what are the odds that if 23 people picked from 4 variables (lets say they're A B C and D) what are the odds that only one person would pick D? i dont know how to do this kind of equation. can someone please tell me? thanks.
odds
- Thread starter austinpetemo
- Start date
You have to determine the probability (p) of only 1 person selecting D, then do p/(1-p) to get the odds.
To find the probability, find the total number of combinations where only 1 person chose D (a very large number, but smaller than the next one.) and divide it by the total number of combinations possible (a very large number).
To find the probability, find the total number of combinations where only 1 person chose D (a very large number, but smaller than the next one.) and divide it by the total number of combinations possible (a very large number).
theres not a simpler way to do it lol? cuz it was just a completely random selection, and i was the only person in a class of 23 to pick the last option. just wondering what the odds were.
was there actually options or just a b c d? if they didn't coincide with anything then i'd say we're pre programmed not to accept a "d" and with b and c being somewhat average in such a question they'd have a higher likelihood. ahhh... the human psyche. it's intriguing to say the least. the funny thing is most people are afraid to shoot for the moon so "a" was not the only choice.
:drool:
Following Zaffy's lead (and assuming we're not dealing with psychological factors per dun), I figure as follows:
If you line up the 23 people in a row and have them each pick silently and then reveal, your total # of combinations is 4^23. The combinations we're interested in are those where exactly one "D" is picked, so we can break that up into: combos where person 1 picked "D" and no one else did + combos where person 2 picked "D" and no one else did + etc.
# of combos where p1 picks "D" alone = 1*3^22 ("D" is fixed, and then the rest pick from "A", "B", and "C"). The # of combinations is the same for all, so the total # of combinations where one person picks "D" is: 23*3^22
Therefore p = (23*3^22) / (4^23) = 0.01026 = 1.026%
or in general, for N people, p = (N*3^(N-1) / (4^N)
Checking for small cases, with N = 1, p = 1/4 as expected,
with N = 2, p = 6/16 as expected (AD, BD, CD, DA, DB, DC)
...woot!
Following Zaffy's lead (and assuming we're not dealing with psychological factors per dun), I figure as follows:
If you line up the 23 people in a row and have them each pick silently and then reveal, your total # of combinations is 4^23. The combinations we're interested in are those where exactly one "D" is picked, so we can break that up into: combos where person 1 picked "D" and no one else did + combos where person 2 picked "D" and no one else did + etc.
# of combos where p1 picks "D" alone = 1*3^22 ("D" is fixed, and then the rest pick from "A", "B", and "C"). The # of combinations is the same for all, so the total # of combinations where one person picks "D" is: 23*3^22
Therefore p = (23*3^22) / (4^23) = 0.01026 = 1.026%
or in general, for N people, p = (N*3^(N-1) / (4^N)
Checking for small cases, with N = 1, p = 1/4 as expected,
with N = 2, p = 6/16 as expected (AD, BD, CD, DA, DB, DC)
...woot!
Wow, I didn't even know what 4dkh is -- had to look it up ---- apparently something to do with pH :thud: :laugh: