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At my work we often run small sweepstakes at big sporting events, e.g. the football world cup, the Olympics, the rugby world cup.

As you may know, the rugby world cup is currently on. There are 20 teams involved, and 11 of us in the office. Using a betting website's odds, we whittled down the 20 teams to the 11 most likely to win outright, and drew them out of a hat.

Of course, each person to draw a team has one fewer potential team than the person who drew previously. The first person to draw has 11 possibilites. The second person has 10. The third has nine, etc.

Of these 11 teams, only three are realistically likely to win. Is it possible to calculate the probability of drawing one of these three at each stage of the draw? If so, is it better to draw earlier or later?

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  • $\begingroup$ It doesn't matter in which order you draw. You could also distribute the tickets face down to everyone randomly and then everyone looks at their ticket at the same time. It has the same effect. $\endgroup$ – Gumeo Sep 24 '15 at 9:09
  • $\begingroup$ It makes no difference. [Consider shuffling 11 cards (A,2-10,J) and then drawing from the deck. Which position in the deck is most likely to have the A? ] $\endgroup$ – Glen_b Sep 25 '15 at 2:50
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Here is a small simulation in R, that might convince you that the order does not affect the probability of drawing a winning team.

# Small simulation for order of drawing from a hat
# Probably an overkill to do it this way...

# Number of iterations
N <- 100000 

# Vector to count number of successful draws
counter <- rep(0,11)

# iterate number of times to estimate probabilities
for(i in 1:N)
{
  # Iterate over each of potential 11 draws
  for(j in 1:11)
  {
    # Create a vector to draw from, i.e. the hat
    hat <- rep(0,11)

    # It has 11 zeroes, now we will sample 3 random
    # places to put the 3 best tickets
    placement <- sample(1:11,3)

    # Put the winning tickets in their place
    hat[placement] <- rep(1,3)

    # Draw from place j
    counter[j] <- counter[j]+hat[j]
  }
}

# Probabilities for each placement of draw
probs <- counter/N
names(probs) <- as.character(1:11)

# print the probabilities
print(probs)

# make a barplot to visualize it
barplot(probs)
title(paste('Probabilities for each placement of draw, 
    iterations: ',as.character(N),sep="",collapse = NULL))

Now we can take a look at the plots for N=10000, N=100000 and N=1000000:

Here are 10K iterations, as we add more iterations the probabilities converge to a constant.

10K iterations

100K iterations

1000K iterations

You can see that the probabilities as converging to the same value. So it does not matter where in the order you draw.

This is an example of a problem that is very easy to simulate if you do not know how to work out the probabilities and get a quick and dirty answer.

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It doesn't matter in which stage you draw : if you are not convinced let's look at what happens for the first and second person that draw : the first who draw as the probability to draw one of these teams equal to 3/11.

For the second : If the first drew a team likely to win, it remains two over ten of them. In this case, the probability to draw it is $\frac{3}{11} \times \frac{2}{10}$. If he didn't, you have 3/10 chance to draw it. So the chance to draw a team likely to win is $\frac{8}{11} \times \frac{2}{10}$. By additionning those two probabilties, you find it equal to $\frac{3}{11}$ also.

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