Three rounds, three bets: how to guarantee a loss of all three rounds Suppose you are buying tickets for three round of some game. Your ticket must have three bets on it before the first round starts. Your three options for each round are for a win, a loss and a draw. How many tickets must be bought to guarantee that one ticket will lose all $3$ rounds. 
I'm not sure about this. I know that the minimum number must greater than $5$ since there can be two tickets with the same first bet and one ticket the third bet which leads to at least $3$ tickets losing the first round. From there, the bets from these three could lie in at least possibilities. That means that two of these tickets could lose. One ticket remains as a loser for the first two rounds but it could win round $3$. I think a similar argument works to refute $6$ though I'm not sure. I am quite confident that $9$ works but I don't know about $7$ or $8$. 
Any help is greatly appreciated!
 A: Consider them separately. You need at least two bets in each round to guarantee at least one bet loses in each round. A bet which wins cannot meet the requirements, and therefore its ticket is terminated.
But you don't know which will be terminated in each round, so you effectively need a way of getting from any bet in each round to any bet in every other round.
So, how many ways of going from a choice of two to a choice of two to a choice of two are there? Just $2^3=8$.
To help you see the equivalence of this solution to its meaning in context, draw three sets of two bets, one for each round. Then draw one line for each possible path from round 1 to round 3. Those lines represent a single ticket. Now if you cover up one bet (to represent it winning) and erase any tickets through it (representing termination) you see you still have at least one ticket which loses on all bets. If you cover up one bet in every round, you have exactly one ticket which loses all bets, and therefore is not terminated, no matter which bets were covered up. 
This also demonstrates minimality. If you don't have even one of those lines to start with, there is a way to cover up bets so that all tickets terminate - by picking exactly the bets that would not terminate the bet you didn't place. Therefore, you need at least all of those tickets, and there are 8 of them, so you need at least 8 tickets altogether.
