## Roll on

Posted by Chris on July 6, 2012 – 5:35 am

Bill and John take turns rolling a fair die. The winner is the first to roll a 6. Bill rolls first. If John wins, what is the probability he did so on his second roll?

Posted by Chris on July 6, 2012 – 5:35 am

July 6th, 2012 at 7:54 am

11/36

July 6th, 2012 at 12:34 pm

Nope.

~~11/36 is simply the probability of John winning.~~oops. 11/36 is involved somewhere.July 6th, 2012 at 12:59 pm

Probability of three unsuccessful rolls followed by a successful one = 5/6 * 5/6 * 5/6 * 1/6 = 125/1296.

Good to see this site come back to life again.

July 6th, 2012 at 1:07 pm

Hi Wiz. It’s good to see you still can’t do probability right LOL.

July 6th, 2012 at 1:39 pm

5/6 * 1/6 = 5/36

July 6th, 2012 at 1:55 pm

Hi SP. ‘nother nope.

So far:

11/36 is a factor that can crop up in the calculations.

125/1296 is the probability that John wins on his second roll. But that’s based on a universe of games where he might also lose.

5/36 is the probability that John wins on his first roll – but again, that’s in a universe of games where he might lose. There are other possible interpretations.

July 6th, 2012 at 7:36 pm

To reword the problem, given that John wins, what is the probability that his win occurred on the second roll?

First, calculate his total winning probability:

(1/6)*(5/6) = win on John’s first roll

(1/6)*(5/6)^3 = win on John’s second roll

(1/6)*(5/6)^(2N-1) = win on John’s Nth roll

Total chance of John’s win =

(1/6)*(5/6)*[1+(5/6)^2+(5/6)^4+...+(5/6)^2(N-1)+...] =

(1/6)*(5/6)*[1+(25/36)+(25/36)^2+...+(25/36)^(N-1)+...] =

(1/6)*(5/6)*{1-(11/36-1)+(11/36-1)^2+…+/-(11/36-1)^(N-1)+…]

The quantity in brackets is the Taylor series for x^-1 where x=11/36. Therefore the probability of John winning is…

(1/6)*(5/6)*(36/11) = 5/11, or about 45.5%.

John’s chance to win on the second roll, including his losses, is (1/6)*(5/6)^3 = 125/1296, or about 9.6%

Thus, given a John win, his chance of winning on his second roll is [(1/6)*(5/6)^3]/(5/11) = 275/1296, or about 21.2%

July 7th, 2012 at 4:02 am

Well done Chuck. That’s correct.

Another way to calculate the probabilities is to note that

P(Bob wins) = (1/6)(1 + (5/6)^2 + (5/6)^4 + …)

P(John wins) = (1/6)(5/6)(1 + (5/6)^2 + (5/6)^4 + …)

Those must add to 1. Let x = 1 + (5/6)^2 + (5/6)^4 + …

Then (1/6)(1 + 5/6)x = 1 => x = 36/11

Hence P(Bob wins) = 6/11 and P(John wins) = 5/11

I’ll also note that 1 + r + r^2 + r^3 + … = 1/(1-r),

when |r| < 1 is well known. In this problem, r = (5/6)^2.

July 11th, 2012 at 7:47 pm

You know two things: 1) the die is fair. 2) John wins.

Therefore, by 2) P(Bob wins) = 0 and P(John wins) = 1

P(John wins on Roll 1) = 1/6

P(John wins on Roll 2) = (1-(1/6))*(1/6) = 5/36

If you are going to include the possibility that Bob wins, then it cannot be given that John wins.

There are ONLY three possibilities:

a) John wins on his first roll. P(a)=1/6

b) John wins on his second roll. P(b)=5/36

c) John wins on a later roll. P(c)=25/36

P(a)+P(b)+P(c) = 1/6+5/36+25/36 = 1

July 11th, 2012 at 9:09 pm

Hi Mark. You’re right in that you can’t predict that John will win any particular game.

The phrase “given that” is misleading, but it is in common use and so we’ve got to live with it. “Given that” really means “only take into account the trials that satisfy the condition that” (or similar). “Given that” doesn’t mean (in context of the problem) that we remove the 6 from Bob’s die to guarantee that he doesn’t win.

If we consider that Bob and John play e.g. a million games, then on average (and rounding to the nearest intger), Bob would win 166,667 on his first roll, John would win 138,889 on his first roll, Bob would win 115,740 on his second roll, John would win 96,451 on his second roll etc.

If we tot that lost up (to infinity) we’d find that Bob would (6/11)*1,000,000 = 545,455 games and John would win 454,545 games. So of the games that John wins, the fraction he wins on his second roll is 96,451/454,545 = 0.21219 = 275/1296 to the limit of accuracy that I’ve used.

July 12th, 2012 at 6:45 am

I’m just going to clarify/extend my last post. “Given that” is often abbreviated to “|” (vertical pipe).

In this problem P(Bob wins | John wins) = 0 and

P(John wins | John wins) = 1.

We want P(John wins on 2nd roll | John wins). NB “John wins” means “John wins on any of his rolls”.

In my last post I tried to make things clearer by considering a million games. But the nature of the term “probability” means that it would be better to consider a single average game.

In a single game, on average, John wins 5/11 of it. In a single game John wins (1/6)(5/6)^3 = 125/1296 games on his second roll. So the fraction of the games that he wins on his second roll of all the games he wins on any roll is (125/1296)/(5/11) = 275/1296 = 0.212191…

—

Aside: Let p = 1/6, q = 5/6, so p + q = 1

P(Bob wins) = p + pq

^{2}+ pq^{4}+ … = p(1 + q^{2}+ q^{4}+ …)= p/(1 – q

^{2}) = (1-q) / ((1-q)(1+q)) = 1 / (1+q) = 6/11Similarly we get P(John wins) = q / (1+q) = 5/11

Note that => P(Bob wins) + P(John wins) = 1

July 26th, 2012 at 12:29 pm

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