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MY ATTEMPT AT SOLUTION: $P(win)=P(draw.at.least.3. red)=P(draw 3 R)+P(draw 4 R) + P(draw 5 R)$

Using the law of total probability:

$=P(bin1)*P(draw.3|bin1)+P(bin2)*P(draw.3|bin2)+P(bin3)*P(draw.3|bin3)+P(bin4)*P(draw.3|bin4)+P(bin5)*P(draw.3|bin5)+P(bin1)*P(draw.4|bin1)+P(bin2)*P(draw.4|bin2)+P(bin3)*P(draw.4|bin3)+P(bin4)*P(draw.4|bin4)+P(bin5)*P(draw.4|bin5)+P(bin1)*P(draw.5|bin1)+P(bin2)*P(draw.5|bin2)+P(bin3)*P(draw.5|bin3)+P(bin4)*P(draw.5|bin4)+P(bin5)*P(draw.5|bin5)$

Factoring this all out I simplify to get:

$P(bin1)*(P(draw.3|bin1)+P(draw.4|bin1)+P(draw.5|bin1)) +P(bin2)*(P(draw.3|bin2)+P(draw.4|bin2)+P(draw.5|bin2)) +P(bin3)*(P(draw.3|bin3)+P(draw.4|bin3)+P(draw.5|bin3) +P(bin4)*(P(draw.3|bin4)+P(draw.4|bin4)+P(draw.5|bin4)) +P(bin5)*(P(draw.3|bin5)+P(draw.4|bin5)+P(draw.5|bin5))$

Since we are given the ratios of the sizes of the bins I tried to find the probability that way. So bin 5 is the smallest.

$P(bin1)=5*P(bin5)$

$P(bin2)=4*P(bin5)$

$P(bin3)=3*P(bin5)$

$P(bin4)=2*P(bin5)$

$P(bin5)=1*P(bin5)$

since the probabilities of all the bins need to add to 1 we have

$1=15*P(bin5) \implies P(bin5)=\frac{1}{15}$

Then I can calculate the remainder probabilities of bins.

$P(bin1)=1/3$

$P(bin2)=4/15$

$P(bin3)=1/5$

$P(bin4)=2/15$

$P(bin5)=1/15$

Then calculating $P(draw.3|bin1)=\frac{{12\choose 3}{13\choose 2}}{{25 \choose 5}}$

So I did these calculations for the remainder probabilities and got something around 0.58.

So the probability of winning 10 is 0.58 and the probability of losing $p$ is 0.42.

So then I set $E[X]=0$ and solved for how much you would lose if playing the game.

$10(0.58)+p(0.42)=0$

$p=-13.8$

This makes no sense. I'm losing more than I would win by playing the game?

I have a feeling I did this question all wrong. Can someone please help me with the solution?

user130306
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  • Please use relevant titles. References to the difficulty of the question are subjective! – FShrike Oct 10 '22 at 14:49
  • okay im sorry, will update – user130306 Oct 10 '22 at 14:49
  • The phrasing implies to me that the net gain on a win would be $10-p$, not $10$. The equation you have implies a winner gets the price for playing back, plus 10 dollars on top of that. So your result says that with a win probability better than $1/2$, the player can afford to bet a bit more than $10$, usually getting it back plus $10$, and less often losing the bet. – aschepler Oct 10 '22 at 15:12

1 Answers1

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Suppose a bin has $r$ red and $w$ white balls. The probability of pulling at least $3$ reds in a draw of $5$ from it is $$\frac{\binom r3\binom w2+\binom r4\binom w1+\binom r5\binom w0}{\binom{r+w}5}$$ Computing this for each bin gives (from most frequent bin to least frequent) $$\frac{739}{1610},\frac{627}{728},\frac{49}{143},\frac12,1$$ We now multiply the first fraction by $\frac5{5+4+3+2+1}$, this being the probability of choosing the corresponding bin; the second fraction is multiplied by $\frac4{15}$ and so on, and the results are added to give the final probability of winning: $$\frac5{15}\frac{739}{1610}+\frac4{15}\frac{627}{728}+\frac3{15}\frac{49}{143}+\frac2{15}\frac12+\frac1{15}\cdot1=\frac{67289}{115115}$$ The expected number of dollars won is this fraction multiplied by $10$, i.e. $$\frac{134578}{23023}=5.8453720\dots$$ which is also the maximum amount you are willing to pay for the game.

Parcly Taxel
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  • wow this makes a lot more sense. thank you so much. so if I understand correctly, we win 10 with a probability of $67289/115115$ and our expected value is $5.84$. so when i was solving for $p$ i.e. the amount we lose, that was totally irrelevant right? – user130306 Oct 10 '22 at 15:26
  • @user130306 Yes. No subtraction at all is needed in this problem. – Parcly Taxel Oct 10 '22 at 15:33
  • actually looking at it, im a bit confused on how you got expected value. since with one X and p(x) we don't have a probability distribution unless p(x)=1 – user130306 Oct 10 '22 at 17:55
  • @user130306 It's OK that the probabilities don't add to one. It is just that the value associated with the omitted probabilities is zero. – Parcly Taxel Oct 10 '22 at 18:04