I buy 8 lucky dip (random number) panels, each panel has 6 numbers chosen at random from 49. I never seem to cover more than 32 different numbers on average, over the total 48 numbers chosen, which seems low? How can I calculate the probability of 32 different numbers or 33 diferent numbers? Hi Andrew, If you buy two tickets, the probability that the numbers on the second ticket are all different from the numbers on the first ticket is ```(# of ways to choose 6 numbers among the 43 that are not on the first ticket) ----------------------------------------------------------------------------- (# of ways to choose 6 numbers among 49) ``` With the combination formula, you find that this probability is ``` 43*42*41*40*39*38/6! -------------------- = 0.436. 49*48*47*46*45*44/6! ``` This means that the probability that two tickets DO have a number (or more) in common is more than one half: 1 - 0.436 = 0.564. Now, your 8 tickets make up 28 PAIRS of tickets (there are 8*7/2 ways to select two tickets among them). So you should expect to have 28*.564 = 16 pairs of tickets with numbers in common. If we assume that all of these pairs of tickets will have only one number in common, and no number appears on three tickest or more, then instead of getting 6*8 = 48 different numbers, you get 6*8 - 16 = 32 different numbers. Now there are two things wrong with my calculation: First of all, two tickets can have more than two numbers in common, so you could get more than 16 repeated numbers. On the other hand, a number could appear on at least three tickets; when this happens, you get more pairs of tickets with repeated numbers involving less actual repetition of numbers. Now, these two errors involve events with small probability, and furthermore they tend to cancel each other out (one leads to an overestimation and the other to an underestimation of the different numbers). This is why my rough estimate of 32 corresponds to what you see when you buy your tickets. Claude Go to Math Central