View Full Version : 1/x chance, odds on favorite?
Roswell
06-20-2005, 08:15 AM
If there is a 1 out of X chance of an event occuring, after how many trials does it become an odds-on favorite to occur?
moomoocow
06-20-2005, 09:08 AM
1/x chance of happening
so (x-1)/x of it not happening
for it to be favourite,
1 - [(x-1)/x]^n > 0.5
[(x-1)/x]^n <0.5
n * ln[(x-1)/x] < ln(0.5)
n > ln(0.5)/ln[(x-1)/x]
So - say it's 1/10
n > 6.6
so you need 7 trials.
moo!
pzhon
06-20-2005, 09:09 AM
[ QUOTE ]
If there is a 1 out of X chance of an event occuring, after how many trials does it become an odds-on favorite to occur?
[/ QUOTE ]
After n*X trials, the probability it has not occured is about e^-n, where e ~ 2.718. e^-(ln 2) = 1/2, so the probability is about 1/2 after (ln 2)*X ~ .693 X trials. (ln 2)*X is a slight overestimate for the number of trials needed.
Some early work on probability was inspired by the observation that you are a favorite to roll at least one 6 among 4 dice, but an underdog to roll 6-6 at least once in 24 attempts.
kiddj
06-20-2005, 11:55 AM
n = #of trials
n > (LN(0.5))/(LN((X-1)/X))
(This is probably the same as what pzhon said, but this seems to work.)
BruceZ
06-20-2005, 12:31 PM
[ QUOTE ]
n = #of trials
n > (LN(0.5))/(LN((X-1)/X))
(This is probably the same as what pzhon said, but this seems to work.)
[/ QUOTE ]
Pzhon's is an approximation of yours which works well for large X (like 10). Note yours is the same as:
-ln(2)/ln(1-1/x) =~ -ln(2)/(-1/x) = x*ln(2).
Also, the chance that it doesn't happen in x trials is:
(1 - 1/x)^x -> 1/e
So, the chance that it doesn't happen in nx trials is:
(1 - 1/x)^nx -> e^-n
kiddj
06-20-2005, 12:53 PM
[ QUOTE ]
[ QUOTE ]
n = #of trials
n > (LN(0.5))/(LN((X-1)/X))
(This is probably the same as what pzhon said, but this seems to work.)
[/ QUOTE ]
Pzhon's is an approximation of yours which works well for large X (like 10). Note yours is the same as
-ln(2)/ln(1-1/x) =~ -ln(2)/(-1/x) = x*ln(2).
[/ QUOTE ]
Yeah, I had to pull the old log rules out of my ass for this one, but it keeps my mind occupied. /images/graemlins/tongue.gif
BruceZ
06-20-2005, 01:23 PM
[ QUOTE ]
Some early work on probability was inspired by the observation that you are a favorite to roll at least one 6 among 4 dice, but an underdog to roll 6-6 at least once in 24 attempts.
[/ QUOTE ]
Also, with 2 dice, you are a 6-5 favorite to roll a 7 before you roll a 6, and you are a 6-5 favorite to roll a 7 before you roll an 8, but you are a 6-5 underdog to roll two 7's before you roll both a 6 and an 8!
AaronBrown
06-20-2005, 01:43 PM
In finance we often solve a related problem, at a given interest rate X, how many years does it take your money to double? You get a decent approximation by dividing X into 70. For example, it takes about 35 years at 2% interest, and about 17.5 years at 4%. Many people know this as the Rule of 72, because 72 works better for interest rates in the 8% range, which were common in the 70's and 80's.
For this problem, you can use 0.69*X as a pretty good approximation for X > 30. It's the same approximation as the Rule of 72. So 69 trials for X = 100 (68.97 is exact).
BruceZ
06-20-2005, 02:07 PM
In case you read the above before I fixed it, try it again now. /images/graemlins/blush.gif
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