For entry into an event, I have 2 entries since I didn’t get selected last year. There are 49 others that were also not selected and have 2 entries. There are 40 others that were selected last year that have 1 entry for this year. There are 56 available spots for the 90 people wanting to enter. If someone with with 2 balls is selected, they cannot be selected again. What are my chances of getting picked in a lottery for an entry?
Not convinced, and I will try a simulation, but initial thought is
spoilertext
Each person with 2 entries is twice as likely to be selected as a person with 1 entry.
Let p = probability that a person with 1 is selected.
Then 50(2p)+40p=56. 56 since 56 are selected.
p=56/140=.4
Or does the lack of independence change it? Tentatively I don’t think so.
So probability that a given person with 2 entries is selected is .8.
Hmmm. Will take more review. Was my calculation wrong or is the simulation wrong? Simulation is consistently getting between .6865 and .687
I think there are commonly developed formulas for picking 56 out of 50 blue + 40 red = 90 total without replacement when any one ball is equally likely to be drawn. Now adjust for blue being twice as likely to be drawn as red. Your chance is avg number of blues drawn/ total number of blues.
Eta: clarifying chance of a single ball being drawn.
56 out of 100 balls get in. so 56%.
he has two balls, so obviously 112% chance
5 Likes
89 people 56 slots
63% get in, 37% will not get in
you have two chances. 13.7% chance of not getting in. .37^2
1 - 13.7% = 86.3%
i like this, since SW .8 is too low, as it ignores ball removal on others with two
can’t comment on the similation
1 Like
The average probability is 56/90 = .6222
50 people have two entries, with probability of winning 2P.
40 people have 1 entries, probability of winning P.
5/9 (2P) + 4/9 P = .6222
14/9P = .6222
P = .4
2P = .8
2 entry people get in 80% of the time, 1 entry people get in 40% of the time.
Double checking EV — 50 * .8 + 40 *.4 = 56.
1 Like
what if 80 get in?
80 / 90 = .89
.89 x 9/14 = .57
the 2p would be 1.14
1 Like
I think I have figured out how to calculate it recursively in Excel. Should be relatively easy. Rows for results after each selection round, 0 selected thru 56 selected. Columns for the number of people starting with 2 tickets who have been selected. Each row should be a function strictly of the row above it, and in particular should depend only on two columns of the row above it (If there a N such selections this round, there must have been N or N-1 in the previous round).
Then the row after 56 selections will give the probability that various numbers of 2 ticket-holders have been selected. Probability that Joe was is 1/50*(probability 1 was) + 2/50*(probability 2 were)+…
Won’t be able to do it until this afternoon or evening.
90 people. 89 plus Kid_Rock.
I think your calculation is correct other than using 89 instead of 90.
I ran 21 “simulations” in Excel using RAND(). It was (un)surprisingly inefficient.
In total, those with two entries got just over 64% of the spots.
The least occurrence was just under 59% and the highest was just over 71%.
edit: now 36 simulations, just under 65%, same mins & maxs.
Wait, no, on second thought that means that the people who only have one chance have a 37.778% chance of not getting picked which is exactly their chance of not getting picked if everyone only has one ball. The people with one ball have to be less likely to get in than if everyone had the same chance, given the addition of the extra balls.
If this doesn’t show up on the CAS probability exam, I’m going to be really disappointed.
I tried setting up a simulation for this in R, but I am getting a different answer (~73.4%) than the other two who previously posted simulation results. Maybe someone can proofread the code to see if we can get a consensus answer.
R code
set.seed(2942)
Iterations ← 5000
OneEntry ← 40
TwoEntry ← 50
Draws ← 56
AvgTESel ← 0
for (loopcount2 in 1:Iterations){
OE_Rem ← OneEntry
TE_Rem ← TwoEntry
OE_Sel ← 0
TE_Sel ← 0
for (loopcount1 in 1:Draws) {
currrand ← runif(1)
if ((currrand > OE_Rem / (OE_Rem + 2*TE_Rem) && TE_Rem > 0) || OE_Rem == 0) {
TE_Sel ← TE_Sel + 1
TE_Rem ← TE_Rem - 1
}
else {
OE_Sel ← OE_Sel + 1
OE_Rem ← OE_Rem - 1
}
if (loopcount1 == Draws) {
AvgTESel ← AvgTESel * (loopcount2-1) / loopcount2 + TE_Sel * (1 / loopcount2)
}
}
}
paste0(“Probability Two Entry Person Selected:”)
AvgTESel / TwoEntry
The problem I was having is there are not necessarily 56 draws. If someone with 2 balls gets picked 2 times, then there are 57 draws. The minimum number of draws is 56. The maximum is all 56 draws getting pulled with people with 2 balls and then 6 more draws occurring. There are a number of combinations between 56-62 draws. In the simulation, you have to essentially have each pick see if it was picked and keep going. Frustrating. Draw is Friday.
The scenario is for our Member-Guest tourney. People that were in last year get 1 ball. People that missed last year get 2. If you miss 2 years in a down then you get in. There were 4 of those to go along w 4 auto entries.
Appreciate all the responses.
I made these assumptions in the simulation I previously posted:
-Balls are drawn, without replacement, until all 56 slots are filled (i.e. 56 distinct people selected regardless of how many draws it takes). Perhaps I should have renamed the “Draws” variable “Slots” for clarity.
-Any 2nd ball for a “two entry” person already drawn is a “dead ball”; just tossed aside if drawn and isn’t used in a probability calc at that juncture for the remaining hopefuls.
Maybe I understand. Maybe I don’t. I thought there would always be 56 players selected, and draw until you get 56. So worst case (max draws) the first 100 tickets drawn are all from the people with 2 tickets. That’s 50 people in, and you still need to draw 6 more. Is that correct?
Meanwhile I remain more confused than ever about the answer. My new “direct calculation in Excel” spreadsheet currently thinks .73342. I’ll try to figure out what’s wrong.
Or maybe the .73342 is correct, and my first simulation is wrong plus my general reasoning solution is wrong.
I’m getting in
I’ve now run 100 simulations.
P(gaining entry|2 entries) ≈ 64.89%
I’m not sure what the standard for full credibility is here, so I ran 212 simulations.
P(gaining entry|2 entries) = 7742/(7742+4410) ≈ 65.21%
My guess as to the official probability is, therefore, 65%.