How does this "wrong" solution, which doesn't explicitly use conditional probability, lead to the correct answer?

Question

AMC, Fall 2021, 10B, Problem 20:

In a particular game, each of 4 players rolls a standard 6-sided die. The winner is the player who rolls the highest number. If there is a tie for the highest roll, those involved in the tie will roll again and this process will continue until one player wins. Hugo is one of the players in this game. What is the probability that Hugo's first roll was a 5 given that he won the game?

Solution 1 takes into account only the first round, and calculates the probability that its highest roll is a 5. This does not take into account the cases possible: single 5, double 5, etc. Hence, the remark that the solution is incorrect. Yet, is there some deeper reason/symmetry/structure in the question because of which this wrong solution leads to the same numerical answer $\dfrac{41}{144}$ as in Solution 2 (which uses Bayes Theorem to correctly arrive at the answer)?

Edit: Based on David K's remark, and what I have been thinking, Solution 1 is correct (based on symmetry). If we were to rephrase the question so that it read "what is the probability that the winner rolled a 5", then solution 1 makes sense. Even if insist on identifying Hugo, the probabilities for Hugo are the same (symmetrical) as the probabilities for the other players, and solution 1 makes sense. However, I am not able to show how Solution 1 and 2 are connected (for example, how the symmetrical probabilities lead back to the same answer). And I think that my symmetry answer is a little "hand-wavy". A more precise answer would be really appreciated.

Answer 1

Solution 1 is correct. Hugo wins, having rolled a five in the first round, if and only if the following two events happen: $A$, the highest score in the first round was $5$; and $B$, Hugo wins. But $A$ and $B$ are independent, since if you know the four numbers rolled in the first round, but not who rolled which one, any given player has probability $1/4$ of winning by symmetry. So $\Pr(A\cap B\mid B)=\Pr(A)$.

Answer 2

Problem 20 In a particular game, each of $4$ players rolls a standard $6{ }$-sided die. The winner is the player who rolls the highest number. If there is a tie for the highest roll, those involved in the tie will roll again and this process will continue until one player wins. Hugo is one of the players in this game. What is the probability that Hugo's first roll was a $5,$ given that he won the game?

Solution 1 Since we know that Hugo wins, we know that he rolled the highest number in the first round. So, the required probability is just the probability that the highest roll in the first round is $5.$

1. Here's a detailed expansion of Solution 1: \begin{align}{}&P(\text{Hugo's first roll} = 5\mid \text{winner} = \text{Hugo}) \\={}&\frac{P(\text{Hugo's first roll} = 5,\; \text{winner} = \text{Hugo})}{P(\text{winner} = \text{Hugo})} \\={}&\frac{P(\text{the winner's first roll} = 5,\; \text{winner} = \text{Hugo})}{P(\text{winner} = \text{Hugo})} \\={}&\frac{P(\text{the winner's first roll} = 5)\times P(\text{winner} = \text{Hugo})}{P(\text{winner} = \text{Hugo})} \\={}&P(\text{the winner's first roll} = 5) \\={}&P(\text{the first round's highest roll} = 5) \\={}&P(\text{the first round's highest roll} \le 5) \\{}&-P(\text{the first round's highest roll} \le 4) \\={}&\left(\frac56\right)^4-\left(\frac46\right)^4, \end{align} as required.

   Note that the third equality is due to the independence of the winner's first roll being 5 and the winner being Hugo (since the probability of the former event is $5$ is unaltered by the knowlege of the latter event), whilst the fifth equality reflects Solution 1's stated observation. While Solution/Argument 1 is not invalid, omitting its key step and justification makes it arguably sketchy rather than pithy.

   On an orthogonal note, it's worth noting that the similar-looking event pair the Hugo's first roll being 5 and the winner being Hugo are not independent (since any knowledge of the former happening increases the probability of the latter).

2. Alternatively, take an arbitrary player $p$ (it doesn't matter whether $p$ is identified); then, by symmetry, \begin{align} {}&P(\text{Hugo's first roll} = 5\mid \text{winner} = \text{Hugo}) \\={}&P(p\text{'s first roll} = 5\mid \text{winner} = p) \\={}&P(\text{the winner's first roll} = 5\mid \text{winner} = p) \\={}&P(\text{the winner's first roll} = 5), \end{align} by the same independence reasoning as above; the working then continues as above.

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Addendum

Solution 2 (which uses Bayes Theorem to correctly arrive at the answer)

Actually, Solution 2 is not less sketchy than Solution 1! Critically, all three instances of the string "Hugo wins on his/the next round" need to be changed to "Hugo subsequently wins".

Besides, Solution 2 should be clearer that its four cases are these conditional probabilities:

1. $P(\text{no player ties with Hugo in the first round},\; \text{winner} = \text{Hugo}\mid \text{Hugo's first roll} = 5)$
2. $P(\text{one player ties with Hugo in the first round},\; \text{winner} = \text{Hugo}\mid \text{Hugo's first roll} = 5)$
3. $P(\text{two players tie with Hugo in the first round},\; \text{winner} = \text{Hugo}\mid \text{Hugo's first roll} = 5)$
4. $P(\text{three players tie with Hugo in the first round},\; \text{winner} = \text{Hugo}\mid \text{Hugo's first roll} = 5)$

Summing these four probabilities gives $P(\text{winner} = \text{Hugo}\mid \text{Hugo's first roll} = 5);\tag*{}$ multiplying this by $\frac16$ gives $P(\text{Hugo's first roll} = 5,\;\text{winner} = \text{Hugo});\tag*{}$ dividing this by $\frac14$ finally gives $P(\text{Hugo's first roll} = 5\mid \text{winner} = \text{Hugo})\tag*{}$ (this is essentially Bayes' Theorem), as required.

Answer 3

Just a note for the key independence part: if $A$ is the event in which the winner first rolls $5$, $B_k$ the event in which the $k$-th player wins, player $1$ is Hugo, then $$P(A)=\sum_{1\leq k \leq 4}P(A\cap B_k)=4P(A\cap B_1)$$ (because no player is any different from another) so $$P(A\cap B_1)=P(A)\cdot\frac{1}{4}=P(A)P(B_1)$$ which shows that $A,B_1$ are independent.

Answer 4

Symmetry arguments always need to be considered carefully, but the argument in this case is fairly strong.

Even more directly, let $W$ be the event that Hugo wins, $H_5$ the event that Hugo rolls a $5$ on the first roll, $R_5$ be the event that the highest roll on the first round is $5$.

Then $W \cap R_5 = W \cap R_5 \cap H_5 = W \cap H_5$. Therefore

$$ P(H_5 \mid W) = \frac{P(H_5 \cap W)}{P(W)} = \frac{P(R_5 \cap W)}{P(W)} = P(R_5 \mid W). $$

To link it specifically to "Solution 2", let $A$ be the event that on the first round, the high roll is a $5$. Moreover, let $F$ be the event that Hugo rolls a $5$ on the first roll, $W$ the event that Hugo wins. Separate the event $A$ into cases according to ties, similar to the cases in Solution 2.

Case 1: The first high roll is a $5$ and no players tie. That is, we have exactly one $5$ and three lower rolls. Call this event $A_1$. The single $5$ could be any of the four players. The unconditional probability of this event is $$ P(A_1) = \dbinom41 \dfrac16\left(\dfrac46\right)^3 = \dfrac{16}{81}. $$

The probability this case occurs and Hugo is the one who rolled $5$ (and therefore Hugo wins) is $$ P(A_1 \cap F) = \dfrac16\left(\dfrac46\right)^3 = \dfrac{4}{81}. $$

The probability that this case occurs and Hugo wins, given that Hugo rolled a $5$, is $$ P(A_1 \cap W \mid F) = \dfrac{P(A_1 \cap W \cap F)}{P(F)} = \dfrac{P(A_1 \cap F)}{P(F)} = \dfrac{4/81}{1/6} = \dfrac8{27},$$ which is case 1 from Solution 2.

Case 2: The first high roll is a $5$ and two players tie. Call this event $A_2$. $$ P(A_2) = \dbinom42 \left(\dfrac16\right)^2\left(\dfrac46\right)^2 = \dfrac{2}{27}. $$

The probability this case occurs and Hugo rolled $5$ is $$ P(A_2 \cap F) = \dbinom31 \left(\dfrac16\right)^2\left(\dfrac46\right)^2 = \dfrac{1}{27}. $$

But if Hugo rolled $5$, Hugo still has only $\frac12$ probability to win, so $$ P(A_2 \cap W \mid F) = \dfrac{P(A_1 \cap W \cap F)}{P(F)} = \dfrac{\frac12 P(A_2 \cap F)}{P(F)} = \dfrac{\frac12\cdot 1/27}{1/6} = \dfrac19,$$ which is case 2 from Solution 2.