The difficulty is that this box needs to have an infinite number of possible outcomes. Nevertheless, we can approximate its behavior just by following the usual procedure.
It will be most convenient to let N be a power of 2, because all the probabilities are powers of 1/2. (Of course you can choose any value of N you like, but then the arithmetic gets a lot messier. Using a power of 2 is another example of the Principle of Mathematical Laziness.) Let's try some small values of N to see what happens.
For the outcome 0, the probability is 1/2. [1/2 * N] = [1/2 * 4] = [2] = 2, so write 0 on two tickets.
For the outcome 1, the probability is 1/4. [1/4 * N] = 1, so write 1 on one ticket.
For the outcome 2, the probability is 1/8. [1/8 * N] = [1/2] = 1, so write 2 on one ticket.
For all the other outcomes, the probability is smaller than 1/8, so 1/8 * N is smaller than 1/8 * 4 = 1/2. It will round to 0. So we don't include any more tickets.
The box in this case contains two 0's, one 1, and one 2.
We proceed as before. The table summarizes the calculations. The box contains eight tickets now.
| Outcome (K) | Probability (P) | P*N | Number of tickets |
| 0 | 1/2 | 4 | 4 |
| 1 | 1/4 | 2 | 2 |
| 2 | 1/8 | 1 | 1 |
| 3 | 1/16 | 1/2 | 1 |
| >3 | <1/16 | <1/2 | 0 |
The table summarizes the calculations. The box contains sixteen tickets now.
| Outcome (K) | Probability (P) | P*N | Number of tickets |
| 0 | 1/2 | 8 | 8 |
| 1 | 1/4 | 4 | 4 |
| 2 | 1/8 | 2 | 2 |
| 3 | 1/16 | 1 | 1 |
| 4 | 1/32 | 1/2 | 1 |
| >4 | <1/32 | <1/2 | 0 |
The pattern should be evident. For N = 2M, half the tickets will have a 0, a quarter of the tickets will have a 1, ..., one ticket will have an M-1, one ticket will have an M, and no tickets will have values greater than M.
We can make a box that approximates the desired distribution very, very well provided we put loads of tickets in it. For example, if M is the largest number that a computer can represent, then using a box with N = 2M tickets should approximate anything the computer can calculate.
Of course this is not practicable, but then nobody actually uses boxes with tickets--this is a conceptual model to help us understand probability and use it correctly.
This box models the number of times one would have to flip a coin before seeing a head: half the time one flip will do, a quarter of the time two flips are required, and so on.
It also models the "gambler's ruin." A popular gambling technique is to double a bet after losing in order to win all the money back on the next bet. The bet can be on the outcome of any repeatable process: a coin flip, the roll of a roulette wheel, whether the Eagles will beat the point spread in the next game. If the odds are even (50% chance of winning), then this box models how many bets have to be made before the gambler wins.
In case you are tempted, don't be. Consider one of those boxes described above with a huge number of tickets in it. Take each ticket and write on it the size of the largest bet you would have to make. For example, if the initial bet is $1, write $1 on all the tickets that have a 0 (representing an initial win). Write $2 on all the tickets that have a 1, $4 on all the tickets that have a 2, and so on. The amount on the largest ticket in the box for N = 2M will be $2M. If M=16, $2M is about sixty-four thousand dollars (a little more actually, but the phrase has a history!). There is a small chance that you would pull this ticket from the box. If you do, you had better have sixty-four thousand dollars! And if you do, here's a sobering thought: your net winnings after risking all that money are a measly $1--the bet you started with.
The conclusion of this analysis is that most of the time you follow the double-up strategy you will wind up winning back your initial bet, but sooner or later you will pull the sixty-four thousand dollar ticket (or worse) and wind up losing everything you have.