So which car he gets is random with a 1\32 probability for any unique car?
We took my grandson to the Santa Cruz Boardwalk. He fell in love with the gondola ride. He wants to go back until he has ridden on all of the 32 "cars".
My probability is too foggy to work out the expected number of rides he would have to take to have a 50% (or 90%) chance of riding on each of the 32 cars.
I am assuming that it is a very large number (thousands?).
Can anyone supply the formula f(N,P) where N is the number of cars and P is the target probability (50%, 90%)?
Thanks
So which car he gets is random with a 1\32 probability for any unique car?
Stop cowardice, ban guns!
hi,
looking at this as a practical problem, by running some simulations it looks like you have a 22% chance to finish in less then 100 rides, 75% chance to finish in less then 150 and 95% chance to finish in less then 200 rides. Now it becomes a question of budget and time I think this is a better indicator then the expected number of rides would be.
Failing to come up with a closed form equation, I also ran a simulation. My results match yours. Mine are below.
But I really want the closed form equation. I'm hoping someone can point me in the right direction.
Code:Number of trials = 1,000,000 Number of balls = 32 Trial Summary Draws Tally Cum Prob 43 1 0.00 44 1 0.00 45 3 0.00 46 2 0.00 47 7 0.00 48 7 0.00 49 6 0.00 50 15 0.00 51 30 0.01 52 26 0.01 53 42 0.01 54 82 0.02 55 112 0.03 . . . 68 1538 0.92 69 1786 1.10 70 2042 1.30 71 2279 1.53 72 2698 1.80 73 2781 2.08 . . . 86 7493 9.04 87 7903 9.83 88 8385 10.67 89 8553 11.53 . . . 122 11671 49.66 123 11596 50.82 124 11361 51.96 . . . 149 7269 74.93 150 6915 75.62 151 6627 76.29 . . . 180 3053 89.93 181 2952 90.22 182 2934 90.51 183 2858 90.80 184 2837 91.08 . . . 202 1663 94.89 203 1556 95.04 204 1562 95.20 . . . 253 349 98.99 254 338 99.02
I think the best "closed form" solution you would get would be in the form of a horrible cascading summation beast.
I don't have emotions and sometimes that makes me very sad.
Although to answer the question in your title we can get the expected value of the number of tries it will take.
(math tags seem broken at the moment but...)
Sum from i = 1 to 32 of (32)/(32 - i + 1) = 129.87184625396864
The simulation I ran also gives a median of 123 which seems reasonable.
I don't have emotions and sometimes that makes me very sad.
N =32 possible rides. Let's say you want the probability of having collected 3 rides after 12 tries. The time before you had either had 3 rides already and you got another old one or you had 2 rides and got a new one. So P(3 after 12) = P(3 after 11)x3/32 + P(2 after 11)x30/32
In general P(r after t) = P(r after t-1)xr/N + P(r-1 after t-1)x(N+1-r)/N. Also P(1 after 1) = 1 All other P(0 after t) and P(r after 0) = 0.
The spreadsheet attached gives the exact probabilities calculated by filling in the t = 0 and r = 0 probabilities, using the general formula for using the general formula for P(1,1) and copying the formula all over the place. The simulation figures agree with the table.
Just wanted to point out that this is known as the Coupon Collector's Problem.
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