Super Mario World 2: Yoshi's Island on the SNES has an interesting roulette bonus game.
The rules are simple: the player bets between one and one less than his or her total number of lives and presses a button to set the two wheels spinning. He or she then presses the button again and the wheels slowly spin down to plus or multiply on the left and zero, one, two, or three on the right. The player wins the number of lives in the expression formed from the bet, the operator on the left wheel, and the value on the right wheel.
I was curious how many lives the user could expect to win in this game, and because I'm taking a statistics course this semester, I had the power to satisfy this curiousity. What follows is an explanation of the calculation I performed and its surprising results:
Givens:
There are 6 multiplies and 8 additions on the left wheel. On the right, there are 7 zeroes, 4 ones, 2 twos, and 1 three. I'm assuming that all values on either wheel are equally likely. Let W be the amount won.
Step One: List All Possible Outcomes
Because there are two possible values on the right wheel and four on the left, it is obvious that there are 2 * 4 = 8 possible outcomes:
- W1 = bet + 0
- W2 = bet + 1
- W3 = bet + 2
- W4 = bet + 3
- W5 = bet * 0 [This result is the worst; the player loses everything]
- W6 = bet * 1
- W7 = bet * 2
- W8 = bet * 3 [This result is the best, and as we'll see in step two, the most rare]
Step Two: Calculate the Probabilities
One can easily see that:
- P(plus) = 6/(6+8) = 6/14
- P(multiply) = 8/14
and:
- P(zero) = 7/14
- P(one) = 4/14
- P(two) = 2/14
- P(three) = 1/14
From these basic probabilities, one can calculate the probabilities of all possible outcomes:
- P(W1) =
P(plus and zero) =
P(plus) * P(zero) = [because the wheels are independent]
8/14 * 7/14 =
2/7 - P(W2) = 8/49
- P(W3) = 4/49
- P(W4) = 2/49
- P(W5) = 3/14
- P(W6) = 6/49
- P(W7) = 3/49
- P(W8) = 3/98
Step Three: Calculate the Expected Value
The expected value is defined as the sum of each value times the probability of getting that value. Basically, we take each outcome from part one and multiply it by its corresponding probability from part two to find the weighted average:
E(W) =
sum(Wi * P(Wi)) =
W1 * P(W1) + W2 * P(W2) + W3 * P(W3) + W4 * P(W4) + W5 * P(W5) + W6 * P(W6) + W7 * P(W7) + W8 * P(W8) =
(2/7)(bet + 0) + (8/49)(bet + 1) + (4/49)(bet + 2) + (2/49)(bet + 3) + (3/14)(bet * 0) + (6/49)(bet * 1) + (3/49)(bet * 2) + (3/98)(bet * 3) =
(89/98) * bet + 22/49
Conclusion
In the long run, the player can expect to win an average of (89/98) * bet + 22/49 lives on each spin.
It surprised me to find that this game doesn't always favor the player. On large bets, the player can expect to lose lives in the long run. For example, if I continually bet 10 lives, I could only expect to win an average of 9.5 lives on each spin. Conversely, if I were to bet only two lives, I could expect to win an average of 2.25 lives. I found it even more surprising that the player can expect the most return, about an additional .35 lives, by betting the minimum one life on each spin. That's a good deal.
Why winnings decreases as the bet increases becomes obvious when we try to calculate the gain rather than the winnings. We'll define the gain G to be the amount above the bet that the player wins. So, we can calculate the expected gain as follows:
E(G) =
E(W) - bet =
(89/98) * bet + 22/49 - bet =
-(9/98) * bet + 22/49
The expected gain is downward-sloping, so as the bet increases, gain decreases.
By setting the expected gain equal to zero, we can determine the point above which the player can expect to lose lives.
E(G) = 0
-(9/98) * bet + 22/49 = 0
bet = -(22/49) * -(98/9)
bet ~ 4.9
This tells us that the player can bet up to four lives and expect to gain lives in the long run. Above that, at a bet of five or more, the player can expect to lose lives.
It makes me wonder if Nintendo planned it that way to help the struggling players but hold back the greedy ones.