Die Race!

Probability Level 3

Andrea and Raleigh are each rolling a fair, six-sided die. They roll their dice simultaneously, individually keeping a sum until someone reaches 100; whoever reaches 100 first wins. (If they reach 100 on the same roll, it's a tie.)

Andrea's die has sides 1, 2, 3, 4, 5, and 6. Raleigh's has sides 1, 1, 1, 6, 6, and 6. Who is more likely to win?

Use your intuition, and then try to prove your result!

Andrea Raleigh They are equally likely

Eli Ross by Eli Ross

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4 solutions

Mark Hennings
Jul 31, 2017

Let A n A_n be the expected number of rolls for Andrea to reach 100 100 or more, given that she is n n away from the target of 100 100 (in other words, her current total is 100 − n 100-n ). Then we have the recurrence relation A n = 1 + 1 6 ( A n − 1 + A n − 2 + A n − 3 + A n − 4 + A n − 5 + A n − 6 ) A_n \; = \; 1 + \tfrac16(A_{n-1} + A_{n-2} + A_{n-3} + A_{n-4} + A_{n-5} + A_{n-6}) with initial conditions A 0 = A − 1 = A − 2 = A − 3 = A − 4 = A − 5 = 0 A_0 = A_{-1} =A_{-2} = A_{-3} = A_{-4} = A_{-5} = 0 . This yields A 100 = 29.0476 A_{100} = 29.0476 as the expected number of rolls that Andrea has to make.

Let R n R_n be the expected number of rolls for Raleigh to reach 100 100 or more, given than he is n n away from the target of 100 100 . This time we have the recurrence relation R n = 1 + 1 2 ( R n − 1 + R n − 6 ) R_n \; = \; 1 + \tfrac12(R_{n-1} + R_{n-6}) with the initial conditions R 0 = R − 1 = R − 2 = R − 3 = R − 4 = R − 5 = 0 R_0 = R_{-1} =R_{-2} = R_{-3} = R_{-4} = R_{-5} = 0 . This yields R 100 = 29.1837 R_{100} = 29.1837 as the expected number of rolls that Raleigh has to make.

Thus Andrea expects to reach 100 100 in fewer rolls, so should win in the long run.

If the problem meant that the two have to meet exactly 100 100 (so that to win from 99 99 a throw of 1 1 is necessary), then we have a similar pair of recurrence relations. Without going into detail, the expected numbers of rolls are now 33.3333 33.3333 for Andrea and 33.4682 33.4682 for Raleigh, so Andrea should still expect to win in the long run.

In response to @Eli Ross 's comment, now let p m , n p_{m,n} be the probability that Andrea wins, given that the current score is 100 − m 100-m to Andrea and 100 − n 100-n to Raleigh. Then p m , n = 1 12 [ p m − 1 , n − 1 + p m − 2 , n − 1 + p m − 3 , n − 1 + p m − 4 , n − 1 + p m − 5 , n − 1 + p m − 6 , n − 1 + p m − 1 , n − 6 + p m − 2 , n − 6 + p m − 3 , n − 6 + p m − 4 , n − 6 + p m − 5 , n − 6 + p m − 6 , n − 6 ] p_{m,n} \; = \; \frac{1}{12}\left[ \begin{array}{l} p_{m-1,n-1} + p_{m-2,n-1} + p_{m-3,n-1} + p_{m-4,n-1} + p_{m-5,n-1} + p_{m-6,n-1} \\ + p_{m-1,n-6} + p_{m-2,n-6} + p_{m-3,n-6} + p_{m-4,n-6} + p_{m-5,n-6} + p_{m-6,n-6} \end{array} \right] for 1 ≤ m , n ≤ 100 1 \le m,n \le 100 , with the boundary conditions p m , n = { 1 m ≤ 0 , n > 0 0 n ≤ 0 p_{m,n} \; = \; \left\{ \begin{array}{lll} 1 & \hspace{1cm} & m \le 0\,,\,n > 0 \\ 0 & & n \le 0 \end{array} \right. From this we can use a computer to calculate p 100 , 100 = 0.4583 p_{100,100} = 0.4583 to 4 4 DP.

If we now let q m , n q_{m,n} be the probability that Raleigh wins, given that the current score is 100 − m 100-m to Andrea and 100 − n 100-n to Raleigh, then q m , n = 1 12 [ q m − 1 , n − 1 + q m − 2 , n − 1 + q m − 3 , n − 1 + q m − 4 , n − 1 + q m − 5 , n − 1 + q m − 6 , n − 1 + q m − 1 , n − 6 + q m − 2 , n − 6 + q m − 3 , n − 6 + q m − 4 , n − 6 + q m − 5 , n − 6 + q m − 6 , n − 6 ] q_{m,n} \; = \; \frac{1}{12}\left[ \begin{array}{l} q_{m-1,n-1} + q_{m-2,n-1} + q_{m-3,n-1} + q_{m-4,n-1} + q_{m-5,n-1} + q_{m-6,n-1} \\ + q_{m-1,n-6} + q_{m-2,n-6} + q_{m-3,n-6} + q_{m-4,n-6} + q_{m-5,n-6} + q_{m-6,n-6} \end{array} \right] for 1 ≤ m , n ≤ 100 1 \le m,n \le 100 , with the boundary conditions q m , n = { 1 n ≤ 0 , m > 0 0 m ≤ 0 q_{m,n} \; = \; \left\{ \begin{array}{lll} 1 & \hspace{1cm} & n \le 0\,,\,m > 0 \\ 0 & & m \le 0 \end{array} \right. and we calculate that q 100 , 100 = 0.4561 q_{100,100} = 0.4561 to 4 4 DP.

These two probabilities do not add to 1 1 , since there is always the probability of a draw.

Since 0.4583 > 0.4561 0.4583 > 0.4561 , Andrea is more likely to win, but (again) not by much.

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Moderator note:

This solution is rigorous. There are some intuitive explanations, but they will not easily lead to a generalized result.

Just because Raleigh's expected value is higher does not mean the probability he wins is lower (at least not without some further justification).

Eli Ross Staff - 3 years, 10 months ago

Your second analysis is correct, but also shows that there is little hope to prove this without a computer.

Here is an argument that would suggest the opposite. What goes wrong here?

A roll of the die has the same expectation value E A = E R = 3 1 2 \mathbb E A = \mathbb E R = 3\tfrac12 for the players, but Andrea's variance is smaller: Var A = 2 11 12 \text{Var}\ A = 2\tfrac{11}{12} but Var R = 6 1 4 \text{Var}\ R = 6\tfrac14 . For the sum of N N rolls, both numbers should be multiplied by N N . For sufficiently large N N , the distribution of the sum becomes approximately normal.

Take N = 28 N = 28 , then E A N = E R N = 98 \mathbb E A_N = \mathbb E R_N = 98 but σ A N ≈ 9.04 \sigma A_N \approx 9.04 while σ R N ≈ 13.23 \sigma R_N \approx 13.23 . Thus R is more likely to have a score significantly greater than 98 (or 100, for that matter) after 28 rolls than A. The same argument can be made for N = 27 , 26 , … N = 27, 26, \dots .

Arjen Vreugdenhil - 3 years, 10 months ago

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... but also more likely to have a score significantly less than 98 98 ... These two options (almost) balance out.

If you look at my comment in Jonathan's submission (in answer to Eli), we can determine the generating function of the p m , n p_{m,n} and of the q m , n q_{m,n} . However, the shape of the generating functions is pretty vile, so I would be surprised if there was a nice closed form for these probabilities.

Mark Hennings - 3 years, 10 months ago
Jonathan Quarrie
Jul 31, 2017

I haven't applied any in-depth mathematics, but here was my intuition.

When either die is rolled twice, 7 is the most likely result. (Albeit, with differing probabilities for each die)

As a multiple of 7, both players are more likely to have 98 after 28 rolls than any of the surrounding values; However, Andrea has a 5/6 chance of reaching 100 from 98, but Raleigh only has a 3/6 chance.

With that, I concluded that A n d r e a \boxed{Andrea} was more likely to reach 100 first.

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Moderator note:

As pointed out in the comments, Raleigh wins with 102. However, if we simply start checking probabilities at 98 as this problem does (by either making a probability tree or making a general argument), Andrea wins at 102.

For reference, here are the winners with the targets from 95 to 110: 

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
 
95, Raleigh
96, Raleigh
97, Raleigh
98, Andrea
99, Andrea
100, Andrea
101, Andrea
102, Raleigh
103, Raleigh
104, Raleigh
105, Andrea
106, Andrea
107, Andrea
108, Andrea
109, Raleigh
110, Raleigh

See Eli's comments to Seb. Your argument would say that Andrea and Raleigh have the same chance of winning if the target was 102 102 instead of 100 100 (both of them would have a chance of 1 2 \tfrac12 of reaching 102 102 from 98 98 ). That would not be the correct answer...

Mark Hennings - 3 years, 10 months ago

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Eli said to Seb that Andrea wouldn't be more likely to win if the target was 102. He didn't say that they wouldn't have an even chance.

But, to be honest, I don't have the capacity to verify or argue against what either of you have said. I just put down my thought process that gave a correct answer. Whether I got there for the right reason is beyond me.

Jonathan Quarrie - 3 years, 10 months ago

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You are a software guy. Implement my recurrence relations in Excel or better and calculate p 102 , 102 p_{102,102} and q 102 , 102 q_{102,102} . You will find that Raleigh has the bigger chance of winning in that case.

The advantage that Andrea has at the 100 100 case, and the advantage that Raleigh has at the 102 102 case, is so slight, that we need a more subtle argument.

Mark Hennings - 3 years, 10 months ago

As a hint, one "intuitive" approach makes this more of a number theory problem than a combinatorics/expected value problem.

Eli Ross Staff - 3 years, 10 months ago

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We could show that the generating function of the p m , n p_{m,n} is P ( x , y ) = ∑ m , n ≥ 1 p m , n x m y n = x y 2 ( 1 + y 5 ) ( 6 + 5 x + 4 x 2 + 3 x 3 + 2 x 4 + x 5 ) ( 1 − y ) ( 12 − ( x + x 2 + x 3 + x 4 + x 5 + x 6 ) ( y + y 6 ) ) P(x,y) \; = \; \sum_{m,n \ge 1} p_{m,n}x^my^n \; = \; \frac{xy^2(1+y^5)(6 + 5x + 4x^2 + 3x^3 + 2x^4 + x^5)}{(1-y)(12 - (x + x^2 + x^3 + x^4 + x^5 + x^6)(y + y^6))} from which we can calculate p 100 , 100 = 3163042987854140309855533129798984880918198654446258624899909963490685 \ 9365600616280099299008449489057390581 6901497876834545854867368632978070816801023432115786962201682200860457 \ 6610843435253147523608071501615464448 p_{100,100} \; = \; \frac{3163042987854140309855533129798984880918198654446258624899909963490685\ 9365600616280099299008449489057390581}{ 6901497876834545854867368632978070816801023432115786962201682200860457\ 6610843435253147523608071501615464448} which gives p 100 , 100 = 0.458313 p_{100,100} = 0.458313 to 6 6 DP. A similar calculation could be performed to evaluate q 100 , 100 q_{100,100} precisely.

If a solution is intuitive, it does not need inverted commas.

Mark Hennings - 3 years, 10 months ago

Let me have another go at explaining why your argument is incorrect, even though it gave you the correct answer.

You say that the most likely total for both Andrea and Raleigh in 2 2 throws is 7 7 . This is true. You then say that the most likely outcome in this game after 28 28 throws is that both have scored 98 98 . This is not true. If p n p_n is the probability that Andrea has scored n n after 28 28 throws, then ∑ n = 6 168 p n x n = [ 1 6 ( x + x 2 + x 3 + x 4 + x 5 + x 6 ) ] 28 \sum_{n=6}^{168}p_nx^n \; =\; \big[\tfrac16(x+x^2+x^3+x^4+x^5+x^6)\big]^{28} If q n q_n is the probability that Raleigh has scored n n after 28 28 throws, then ∑ n = 6 168 q n x n = [ 1 2 ( x + x 6 ) ] 28 \sum_{n=6}^{168}q_nx^n \; = \; \big[\tfrac12(x + x^6)\big]^{28} Using a computer, the probability that the score after 28 28 pairs of throws is 98 : 98 98:98 is therefore p 98 q 98 = 0.00655995 p_{98}q_{98} \; = \; 0.00655995 On the other hand, the probability that Andrea has not yet reached 100 100 , while Raleigh has already reached 100 100 , is ( ∑ n = 6 99 p n ) ( ∑ n = 100 168 q n ) = 0.240529 \left(\sum_{n=6}^{99}p_n\right)\left(\sum_{n=100}^{168}q_n\right) \; = \; 0.240529 Thus it is much more likely that Raleigh has already won. On the other hand, the probability that Andrea has already won after 28 28 pairs of throws is ( ∑ n = 100 168 p n ) ( ∑ n = 6 99 q n ) = 0.249671 \left(\sum_{n=100}^{168}p_n\right)\left(\sum_{n=6}^{99}q_n\right) \; = \; 0.249671 so Andrea is more likely to have already won by this stage.

It is in fact true that of all the possible scores a : b a:b that could occur after 28 28 pairs of throws, for 28 ≤ a , b ≤ 168 28 \le a,b \le 168 , the most likely is 98 : 98 98:98 . That does not help us in the context of this question, however. Even though 98 : 98 98:98 is the most likely score (if we ignore the "win if you reach 100" bit) the probability of its happening is still only 0.66 0.66 %, and outcomes involving a collection of other, less likely, scores become much more important.

Probability is frequently counterintuitive.

Mark Hennings - 3 years, 10 months ago

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I didn't really need any more convincing that my intuition was wrong. Maybe I didn't make the appropriate signals for this. But then I'm probably not who you're trying to convince. I'd hazard a guess that most of the upvotes for my 'solution' are from people (like me) that don't have the strongest grasp of probability and don't understand your solution or annotations.

Jonathan Quarrie - 3 years, 10 months ago
Seb Wilkes
Jul 31, 2017

We use can use some simple statistics; using a theory called "Expected Value" which can be written as E(X). Andrea has a discrete uniform distribution so E(A)= ( 6 + 1 ) 2 \frac{(6+1)}{2} = 7 2 \frac{7}{2} while E(R) is a little tougher to calculate. 3( 1 6 \frac{1}{6} )+3( 6 6 \frac{6}{6} )= 19 6 \frac{19}{6}

If we divide 100 by the E(X) values this will give us a value that will tell us how many throws arriving at the Expected Value will take. This shows that Andria will get there first because 100/3.5 > 100/3.17

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3 1 6 + 3 6 6 = 21 6 3\frac{1}{6}+3\frac{6}{6}=\frac{21}{6}

Brian Riccardi - 3 years, 10 months ago

This does not mean that the probability is greater. For example, this logic would imply that Andrea is also more likely to reach 102 first -- but this is actually not the case.

Eli Ross Staff - 3 years, 10 months ago

i underestimated the problem. nice solutions.

Srikanth Tupurani - 3 years, 2 months ago

The answer is Andrea because she has #'s 1, 2, 3, 4, 5, and 6. That means that Andrea has a 1 out of 6 chance to get 1 while Raleigh has sides 1, 1, 1, 6, 6, and 6 which means that she gets a 1 out of 2 chance to get 1. IF YOU COULDN'T DO THIS PROBLEM, IT'S OK. YOU ARE THE ONE WHO GET'S TO LEARN.

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