| A | B | C | D | |
| 1 | 1000000 | [A,B,C] |
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| 2 | 0 | 0 |
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| 3 | for A1 | =int(rand(3))+1 | =int(rand(3))+1 |
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| 4 | =B1(B3) | =B1(C3) |
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| 5 | =B1\C4 | =B5\B4 |
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| 6 | =int(rand(C5.len()))+1 | =C5(B6) |
|
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| 7 | =(B5\C6)(1) |
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||
| 8 | if B4==C4 | >A2+=1 |
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| 9 | else if B4==B7 | >B2+=1 |
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| 10 | =A2/A1 | =B2/A1 |
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A10中不更改答案获奖的概率如下:
B10中更改答案后获奖的概率如下，可以看出，在这种游戏中，更改答案后的获奖可能性差不多提高1倍:
