# Probability of Winning a Prize from a raffle with Multiple Winners with Multiple Entries

So let's say there's a raffle to win 500 prizes for a raffle and you can have multiple entries. There are currently 200,000 entries and I have 200 of those entries. Each person can only win once as in if I win with any of those 200 entries, I won't be able to win with the rest of the 199 entries.

• Scenario A: What are the chances of me winning using those 200 entries?

• Scenario B: What are the chances of me winning in one of my 4 accounts with 50 entries each?

Which would be the best way to enter for the best chance of winning?

• How many entrants are there? If each person can only win one prize, no matter how many times they enter, it makes a great difference if there are 499 other entrants, or 199,800 other entrants!
– Lynn
Commented Jul 10, 2022 at 7:43
• That's a good point! However, it's undisclosed how many people entered. Let's just there's an average of 25 entries per person, so 8,000 entrants. Commented Jul 10, 2022 at 9:43
• Even that information is inadequate, because the answer changes tremendously if 7,999 entrants have one ticket apiece and the remaining entrant holds the remaining 200000-500-7999 tickets. // Please explain what it means to "enter." Are you suggesting you have the option of establishing multiple accounts? If so, isn't it obvious that your expectation is maximized by employing 500 accounts? After all, the chance of winning depends only on how many tickets you control, but the expectation is lowered every time a non-winning ticket is excluded because some other ticket in an account wins.
– whuber
Commented Jan 12 at 13:09

### Scenario B versus Scenario A

Scenario B is definitely better. It gives the same number of entries, and thus the same probability to win at least once. But in addition you also have the probability to win two, three or four times.

Potential disadvantage with scenario B is that it might come with some costs. Are you allowed to use 4 accounts? If not you might get disqualified. Are you bypassing this by making a deal with friends? But then you might need to give up a share of the win.

### Exact probabilities

To compute the exact probabilities is difficult. For this you need to know more about the distribution of entries and number of competitors.

• In the worst case you have 199800 competitors with one ticket each and you probability is related to a hyper geometric distribution $$1-\frac{199800\cdot199799\cdot199798\cdot\dots\cdot199301}{200000\cdot199999\cdot199998\cdot\dots\cdot1999501}\approx1-\left(\frac{199800}{200000}\right)^{500}\approx 0.3936211$$

• In the best case you have 499 competitors who have together the remaining 199800 tickets (a bit more than 400 tickets each) and your probability to win is equal to $$1$$.

If you have some estimate for the distribution of the tickets then you could make a better estimate.

• For instance, if everyone has 200 entrees, then there are 1000 competitors and your probability to win is $$0.5$$.

• It might be that not everyone has exactly 200 entrees. Then, you might consider this as a statistical physics problem. But nothing comes to my mind directly (I am currently thinking about turning all contestants into molecules where only 500 can be 'on' and the rest is 'off' and compute some sort of Boltzmann distribution or entropy).

Alternatively you simulate this with a computer which can estimate your probabilities very quickly.

### Estimate with exponential law for complex case

Edit: I went a bit further on the last case and did a simulation for a case with some randomly generated entries distribution. It appears like the probability to get no prize is an exponential function of the number of entries.

$$P(\text{no prize}) = \exp(-\alpha k_\text{entries})$$

Where $$k_\text{entries}$$ is the number of entries for the specific person for which we want to compute the probability.

The parameter $$\alpha$$ can be found by solving the following equation

$$\sum_{i}^n P_i(\alpha) = \text{total entrees} -500$$

Where $$P_i$$ is the probability for the $$i$$-th person.

This equation stems from the fact that the expectation value for all entrants must be such that it equals the total number of places that are not gonna win.

A heuristic motivation for the formula above is that we can regard the problem as a randomly drawing entries (also from people that have already won) untill 500 people have won. Say that the number of draws is $$M$$ (and this number is typically distributed around some number above 500) then the number of entries drawn is approximately Poisson distributed with the probability for zero entries being $$\exp(-\lambda) = \exp(-M \cdot k/K)$$, where $$k/K$$ is the fraction of entries that the person has.

set.seed(1)

# generate 200000 entrees among people
# with a random distribution for entrees
# first we generate a surplus
# which later we cut down to 200000
#
# np is the number of people
# k is the number of lottery tickets (we order this list)
# is is a vector of size np giving id's for the people
k = rpois(10^4,10)*20
np = which.min(abs(cumsum(k)-200000))
k = k[1:np]
k = k[order(k)]
id = 1:np
p = k/sum(k)
counts = rep(0,np)

np

### simulate some sampling
### repeat 10000 times
### the vector counts contains how often person wins
n = 10^4
for (i in 1:n) {
smp = sample(id,500, prob = p, replace = FALSE)
counts[smp] = counts[smp]+1
}

### plot simulation
### for each person we plot as a function of k,
### the observed probability of loosing
### this is 1-counts/n
plot(k[id],1-counts/n, log = "", xlab = "number of entrees", ylab = "probability no prize")

### compute parameter for exponential model
f = function(r) {
(sum(exp(-r*k))-(np-500))^2
}
mod = optim(0.01,f)

x = 1:max(k)
lines(x,exp(-mod\$par*x))