c4countdown

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?

25/216 ?

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

Edit: scrap that, how about 5/36?

Junaid Mubeen wrote: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.

275/1296??

Gary Male wrote:

Junaid Mubeen wrote: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

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.)

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...

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.

Charlie Reams wrote: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?

Gary Male wrote:

Junaid Mubeen wrote: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.

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

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...

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

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.

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

Therefore all dice are blue.

Gary Male wrote: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?

Ok, it's 0. By induction.

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.

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

Paul wins!

Why doesn't David win?

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

Paul Howe wrote: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.

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).

My head hurts.

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.