# The Good, the Bad and the Ugly

Tuco, Blondie and Sentenza play a game with a single die. Tuco has the first roll, then Blondie, and then Sentenza, and so on. As soon as a player tosses a six, that player drops out of the game and the remaining players continue rolling the die, until everyone has rolled a six.

What is the probability that Tuco rolls the first six, Blondie rolls the second six, and Sentenza rolls the third six?

Hint:

The answer is not $\frac{ 1 }{ 6 }$.

• @dpwilson why die instead dice? – Vinicius Monteiro Mar 29 '16 at 18:51
• "Die" is the singular form of "dice". – dpwilson Mar 29 '16 at 18:59
• @dpwilson both "dice" and "die" is correct for the singular of plural "dice". In my opinion the edit wasn't necessary. see also oxforddictionaries.com/definition/english/dice, oxforddictionaries.com/definition/english/die#die-2 even suggests that "die" is more uncommon – Ivo Beckers Mar 29 '16 at 19:38
• @Ivo -That's what happens when you look up English words in a British dictionary. According to ELU Stack Exchange, the plural is dice and the singular is die. – Mazura Mar 29 '16 at 21:14
• I don't suppose the answer is, "0 probability since they will stop rolling when only Sentenza is left, as there is no point in continuing to roll when only one person is left." ;) – jpmc26 Mar 30 '16 at 2:38

Let $P$ be the probability that the player to start throws the first six.

The probability that Tuco throws the first six is, per definition, $P$.

If Tuco does not throw a six (chance $5/6$), Blondie becomes the first to play. So, his chance to throw the first six is $(5/6)P$.

If both Tuco and Blondy do not throw a six (chance $(5/6)^2$), Sentenza becomes the first to play. Hence, the probability that Sentenza throws the first six is $(5/6)^2P$.

The three probabilities have to add up to unity: $$P + (5/6)P + (5/6)^2P = 1$$

which yieds $P=36/91$.

Repeating the same argument for two players, yields a first player chance to throw the first six of $6/11$.

Therefore, the probability for sixes to be thrown by the players 'in order of first throw' is $36/91$ times $6/11$ which equals $216/1001$.

• Very clear explanation. – Fimpellizieri Mar 29 '16 at 18:25
• FYI - Blondie was played by Clint Eastwood, so that should be "his" chance. – Paul Sinclair Mar 29 '16 at 20:11

$\frac{216}{1001}$

Reasoning

First we find the probability that Tuco rolls the first six.

The probability that the first six appears on Tuco's $k$th go is $\frac{1}{6}\left( \frac{5}{6} \right)^{3k-3}$. Hence, the probability that he rolls the first six is $\frac{1}{6} \sum_{k=0}^{\infty} \left(\frac{5}{6}\right)^{3k} = \frac{36}{91}$.

Given that Tuco rolls the first six the probability that Blondie rolls a six on his $k$th subsequent turn is $\frac{1}{6}\left( \frac{5}{6} \right)^{2k-2}$ and so the probability that he rolls the next six is $\frac{1}{6} \sum_{k=0}^{\infty} \left(\frac{5}{6}\right)^{2k} = \frac{6}{11}$

Overall, we find that the probability of Tuco followed by Blondie is

$p = \left(\frac{36}{91} \right)\left(\frac{6}{11}\right) = \frac{216}{1001}$

216/1001 ≈ 21.58%.

For Tuco to roll the first six, it must be the case that one of the following cases holds:

Tuco rolls a six on the first roll; for this, $P = 1/6$. Or, all three roll a non-six in the first round, and Tuco rolls a six on his second roll. For this; $P = (5/6)^3 * (1/6)$. Or, all three roll a non-six in the first two rounds, and Tuco rolls a six on his third roll. For this $P = (5/6)^6 * (1/6)$. And so forth up to infinity.

Thus, the probability for Tuco to get the first six is

$P_\mathrm{Tuco} = \sum_{n = 0}^\infty \left( \frac{5}{6} \right)^{3n} * \frac{1}{6} = \frac{1}{6} \left( \frac{1}{1 - (5/6)^3} \right) = 36/91.$

Once Tuco has rolled his six, the game is down to Blondie and Sentenza.

By the same logic, the probability for Blondie to roll a six before Sentenza is $P_\mathrm{Blondie} = \sum_{n = 0}^\infty \left( \frac{5}{6} \right)^{2n} * \frac{1}{6} = \frac{1}{6} \left( \frac{1}{1 - (5/6)^2} \right) = 6/11.$

Thus, the total probability that the three will roll their sixes in the given order is

$(36/91) * (6/11) = 216/1001 \approx 21.58\%,$

which is, as noted in the hint, not equal to 1/6.

EDIT: To generalize the problem, if we have $n$ players playing this game, the probability that they will roll their sixes in the order in which they play is

$$P = \prod_{m = 1}^n \frac{1}{6}\frac{1}{1 - (5/6)^m} = \frac{6^{-n}}{\prod_{m = 1}^n (1 - (5/6)^m)}.$$
The product in the denominator can be expressed in terms of q-Pochhammer symbols:
$$P = \frac{6^{-n}}{(5/6; 5/6)_n}.$$

If we calculate this for various values of $n$, we find that:

The probability falls off rather quickly; by the time you have 6 players, the probability is less than 0.5%. However, perhaps surprisingly, the probability is always substantially greater than $1/n!$. In other words, among all the possible orders in which the players could roll their first sixes, the "canonical" order of the players always seems to be favored, being more and more strongly favored as $n$ increases. • Each term in the product is greater than $\frac{1}{6}$, so it's obvious that it will decrease slower than the inverse factorial. – f'' Mar 29 '16 at 20:25

Let $p$ be the probability that Tuco rolls the first 6. Then $p$ is also the probability that each of the others rolls the first 6 if the game started with that player.

Then for Tuco to roll the first 6, Blondie the second 6 and Sentenza the third 6, we need Tuco to start and roll a 6 first; then the game 'resets' with Blondie starting and rolling a 6 first; then again, this time with Sentenza. Each of these probabilities is $p$ and happens in sequence, so the required probability is $p^3$.

Now, starting with Tuco, the probability of Tuco rolling a 6 on the first roll is $\frac{1}{6}$, and the probability of all 3 not rolling 6 on the first roll is $(\frac{5}{6})^3$. If all 3 didn't roll a 6 on the first roll, the game 'resets' and the probability of Tuco rolling a 6 from his second roll onwards is again $p$.

We then have $p = \frac{1}{6} + {(\frac{5}{6})^3}p$, i.e. $p = \frac{36}{91}$.

So the required probability is $p^3 = (\frac{36}{91})^3 \approx 6.2\%$.

• Tuco drops out of the game once he rolls a six, though. So once the game "resets" for the first time, there are only two players, and the probability is different. – Michael Seifert Mar 29 '16 at 17:28
• Though Michael Seifort is correct that you mishandled what happens after Tuco is eliminated, I am still giving a +1 for the different (simpler?) method for calculating Tuco's probability. – Paul Sinclair Mar 29 '16 at 20:09
• @MichaelSeifert Oops, I missed the dropping-out part in the question. – Lawrence Mar 29 '16 at 23:46
• @PaulSinclair Thanks Paul. I'm not sure I can work that in nicely, but I'll leave the answer up as the approach, at least, was quite different from that of the existing answers at the time I posted this. – Lawrence Mar 29 '16 at 23:49
• Johannes gives the corrected argument, but since you posted this approach first and the error was only later, I thought it deserved an upvote. – Paul Sinclair Mar 30 '16 at 0:36

First, let me shorten their names by referring to them by their first letter, $T$, $B$ and $S$.

Let us solve this problem using states.

For those of you who have not come across this concept before, I have posted a link below in the comments which is the wiki from which I learnt this problem solving method. I really recommend it as it has been very helpful for me!

Nevertheless, I'll try to explain the concept in this spoiler for those of you who are new to it.

In this game, there are different 'states'. For example, it might be $T$ to roll. For each of these different states, I work out the probability that an event occurs in that state, like $T$ getting the next 6. Often, I will write this in terms of $T$ getting a 6 in other states. Eventually, I can combine the equations to give me a numerical answer.

States are especially useful in cases when you have no idea when the game will end (i.e. in games where it is a constant loop where there is some chance of the game ending on each round).

Let $E(X_Y)$ denote the probability $X$ will roll the next 6 when $Y$ is the next person the roll the die.

Let us first consider the probability that $T$ will roll the the first 6:

$$E(T_T)=\frac{1}{6}+\frac{5}{6}E(T_B)$$ $$E(T_B)=\frac{5}{6}E(T_S)$$ $$E(T_S)=\frac{5}{6}E(T_T)$$

The game starts with the state where $T$ is about to roll, and we are interested in the chances that $T$ will roll the next 6, so we want to find $E(T_T)$.

Using the above 3 equations:

$$E(T_T)=\frac{1}{6}+\frac{5}{6}\left(\frac{5}{6}\left(\frac{5}{6}E(T_T)\right)\right)$$ $$E(T_T)=\frac{1}{6}+\frac{125}{216}E(T_T)$$ $$\frac{91}{216}E(T_T)=\frac{1}{6}$$ $$E(T_T)=\frac{36}{91}$$

We are only interested in what happens when $T$ rolls the first 6. Any other cases are not relevant.

So now we need to find the probability that $B$ rolls a 6 given that $T$ has already rolled a 6. The game is now reduced to 2 players with it being $B$'s turn (as $T$ just rolled a 6).

Note that this is a different game from before as it has changed from 3 to 2 players, so $E(X_Y)$, in general, has a different value now, since we assume $T$ is out. For example, $E(T_B)=0$ now, since $T$ cannot roll a 6, of course (as $T$ is out of the game).

Now we have the new equations:

$$E(B_B)=\frac{1}{6}+\frac{5}{6}E(B_S)$$ $$E(B_S)=\frac{5}{6}E(B_B)$$

So:

$$E(B_B)=\frac{1}{6}+\frac{5}{6}\left(\frac{5}{6}E(B_B)\right)$$ $$E(B_B)=\frac{1}{6}+\frac{25}{36}E(B_B)$$ $$\frac{11}{36}E(B_B)=\frac{1}{6}$$ $$E(B_B)=\frac{6}{11}$$

Now if $T$ gets the first 6, and then $B$ gets the next 6, then $S$ cannot fail to get the 6 after that ($S$ will eventually roll one, even if it takes over 9000 tries).

So therefore, the probability that order will happen is:

$$\frac{36}{91}\times\frac{6}{11}=\frac{216}{1001}$$

• I use states to solve this problem. However, this concept might be new to some. I actually learnt about it from reading this really good wiki, and I suggest reading up the section on indicator variables the next on states to learn about this way of solving probability problems: brilliant.org/wiki/linearity-of-expectation $.$ States are really helpful when it comes to solving probability problems like this - it has surprised me a lot of the time including this one!! – Shuri2060 Mar 29 '16 at 21:31