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an urn contains 25 balls 40% of which are green. a contestant reaches in the urn to choose three balls the contestant will win 200 if he or she selects a green ball but will lose 120 for any other colour. is this a fair game

A) the ball is replaced after each draw B) the ball is not replaced after each draw

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    Sorry I meant 40% – 2017-02-28

2 Answers 2

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Whether balls are replaced or not is irrelevant in terms of expected winnings, since the expectation of what you replace obviously coincides with the expectation of what you draw (note that replacement does increase the variance, however). So, whether you replace balls or not, the expected winning per ball is $200\cdot 0.4 - 120 \cdot 0.6=80-72=8$, yielding an expected win of $3\cdot8=24$ over three balls. So it's not a fair game.

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    oh.... :P upvote – 2017-02-28
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For part A:

Let's calculate the expected value (aka winnings). There are $25\cdot \frac{2}{5}=10$ green balls.

If the balls are replaced after each draw the expected value remains the same for each drawing (so we take it for one drawing and multiply by 3).

The expected value for one draw is $\frac{2}{5}\cdot 200-120\cdot \frac{3}{5}=80-72=\boxed{8}$, so the game is unfair. In fact, it's in your favor; you have an expected winning of $24$! (though obviously the value itself is impossible)

Part B:

We have to brute-force the cases we have here. The chance of all three flipping green is $\frac{10\cdot 9\cdot 8}{25\cdot 24\cdot 23}=\frac{6}{115}.$ The chance of two flipping green and one something else (let's assume the other 15 balls are red) is $3\cdot\frac{10\cdot 9\cdot 15}{25\cdot 24\cdot 23}=\frac{27}{92}.$ (Pay special attention to the 3; we can draw RGG, GRG, or GGR)

For 2 red and one green it's $3\cdot\frac{15\cdot 14\cdot 10}{25\cdot 24\cdot 23}=\frac{21}{46}.$ For all red it's $\frac{15\cdot 14\cdot 13}{25\cdot 24\cdot 23}=\frac{91}{460}.$

So the expected value here is $600\frac{6}{115}+280\frac{27}{92}-40\frac{21}{46}-360\frac{91}{460}.$ This gives $24.$

Yeah, there's definitely a better solution that I'm missing. But anyway this one is also rigged in your favor. Does anyone know why the second one also has an expected value of 24?

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    Where did you get 24 – 2017-02-28
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    Each round (draw) gives you an expected value of +8, so it's just 8*3=24. – 2017-02-28
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    For part b where did you get all the coefficients like 600 – 2017-02-28
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    if three green balls are drawn you win 200+200+200=600. If two green/1 red, then 200+200-120=280. And so on. – 2017-02-28