A Math Problem

Courtesy LispClub.com:

Sue and Bob take turns rolling a 6-sided die. Once either person rolls a 6, the game is over. Sue rolls first. If she doesn’t roll a 6, Bob rolls the die; if he doesn’t roll a 6, Sue rolls again. They continue taking turns until one of them rolls a 6.

Bob rolls a 6 before Sue.

What is the probability Bob rolled the 6 on his second turn?

I love puzzles which are simple to state but have a fiendishly tricky or counterintuitive answer. I just threw up a page on the xkcd IRC wiki to hold some of the better ones I’ve found over the years. I’ll be adding more over the next few weeks as I remember or find good ones. Feel free to add some of your own!

Edit: Buttons and then Daniel Barkalow got the correct answer first.  Here it is, rot13‘d.  Check your answer against this before posting smugly or people (I) will tease you:  gjb friragl svir bire gjryir avargl fvk, be nobhg gjragl-bar cbvag bar creprag.

0 thoughts on “A Math Problem”

1. I suppose the most confusing part people have is that the last statement could be considered “incomplete.”

At the most basic “interpretation”:

Q:What are the odds Bob will roll a six on his die?
A: 1/6

Now taking into the previous paragraph:

Q: How many out of all his sixes does Bob get on his second roll when only he is playing the game? OR If Bob always wins, what is the probability THE victory will be on his second throw?
A: 5/36 (every preceeding throw decreases the probability this will happen on the next roll)

Making the big assumption:

Q: What is the ratio of victories Bob gets on his second roll when taking into account ALL possible victories?
A: 125/1296 (1/6 x (5/6)^3)

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2. Just to say for that one guy, 5/11 is just roughly the amount of games Bob’d win overall. It’s not useful to know it for winning a specific roll.

Think if it this way, if you knew Alice rolled first, she’d win 6/11 of the times. In a game where they payed you 2:1 | them:you you’d make money betting on Alice.

P.S. I though of one situation it’d be useful: If you know the ratio of how many games (in all) are won on the second roll, THEN you could multiply by 5/11 to get Bob’s victories on the second turn.

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3. P.P.S. Or dividing 125/1296 by 5/11 gives you the TOTAL victories on the second round. (275/1296) Hope the makes it clearer why you kept getting that number.

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4. There’s another way to (trivially) see that the answer can’t possibly be 1/6, as several posters have claimed.

If it’s true that the odds of Bob winning on turn 1 are 1/6, and the odds on turn 2 are 1/6, and on turn 3, and 4, and 5, and 6, and 7 (please note: this isn’t true)… then, what are the odds he wins in the first 7 turns?

1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 7/6 ???

Clearly, this cannot be.

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5. but the question asks in relation to his second turn and not necesarily to the game in general.
the odds of him rolling a 6 on any of his turns is going to be 1:6, but the odds of him getting to that turn and then rolling a 6 are going to be vastly different

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6. how many different sets of 5 numbers you can get from 1-25

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7. The chance of rolling a six on a six sided die is always 1/6. No matter how many times you’ve rolled it or what game your playing or what the hell you’re doing.

If you roll a die to get a six the chance is 1/6.

Simple.

You can make it less simple though but 1/6 is actually correct since it is a D6 that is rolled by a person. Therefore I must say that most statements here are rather irrelevant since you make a mathematical problem out of something simple.

I must say that it is fun to read through all sollutions though^^

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8. Jake says:
February 11, 2009 at 12:57 pm

“Bob rolls a 6 before Sue” just says he ended the game.

^^

I agree with Jake. What’s the point of solving a non-existent problem? The rules clearly state that the game ends when either of them rolls a six.

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9. I really liked this problem. Not being familiar with probabilities, I went through several iterations of the problem before really coming to an understanding.

My first attempt summarized the problem as: What are the odds of Bob rolling a 6 on his second turn? In that case, it would be 1/6, since each throw is an independent event. Looking more closely, though, that’s not what we’re asking. “What is the probability Bob rolled the 6 on his second turn” is not equivalent to “what is the probability Bob rolled *a* 6 on his second turn.”

My second attempt stated the problem as: what are the odds of Bob winning the game by rolling a 6 on his second turn? In that case, we have Sue’s odds of rolling a non-6, Bob’s odds of rolling a non-6, Sue’s odds of rolling yet another non-6, and finally, Bob’s odds of rolling a 6. (5/6)*(5/6)*(5/6)*(1/6)=125/1296. Ah ha! Except… that’s still not what we really want to know. That tells us the fraction of games that Bob wins on his second turn, from the set of all games.

The correct summation is: of the games that Bob wins, what is the fraction that Bob wins on his second turn? It is tempting to say that Sue cannot win, and thus discount her throws completely, but she can, in fact, still affect the outcome. We have to take into account the fact that in the set of all possible outcomes, not all games can be sorted into the “Bob wins” category. Sue will win some games, making it less likely for Bob to roll the die several times and win later on down the line. So, even though we are not looking at games that Sue wins, we are looking at the fraction that she doesn’t. That doesn’t affect Bob’s overall odds of winning in the set of games where he wins, but it does affect the probability of winning on a given turn.

So, knowing this, the only missing piece of the puzzle is knowing the fraction how many games Bob wins. Trying not to get bogged down in infinite series, which I don’t enjoy, I just reasoned it out as this: the probability of Bob winning on any given turn is B = (5/6) * S, where S is Sue’s chance of winning. Sue has a better chance of winning because she rolls first. We also know that S = 1 – B, since any game that Bob doesn’t win, Sue does. The system of equations can be solved as B = 5/11.

Knowing, then, that Bob wins 5/11 of the games overall, how many does he win on his second turn? He wins 125/1296 games overall on his second turn, divided by 5/11 games overall, for a grand total of 275/1296.

A great problem. It really challenged me to read the problem and figure out what’s going on, and what we really want to know. I love that moment when everything clicks and finally makes sense.

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10. surely its just 5/6 (prob. of not getting a six) *5/6 *5/6 (sue’s go, bob’s first go then sue’s second go) * 1/6 (prob. of bob then getting a six)

so:

it’s (5/6)^3 *1/6

(5^3/6^3)* 1/6 = 125/216 /6 (or *1/6)

=125/1296

not that tricky.. or am I doing something stupid like using a calculator :-S

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11. >surely its just 5/6 (prob. of not getting a six) *5/6 *5/6 (sue’s go, bob’s
>first go then sue’s second go) * 1/6 (prob. of bob then getting a six)

That will tell you the probability that Bob wins on his second turn, from all possible outcomes. That’s not what the problem is really all about — as far as we’re concerned, Sue cannot win. We want to know what the odds are that Bob will win on his second turn, if we know in advance that Bob must win.

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12. I feel bad that I figured out method to find the answer immediately but somehow failed to correctly multiply numbers together, even with a calculator.

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13. Having not looked at the answer (yet) I would say that it would be about 9.6%.
This was my math:
(5/6) (5/6) (5/6) (1/6) because the probability of either of them NOT rolling a 6 on any turn is 5/6 and there are three turns taken before Bob reaches his second turn, wherein his probability of rolling a 6 is 1/6. Feel free to ridicule if this answer is not correct, I will now go look at the actual answer.

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14. Hmm I was wrong… What a surprise! NOT… Tricky tricky! Thanks to Chris for the full explanation, without that I would have never figured it out!
Just sent it to my stats professor (who is actually a spatial/cognition psychologist) to torture him a bit, we’ll see how he does.

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15. Wait a minute… we know that Bob wins… it’s in the problem. Why do we factor in Bob’s chance of winning in games overall when we are only being asked the probability that he won on his second turn in an isolated game which we already know that he won? Chris? Can you tell me this? I’m getting a glimmer of hope that maybe I was right… Could someone please quash that quickly?

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16. Well, as I said above, the question isn’t “what are the odds that Bob will win on his second turn?” That’s what p = (5/6) (5/6) (5/6) (1/6) gets you — his odds of winning, assuming that Sue could as well. In that equation, we’re still allowing for the possibility that Sue could win. That’s what the first and third 5/6 terms represent.

In reality, we know that Sue cannot win. From there, it’s tempting to toss Sue out on her bum and say that Bob’s odds of winning on any given roll are 1/6, but that’s just answering the question “what are Bob’s odds of rolling a 6 on any given turn?”

The real question being asked here is “if we know that Bob wins, what are the odds that it will happen on his second turn, rather than his first, third, or any other?” To answer that, we have to consider the set of all possible outcomes. We can split that set into two, the games Bob wins, and the games he doesn’t. We can also split that set into games that end on the second turn, and games that don’t. Ultimately, we’re interested in the intersection of two sets: games in which Bob wins, and games that end on the second turn.

The part that really seemed to be a stumbling block for me and many others is the first part of the question. Figuring out the odds that Bob will win on his second turn is pretty trivial. Unfortunately, a close examination reveals that “what are the odds that Bob will win on his second turn” is not an equivalent question to “given that Bob wins, what are the odds that he will do so on his second turn?”

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17. I like to think of it using actually probability symbols, it makes it clearer for me.

We are looking for the probability that Bob wins on this 2nd turn, given that he wins.

2=wins on second turn
W=wins

P(2|W)=P(2&W)/P(W)

Which translates to, the probability of winning on his second turn given that he wins is equal to the probability [that he wins on his second turn and that he wins] divided by the probability that he wins.

Obviously P(2&W)=P(2) because winning on his second turn implies winning.

Thus you are looking for P(2)/P(W), which is exactly what Chris calculated above.

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18. Geez people, you make this extremely simple problem way too complicated. Here’s how ya do it:

P(1st roll, Sue’s 1st roll: Sue rolls anything but a 6)*P(2nd roll, Bob’s 1st roll: Bob rolls anything but a 6)*P(3rd roll, Sue’s 2nd roll: Sue rolls anything but a 6)*P(4th roll, Bob’s 2nd roll: Bob rolls a 6)

= (5/6)*(5/6)*(5/6)*(1/6)

=125/1296

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19. That is how you do it incorrectly, yes. You’re answering the question “what are the odds that Bob will win on his second turn,” which is not the same as asking “if we know that Bob must win, what are the odds that he will do so on his second turn?”

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20. The best way to rephrase the question is, “What fraction of the games that Bob wins would we expect him to win on his second turn?”

“What fraction of the games played would we expect Bob to win on his second turn?” is a completely different question, because it also considers the sequences of rolls in which Sue wins.

The key, after reading this question correctly, is to realize that even if we are only thinking about games Bob wins, the frequencies with which any two winning sequences occur still occur in the same ratios. That is essentially why the hint (answer is not 5/36) is given.

I would say that 5/36 is obtained by a more reasonable (though still misguided) approach than 125/1296. There’s a good chance that if you never thought about why anyone would think the answer is 5/36 that you never thought about the problem correctly.

The answer 5/36 comes from the false assumption that the ratio between the probabilities of Bob winning on turn n and on turn n+1 is 6:5 instead of 36:25. Read the solution of any of the people who came up with 275/1296

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21. This is not a mathematical or stats problem. But a language problem..

The only problem here is the “Bob rolls a 6 before Sue.”
Given that, we can assume Sue’s never rolls a six so each time she rolls, you have 5/5 to go the next turn.
That would make :
5/5 * 5/6 * 5/5 * 1/6 = 5/36

That wasn’t my first understanding of the problem
My first one was the 125/1296 solution.

When Chris says that : “The correct summation is: of the games that Bob wins, what is the fraction that Bob wins on his second turn? It is tempting to say that Sue cannot win, and thus discount her throws completely, but she can, in fact, still affect the outcome”

You can disagree for a simple a reason :
“of the games that Bob wins” => Bob always wins, it’s a fact.
“still affect the outcome” => no she can’t, if she can win she can’t affect the outcome since she will never roll a six before bob.

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22. To add to David and Chris’ explanation, one can calculate the chance for Bob to win the game at a given turn n as: (5/36)*(25/36)^(n-1). If you sum this for n=1 to infinity, you find the 5/11 that Chris found, the chance for Bob to win. After that, use David’s explanation, which I find the clearest, since it boils down to basic conditional probabilities.

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23. NicKy: The rules are all normal, with no weird biases. The next paragraph states that Bob won a game. The third paragraph then asks for the probability that bob rolled “the 6” (referring to the previous paragraph as the outcome of one game for which you’ll be finding the probability, rather than the puzzle contradicting itself) on his second turn. Show me how any other interpretation is valid if you want to attribute it to language. For now I can only think the puzzle legitimately pwned you.

Hopefully my comment will get through now. The blag likes to drop them about 1021 times out of 1296.

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24. I keep reading the same things, and keep thinking they are wrong. Which means I probably am.

The formulas that bring about the “correct” answer all take the probablity that Bob rolls his 6 on his second role as a ratio of the probability that he wins at all. However, this seems to be ignoring the fact that, in the game we are analyzing, we know he wins. The probability that he wins (or has won) is 1. The probability that Sue wins (or has won) is 0.

If the question was: What is the probability that Bob won the game? The answer would be 1. This is simple. Sue did not win.

Given this, I see no reason to take Sue’s rolls into account. She did not roll a 6. Therefore the probability she rolled a 6 before Bob got to roll is 0, so she can be discounted.

I keep reading other people say that this problem is so simple, and that everyone is making it complicated, and then they give a different answer to me! So something is amiss.

I think the most telling thing is that if we kept the information the same, and changed the question to: What is the probability that Sue rolled a 6 on her second roll? The answer is simply, and unequivocally, 0. She did not roll a 6 in this game, so she could not have rolled on one her second roll.

In order for the answer to be anything other than 5/36, the answer to the above question needs to not be 0.

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25. The probability Bob has won is 1. The probability Sue has won is 0. But you can’t just toss Sue, at least not that thoroughly. Let that be a hint.

Try running a simulation without putting any of your assumptions in it. Just throw dice, and count how many times Bob wins, and how many times he wins on his 2nd. If you assume anything you might as well jump straight to 5/36 again.

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26. What is the probability Bob rolled the 6 on his second turn?

i dont see how the answer is such an obscure percentage… 1/6*1/6 seems more logical to me

in fact, having done probability in highschool and probably covering this exact problem, it makes no sense to me 😦

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27. If I run a simulation and see what percentage of times Bob wins on his second roll, I am answering the question: “What is the probability that Bob wins on his second roll”. without knowing that he has won the game. You could run the same simulation and see what the probability that Sue won on her second roll. You will get a very similar answer, and it is wrong. Sur did not win on here second roll. This is known. Therefore the probability that Sue rolled a six on her second roll is 0.
Let me add another wrinkle. If the information said:
“Bob rolls a 6 first”
“Bob rolled a 6 on his third roll”.
If I now answer the question “What is the probability that Bob rolled a 6 on his second roll”, the answer is 0. There is no chance that Bob rolled a 6 on his second roll, as we know he rolled it on his third, so could not have rolled on second (otherwise there would not have been a third roll!).
However, the original simulation will still give the original answer, and will still be wrong.

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28. A good simulator would look like this (excuse my crude C, just first thing that came to mind):

unsigned long samples = 0, true_samples = 0, iterations = 1000000;
while(iterations–) {
unsigned long turn = 0; // 0=Sue1, 1=Bob1, 2=Sue2, 3=Bob2, …
while(random() % 6) ++turn; // close enough to 1/6 prob of ending; feel free to substitute if you’re paranoid about % 6 or PRNGs
if(turn & 1) { // the precondition is that Bob wins; if true we count the game, if false we disregard it
++samples;
if(turn == 3) // we know the precondition is true; now if Bob won on his second as well, we increment that counter
++true_samples;
}
// you can also use turn & 1 and then turn == 2, or turn == 5 and then turn == 3
}
printf(“%lu/%lu or %g%%n”, true_samples, samples, 100.0 * true_samples / samples);

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29. Make the simulation handle things exactly as described. You should have two variables to keep track of the number of games that satisfy the precondition, and the number of games that satisfy the precondition (Bob wins) as well as the number of games that satisfy the precondition as well as anything else you want (Bob wins on second). Have an outer loop that loops, say, a million times. In that loop, run a new loop that runs the game and keeps track of who ended on which turn. Now check if the outcome satisfies the precondition: if not, disregard the game; if yes, count the game and also check for your other condition.

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30. Ugh, sorry, at first my comment with code didn’t appear at all for a few mins for whatever reason (timed-out CAPTCHA or gibberish detection or something). Now that I posted a new one it says awaiting moderation for the one that didn’t appear at all at first.

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31. If you don’t get it, read the post from BertBert on February 20th and the following. I like his illustrating third example.

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32. You could model tha game as a markov chain, and then just make the necessary computations(which i dont really want to).

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33. >the probability of Bob winning on any given turn is B = (5/6) * S, where S is >Sue’s chance of winning. Sue has a better chance of winning because she >rolls first.

Sorry for being dense. This looks intuitively right to me, but I for some reason cannot find a simple explanation for it. Can someone prove this quickly ?

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34. @Jean-Marc: If we model the game as Bob and Sue rolling at the same time, then If Sue rolls a six, she wins on that turn, and if Bob rolls a six and Sue does not, he does, but if both Bob and Sue roll a six, Sue wins. That means Sue wins 6/36 of all rolls, and Bob only wins 5/36.

The ratio (6/36):(5/36) is the ratio of Sue’s wins to Bob’s wins, overall.

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35. I got the right answer (275/1296) without any problem, by just treating the whole thing as a semi-basic conditional probability question.

So, instead of trying to add further confusion by trying to explain conditional probability to everyone, I shall simply rephrase the question in a fun way:

You are talking to Sue. She said that she and Bob had decided to play a game of Russian Roulette, with one bullet in a standard 6-slot revolver, except each person spins the revolver before pointing the gun at their head and shooting. They made a second bet just for fun: if the person who lost died on their second turn, the other person would get \$1,000,000. Sue took her turn first. During this game, Bob died.

What is the probability you can steal a million dollars from Sue?

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36. The answer is 5/6 * 5/6 * 5/6 (the probability that a six is NOT rolled on either of the first two rolls) times 1/6 or 125/1296. At any rate, as an economist I thank you from the bottom of my heart for a little perspective on the still upsetting and outrageous bailout bonuses. Our province recently gave out similar bonuses… you know what the amount per person was? About seven thousand dollars… not 1/3 of the average income as my local paper claimed, since that would mean we were all working for a little over minimum wage, but seven large… whereas these bonuses, given to the architects of the credit default swaps, were more than a quarter million dollars, and distributed on an annual basis, both important facts… anyway, this concludes your division for the day, until lunch time.

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37. Hmm. I can arrive at the right answer, but apparently only by asking the wrong question. My intuitive answer was 5/36

If I ask: What’s the probability that Bob wins on his 2nd roll?

…but why do I _need_ the information that Bob wins to arrive at that result? I don’t see how that piece of information enters the equation.

Can anyone explain to me what I am missing?

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38. Excuse my poor diagramming skills and hosting, but try my linkified name. It explains the right answer in detail, and includes some other answers.

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39. Ouch, and ouch once more.

For everyone that did not ‘get it’:

We have the game defined – players alternatively rolling 1D6 with Sue starting, and the players who first rolls a 6 (P=1/6) wins.

we know that Dick won, and we are asked to compute probability it was on his 2nd turn.
In other words the question asks – what is the ratio of probability Dick has won on his second turn to the probability that dick has won on any of his turns.
Now, we consider dick’s won games
5/6 * 1/6 (dick won on his first turn) 5/6*5/6*5/6*1/6 (Dick won on his second turn) 5/*6*5/6*5/6*5*6*5/6*1/6 (dick won on his 3rd turn)
Overall we get this geometric series : 5/6*1/6 (1+ (5/6)**2 + (5/6)**4 ….)
= 5/36( 1 + 25/36 + (25/36)**2 …. )
to compute the limit we can use the formula 1/(1-n) where n is the multiple of this convergent series. so we have 1/(1 – 25/36) = 1/(11/36) = 36/11
multiplying by the constant : 5/36 * 36/11 = 5/11 his overall prob. of winning

then we take the probability he won on his 2nd turn (5/6*5/6*5/6*1/6) (translated into words : sue did not roll a six, dick did not roll a six , sue did not roll a six , dick rolled a six and won) which is 125/1296
now (125/1296)/(5/11) = (125*11)/(5*1296) = 25*11/1296 = 275/1296 as the result says.

(note for fun – i myself made a mistake – i misread the question as Dick winning on THE second turn instead of HIS second turn)

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40. This can be done with slightly less arithmetic if you use a bit of probability theory (Bayes’ rule).

Define the two events we care about as follows:

T2 = the game ends on someone’s second turn
B = Bob wins the game

The problem asks to find Pr[T2 | B] (the probability that the game ends on the second turn *given* that Bob wins).

First off, we need the fact that Pr[B | T2] = Pr[B]. This is saying, essentially, that if I tell you the game ends on the second turn (either Sue’s second or Bob’s second), this doesn’t give you any additional information about whether or not Bob wins. This makes intuitive sense, but can also be formally shown in a couple of lines.

Then, Bayes’ rule says Pr[T2 | B] = Pr[B | T2] * Pr[T2] / Pr[B]. The right hand side simplifies to just Pr[T2] because Pr[B | T2] = Pr[B]. So, the probability that we’re interested in, namely Pr[T2 | B], is equal to the probability that the game ends on the second turn, which is
(5/6)(5/6)(1/6) + (5/6)(5/6)(5/6)(1/6) = 275/1296.

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41. For anyone still unconvinced of the correct answer, here is a simple python program simulating 100 000 games. This usually comes closer than .5% to the analytic answer.

“””
import random
from itertools import count

def game():
for turn in count(1):
if random.randint(1,6) == 6:
return ‘sue’, turn
if random.randint(1,6) == 6:
return ‘bob’, turn

def main():
N = 100000
bob_wins = 0
bob_wins_on_second_turn = 0
for i in xrange(N):
winner, turns = game()
if winner == ‘bob’: # we know that bob wins, so only count those games
bob_wins += 1
if turns == 2: # of these, how many does he win on the second turn?
bob_wins_on_second_turn += 1

print 100. * bob_wins_on_second_turn / bob_wins # as a percentage

main()
“””

bye, David.

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42. Sorry, the comment messed up the indention, I don’t know how to fix that quickly… it should be pretty straightforward to indent that yourself (the only important bit is that the second “if” (if turns == 2) is below the winner == bob “if”).

Interesting sidenote (don’t know if anyone has mentioned that yet): Sue has the same chances to win on the second turn, given that Sue wins.

bye, David.

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43. For the skeptics who prefer R, here is the R code. The sim is sloppy, but it only takes like 3 minutes to write down. It isn’t as fast as python, but, at least to my eyes, it is a bit more transparent:

### Start R code
win1stfunc <- function(x, odd = TRUE) {
return(which(x == 6)[1])
}

set.seed(1234567)
rolls <- matrix(sample(1:6, 100*100000, replace = T), nc=100)
win1st <- apply(rolls, 1, win1stfunc)
bobwin <- win1st[win1st %% 2 == 0]
sum(bobwin == 4)/length(bobwin)
### End R code

If you use the random seed I did, you get the answer 0.213. If you comment out the set.seed() call or use a different seed, you’ll get a different, but close, answer.

The “trickiness” of the problem comes from the “misdirection” conferred the conditional probability. In this case you have to condition on Bob’s victory, which requires a different calculation than the naive answer, as many at least one commenter notes above.

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44. Actually, the most interesting part is that everyone assumes Bob “wins” when he ends the game. The puzzle makes absolutely no reference to winning. Not that it matters, but it’s almost like it was worded like that just to see if anyone would *NOT* call it a “win” 275/1296 of the time (besides 5/36, 125/1296, 1/6, etc.). And because I noticed I get a win certificate from Randall in my mailbox. Or something; must be my imagination going crazy. Oh, and I do call it winning. 🙂

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45. This is just straightforward Bayes rule, with a bit of geometric series, isn’t it?

275/1296

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46. I think I am (slowly) being convinced. Eric’s discussion of Bayes’ rule helped a lot (and sent me off looking it up!). To convince myself, I changed the question to:
What is the probaility that Sue won on here second roll, given that Bob won. The answer is pretty obvious (0), but I wanted to see if the maths held up. Here is some the maths (I think):

Define the two events we care about as follows:

S2 = the game ends on Sues second turn
B = Bob wins the game

Bayes’ formula is:

Pr[S2 | B] = Pr[B | S2] * Pr[S2] / Pr[B]

Pr[S2 | B] is the probability that Sue wins on her second roll, given that bob wins.
Pr[B | S2] is the probability that Bob won, given that sue wins on her second roll.
Pr[S2] is the probability that Sue wins on her second roll.
Pr[B] is the probability that Bob wins.

We can quickly deduce that Pr[S2 | B] is 0, as the probability of Sue winning on her second roll is 0 if we know Bob wins. We also know that Pr[B] is non-zero, so the calculation yields:

Pr[S2 | B] = 0. Which is what we expected.

Okay, I believe the maths. And I’m sure the simulations work. I now need to re-read WHY this is the case. Why 5/36 is wrong (which I am now convinced it is).

This has been a fun problem!

The problem asks to find Pr[S2 | B] (the probability that the game ends on Sue’s second turn *given* that Bob wins).

First off, we need the fact that Pr[B | S2] = 0. This is saying, essentially, that if I tell you the game ends on the second turn (either Sue’s second or Bob’s second), this doesn’t give you any additional information about whether or not Bob wins. This makes intuitive sense, but can also be formally shown in a couple of lines.

Then, Bayes’ rule says Pr[T2 | B] = Pr[B | T2] * Pr[T2] / Pr[B]. The right hand side simplifies to just Pr[T2] because Pr[B | T2] = Pr[B]. So, the probability that we’re interested in, namely Pr[T2 | B], is equal to the probability that the game ends on the second turn, which is
(5/6)(5/6)(1/6) + (5/6)(5/6)(5/6)(1/6) = 275/1296.

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47. Cool, I learned something about statistics today. (That was the one math class that gave me any trouble back in high school. I think I finished with a B-.)
I got to 125/1296, but I couldn’t think how it would get from there to 275/1296. I never would’ve thought to approach “What are the odds that Bob wins on his second turn?” as “What are the odds that Bob wins at all, and what’s the chance of it happening on his second turn?’

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