Sue and Bob

[Charlie Reams]

After the fun of Making Change, here's another puzzle, this one stolen from xkcd:

Sue and Bob take turns rolling a normal die. Once either person rolls a 6, the game is over. Sue rolls first.

In this particular game, Bob rolls a 6 before Sue. What is the probability Bob rolled the 6 on his second turn?

[Junaid Mubeen]

25/216 ?

Seems too simple so I reckon I've missed something.

Edit: scrap that, how about 5/36?

[Gary Male]

25/216 ?

Seems too simple so I reckon I've missed something.

I'm looking at 5/36. As defined in the terms of the puzzle Sue definitely doesn't roll a 6, so we can effectively ignore her rolls. I think.

[David Williams]

275/1296??

[Junaid Mubeen]

25/216 ?

Seems too simple so I reckon I've missed something.

I'm looking at 5/36. As defined in the terms of the puzzle Sue definitely doesn't roll a 6, so we can effectively ignore her rolls. I think.

yep, i agree with gmail

[Charlie Reams]

This puzzle has hidden depths.

Incidentally if you think you've solved it then try solving it for general probability of success (i.e. abstract out the 1/6.)

[Neil Zussman]

Is it not 5/6 * 5/6 * 5/6 * 1/6?
The only way the fourth turn was the first 6 is if the first three turns weren't a 6...

[Junaid Mubeen]

Are we equipped with the assumption that Sue and Bob both have a 1/6 chance of landing a 6 (or 1/n of landing an n) and that each throw is independent?

If not, then the fact that Bob won despite Sue going first gives us more info.

[Martin Gardner]

After the fun of Making Change, here's another puzzle, this one stolen from xkcd:

Sue and Bob take turns rolling a normal die. Once either person rolls a 6, the game is over. Sue rolls first.

In this particular game, Bob rolls a 6 before Sue. What is the probability Bob rolled the 6 on his second turn?

I think the obvious one is 5/6 * 5/6 * 5/6 *1/6, but if it is that, isn't it not very difficult? 125/1296. I think the Bob rolls a six before Sue might somehow be the key to this.

OK what I've written above is definitely wrong, because he might roll a six on his first turn, there's nothing in the question that prohibits it. And if he rolls one after his second turn, it's irrelevant to the question. So right, I think I can have a go at this:

OK, for Sue's rolls, the probability of her getting a six is zero, because that's in the question (Bob rolls a six before Sue). On his first turn, it's 1/6 that he rolls a six, and 5/6 that he doesn't. If it's a six, there's a chance of 0 of him doing it on his second turn, because the game's over. Now Sue rolls and it's not a six, because that's what it says in the question, so now his chances of rolling a six are 1/6. Therefore I suggest that it's 5/6*1/6 = 5/36. How did I do?

[Martin Gardner]

25/216 ?

Seems too simple so I reckon I've missed something.

I'm looking at 5/36. As defined in the terms of the puzzle Sue definitely doesn't roll a 6, so we can effectively ignore her rolls. I think.

Sorry I hadn't noticed this. Still, I've tried to explain how I got the answer, as well.

[Nicky]

Yep. I'm in the 5/36 camp too.

[Martin Gardner]

I've just Googled this (maybe you should have changed the names, Charlie) and it doesn't give 5/36, in fact it says it is NOT 5/36 at the end of the question, presumably because people keep coming up with that and thinking it's right. I'm not entirely convinced by the answer they give either. I won't post it here though, you all know what Google is...

[Nicky]

D'oh! Yeah, the 5/36 reasoning should give us 1/6. Google is my friend.

[Gary Male]

Hmm, alright. Time to break this down and see what's really going on.

Say Bob and Sue roll at the same time, and have distinctly coloured dice. Of the 36 rolls, 5 contain a Sue six alone, 5 contain a Bob six alone, and one roll has both Bob and Sue sixes which under the terms of who goes first is a win for Sue. If neither won then subsequent turns will have the same probability for each side so to infinity so Bob wins 5/11 games and Sue wins 6/11 games.

Hmm. I'll come back to this when drunk.

[Jon O'Neill]

By induction, if n=1 then n=(1+1)=2, so 1=2.

Therefore all dice are blue.

[Martin Gardner]

Hmm, alright. Time to break this down and see what's really going on.

Say Bob and Sue roll at the same time, and have distinctly coloured dice. Of the 36 rolls, 5 contain a Sue six alone, 5 contain a Bob six alone, and one roll has both Bob and Sue sixes which under the terms of who goes first is a win for Sue. If neither won then subsequent turns will have the same probability for each side so to infinity so Bob wins 5/11 games and Sue wins 6/11 games.

Hmm. I'll come back to this when drunk.

They came up with 5/11 as well, which looks fine to me, but as you say that's the general case and we're looking for a specific case set out by what Charlie has said. Say if we said 'what's the chance of Bob winning on his 100th roll" it wouldn't be the same, would it?

[Kai Laddiman]

Ok, it's 0. By induction.

[Junaid Mubeen]

Conditioning on Bob winning does indeed give 5/11, which I think makes David Williams' answer correct...but intuitively it's difficult to see why the 5/36 argument fails. I guess that's what makes the problem interesting, as is often the case in probability.

[Paul Howe]

Right, after puzzling over this for a bit I think I agree with David.

We're looking for P(Bob rolls a six on his second turn | Bob won). Such an expression is begging for Bayes rule to be applied to it, which gives

P(Bob won| bob rolled a six on his second turn)*P(Bob rolls a six on his second turn) / P(Bob wins)

P(Bob won| bob rolled a six on his second turn) = 1 because if Bob rolls a 6 he's won by definition
P(Bob rolls a six on his second turn) = (5/6)^3*(1/6)

And, as Gary pointed out, the probability of Bob winning in the absence of any extra information is 5/11 (which I severely overcomplicated by expanding geometric series and figuring out their win ratios were 5/36:6/36)

Plugging all this in gives 1375/6480, or 275/1296

[Charlie Reams]

Paul wins!

[David Williams]

Why doesn't David win?

[Charlie Reams]

You got 1 mark for the correct answer, but you lost a mark for not replying to my PM.

[Gary Male]

Right, after puzzling over this for a bit I think I agree with David.

We're looking for P(Bob rolls a six on his second turn | Bob won). Such an expression is begging for Bayes rule to be applied to it, which gives

P(Bob won| bob rolled a six on his second turn)*P(Bob rolls a six on his second turn) / P(Bob wins)

P(Bob won| bob rolled a six on his second turn) = 1 because if Bob rolls a 6 he's won by definition
P(Bob rolls a six on his second turn) = (5/6)^3*(1/6)

And, as Gary pointed out, the probability of Bob winning in the absence of any extra information is 5/11 (which I severely overcomplicated by expanding geometric series and figuring out their win ratios were 5/36:6/36)

Plugging all this in gives 1375/6480, or 275/1296

Ok, I won't come back to this when drunk...

Good work.

[David Williams]

Sorry, never had an unsolicited PM before!

Chance of the game finishing on any particular numbered pick is 5/6 times chance of it finishing one pick earlier. Let k = 25/36, and assume over many games Bob wins n on his first pick. So he wins nk on his second pick, nk^2 on his third, n/(1-k) altogether. The proportion of these that finish on the second pick is nk/(n/(1-k), or k*(1-k).

[Martin Gardner]

My head hurts.

[Ian Volante]

I hate probability. I worked that out by induction. I know that joke's been made twice in this thread now, but I don't care.