Although, seriously, life would be so much simpler if we could just go with 5/36.. too bad it’s wrong :(

Love the blog btw :)

I’m sure that this takes more than my meager high-schools math, though.

We want to find the probability that Bob wins in his second turn, given that he wins in first place:

P(C|B) = P(C^B) / P(B)

Since B follows logically from C:

P(C|B) = P(C) / P(B)

The probability that the winner is determined in the fourth die toss (Bob’s second) can be easily calculated:

P(C) = (5/6) * (5/6) * (5/6) * (1/6) = 125/1296

Calculating P(B) is a little bit trickier, but not so much. First, we note that, with probability 1, either Sue or Bob must win the game at some point. (The probability that the game hasn’t before the nth die toss converges to zero as n grows. This doesn’t mean that it’s impossible that the game never ends.) And, since they can’t both win at the same time:

P(A) + P(B) = 1

Given that Sue doesn’t win in her first turn, Bob’s chances are exactly the same that they were for Sue before her first turn. So we have:

P(A) = P(B|D) = P(B^D)/P(D)

Since D follows logically from B:

P(A) = P(B|D) = P(B)/P(D) = P(B) * (6/5)

Replacing for P(A) in P(A) + P(B) = 1 gives:

P(B) * (6/5) + P(B) = 1
P(B) = 5/11

(Kudos to Nathanael Nerode for calculating the probability that Bob wins using geometric series!)

So we can calculate the original conditional probability:

P(C|B) = P(C) / P(B) = (125/1295) / (5/11) = 275/1296

Now I guess I’ll have to simulate it.

Define events:

A1: Bob wins in the second throw.
A2: Bob wins.

We need, P(A1|A2) = P(A1,A2)/P(A2) = P(A1)/P(A2)

Now, P(A2) = 5/6*1/6 + 5/6*5/6*5/6*1/6 + …. = 5/6 (Infinite GP).

P(A1) = 5/36.

So, P(A1|A2) = (5/36)/(5/6) = 1/6.

What’s wrong with this?

Answer = 0.212191358

So using the formula in my previous post;
The numerator is just; [(5/6)^3]*(1/6)
The denominator is just; {SUM[(5/6)^(2n-1)]}*(1/6)
which equals; (30/11)*(1/6)

Numerator over denominator gives you the answer!

As for an explanation of the denominator; the denominator stands for the probability that Bob wins (Regardless of how many turns it takes him, so it covers ALL possibilities). In order for Bob to win, the dice has to roll a six on a EVEN turn (Since Bob always goes after Sue). Hence, the probability of Bob wining is the probability of not rolling a six ‘2n-1′ times, followed by the rolling of a six. It is ‘2n-1′ in order to form an odd number regardless what ‘n’ is.
Hence, you sum over all possible values of ‘n.’ The limiting sum formula is a/(1-r) where ‘a’ is the first term in the series and ‘r’ is the ratio in which each term increases by. So hence, the limiting sum has the first term being (5/6) and the ratio being [(5/6)^2].
Following this will get you to the denominator!

Hope this helps!

So the formula for conditional probability is the intersection of the two events divided by the condition.
I.e. Pr(2 turns ‘intersect’ Bob-wins)/Pr(Bob-wins)

I’m not sure what the answer is because I don’t remember how to calculate infinite series but yeah… hope this helps clarify things!