# 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. Didn’t take too long for me. My quick working was as follows:
Once Sue does not roll a 6 the first time, the chance that Bob wins is A.
5/36+25/36A=A, so A=5/11
Thus the chance that Bob wins eventually is 5/6 of 5/11 which is 25/66.
The chance that he wins on his second turn is (5/6)(5/6)(5/6)(1/6)/(25/66) which is just 275/1296.

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2. So, as I understand it, it’s a two-level question.

What are the odds that:

a) Bob will win [excuse me, ‘end the game’], and

b) he will do so on his second turn.

Which means that you have to multiply a * b.

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3. bob and sue rolling dice. sue misses the 6 on her first roll (5/6), bob misses his first roll(5/6). sue misses second roll (25/36), bob hits second roll (5/36).
5/36+25/36a=a, solves to a=5/11, odds he wins then is 5/6 of 5/11 so 25/66. then plug that int to the odds he wins on turn 2 (5/6)^3(1/6)(25/66) comes to 275/1296

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4. Is it 0 because he says Sue rolls, Sue rolls again meaning she got a six? Ps. How do you work the rot13 thingy it just says cypher? ?

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5. the only thing I find odd at all is that the odds of Bob winning on his second turn was 9.6% when we say that he always wins…

and when we add in Sue’s chance of winning, using Chris’ numbers, he jumps to a 21.2% chance of winning on his 2nd turn….

am I just being dumb?

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6. Its a trick question, and I think you’re all wrong.

If you flip a coin 50 times and it lands on heads each time, what is the chance it will land on heads the 51st time?

1/2. Disregard all of the previous flips, because the question is only asking what will happen on the 51st flip, and regarding this one flip, its a 50/50 chance of it landing on either heads or tails.

If the question stated “what are the chances that you will flip heads 51 times in a row?” then it would be an insane, highly improbable skewed number. But it didnt, it just asked what would happen on the 51st flip.

To translate, this question is asking if Bob will roll a 6 on his second roll, which is not taking in to account any of the other rolls. During this one, singular roll, he has a 1/6 chance of getting a 6.

If the question had stated “what is the probability Bob can avoid rolling a 6 until his second roll”, then it would be a different question entirely. But it wasnt, it was just asking “if Bob rolls a dice, what chance is there that it will be a 6?”

If you want to read in to the questions themselves, anyway. If you want to go with the implied meaning, do it your way.. but I think its a trick question.

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7. Mason135: 21.2% all the way. 9.6% is just kinda like 5 would be in 2+3+4; just an intermediate answer, before you get to add the 4. Only the puzzle is a little more complex than that, so it’s harder to grasp.

Nokarot: It doesn’t ask, “Bob made it to his second turn. Now, will he end the game?” It asks “Bob ended this one game, but I won’t get more specific and tell you on which turn. Did he end it on his second turn?”

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8. This is a bit disorganized. It’s late and I’m tired. But it’s correct.

You’re all wrong! The problem does *not* state “out of all possible games, what is the probability that the one they *actually* played *happens* to be the one that he rolled a 6 on his second turn.” at which point 21.9% or whatever would still be incorrect, for reasons I will go into below. If you’re trusting the guy who says “infinite series, which bother me,” stop right there. They’re rather elementary things in most cases, especially this one. Basic computability theory.

Rephrasing, the problem states, these are the rules of a the game. Sue goes first. Bob wins. Thus he rolled a 6, at some indeterminate point. Sue did not. What are the odds the 6 was rolled on his second turn?

Let’s ask a different question. What are the odds for ANY turn? What are the odds that he did it on his two hundred thousandth turn?

They’re all the same. Bob wins, so Sue never rolls a 6. She is excluded explicitly in the problem. Considering her is wrong-headed. 1/6. Independent events, people. You can’t find the limit of an infinite series on this question, or do clever math like was done in the comments. *the* 6 is equally probable to pop up at any turn, and that probability is 1/6. plain and simple. I’ll stake my semi-anonymous reputation on it.

The 21.9% assumes that Sue *could* roll a 6. But she DOES NOT.

The reason is, there’s an *alternate interpretation* you all missed that leaves an ambiguity. It was elucidated in my rephrasing (without changing the actual problem). Here’s a further hint: what are the odds that Sue doesn’t ever roll a 6 for eternity? That Bob doesn’t either?

Bob won, it is stated, it is no longer a possability but a fixed event; his winning was 1/1, encompassing the entire game state-space to search.

Now suppose that the game goes on for 500,000 turns before Bob finally gets a 6?

Suppose it goes on infinitely long before Bob rolls a 6? There was no time limit set.

But anyway, what are Bob’s odds on that 500,000th turn?

Bob doesn’t win 53/131 (or whatever) games, he wins 1 of 1. It states that explicitly. We’re NOT searching the entire game state-space, which you actually failed to do properly anyway. It is CERTAIN that Bob wins this game, and thus it is CERTAIN that Sue does not. Thus her probability of ever rolling a 6 is 0. Her win percentage in the correct game state-space is 0. Bob’s is 1. Using your logic, that makes 1/0 over 5/36 doesn’t it? I forget, it was so wrong and it’s late, the answer either comes out to undefined or infinity with that math.

Because technically speaking, by a strict interpretation, without knowing the length of the game, the answer to that incorrect question is actually uncomputable, because they could’ve played 500,000,000,000 rounds, or just 2. Bob’s odds for winning a game that happened were 0 or 1. He won or he didn’t. In this case it was 1. It is an event that happened, it is no longer a probability. He won it. How often is that the case is then asked. (This is the intuitive leap people make that is incorrect, but clever. It assumes that the game always eventually ends, but it provably DOES NOT ALWAYS END. In an infinite number of cases, the game goes on forever, with neither winning. This is the second critical flaw, after reading in too much.) Thus, if bob wins a only a certain percentage (due to Sue going first), and he won this time, what are the odds it happened to be on his second roll? That is equivalent to, what percentage of the games that Bob wins does he win on his second roll?

That answer to that problem is not 21.9%. Or at least, not for the reasons given. I have not done the actual math for that question, since it wasn’t asked, but the logic is flawed. As I said, the poster made an assumption that all games eventually end. This was not stated in the rules. Nowhere did it say “eventually one will definately roll a 6.” Heck, it didn’t even say that the 6-sided die was standard and even HAD a 6, but that’s beside the point, getting into real mathematical pedantry. That’s outside of the scope of this.

Because that’s not the question. The question is, quite simply: Sue does not roll a 6. Ever. Bob does, eventually. What is the probability that bob’s second turn rolled a 6. Give no more information, the infinite series HAS no solution, because it won’t converge before infinity. You’d have to simulate infinite games to compute it. The question is more ambiguous than intended. Likely an attempt to be as clever as the whole what’s behind door number 3 probability fun. But the questioner wasn’t quite so clever as (s)he thought, and ends up being a double-trick, or deceptively simple, because the answer is that each and every time he rolls the die, even if he does it 2^70000 times, Bob’s chance is 1/6. Including the second time he rolled it.

(Quick question. How could you even mistakenly answer 5/36? What crack were you smoking when you thought that was a plausible incorrect answer?)

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9. I had to re-post to clarify, I posted on a bb board about the same problem much more clearly.

Quoting myself,

You’re all wrong, because the puzzle is flawed, or intentionally much more deceptive than anyone anywhere seems to get.

Even the people who head down the clever, but incorrect road to arrive at 21.21% haven’t taken into account the fact that there are an infinite number of cases in which neither one ever, ever rolls a 6, for all of eternity. Nowhere in the rules was it stated that eventually someone HAS to roll a 6 in general play and end the game. It didn’t even state that the 6-sided die had a side denoting ‘6’, nor did it state if the game could move through an infinite number of iterations before one person finally rolls a 6 or not, but that’s just getting into set theory pedantry that’s unnecessary. Games can be infinite and never resolve, or infinite AND resolve, and you didn’t search “all possible games” since you didn’t search infinity. Except that in this SPECIFIC instance, it was stated that Bob *DID* eventually roll a 6. Sue did not, her odds of ever doing so are 0 over the whole of the game space, and his are 1, even over an infinitely large game space. The odds that it was on his second roll are uncomputable because of this basic fact: we don’t know how long the game was. The odds are either 0 or 1, but we don’t know. This problem is not possible to solve. You could say the odds it happened on the 2nd roll are infinitely small, I suppose. I.e. 0. This presents a paradox, since the puzzle explicitly stated that he did in fact roll a 6, eventually. This means the puzzle is actually a paradox with no resolution.

Or is it? Let’s look at it another way. The only question we *can* answer is what is the probability that he rolled *a* 6 on his second turn. That probability is 1/6. This is equivalent to asking *the* 6, because it is game-ending; in an infinite number of games which Sue does not roll a 6 before him, in which he does eventually roll a 6 and the game terminates, he will roll a 6 on his second turn with a probability of 1/6. No more rolls are made, and the sequence is made non-infinite, and thus computable. Thus, the probability that he rolled *the* 6 on his second roll is 1/6, because that’s the probability he will roll *a* 6 on *any* roll. This is seriously basic computability and probability, messed up by attempts to be clever that just aren’t clever enough, because of the deceptive nature of the puzzle. Sorry guys.

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10. Hmmm. Given that it is stated that the game ended on the second turn, and it’s asking for the chance of that occuring in that fashion, I’d say it’s fair to count all the rolls together. (Sometimes, people think too much.)

so. . .

Sue rolled twice, 5/6 on each since she rolled something other than a 6, and 5 possibilities other than that.

Bob rolled twice, one is 5/6 from his first time that wasn’t a 6, and then 1/6, for the second turn where he did.

so it’s (5^3) / (6^4), or aprox. 9.65%

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11. portrays more:
‘Bob rolls a 6 before sue’ implies that the game does end.
In any case, the longer the game goes on, the less likely it is to continue, so these infinite number of games which do not end have a probability to actually occur of 0. The probability of not ending by turn N is (5/6)^N, which tends to 0 as N tends to infinity.

This means that although the infinite series only converges at infinity, it does converge TO something, so has an answer.
For Example, the probability of SOMEONE winning, by rolling a 6, = 1/6 times the sum of the probabilities of all possible outcomes before a 6 is rolled.
I.E.
1/6 * (sum k=0 to infinity (5/6^k)). This is a geometric sequence, so the sum is given by a simple formula, = 1/(1 – 5/6) which = 6. So the total probability of someone winning, and ending the game, is 1/6 * 6, which is 1. Thus the game WILL end, at some point.

“The question is, quite simply: Sue does not roll a 6. Ever. Bob does, eventually. What is the probability that bob’s second turn rolled a 6”

So, this is the probability of Bob rolling a 6 on his second go, given that he does win at some point. In order to find out what the probability of bob rolling a 6 on his second go, we have to look at ALL cases, even ones where Sue wins, and then normalise this probability in light of the fact that we know Bob wins. We do this by dividing the probability bob wins on go 2 by the probability bob wins overall. (Note that this gives the normalised probability that Bob wins to be P(B) over P(B) = 1 meaning that he does win). The point is that if we ignore Sue from the beginning we get the wrong answer.
And that the “right” answer is indeed right.

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12. My answer late last night was incorrect, but on the right path. What can I say, it was 4am and I was on sleeping pills, my logic wasn’t 100% watertight. I have posted a revised and correct version to the wiki, at

http://wiki.xkcd.com/irc/Talk:Puzzles#Dice_Rolls

You end up needing set theory and basic calculus and that’s about it. I’m the one down at the bottom of that particular discussion. It’s much easier to discuss there if I’m wrong, because things can be revised and nested, etc., so please take the discussion there 😛

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13. I really liked this problem. I thought about for a few days, then I had an epiphany suddenly 10 minutes ago.

Okay, first of all, let’s modify the rules slightly. The modification only serves to help us better see the probabilities and will be explained later.

Instead of Sue rolling and then Bob rolling, they now roll at the same time, each with a fair 6-sided die. Sue wins when she rolls a 6, and Bob wins when he rolls a 6 and Sue doesn’t. Now we can see Bob’s roll, regardless of Sue’s

Now, with this game, there are 36 possible outcomes (1:1, 1:2, etc.) Of these, Sue wins 6 of them: (6:1, 6:2, 6:3, 6:4, 6:5, and 6:6) and Bob wins 5 (1:6, 2:6, 3:6, 4:6, and 5:6). With any other outcome, the game continues

Notice that Bob does not win in the instance when they both roll a six, since sue rolls a 6 and she wins regardless. Also, notice that there are only 3 possible ways for that turn to come out (otherwise known as events):

S: Sue wins
B: Bob wins
I: nobody wins (indecisive)

The probability of Event S, or P(S) is 6/36, or 1/6. P(B) is 5/36, because it only happens 5 times, and P(I) is 25/36, the other 25 possibilities. Now, let’s add in a new event, which we will call W. W is when somebody wins, and it doesn’t care who. In other words, W=S union B. Since S and B are disjoint events (i.e., there is no possible way for both events to occur simultaneously), we can apply the Probability Addition Rule:

P(W)=P(S)+P(B), or 11/36

Now we know the probability of someone winning on a given turn, or, in other words, the probability of a 6 appearing in that turn. It is tempting at this point to think that this means that only 11/36 of games played have a victor, but this is not true. Instead, what P(W) is saying is that 11/36 TURNS played end the game. The outcome of one turn only has bearing on the next turn if and only if that one turn ends the game, in which case the next turn doesn’t exist.

However, in a full-blown game, somebody has to win eventually, and that means that in a game, W is a given. Now we can use conditional probability. Conditional Probability goes something like this: The probability of B given the information that W is true (this is denoted as P(B|W)) is equal to:

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

Because B implies W, we can replace P(B&W) with simply P(B). Plug in the numbers:

P(B|W)=(5/36)/(11/36), or 5/11
The probability that Bob will win any game is 5/11. Let’s define this event as Event C, to avoid any confusion.

Now, we want to know, what is the probability that Bob wins on his second turn, or with our modified rules, *the* second turn? This is the sequence of events denoted as I, B. Since we want both of these events to happen, we can apply the Probability Multiplication Rule:

P(I&B)=P(B)P(I), or (5/36)(25/36), or 125/1296

This is the probability that, out of all games, Bob wins on the second turn (event 2). However, the statement “Bob rolls a 6 before Sue.” gives the assumption that Bob wins. This means that we must apply some more Conditional Probablility:

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

2 implies C, so P(2&C)=P(2):

P(2|C)=P(2)/P(C), or (125/1296)/(5/11), or 275/1296, which is approximately equal to 21.2%

That’s how I figured it out, anyways.

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14. I’m having trouble applying Bayes’ Theorem here. I would appreciate some help with the maths/definitions/concept.

As I understand it the relevant values to calculate P(T2|B) are:

P(B|T2) = 125/1296
P(T2) = 275/1296
P(B) = 5/11

which gives us the equation of:

P(T2|B) = (125/1296)*(275/1296)/(5/11)

I make that 75625/1679616 or 22.2%. I don’t see how the answer can be 275/1296 when that’s one of the inputs and the other two don’t cancel out.

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15. Ok never mind I got it – I was misunderstanding what P(B|T2) meant. P(B|Tx) will always be equal to P(B) in this scenario, because in any given round (or pair of rolls if you prefer to think of it that way), the chances of winning remain the same for both players, ie of the 36 possible outcomes of a round, 6 will be Sue wins, 5 will be Bob wins and 25 will be no result. Hence, if you know the game finished in round x, you’ve narrowed down the possibilities of the game to those 11. This is as true in round 1 as it is in round 5562538576.

Similarly, knowing who wins imparts no information about when the game ended, so the proportion of Bob’s victories that happened in turn 2 is purely the proportion of games that ended in turn 2.

(As a sidenote, I don’t think that’s strictly true, because knowing Bob won means that Bob rolled the dice the same number of times that Sue did. If Sue won it means she rolled once more. There’s probably a more rigorous way of calculating it, but you’d need to model it on just rolling until a 6 came up and then asking whether you’d rolled an odd or even number of times.)

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16. Wow, good problem! I was firmly entrenched in the theoretical “Sue doesn’t matter, just ignore her” camp until I actually wrote a little simulation and found that, son of a…. the answer really is about 21.2%.

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17. Hi all.

The logic of the question stands up to mathematical analysis. English interpretation can blur the lines, but mathematically it is entirely correct and accurate.

Here is how I can prove the answer is right without fractions:
Do a montie carlo simulation where chance of a roll is even between 1 and 6. Each turn the “player” switches from Bob to Sue. Any win is recorded for the turn, and the player.

I ended up with something like this:
Player Round
Sue 1
Sue 4
Bob 4
Sue 5
Bob 2

The aggrigate of which is (over 100,000 rounds):
Player Round Count
Sue 1 8.335e4
Bob 2 6.956e4
Sue 3 5.792e4

The % of each round over all rounds becomes:
Player Round % of wins
Sue 1 16.7
Bob 2 13.9
Sue 3 11.6

Yes, this roughly matches with the values:
1/6, 5/36, 25/216 etc.

Now – remove all Sue values from that smulation. Then you have values like this:
Player Round Count
Bob 1 6.956e4
Bob 2 4.822e4
and so on.

Now – express each rould of Bob wins by all Bob wins.
Player Round Count
Bob 1 30.7
Bob 2 21.3

LOOK!!!! round 2 is 21.3%, which is just off the expected 21.2% (or 275/1296). What if the question had asked “Sue won.” Well, the count of her round wins divided by all her wins we come up with the same values (within a small margine of error).

Now that we know the correct answer it’s up to us to work out the reasoning.

My reasoning is that each round has a probability of winning.

Each “round” actually being Player 1 rolls then Player 2 rolls.

The chance of a win from ANY round is 1/6 + 5/36 (the chance that Player 1 wins, or the chance that Player 2 doesn’t, but Player 2 does). That = 11/36. Since we know that only Bob won, then this is now his probability. So the probability that a player won in the first round of the game is 11/36.

The probability that nobody won the first round (and therefore Bob didn’t win the first round) is:
1 – 11/36 = 25/36

The chance that somebody won the second round (and therefore Bob won the second round) is :
25/36 * 11/36 This is the probability that nobody has won to this point AND the proability that somebody won in this round.

Great question though – I like “the game has been played” questions because it stops you thiking like a frequentist and switches you into bayesian thinking.

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18. If Bob play all alone rolling dice, and stop to play when he have a 6, the probability that he stop to play at the fourth throw, are they of 275/1296 ?

If no, I’ll start becoming crazy.

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19. Sylario: That would be 125/1296; you didn’t prevent sixes on odd turns from reaching your stats. This is really the first of the two key points of the puzzle:
– The fact that Bob won does affect results.
– You might not ever see Sue, but she’s still there, hitting your probabilities.

If you didn’t get either, you’re likely to end up with 125/1296 (if not one of the others that show an overall lack of understanding). If you only grasped the first, you’d end up with 5/36. If you only grasp the second, it doesn’t seem too different from not grasping either, and you’ll probably get 125/1296. It’s fairly easy to switch back and forth between the latter two. Only if you account for both at the same time, you get 275/1296.

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20. The answer is 1/6. If you take a large enough sample size, each number should come up approximately 1/6 of the time. Regardless though, each roll’s probability is independent of the last. So even if you rolled your die 27,000 times and never got a six, you still only have a 1 in 6 chance of rolling a six on the 27,001 roll.

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21. So, let’s take a huge green casino table, and two camera, mr White is viewing the widescreen camera with bob and sue, he know the normal rules, for him probability are of 275/1296.

Then we have Mr Black, he only see the part of the table where dices roll. For him the rules are : at each roll if there is a six, we stop rolling.

How the hell is it possible that the probability is different for mr white and mr black to reach the 4th dice roll?

How do you reach 275/1296 with the knowledge of Mr Black ?

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22. OK, it is given that Sue never rolled a 6. We can ignore her.

Now, given that Bob has won, what are the odds that he did so on the second turn?

The odds for the first turn: 1/6
The second turn: (5/6) * (1/6)
The third turn: (5/6)^2 * (1/6)
Nth turn: (5/6)^N * (1/6)

But these are all the possibilities, aren’t they? So out of all of them, the second one happening would be:
[(5/6) * (1/6)] / [The infinite sum]

I’m not good at maths, but AFAICS, we can cancel out the 1/6′ and use geometric progression:

(5/6) / [ 1 / (1 – r) ]
=(5/6) / [1 / (1/6)]
=(5/6) / 6
=5 / 36 ??

But this is not the answer. What am I doing wrong?

And, say, how is it different from:
“The game draws after two turns. Given Bob wins, what are the odds it was on the second turn?”

t2 / t1+t2

“Draws after four turns; odds that Bob wins on the third”

t3 / t1+t2+t3+t4

Sometimes I need my mistakes pointed out. A little help, please?

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23. After reading some clues in the comments, the first solution I tried worked out. Go me, I guess, although I don’t think I would’ve gotten it that fast without the comments.

The question is asking “In this game with this set of rules, Bob happened to win this round; what’s the likelihood that he won on his second turn?” (I’m interpreting ending the game as winning, just to clarify.) That’s the same as asking, “Out of all games played with these rules in which Bob wins, what’s the percentage of games he wins on his second turn?” Which again is the same as, “Out of games played by these rules, what’s the ratio of Bob-second-turn victories to total Bob victories?”

So, take the probability that Bob will win on his second turn, and divide it by the probability that he wins at all. Say B is the probability that Bob wins, and bn is the probability he wins on the nth turn.

The total probability of a Bob win is the sum of the probabilities of him winning on each given turn.
B = b1 + b2 + b3 + b4 + …

For him to win on his first turn, Sue has to roll one non-6, then he rolls a 6. For him to win on his second turn, Sue rolls a non-6, then Bob does, then Sue does, then Bob rolls a 6.

b1 = (5/6) * (1/6)
b2 = (5/6) * (5/6) * (5/6) * (1/6)
b3 = (5/6) * (5/6) * (5/6) * (5/6) * (5/6) * (1/6)

The probability of Bob winning on turn n is the probability that Bob *makes it* to turn n (i.e. a 6 was not rolled on a previous turn), times the probability that he then rolls a 6. As we can see, he has a (5/6) ^ (2n-1) chance of making it to turn n, and a 1/6 chance of then rolling a 6. So bn = [ (5/6) ^ (2n-1) ] * (1/6), and B = the sum of [ (5/6) ^ (2n-1) ] * (1/6) from n=1 to infinity.

I didn’t remember how to calculate infinite sums, so I just wrote a program that took the sum up to 200 turns, saw it converged nicely, and did the math from there. I came up with b2/B = 21.2%, which is close enough.

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24. Sorry to double post!

Another explanation of why Sue can’t be ignored; hope this helps:

What we’re doing is looking at a set composed of rounds of this game played by Bob and Sue, and tossing the Sue wins out of the set, leaving only Bob wins. The longer Bob goes without rolling a 6 in a given round, the more likely Sue will win and that round will be tossed out of our set. The sooner he rolls a 6, the more likely the round will “count” for our purposes.

That means there’s a significant difference in Bob’s rolling behavior between rounds he wins, which count, and rounds Sue wins, which don’t. His rolls on winning rounds are better than average, and his rolls on losing rounds are worse than average (“better” and “worse” defined by how soon he gets a 6). That’s why simulating the game as described and counting only Bob wins gives you a different result than just having Bob roll repeatedly by himself.

An example using fewer numbers and more raptors:

One hundred people are sprayed with pheromones and let loose in a raptor pit. Ninety of them got sprayed with Formula 1, which makes raptors hungry, and ten got Formula 2, which makes raptors sleepy. Bob survived. What’s the likelihood that he was sprayed with Formula 2?

Answer: More than 10%, that’s for sure!

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25. Wait What? The problem is what exactely ?

1)What are the probability that bob win at the second tour considering all possible game scenarios ?

OR

2) We know bob has won but we do not know when, what are the probability that he won at second turn ?

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26. >Wait What? The problem is what exactely ?

>1)What are the probability that bob win at the second tour considering >all possible game scenarios ?

>OR

>2) We know bob has won but we do not know when, what are the >probability that he won at second turn ?

The second one.

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27. It’s 21.2%, as people above me have so ably demonstrated in math and with sample programs. There’s no real question, the mathematics are simple once you remove your intuition and just lay out the variables.

I just want to add that a very quick way of disproving many of the answers given, like 1/6 or 5/36, is to just add up the first ten turns or so, and you’ll see that the odds of Bob winning in 10 turns are more than 100%, which is clearly ridiculous.

You can do a similar, though more complicated, thing to disprove the underestimates. Just note how it’s monotonically decreasing (if your probability isn’t monotonic you’re beyond hope, and if it isn’t decreasing then you’ll come to more than 100% in short order). Infinite sums are tricky, but everybody realizes that a car that drives halfway up the driveway to the garage, then half of the remaining distance, etc., will never drive past the garage door. Unless you manage to come really close to the real solution, you should see quite clearly that it never comes even close to 100%.

But we’re given that Bob wins. The sum of the probabilities of Bob winning on turn N for all N is 100%; it’s given. Whatever your answer, it has to converge on 100% to be right, and most answers given very obviously never will, either moving to infinity% chance of winning, or plunging well below the 50% mark.

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28. The 5/36 train of thought does converge on 100%. Only it converges half as fast. Others tend not to converge on 100%, but see if you can convince them that you’re making sense. Remember they probably weren’t thinking properly about this problem. Still a nice try, and one that has some hopes with the 125/1296 flood.

Sylario: 2nd one indeed. I’m not sure if I like the raptors Dana brought though.

I wonder what I’m still doing here in this comment thread. I guess I just like this puzzle because while it’s simple, it manages to confuse people in a gazillion different ways. Cherry pickers are much more monotonic, mathematically speaking.

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29. Sorry if someone already posted a similar explanation, but here’s how I did it from a frequentist perspective–without resorting directly to Bayesian notation:

First consider the entire frequency space, and call the overall probability that Bob wins on turn n “Bn.” (By “overall” I mean not conditioned by our knowledge that Bob wins.) Then, given that we know Bob wins, the probability that Bob won on his second turn is B2/(B1 + B2 + B3 + … ) where the sum in the denominator extends to infinity. We don’t have to include probabilities for Sue winning in our denominator since we know Bob wins.

But we cannot ignore Sue because to calculate the overall probabilities Bn we need to include the probability that the die passes to Bob each turn–that Sue did not roll a six. This is of course (5/6). Thus, the overall probability that Bob won on his first turn is (5/6)*(1/6) = (5/36). Let’s call this fraction “w” (for win).

The probability that Bob did NOT win on his first turn is (5/6)*(5/6) = (25/36). Let’s call this fraction “p” (for pass).

If the die passes to the second round the probability that it then passes to Bob is (5/6). If it passes to Bob, the probability that he then wins is (1/6) and the probability that he then loses is (5/6). Thus, the overall probability that Bob wins in the second round (B2) is (5/6)*(1/6)*p = w*p. The overall probability that the die will pass to a third round is (5/6)*(5/6)*p = p^2.

Extending this reasoning to later rounds, the probability of Bob winning in round n (Bn) is w*p^(n-1). To get to the nth round Bob must have passed n-1 times. Thus: B1 = w, B2 = w*p, B3 = w*p^2, etc.

This leads to a simple expression for our final probability:
P = (w*p)/{Sum[n = 0 to inf](w*p^n)}. I was too lazy/incompetent to solve this infinite sum analytically, but numerically it does converge to 0.45…, or 5/11. So P = (w*p)/(5/11) = (11*5*25)/(5*36*36) = 275/1296. Yay!

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30. Having said all that, the Bayesian formalism is SO much easier to use!

Also, I just read through all the comments and I see that many others explained this much like I just did, with or without the infinite sum. I’m grateful for the people who pointed out the simpler way of calculating Bob’s overall odds of “winning.” And kudos to the person who pointed out that the game could be Russian roulette!

Also, (except for the immediately preceding post) I am not the same person as any earlier Brian/Bryan.

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31. For some reason I was stuck thinking the problem said they only played one game, and was frustrated about the infinite part since if we were certain they only played one game it shouldn’t matter. Am I correct in thinking that?

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32. I wrote a couple of methods in Java to model the situation and with 100 000 trials, the percentage of times that bob wins on his second turn out of the times that he does win is 21.7%

Extending this to 100 000 000 trials it came up with 21.224%

Sue can’t be ignored because if she ends up winning, that game is not included in the total of games that you consider in this calculation. This is a conditional probability problem! The probability of bob winning on his second turn given that he wins eventually (ie considering only the times when he wins, roughly half of the time) is not equal to the probability that he wins on his second turn.

“They?re all the same. Bob wins, so Sue never rolls a 6. She is excluded explicitly in the problem. Considering her is wrong-headed”

Clearly sue *can* roll a 6, so the fact that she *doesn’t* makes her very much included.

Even if you don’t trust the maths, trust in Math.random()! Actually my results technically aren’t trustworthy, as Math.random() doesn’t generate truly random numbers. But they’re pretty close!

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33. Ah, thank you Lee! Out of all the attempts of explaining why Sue needs to be included in the problem, your explanation did it. For someone like me, who is matematically able, but without any real education in advanced maths (lower than average high school standards), explanations like these give inspiration to go get that education, and it lets me understand problems and solutions without really knowing probability theory. I like that 🙂

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34. Thanks xkcd for this! Absolutely love counter-intuitive probability puzzles.

Conditional probability is definitely the way forward; the trick to spotting this is in the wording of the question.

The question is:
“What is the probability Bob rolled THE 6 on his second turn?”

That’s -the- six with which he wins the game, not any old 6 that would win him the game. It is the difference between the existential and universal where people are going wrong and getting 125/1296 (the universal).

Definitely P(Bob wins on 2nd turn) / P(Bob wins)
= (125/1296) / (5/11)
= 275/1296

When computing P(Bob wins), it helps to use 1+x+x^2+… = 1/(1-x)

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35. as agreed earlier, there are 11/36 outcomes in which the game ends each turn

THIS game ends turn 2

25/36 passes round 1
x
11/36 ends round 2

THIS gives 21.2%. but this is clearly not the chance of player 2 winning round 2, its the chance of the game ENDING in round 2.

player 2 wins 5 of the 11 “win” outcomes, so x 5/11

= 9.6%

some people seem completely convinced the answer is 21.2%, so where is the flaw in my logic here?

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36. Dave – The flaw in your logic is that you are including the games that sue wins in your sample space. You have calculated the probability that bob wins in round two, which is different to this probability given that sue does not win. If you do not consider the games in which sue wins (which is about half of them) then you will get the correct answer. This conditional probability is roughly, but not exactly equal to 2 times the probability that bob wins on his second turn, as you ignore roughly half of the games. I say roughly, because sue will win slightly more often as she rolls first.

To calculate this conditional probability you need to:

1. Consider all possible games
2. Ignore/disregard the games in which sue wins
3. Find the proportion of remaining games that end on the second turn

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37. @portrays:

Wow. You’re an idiot. I can see where you’re coming from, but still…

What we’re given:
* Sue can’t roll a 6; has no effect
* Bob can roll a 6, but we are looking for the likelihood that he rolls it SECOND TURN

In order for Bob to roll it on any given turn:
* Sue mustn’t roll a 6 (already stated that she can’t, so no effect)
* Bob mustn’t have rolled a 6 yet, either.

Therefore, assuming that for each roll, Bob has a 1/6 chance of rolling a six and a 5/6 chace of not rolling a six, the probability for Bob to win on any turn “n” is:

((5/6) ^ (n – 1)) / 6

Of course, I obviously messed up somewhere, because for the second turn this yields 5/36 Like

38. HOW TO SOLVE THIS PROBLEM IN THREE SIMPLE STEPS:

-Figure out Bob’s chances of winning in his second round, which are (1/6)*(5/6)*(5/6)*(5/6). (i.e. three misses, then a six)
-Figure Bob’s total chances of winning this game, no matter which round he wins in. This can be kind of tricky since you have to make a sum that goes into infinity… but anyway, the result is 0,45454545454545…. (sue chances are 0,5454545454….).
-Divide the first number with the second. Why? cause you already know bob has won, and the first number is the probability of bob winning in the second round ALL THINGS BEING EQUAL, but you already know that bob has won. So you have to “filter out” sue winning chances.

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39. This is a simple conditional probability problem. First, define “rolling a 6” as “winning” for convenience. Now, for my convenience, define each person’s turn as a roll; Bob’s second turn is then roll 4.

According to the conditional probability law, P(Bob wins on roll 4 given that Bob wins) = P(Bob wins on roll 4 and Bob wins)/P(Bob wins). But if Bob wins on roll 4 then Bob wins, so this is equal to P(Bob wins on roll 4)/P(Bob wins).

P(Bob wins on roll 2) = (5/6)*(1/6).
P(Bob wins on roll 4) = (5/6)*(5/6)*(5/6)*(1/6)
P(Bob wins on roll 2 + 2n) = [(5/6)^(2n)]*(5/6)*(1/6)
Bob can only win on an even-numbered roll, so
P(Bob wins)= (5/6)*(1/6)[1 + (5/6)^2 + (5/6)^4 + (5/6)^6 + (5/6)^8 +….]
Or, more formally,
P(Bob wins)= (5/6)*(1/6)*[Sum from 0 to infinity of (5/6)^(2n)]

Now, the sum from 0 to infinity of (5/6)^(2n) is a geometric series, better represented as the sum from 0 to infinity of (25/36)^n. Therefore it converges, and by standard formula (1/1-r) it converges to 1 / [1 – (25/36)] = 1 / (11/36) = 36/11.
P(Bob wins) = (5/36) * (36/11) = 5/11

P(Bob wins on roll 4) = (25/36)*(5/36)

P(Bob wins on round 4 given that Bob wins)= [(25/36)*(5/36)]/(5/11) = (25/36)*(11/36)=275/1296.

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40. Note, however, that this is an essentially Bayesian style of question; treating probability as related to personal knowledge. The assumption is that it is meaningful to discuss conditional probability after the fact — the probability that something “did happen” in the past.

A hardcore frequentist might say “Whoever told me that Bob won knows whether Bob won on the second turn or not. I don’t need to use probability — I’ll shake it out of him.” 🙂

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41. I apologize for cluttering up the comment page by leaving a comment before reading them all, but the answer I arrived at was 11/36. My reasoning was (after being warned off the simplistic approach that was my first thought): We know absolutely that Sue’s first roll was irrelevant. Therefore we can ignore it. (Or in other words, since Sue’s first roll affects all subsequent rolls equally, it factors out, unlike say, Sue’s hypothetical second roll which affects the chances of Bob’s getting a third roll, but not of his getting a first roll.)

So after Sue misses her first roll, There is a 1/6 chance that Bob wins with his first roll. There is a (5/6)(5/6)(1/6) chance that neither wins on the first roll and Bob wins on the second roll. And so on: the probability that Bob wins on the n’th roll is (25/36)^(n-1) * (1/6). This is an infinite series, which, if I did the math right, converges on (6/11). So when Bob rolls first he wins in (6/11) of the cases. Since (25/256) of these are accounted for by wins on the second roll, wins on the second roll account for [(6*256)/(25*11)] of wins by Bob. Since we are given that Bob won, there is a [(6*256)/(25*11)] chance that it happened on his second roll. This is around 5.6%, which is not what the rot13 answer says. I will have to read the comments now.

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42. PS I inverted the fraction by mistake; my answer should have been ~18% (instead of 560%) which is more reasonable, but still not canon.

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43. OK, after checking, it seems that my mistake was in mentally calculating the cube of 6 to be 256. Oops.

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44. Oh, and when I first wrote (11/36) I was thinking that Bob had won on his first roll, and I forgot to change that in my post. So my sins were:
[Thinking Bob won on the first roll]–>[Messing up the cube of 6]–>[Inverting the fraction that gave the correct answer]–>[Moving the decimal point to turn my ridiculous answer into something less than 1]–>[and now, quadruple posting].

sorry it’s late and I am careless

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