Sue and Bob
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Sue and Bob
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?
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?
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Re: Sue and Bob
25/216 ?
Seems too simple so I reckon I've missed something.
Edit: scrap that, how about 5/36?
Seems too simple so I reckon I've missed something.
Edit: scrap that, how about 5/36?
Last edited by Junaid Mubeen on Mon Feb 16, 2009 4:47 pm, edited 1 time in total.
Re: Sue and Bob
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.Junaid Mubeen wrote:25/216 ?
Seems too simple so I reckon I've missed something.
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Re: Sue and Bob
275/1296??
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Re: Sue and Bob
yep, i agree with gmailGary Male wrote: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.Junaid Mubeen wrote:25/216 ?
Seems too simple so I reckon I've missed something.
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Re: Sue and Bob
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.)
Incidentally if you think you've solved it then try solving it for general probability of success (i.e. abstract out the 1/6.)
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Re: Sue and Bob
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...
The only way the fourth turn was the first 6 is if the first three turns weren't a 6...
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Re: Sue and Bob
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.
If not, then the fact that Bob won despite Sue going first gives us more info.
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Re: Sue and Bob
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.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?
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?
If you cut a gandiseeg in half, do you get two gandiseegs or two halves of a gandiseeg?
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Re: Sue and Bob
Sorry I hadn't noticed this. Still, I've tried to explain how I got the answer, as well.Gary Male wrote: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.Junaid Mubeen wrote:25/216 ?
Seems too simple so I reckon I've missed something.
If you cut a gandiseeg in half, do you get two gandiseegs or two halves of a gandiseeg?
Re: Sue and Bob
Yep. I'm in the 5/36 camp too.
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Re: Sue and Bob
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...
If you cut a gandiseeg in half, do you get two gandiseegs or two halves of a gandiseeg?
Re: Sue and Bob
D'oh! Yeah, the 5/36 reasoning should give us 1/6. Google is my friend.
Re: Sue and Bob
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.
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.
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Re: Sue and Bob
By induction, if n=1 then n=(1+1)=2, so 1=2.
Therefore all dice are blue.
Therefore all dice are blue.
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Re: Sue and Bob
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?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.
If you cut a gandiseeg in half, do you get two gandiseegs or two halves of a gandiseeg?
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Re: Sue and Bob
Ok, it's 0. By induction.
16/10/2007 - Episode 4460
Dinos Sfyris 76 - 78 Dorian Lidell
Proof that even idiots can get well and truly mainwheeled.
Dinos Sfyris 76 - 78 Dorian Lidell
Proof that even idiots can get well and truly mainwheeled.
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Re: Sue and Bob
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.
Re: Sue and Bob
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
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
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Re: Sue and Bob
Paul wins!
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Re: Sue and Bob
Why doesn't David win?
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Re: Sue and Bob
You got 1 mark for the correct answer, but you lost a mark for not replying to my PM.
Re: Sue and Bob
Ok, I won't come back to this when drunk...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
Good work.
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Re: Sue and Bob
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).
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).
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Re: Sue and Bob
My head hurts.
If you cut a gandiseeg in half, do you get two gandiseegs or two halves of a gandiseeg?
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Re: Sue and Bob
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.
meles meles meles meles meles meles meles meles meles meles meles meles meles meles meles meles