this needs to be on exam P. The pass rate is too high on that exam!
Also we need to tell them that a bunch of credential actuaries can’t solve this problem because exams don’t matter
which is reasonably close to ahj911211’s simulation (73.4%) and to my Excel .7332
Yes, from the beginning I said my direct calculation had matched ahj911211’s simulation
I don’t think 90 choose 56 is the right denominator. That would be the right denominator if everyone had an identical chance of getting picked.
I mean that’s the right number of outcomes, but it’s not the right number of ways to arrive at those outcomes.
And 56 choose 50 is clearly not right as it assumes that all 50 of the two-ballers get in.
Basically, I can tell you why you’re wrong, but I can’t come up with the right answer, so I’m mostly useless.
Let’s see if I can improve on this…
Probability of 1 ticket getting in
(1/5)(2/4) + 2(2/5)*(1/3)
Okay, I really have no idea how to generalize this to greater numbers
sorry, only looked at numbers & skipped words
I claim this to be solved
I would think that a successful person’s second ball being drawn would be ignored. So, it’s effectively “no longer in the pot.”
Not sure what to do with that info, though.
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Sig lines! Sig lines for sale!!
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Yes, after looking at SteveWhite’s example of picking 2 instead of 56, I convinced myself that after a draw of the first “blue” the results of the next draw are the same either by modeling the duplicate as an option or just ignoring possibility of a duplicate (you do still have to adjust denominator, which is like removing all duplicates). Hadn’t gone farther to see if it works with multiple blues drawn (it probably will).
What to do with it? It would simplify my tree as each draw has 1 of 2 possible outcomes (instead of sometimes having 3).
Continuing to try to do this without resorting to simulation. I’ve gotten much closer to the simulation results (with the bonus that regardless of winners the probability never goes above 1 – I’m getting 77.33% for the 56 winner case.
Here is my thought process.
If there are M 2 chance entrants and N 1 chance entrants, and we select W winners:
There are guaranteed to be M* = max(0, W-N) 2 entrant winners and N* = max (0, W-M) 1 entrant winners. Each 2 chance entrant has an M*/M probability of winning one of those slots, and each 1 chance contestant has a N*/N chance of winning one of those slots.
Now let m = M - M*, and n = N - N*.
We have m+n people left to win W - M* - N* prizes. Let A = (W-M*-N*)/(m+n) (the proportion of people left that will win)
The probability that a 1 chance person wins one of the non guaranteed spots is:
n/N* (A (m+n) / (2m+n))
derivation
A = m/(m+n)(2P)+n/(m+n)(P)
A(m+n) = m(2P) + n (P)
A(m+n) = (2m+n) P
P = A(m+n)/(2m+n)
Multiplying that by n/N, the probability that they are still around after the guaranteed winners are selected, gives the final answer.
The probability that a 2 chance person wins one of the non guaranteed spots is:
m/M * 2 * A * (m+n) / (2m+n)
The final probability for the 2 chance person is M*/M + m/M * 2 * A * (m+n) / (2m+n).
For 56 winners from 50 2 chance and 40 1 chance people, I get 16/50 + 34/50 * 2 * .50 * (34+34) / (68 + 34) = 77.33%.
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Nice.
Another approach might be to determine the probability of getting picked if he were the ONLY one with two chances. Then, increase that by one, see if a pattern or some kind of recursion emerges.
I think your formatting here is misleading. I read this as n divided by N* times (A…). But I think you mean this to be n divided by N times (A…). I’m assuming this latter.
If I followed correctly, you have here the Prob of winning a guaranteed slot OR a non-guaranteed slot.
This could work for calculating the 1-entrant scenario (since it’s impossible for that entrant to be in both situations). I get this probability to be ~0.28333. Using your calculations above, this gives a sum of ~1.05633333.
It seems that you would still need to subtract the prob of a 2-ticket entrant being in both scenarios. Based on the calculation in the preceding paragraph, this should be around .05633. Equivalently, using PZ’s method of using the complement for an indirect calculation, the probability of a 2-ticket entrant getting a prize is ~0.71666667
Sorry, that was misleading. It was the latter.
I was trying to segregate the people into ‘guaranteed winners’ and ‘non-guaranteed entrants’
Number of entrants: M* + m = M
Guaranteed winners: [M*/M] * 1 = M*/M total probability
Non-guaranteed entrants: [m/M] * 2 * A * (m+n) / (2m+n) total probability
It’s essentially a weighting of the chance they win depending on which of the groups they fall in.
I haven’t read through detailed solution ideas, sorry. so maybe somebody suggested this already.
But what about doing it in two parts. First find the probability that X winners are double ball holders. This would involve taking the number of permutations of double ball winners/ single ball winners.
Then calculate the probability that you win given the number of double ball winners.
you’d have to sum over the number of double ball winners but that too should be tractable.
And I like this idea; but you also need to consider the possibility that an entrant could potentially be in both (if they had two tickets). That is, your work only considers each group separately, but not jointly.
For the 1-ticket entrant, it’s impossible to be in both groups; so it’s not needed to look at the two scenarios jointly.
A good test of the methodology is to see if the probabilities of the 1-ticket entrant and the 2-ticket entrant sum to unity.
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I don’t think they should sum to unity. M P(2 entrant) + N P(1 entrant) should sum to W.
I think the premise is that they can’t.
Also, I’m assuming a 2nd draw of a entrant is a null draw and doesn’t matter. That ball is effectively removed by my calculations, so if it gets drawn, no it didn’t. 