iirc it’s credible as soon as it boils, so 212 seems like the right number. In Europe you only need to do 100.
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if the probability is 65%, the credibility standard is 2, since 65%*2 > 100%
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You are correct. I wrote it wrong. There could be 106 selections. I had that and then blanked when I wrote it. I was getting myself confused while doing this which is why I posted. Haven’t done enough real math work in a while.
The 65% figure would seem a little low since if we all had 1 ticket, it would be 56/90=62%. Having second ball doesn’t appear to help then.
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Well, those with only one ticket have only a 35% chance, so having a second ball hurts them considerably, imo.
My simple simulation is showing the same .73 range that others are seeing.
I’m wondering if you are actually simulating P(2 entries | gaining entry). It’s been a while since I’ve done Bayes Theorem, but if I use
P(gaining entry | 2 entries) = 0.73
P(gaining entry) = 56/90
P(2 entries) = 50/90
I get P(2 entries | gaining entry) = 0.65, which is close to what you are getting from your simulation.
Of course, it’s been years since I’ve done much of this stuff, so take it for what it’s worth.
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yeah…I think you’re right…
65% is the “number of who gained entry and had two tickets” / “number who gained entry”.
…but I don’t agree with the 70% number either. Let me explain…
On average & rounded to the whole person, 20 people with one ticket and 36 people with two tickets will gain entry. There are 90 people trying to get in.
Pr(entry|2 balls) ≈ 36/90 ≈ 40%, Pr(entry|1 ball) ≈ 20/90 ≈ 22%
No, if having two balls equates to a 65% chance of getting in then you would expect to have
65% * 50 = 32.5 two-ball folks get in (on average)
Which leaves 56 - 32.5 = 23.5 slots for the one-ball folks, of which there are 40. 23.5 / 40 = 58.75%.
So if 65% is the correct chances for getting in when you have two balls, then 58.75% is the chances for getting in if you have only one.
On further reflection, I don’t think my description matches how things will actually happen. If the second ball for a particular person magically disappeared (or is otherwise removed) from the pot when the first was drawn then my description would be valid.
I then thought, figure how many reds and blues get picked from 100 blues and 40 reds, then take the number of blues and figure how many distinct blues would be in the total. But since the number of red + blue drawn depends on how many are distinct blues, that won’t work like i thought.
My current thinking is building a tree where drawing a " new" blue creates a third group, call it green, that enters the tree.
My intuition is not good enough to say whether the green tree is mathematically equivalent to ignoring the green branch entirely.
Since there are at least 16 blues drawn, floor has to be 16/50.
Still not sure why my first simulation got an answer different than .733, and sill struggling with why this logic is wrong, but it demonstrably is wrong.
Suppose we were just choosing 2 people instead of 56.
The probability both are from the 2-entry group is (100/140)(98/138) = .50724
The probability both are from the 1-entry group is (40/140)(39/139)= 0.08016
The probability one from each is 1-.50724-.08016=.41260
Or (100140)(40/138)+(40/140)(100/139)=.41259
Let Joe be a specific entrant with 2 tickets. Chances he was selected is (2*.50724+,41260)/50 = 0.02854
Let Fred be a specific entrant with 1 ticket. Chances he was selected is
(2*.08016+.41260)/40=.01432
So chances of Joe being picked is less than twice the chances Fred is picked. As the number picked increases, Joe’s chances and Fred’s chances each increase, but the ratio of Joe’s chances to Fred’s chances continues to decline.
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Yeah, my logic is clearly wrong. Probability is really good at messing with you.
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I messed up my quote. I was intending to acknowledge I did exactly the same as you. I’m about to fix that.
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Yeah, this much was obvious in the “what if 80 are picked” question posed by (I think) PatientZombie. And I illustrated it too in my “if 65% is right for two balls then it’s 58.75% for one ball” post.
So it’s some summation of all the different possible outcomes.
Seems like there should be a way to count the possible outcomes and then count the outcomes where KidRock wins and divide. But my A in Combinatorics is failing me now. My professor would be abjectly disappointed in my participation in this thread.
I agree my initial was off (in addition to the 89 instead of 90).
But for these types of calcs I still like looking at not getting in and counting people, not tickets (too adolescent to continue to say balls).
Using Excel and circular calculations. If we agree two tickets = probability of 1 ticket squared, to NOT get in.
And 34 people will not get in
Setting a probability of 1 ticket at 51.65% (circular logic) to NOT get in, gives 26.68% for 2 tickets
so with 50 people at 26.68% = 13.34 & 40 people at 51.65% = 20.66 — 13.34+20.66= 34 people who will not get in.
so the odds to get in is 1-.5165 = 48.35% for 1 ticket & 1- .2668 = 73.32% for two tickets
This is very close to ahj911211’s simulation
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Bumper sticker.
Directly:
56 (winning) slots; so the denominator would be {}_{56}C_{90}.
How many ways to select 50 two-ticket players for 56 slots? This will be your numerator.
Indirectly:
PatientZombie’s method of looking at the complement.
56 choose 90? lol
I feel like the right approach here might be to calculate the probability of the number of draws it takes to get the 56 winners (somewhere from 56 to 106), and then calculate the chance a single entry had their ball selected in each of those scenarios.
I suck at combinatorics.
So here’s my simplification
2 people, 1 has two tickets, 1 has 1 ticket. And we’re choosing 1 person to get in
Probably of 1 ticket getting in, 1/3
I don’t know how to generalize that. But I tried.
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I figured out where my simulation went wrong. :duh: It was starting with 56 people with 2 tickets and 40 people with 1 ticket. That increased the average number of 2-ticket winners, but decreased the chances for each specific 2-ticket entrant.
Now each of 10 sets of 1 million trials gives a probability of .734, to 3 decimals. My “exact” calculation still has .7334, to 4 decimals.