Seemingly similar but different probability games

Burger King is currently running its "family food" game in which each piece can be modeled as a scratch off game where exactly one of three slots is a winner and you may only scratch one slot as your guess. As I was standing in line the other day I realized that their advertisement of "make it a large to double your chances of winning" (where large drinks/fries have two pieces on them) was actually not exactly true. The real probability of having at least one winning with two tickets is $\frac{5}{9}$ rather than $\frac{2}{3}$ which would really be double the chance. This is because the probability of both losing is $\frac{2}{3}\times\frac{2}{3}$ so the probability of at least one winning is $1-\frac{2}{3}\times\frac{2}{3}=\frac{5}{9}$.

Now this was all very clear to me but while I sipped my strawberry banana smoothie, I wondered why this game of two three-slot tickets where each one has one winner and you can scratch one from each is different from the game of a single six-slot ticket where there are two winning slots and you get two scratches. The games must be different because the probabilities of winning are different. The six-slot game has $1-\frac{4}{6}\times\frac{3}{5}=\frac{3}{5}$ chance of getting at least one win. The two games seem the same to me intuitively. Can anyone explain how they are different?

EDIT: I had noticed that in the two tickets, by making one guess, you actually eliminate 2 other possibilities with it so maybe this is the core of it, but I am still trying to see it more intuitively.

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Interesting fact: The two cards with one winner each, and the one card with two winners do have something in common. If you win a dollar for each "winner" then the expectation (average) of the number of dollars you win is the same for each game, namely $\frac{2}{3}$. Another interesting question would be the average number of minutes one shortens one's life with each sip. – André Nicolas Aug 24 '12 at 23:26

Your probability of winning on the first scratch is the same in each instance (this is the intuitive sense in which the two games are the same), so let's look at what happens if you don't win on the first scratch and want to win on the second scratch. This is where the games have to start differing.

In the six-slot case, you know there are $2$ winners left out of a possible $5$ slots, so your probability of winning on the second scratch is $\frac{2}{5}$. In the pair of three-slot case, you still know that there are $2$ winners left out of a possible $5$ slots, except that the $5$ slots are divided into two groups:

• $2$ slots that you aren't allowed to scratch, and that contain $1$ winner,
• $3$ slots that you are allowed to scratch, and that contain $1$ winner.

In other words, you aren't allowed to take advantage of the fact that there are $2$ winners left because you can't scratch the slots corresponding to one of them, and those slots are more likely to contain a winner than the ones you are allowed to scratch. This is where the games differ.

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There are ${{6}\choose{2}}=15$ ways to choose two slots out of six. If there are two winners among the six slots, then $1$ of these choices has two winners, $8$ have one winner, and $6$ have no winners. So your chance of getting at least one winner is $$\frac{1+8}{1+8+6}=\frac{9}{15}=\frac{3}{5}=0.6.$$ In the game with two three-slot tickets, you are constrained in which pairs of slots you can choose: you must choose exactly one of $\{1,2,3\}$ and one of $\{4,5,6\}$, eliminating $6$ of your original $15$ choices. These eliminated choices are known to consist of $4$ with one winner and $2$ with no winner: the choices you aren't allowed to make have a $2/3$ chance of winning, which is greater than $3/5$! The choices that are left are obviously worse on average: your chance of winning now is reduced to $$\frac{1+8-4}{1+8+6-6}=\frac{5}{9}=0.555\dots.$$

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Here's an attempt at an intuitive explanation.

Imagine a variation on your game with six-slot tickets with two winning slots. In this variation you have a choice after one scratch, assuming that scratch failed:

choice A: you can take your second scratch on a new unused six-slot ticket

choice B: you can take your second scratch on the same six-slot ticket you used for your first scratch.

You would prefer choice B, because one of the bad choices has already been eliminated on that ticket, enhancing your chances of winning. Note that choice A has exactly the same probability of winning as the original three-slot game you described, while choice B is your six-slot game.

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