# Ticket selling probability puzzle

I was talking with a friend of mine about a simple statistical puzzle that we weren't able to solve due to our ignorance about the statistics world.

Consider a ticket selling web. A musical event is coming up, and there are only 400 tickets available for a total of 800 solicitants. Since the web is not optimized at all for heavy internet traffic, there is a high chance of if getting freezed and unusable. In order to lighten this internet traffic, there is the possibility of getting multiple tickets with a single purchase.

There is a group of 10 friends wanting to attend this event. When purchasing all the tickets at once, I understand there would be a probability of 50% of getting them in the first attempt (being 10 tickets it should be even less, but for the sake of simplicity let's keep it at 50%).

One of the friends, in an inspirational moment, talks about splitting the group in two groups of 5 members, so that the probability of getting all the tickets would increase substantially.

Is this idea correct?

My very basic understanding of statistics makes me think that even if the probability of getting tickets for at least one group would increase substantially, there would be a sligthly lower probability of getting all the 10 tickets.

If tickets are booked by splitting in two separate groups (Say Group A and Group B) of 5, there are 4 possible cases

1. Group A gets 5 tickets, Group B books 5 tickets
2. Group A gets 5 tickets, Group B books 0 tickets
3. Group A gets 0 tickets, Group B books 5 tickets
4. Group A gets 0 tickets, Group B books 0 tickets

Assuming that both ticket booking events are independent with each having probability of 0.5, following are the probabilities for each case

Conclusion :

• Probability that you will be able to book 5 or more tickets is increased to 0.75 i.e. P(Case 1) + P(Case 2) + P(Case 3) , so if you are OK with 5 or more tickets, splitting in two groups is better strategy.
• If you want 10 tickets or nothing , then single group is better strategy as splitting in 2 groups will reduce the probability of getting 10 tickets to 0.25 (Case 1) as compared to 0.5 in single group.