Odd or probability question in gaming theory.

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I have two games with different probabilities of winning. Let's say Game A has 80% of return to player percentage while Game B has 100% (which means a player will win no matter what). And there is a setting that I can adjust percentage to select game A or game B. Bpth Game A and Game B have 1/10 million odds to win Grand Jackpot.

  • Scenario 1: When I set up the game, I set up 50%/50% so I have 50% chance of selecting Game A and another 50% chance of selecting Game B. As it was mentioned above, the game A has 80% chance to win while Game B has 100% probability to win. So overall, I get 90% chance of winning (0.5 * 0.8 + 0.5 * 1 = 0.90).

  • Scenario 2: I set up my game as: 40% chance of selecting Game A and 60% chance of selecting Game B. I get total 92% chance of winning from both game packages now. (0.4 * 0.8 + 0.6 * 1 = 0.92).

So in this case, is my overall probability or odd of winning Grand Jackpot going to change between scenario 1 and scenario 2?? Does it matter when the odd for Jackpot is calculated separately for Game A and Game B? Hopefully this is clear question.

Any comments will be appreciated. Thanks

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Yes, the overall probability of winning the Grand Jackpot is the same if you assuming that both games have a $1/10000000$ of winning the Grand Jackpot. You can compute it directly. Say $\alpha \in [0,1]$ is the probability of playing Game A.

Then the probability of winning the Grand Jackpot is $$ \alpha \times 1/10000000 + (1-\alpha) \times 1/10000000= 1/10000000.$$

In other words, it is still $1/10000000$ regardless of what $\alpha$ is.

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Cutting to the nitty gritty, suppose you have bought one ticket in $10$ million issued tickets in Scenario $1$

There are two possible questions, what you have in mind isn't clear.

  • If you just want the absolute probability of winning the Grand Jackpot, it will clearly be the same for options $A$ and $B$ at $1$ in $10$ million

  • If you want the probability of winning the Grand Jackpot given that you have won a prize, it will be different for game $A$ and game $B$ at $1$ in $5$ million, and $1$ in $10$ million respectively.