Lottery and win with less numbers - Hypergeometric Distribution I got a problem in showing this proof in paper.
One of my coworkers insists that using a sequential number in a Brazilian lottery increases the chances of winning the prizes with less odds I pretty certain it's not true but I can't come up with an adequate proof.
Here we have a special lottery that works as the following. One must choose 15 numbers from 25. With this ticket, you'll be able to win the biggest prize (hitting the 15 numbers from 25) but you can also win the smaller prize with 14. This continues until 11, where you get the smallest.
His theory is that betting with a sequential ticket from 1-15, 2-16 and 3-17 he'll have more chances to win instead of just picking 3 random sequences. The problem gets worse when others coworkers go gambling together. They insist that combining sequences like that in different tickets also increases their chances.
I'm thinking about using Hypergeometric distribution with conditional probability but this is not working since they don't have this mathematical background.
Thanks!
 A: I don't think you need to bring up the hypergeometric distribution.
You can  try illustrating with a smaller example. Assume 3 winning numbers are chosen out of 5. The possible winning combinations are:
$$\{123, 124, 125, 134, 135, 145, 234, 235, 245, 345\}$$
If you choose $123$ then you would get exactly 2 numbers right when the winning numbers are one of $\{124, 125, 134, 135, 234, 235\}$, which is 6/10's of the time.
If you choose $124$ then you would get exactly 2 numbers right when the winning numbers are one of $\{123, 125, 134, 145, 234, 245\}$ which is 6/10th's of the time.
There's no difference between consecutive and non-consecutive when the numbers are randomly drawn. Each combination of numbers is equally likely to show up.

Not sure if this more general explanation is suitable for your colleagues:
Say you have chosen your numbers and want to know the probability of a specific set of 11 of them showing up in the winning numbers. 
Out of the original 25 numbers, there are 14 left and 4 of them are 'invalid' since we don't want anymore of our numbers to appear.  This leaves ${10} \choose {4}$ ways of combining 15 numbers. Notice the number of ways doesn't depend on the numbers being consecutive or not. 
If you want the number of combinations for any set of 11 out of the 15 numbers you picked, it's ${15\choose 11}{10 \choose 4}$ since there are ${15\choose 11}$ unique sets of 11 numbers. 
So no matter which 15 numbers you pick, the probability of 11 of them showing up in the winning numbers is $\frac{{15 \choose 11}{10 \choose 4}}{{25 \choose 15}}$.
You can do similar reasoning for 12, 13 or 14 numbers. This is just the hypergeometric distribution but maybe deriving it in this conversational way would make it less intimidating.
