Wherein I shall steal a math problem from XKCD:

[theweaselking]

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?

Hints: The answer is not 1/6. It is not 5/36. It is not 125/1296.

Comments

[kadath.livejournal.com]

In theory I know how to do this, but my brain is tofu. Okay:

Probability that Bob does not win on his first roll: 5/6

Probability that Sue does not win on her first roll: 5/6

Probability that Bob wins on his second roll: 1/6

So, 5/6 * 5/6 * 1/6 = 25/216

Nobody quote me I can't thiiiiiink today.

[theweaselking.livejournal.com]

You forgot that Sue goes first.

[argonel.livejournal.com]

I will spoil the game and point out that the probability of things that have already occured is by definition 1. Therefor the probability that bob rolled a 6 on his second turn is either 1 or 0. It either happened or if didn't. If you ask the probability that Bob will roll a 6 on his second turn next game I'm not sure I can figure out the answer.

[theweaselking.livejournal.com]

But we don't know that Bob won on the second turn. We just know that Bob won.

We want to know what the odds are of it having been on the second turn.

[chaosrah.livejournal.com]

Game is over! I are smartz.

[chaosrah.livejournal.com]

Oh, nm, I read it wrong. But I don't care. Game is still over. So who cares? lol.

[ikkarus01.livejournal.com]

0. There is no Sue or Bob, only Zul.

[drjon.livejournal.com]

1/36. But yes. No Sue or Bob, only Zul.

[silmaril.livejournal.com]

185/1296.

(Probability that the game did not end in the third turn overall, i.e. in Sue's second turn, times the probability that Bob rolled a six in his second turn, the fourth turn overall.)

( Moar explanations: P(game ends in first turn)=1/6.
P(game ends in second turn)= (1-1/6)*1/6 = (5/6)*1/6 = 5/36.
P(game ends in third turn)= (1-5/36)*1/6 = (31/36)*1/6 = 31/216.
P(game ends in fourth turn)= (1-31/216)*1/6 = 185/1296. )

ObVious: Posted without looking at the earlier answers.
Disclaimers: Ready and willing to eat crow with mustard. Was not likely to think deeply enough about it to come up with this reasoning without the hint that it was not 125/1296. Arithmetic error entirely possible in the above, as I did not use a calculator and I'm already rather tired, but I stand by my reasoning.

[theweaselking.livejournal.com]

You've got an arithmetical error in there somewhere, probably caused by calculating "does not end" as "1 - does end" at each stage.

Were I using your method, I would write it:

Sue doesn't win: 5/6
Bob doesn't win: 5/6
Sue doesn't win: 5/6
Bob wins: 1/6

Multiply all four together to make them happen in a row, and you'll get 5^3/4^6, which is 125/1296, which is wrong.

You've just calculated the odds of Bob winning in the second turn.

What we want are the odds that, given that Bob has won, his win came in the second term.

[dolston.livejournal.com]

125/1296

[dolston.livejournal.com]

And to be clear I didn't put much thought into figuring this out.

[cantkeepsilent.livejournal.com]

All wrong. It's 275/1296.

The probability that Bob wins on the second turn is the 125/1296 that everyone is coming up with. But Bob only wins 5/11 of the time. Therefore, among the trials that Bob won, the probability that he won on the second turn is (125/1296)/(5/11) = 275/1296.

[theweaselking.livejournal.com]

Interesting path of reasoning, but it did come to the right answer.

(how do you get "Bob only wins 5/11 of the time"? It's the sum of the infinite series of 5/36 + 125/1296 + etc etc etc, but I don't remember how to do that.)

[dwaleberry.livejournal.com]

[opaqueplanet.livejournal.com]

bwahahaha

I love it!

[autobotsrollout.livejournal.com]

On the second turn, Bob pulls a gun and says "it's a six, bitch."

Therefore the probability is 1.

[theweaselking.livejournal.com]

That doesn't address the possibility of Bob losing on turn 1.

Unless he pulls the gun earlier, and starts swearing. Swearing changes the odds.

"KEEP ROLLING! KEEP FUCKING ROLLING OR I WILL BLOW YOUR FUCKING HEAD OFF! SHUT YOUR WHORE MOUTH YOU FAT COW AND ROLL!"

[opaqueplanet.livejournal.com]

I hated probability in highschool. Mostly because I hated fractions in elementary school. I see them and my brain goes "graaaah" and rolls over towards geometry or calculus.

So what career path did I take straight outta high school? Psychology! Until I got to Stats... graaaaaah.

[theweaselking.livejournal.com]

I was an electrical engineer until I realised that I hated EM, power, and semiconductor physics.

[glitteringlynx.livejournal.com]

I did it wrong.

Online help says the formula is: 1 - (5/6)^n where n = number of rolls it took to get the 6. Therefore:

1 - (5/6)^4
1 - 5^4/6^4
1 - 625/1296
= 671/1296

[glitteringlynx.livejournal.com]

Source: http://mathforum.org/library/drmath/view/56502.html

The obvious answer is

[mcpino.livejournal.com]

42.

[heraldofchaos.livejournal.com]

sorry...

it doesnt matter who rolls first. previous rolls do not affect probability.

every time you roll a dice it has a 1:6 chance. just because ive flipped 1000 heads doent not affect the probability of a tail.

i would ask these questions.

is it a perfect cube? is it being rolled the same way every time? are we dealing with a janky "random" number generator?

the solutions that are give are based on pure math.

Dice (a physical object) ISNT.

and i should know. ive delt with more than enough slight of hand and loaded dice.

i have in my possession a set of dice that will roll 9-12 for me but not for anyone else... unless you know the trick to them.

[theweaselking.livejournal.com]

We're assuming perfect 1/6 dice, yes. These are magical hypothetical dice that produce magically perfect results.

But the question isn't "what are the odds that the next roll will be a 6". That's too easy.

(Also: Honest dice can be made such that their results are functionally indistinguishable from random, for our purposes. Even if we were using physical dice and not magic hypothetical dice, we aren't stuck using Parker Brothers or WotC "vaguely cube-like" D6en, here.)

[opaqueplanet.livejournal.com]

Sue rolls. Chance of not-six: 1 (since we know Bob wins)
Bob rolls. Chance of not-six: 5/6
Sue rolls. Chance of not-six: 1
Bob rolls. Chance of six: 1/6

1 x 5/6 x 1 x 1/6 = 5/36

I'm sure this is wrong, but why?

[theweaselking.livejournal.com]

Because that is the chance that, out of two rolls, the first is not a 6 and the second is a 6.

The problem is that Bob rolling a not-6 and then a 6 does not guarantee him victory in turn two. Sue could have rolled a 6 in either of two cases before then - meaning, for Bob to win by rolling not-6 then 6, you have to ALSO consider that Sue will miss both times.

The real catch, here, is that the fact that Bob wins tells us we need to look at a subset of all results, but the fact that Bob *can lose before rolling* affects that subset. He doesn't win 1/6 of the first round - he wins 1/6 *if Sue loses*. This means that his odds of winning in round 1 are lower, even when we consider only his wins, because him winning in the first round will come up less often in the list of total results.

[dscotton.livejournal.com]

Not having read the comments yet, I think it's 11/36.

[dscotton.livejournal.com]

Whoops, this is not right. I was trying to do (odds Bob wins on the second turn) / (odds Bob wins) but my numerator was completely wrong.

275/1296 is my new guess... let's see.

[neko-san.livejournal.com]

My current calculation is 25/186, but I'm afraid I may still have missed something in there. Beh.

[theweaselking.livejournal.com]

Indeed, you have.

There's a full solution in the comments above, in various places.

[undeadbydawn.livejournal.com]

well, that was an education. Which has made me seriously reconsider my oncoming career option in engineering.