AIME 2026 II · 第 7 题

Solution 1

Denote $A$, $B$, $C$ for Alice, Bob, and Carol respectively. Consider the situation after three rolls.

Case 1. One of $A$ or $B$ received two coins and the other received one, which occurs with probability $\frac{6}{27}$ since there are $6$ ways to permute $AAB$ or $BBA$ and $27$ ways to choose three letters in total. Then there is a $\frac{1}{2}$ that the player who received one will receive another before $C$ does while ignoring the third player.

Case 2. One of $A$ or $B$ received three coins and the other received none, which occurs with probability $\frac{2}{27}$ due to there being only two cases, namely $AAA$ and $BBB$. Then, there is a $\frac{1}{2}\cdot\frac{1}{2}$ chance of the second player receiving both coins before $C$ does while ignoring the third player.

The total probability is $\frac{6}{27}\cdot\frac{1}{2}+\frac{2}{27}\cdot\frac{1}{4}=\frac{1}{9}+\frac{1}{54}=\frac{7}{54}\implies\boxed{754}$.

~ eevee9406

Solution 2

Let $P(a,b)$ be the probability that Alice and Bob recieve two coins before Carol receives any, given that Alice and Bob respectively start with $a$ and $b$ coins. Let $2$ be a placeholder for all integers greater than two. Also, by symmetry, we have $P(a,b)=P(b,a)$.

We find that $P(2,2)=1$ and $P(2,1) = \frac{1}{2}$ by symmetry. Then

$$P(2,0) = \frac{1}{3}P(2,0) + \frac{1}{3}P(2,1),$$ and we find $P(2,0)=\frac{1}{4}$. Next, we have \begin{align*} P(1,1) &= \frac{2}{3}P(2,1) = \frac{1}{3},\\ P(1,0) &= \frac{1}{3}P(2,0)+\frac{1}{3}P(1,1) = \frac{1}{12}+\frac{1}{9} = \frac{7}{36}. \end{align*} Finally $P(0,0) = \frac{2}{3}P(1,0)=\frac{7}{54}$ and the requested sum is $\boxed{754}$. ~TThB0501

Solution 3 (Less Convoluted Solution 4)

Consider the ordering of the first four draws. We must have $AABB$, or some permutation of such, giving us $\frac{4!}{2!2!} = \binom{4}{2} = 6$ total configurations. The string has a $(1/3)^4$ probability of occurring, multiplying by $6$ total strings gives us the index probability $6/81$ or $2/27$.

Now, from this index, we have four states. Given that Carlos rolls next, for more than five strings (four from the $AABB$ rolls and one from the $C$ roll),

- Either we draw $A$ and then $B$ (or vice versa)

- We draw $A$

- We draw $B$

From this, we can easily continue with series. To solve the first point, notice that there are $2!$ orderings to $AB$. Multiplying by the probability of this $AB$ string, which is $1/9$, to the index and every other plausible index with the inclusion of the roll Carlos pertains results in:

$$2! \cdot \sum_{k=0}^{\infty} \left(\frac{2}{27}\right)\left(\frac{1}{3}\right)\left(\frac{1}{9}\right)^k$$ The sum is an infinite geometric series with first term $a = 2/81$ and common ratio $1/9$, through the geometric series formula is a sum:

$$\frac{2/81}{1 - 1/9} = \frac{2/81}{8/9} = \frac{2}{81} \cdot \frac{9}{8} = \frac{2}{9} \cdot \frac{1}{8} = \frac{2}{72} = \frac{1}{36}$$ Multiplication by $2$ gives us $\frac{1}{36} \cdot 2 = \frac{1}{18}$.

Then, for the second and third points, notice the symmetry between the two. Thus, we solve for $A$, and multiply by $2$ to account for $B$. The chance of $A$ happening is $1/3$, and we apply this to the index, along with Carlos' probability, obtaining:

$$\sum_{k=0}^{\infty} \left(\frac{2}{27}\right)\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)^k$$ This is an infinite geometric series with first term $2/81$, and common ratio $1/3$, and solving via the formula obtains:

$$\frac{2/81}{1 - 1/3} = \frac{2/81}{2/3} = \frac{2}{81} \cdot \frac{3}{2} = \frac{3}{81} = \frac{1}{27}$$ Multiplication by $2$ gives us $2/27$.

The answer is $\frac{1}{18} + \frac{2}{27}$, which we can solve through LCD, obtaining a probability of $\frac{3}{54} + \frac{4}{54} = \frac{7}{54}$, which is in the form $m/n$, where $\gcd(m, n) = 1$. Thus the answer $100m + n$ is nothing but $700 + 54 = \boxed{754}$.

~Pinotation

Solution 4 (Infinite Geometric Sums)

The problem is asking for the probability that both Alice and Bob have 2 coins before Carol gets her first coin. Thus, we only care about what happens before and up to the point where both Alice and Bob have 2 coins.

So, once we see that both Alice and Bob have two coins, we stop checking.

We can check all series of coin flip outcomes from length 4 to infinity. Let's first assume Alice gets to 2 coins first, then Bob. Length 4: Must be some form of ABAB -> last B is fixed, there are 3 ways to arrange the ABA, this has a $\frac{1}{3}^4=\frac{1}{81}$ chance of happening -> chance that it takes 4 tries for both Alice and Bob to have at least 2 coins is $\frac{3}{81}.$

Length 5: Must be some form of ABAAB -> Last B is fixed, there are 4 ways to arrange the ABAA, this has a $\frac{1}{3}^5 = \frac{1}{243}$ chance of happening -> chance is $\frac{4}{243}.$

In general, for a sequence of length $n,$ the probability is $\frac{n-1}{3^n},$ so our final probability sum will look something like this:

$$P = \frac{3}{81}+\frac{4}{243}+\frac{5}{729}+\frac{6}{2187}+...$$ We can break the individual components of P after the first term down into 3 and several "1 bits":

$$\frac{4}{243} = \frac{3}{243}+\frac{1}{243}$$ $$\frac{5}{729} = \frac{3}{729}+\frac{1}{729}+\frac{1}{729}$$ $$\frac{6}{2187} = \frac{3}{2187} + \frac{1}{2187} + \frac{1}{2187} + \frac{1}{2187}$$ So, from this, we can take first the sum $S_1 = \frac{3}{81}+\frac{3}{243}+\frac{3}{729}+\frac{3}{2187}+... = \frac{3}{81}\cdot\frac{3}{2} = \frac{1}{18}.$

Then, we can do

$$S_2 =(\frac{1}{243}+\frac{1}{729}+\frac{1}{2187} + \frac{1}{6561}+...) + (\frac{1}{729}+\frac{1}{2187}+\frac{1}{6561}+...) + (\frac{1}{2187} + \frac{1}{6561}+...)+...$$ $$S_2 = \frac{1}{243}\cdot\frac{3}{2} + \frac{1}{729}\cdot\frac{3}{2}+\frac{1}{2187}\cdot\frac{3}{2}+...$$ $$S_2 = \frac{1}{162} + \frac{1}{486} + \frac{1}{1458} + ...$$ $$S_2 = \frac{1}{162}\cdot\frac{3}{2}$$ $$S_2 = \frac{1}{108}.$$ Thus, $P = S_1+S_2 = \frac{1}{18} + \frac{1}{108} = \frac{7}{108}.$

Note that we just talked about the case where Bob gets the last coin. But note that the case where Alice gets the last coin is symmetrical to this, and thus our answer is simply $2P = \frac{7}{54},$ and thus our requested answer would be $\boxed{754}.$ -DuoDuoling0

Solution 5 (basically steps of solution 4 but different math)

Like solution 4 we will stop the sequence after getting two of each $A$ and $B$. $A$ and $B$ are symmetric in terms of probability so lets first tackle the case where it ends in $A$.

• Case 1: String of $B \dots A B \dots A$ where the strings of $B$'s are of length 1 or more.

The chance of this is $(P(B \dots)) \cdot (P(A)) \cdot (P(B \dots)) \cdot (P(A))$.

The chance of a string of $B$'s can be evaluated using a geometric series:

$P(B) = 1/3, \quad P(BB) = 1/9, \quad P(BBB) = 1/27 \dots$

Add all of these up and you get $1/2$.

The total probability for this case is:

$(1/2) \cdot (2/6) \cdot (1/2) \cdot (2/6) = 1/36$

• Case 2 & 3: There are 2 more cases, $ABB \dots A$ or $BB \dots AA$ where the strings of $B$'s is of length 2 or more.

$P(BB) = 1/9, \quad P(BBB) = 1/27 \dots \implies P(BB \dots) = 1/6$

So the total probability is:

$(2/6)(1/6)(2/6) + (1/6)(2/6)(2/6) = 1/54 + 1/54 = 1/27$

Final Calculation:

• Chance of string ending in $A$ is equal to:

$1/27 + 1/36 = 7/108$

• Chance of ending in $A$ or $B$ is:

$7/108 \cdot 2 = 7/54$

The requested answer is $100 \cdot 7 + 54 = \mathbf{754}$.

-Olecram(hopefully someone can make it into latex and format it)

-LI,CHENXI (no problem bro )