Originally Posted by
Humots
I believe that in .../ACT2Scene1.php Jabba miscalculated the probability of having drawn a card from the All-Ace deck.
This is an example of conditional probability: the probability that an event would have, given that another event has occurred. In this case, the probability we want is the probability that, given that an Ace has been drawn, the Ace was drawn from the all-Ace deck.
Let's represent this probability by the expression P(All-Ace deck|Ace drawn).
Bayes’ theorem (see …/wiki/Bayes’_theorem) states that for conditional probabilities A and B,
P(A|B) = ( P(B|A) * P(A) ) / ( P(B) )
In this case, for A = All-Ace deck and B = Ace drawn,
P(All-Ace deck|Ace drawn) =
( P(Ace drawn|All-Ace deck) * P(All-Ace deck) ) /
( P(Ace drawn) )
For Jabba’s example,
P(Ace drawn|All-Ace deck) = 1.0 since the All-Ace deck contains Aces only;
P(All-Ace deck) = 1 / 50 = 0.02;
P(Ace drawn) = P(Ace drawn | All-Ace deck) * P(All-Ace deck) +
P(Ace drawn | not All-Ace deck) * P(not All-Ace deck)
= 1.0 * (1/50) + (1/13) * (49/50)
= 0.02 + 0.075385
So, P(All-Ace deck|Ace drawn) = (1.0 * .02) / (0.02 + 0.075385)
= .02 / 0.09538
= 0.2097,
which is slightly better than 1 chance in 5.
While this is not that far from Jabba’s result ( 0.07538 / .02 ) which he called 1 in 4, it indicates to me that Jabba misapplied conditional probability, and probably Bayes’s Theorem, by leaving out one term and getting the expression upside down.
This does not support Jabba's claim to be a certified Statistician (whatever that means). It looks to me like Jabba is cutting and pasting a few statistical arguments he understands poorly if at all.
Humots,
- This is going to take me awhile, but so far, I can't figure out why we can't just compare the 2 "combined" probabilities -- i.e. combined probability #1) the probability of randomly selecting the ace deck from the total number of decks (.02), times the probability of drawing an ace, once the ace deck has been chosen (1.00), and #2) the probability of randomly selecting a normal deck from the total number of decks (.98), times the probability of drawing an ace, once the normal deck has been chosen (.076923077).
- Consequently, before we get started, the probability of chosing the ace deck and then drawing an ace is .02*1.00, or
.02, while the probability of chosing a normal deck and then drawing an ace is .98*.076923077, or
.075384615. And, the probability of drawing an ace via the second route is almost 4 times as large as the probability of doing it via the first.
--- Jabba
- The smilie at the top is an accident, but I don't know how to get rid of it.