would be:
p(a and b and c) = p(a) x p(b|a) x p(c|b|a)
I'm not sure how to interpret the "|" symbol arithmetically... is it just a reminder that the actual calculation must be without replacement?
The story... A long time ago someone grabbed a deck of cards and asked me to guess the first one. Amazingly I guessed right, and then guessed correctly the second and the third. She wanted to pull a fourth card, but intuitively I felt that guessing three out of three was more improbable than guessing three out of four if I missed the last one, so I declined and said let's stay with the improbability of just three tries...
Recalling this event, I decided to check if my intuition was right, but am a bit confused on the calculation... and it may be that the calculation is additionally complicated by the guesser being allowed to halt the series of card guesses whenever he wants - another aspect of conditional dependent?
Any help on untangling the process to get a calculation is appreciated.