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Assume A and B have access to the set $\{1,\dots,9\}$ and $\{1,\dots,8\}$, respectively. They choose three numbers from each's set without replacement and form the largest 3-digit number accordingly. For example, 4,2,5 means 542. What's the probability for A getting a larger number than B?

My attempt: $P(A\ win) = P(A\ win|A\ with\ 9)P(A\ with\ 9) + P(A\ win|A\ without\ 9)P(A\ without\ 9)=1\times 3/9 + 1/2\times6/9 = 2/3$.

I am not sure if the above works. Specially, it seems $P(A\ win|A\ without\ 9)$ is not $1/2$ given the possibility of tie.

Any hint and suggestion?

Thanks a lot!

WWSS
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  • In the "A without $9$" case, if A chose the 3-digit number $a$, what is the probability that B have chosen the same number? How would that probability change for small or large $a$? – peterwhy Aug 14 '22 at 03:29

1 Answers1

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The case $\{$A without $9$ $\}$ is the same as changing $A$'s drawing set to $\{1,2\dots8\}$. Since $B$'s drawing set is the same, we can say that $$P(A\text{ wins }|A \text{ without } 9)=P(B \text{ wins}|A \text{ without } 9)$$

Since the only possible outcomes are $A$ winning, $B$ winning or a tie $$P(A\text{ wins }|A \text{ without } 9)+P(B \text{ wins}|A \text{ without } 9)+P(\text{Tie}|A \text{ without } 9)=1$$

$P(\text{Tie}|A \text{ without } 9)$ is the probability that $B$ chose the 3 particular digits $A$ chose from $\{1,2\dots8\}$. Since the probability of choosing 3 particular digits is$$\frac1{\text{Total Possible Choices}}=\frac 1{\binom83}=\frac1{56}$$

we therefore have $$ 2P(A\text{ wins }|A \text{ without } 9)+\frac1{56}=1\implies P(A\text{ wins }|A \text{ without } 9)=\frac{55}{112}$$

You can now use this in your mentioned formula to calculate the total probability of $A$ winning as $$P(A \text{ wins})=1\times\frac39+\frac{55}{112}\times\frac69=\boxed{\frac{37}{56}}$$

Cathedral
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