A maths problem.

Heggy

Just thought you might like to try this. (You may have seen it elsewhere.)
This is just one of those seemingly counter-intuitive ones, though perhaps that not the right word.

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?

Bear in mind that the answer is not

125/1296.

If you get it don't forget to explain it.

LeixlipRed

Is this a trick question? As in the answer is 1/6?

Heggy

It's a not a trick question, it's just something that needs to be read carefully. I guess you would also need to have across the elements necessary to calculate it too though.

Heggy

I saw someone had an answer there but deleted it because I had mentioned the answer that it wasn't in the first post, but I hadn't it clearly marked so now it's in spoiler tags.
If you want a hint:

It definitely is a number over 1296

Thomas_S_Hunterson

Heggy wrote: »

I saw someone had an answer there but deleted it because I had mentioned the answer that it wasn't in the first post, but I hadn't it clearly marked so now it's in spoiler tags.

Don't mind me, bit of a late one last night;)

Off to bed before I do any damage.

Heggy

Maybe I should have posted it earlier.

TommyGunne

EDIT: Sigh I'm a little slow. We can just ignore Sue completely because Bob rolls a 6 first. So the answer would be

5/6*1/6 = 5/36

jhn_noln

damn you and your fast typing!!!!:mad::p

Heggy

You can't ignore Sue because you have to take into account the fact that it's possible to roll a 6.

Andrew76

For Bob to roll a 6 on his second turn means:

Turn 1 - Sue and Bob both roll something other than a six = 5/6 * 5/6
Turn 2 - Sue rolls something other than a six and Bob rolls his six = 5/6 * 1/6

Probability of that happening is turn1 * turn2 = 5/6 * 5/6 * 5/6 * 1/6 = the answer.

hivizman

I suspect that this is a "conditional probability" problem. We are not looking merely for the probability that Bob throws a 6 on his second throw, which is what Andrew76 has calculated, but for the probability that Bob throws a six on his second throw given that Bob throws a 6 before Sue. The reason we can't ignore Sue is that, if she wins on her first or second throw (or indeed if Bob wins on his first throw), then Bob doesn't get a second throw. The probability quoted in the first post is a somewhat different conditional probability (the probability that Bob throws a 6 on his second throw conditional on (a) Sue doesn't throw a 6 on her first throw, (b) Bob doesn't throw a 6 on his first throw, and (c) Sue doesn't throw a 6 on her second throw).

The conditional probability required is the simple probability that Bob throws a 6 on his second throw, divided by the probability that Bob wins (that is, Bob throws a 6 on any of his throws, conditional on Sue not having already thrown a 6). If this analysis is correct, I'm happy to leave the arithmetic to others.

A simpler example of conditional probability is this: suppose you have a standard pack of 52 playing cards, and draw a card at random. The probability that this is the Ace of Spades is 1/52. But suppose that you draw a card at random and I tell you that it's a black card. Now, the probability that the card is the Ace of Spades is 1/26, because there are only 26 possible cards that you could have drawn, and one of them is the Ace of Spades. On the other hand, if I tell you that the card is red, the probability that the card is the Ace of Spades is zero, because it's a black card. Similarly, if I were to tell you that Sue won the dice game, the probability that Bob threw a 6 on his second throw must be zero, because either Sue had already won the game and Bob didn't get a second throw, or Sue won on a later throw, which could only happen if Bob hadn't already thrown a 6.

Heggy

You're on the right track hivizman, keep going, it is conditional.

Andrew76

hivizman wrote: »

The reason we can't ignore Sue is that, if she wins on her first or second throw (or indeed if Bob wins on his first throw), then Bob doesn't get a second throw.

The OPs quote says Sue rolls first, Bob rolls a 6 before sue, whats the probability he rolls 6 on his second throw - does that not mean Sue cannot have won on her first or second throw and Bob cannot have won on his first throw - if he is to have a chance of winning on his second throw? = 5/6 * 5/6 * 5/6 * 1/6?

Totally confused now.

TommyGunne

Yeah I think I got it now. In work so don't have time to write out my method but it gives

275 / 1296

. Please tell me if that is right! Similar method to hivizman's.

hivizman

I think that's why the OP said that the result is "counter-intuitive".

Also, the OP said that the problem is "something that needs to be read carefully".

If the problem had been worded without the sentence "Bob rolls a 6 before Sue", then your calculation would be fine (and we need to do your calculation anyway in order to get what I think is the correct answer).

But we are given this extra piece of information, that "Bob rolls a 6 before Sue", so we already know that all the possible situations where Sue rolls a 6 before Bob can be ruled out. The probability that Sue rolls a 6 before Bob is

6/11

[this can be calculated by summing a geometric progression or using a number of other approaches), which means that the probability that Bob rolls a 6 before Sue is

5/11

(these are mutually exclusive and exhaustive events so the probabilities must add up to 1). You can show that this second probability is equal to the sum of the probabilities that Bob rolls a 6 on his first throw (assuming that Sue doesn't roll a 6 on her first throw), which is

1/6 * 5/6 = 5/36

, that Bob rolls a 6 on his second throw (assuming that neither Bob nor Sue has already rolled a 6), which is your figure of

5/6 * 5/6 * 5/6 * 1/6 = 125/1296)

, that Bob rolls a 6 on his third throw, etc.

So, given that we are told that Bob has rolled a 6 before Sue, we can ignore all those "events" where Sue rolled a 6 before Bob. That leaves only

5/11

of all the possible outcomes of the game as relevant, and the conditional probability we are looking for is

125/1296 divided by 5/11

, which is

275/1296

.

To reiterate, the key point is that we know already that Bob rolls a 6 before Sue, so all those possibilities where Sue rolls a 6 before Bob are irrelevant. Similarly, in my playing card examples, if we already know that we have drawn a black card, then we can ignore the possibility that we drew, say, the Queen of Hearts, because that's simply not a black card.

I see that TommyGunne has just posted the same answer, so if I'm wrong, at least I'm in good company.

Heggy

You've got it right hivizman and TommyGunne, don't worry, (I don't like being this scary nameless OP ). And avery good explanation too.
I think there's a theorem, Bayes' theorem isn't it?

TommyGunne

Heggy wrote: »

You've got it right hivizman and TommyGunne, don't worry, (I don't like being this scary nameless OP ). And avery good explanation too.
I think there's a theorem, Bayes' theorem isn't it?

Yup. Just conditional probability though. You don't really need the theorem to work it out logically. P(A|B) = (P(B|A)*P(A))/P(B) (where | means given), is the form its usually expressed in.

hivizman

TommyGunne wrote: »

Yup. Just conditional probability though. You don't really need the theorem to work it out logically. P(A|B) = (P(B|A)*P(A))/P(B) (where | means given), is the form its usually expressed in.

Yep, it's Bayes' Theorem. In this case, if A is the event that Bob rolls a 6 on his second turn and B is the event that Bob rolls a 6 before Sue, then P(B|A) has to be one (if Bob actually gets to have his second turn and rolls a 6 on this turn, then he must roll a 6 before Sue does), so the formula simplifies to P(A|B) = P(A)/P(B), which was how I did the calculation.

Thanks, Heggy, for posting this problem.

Heggy

Well thanks for putting so much effort into it. I got it off the xkcd 'blag' where they all had extensive discussions about the whole problem. I have nothing more than Leaving cert probability to my name, I was just curious to see how long it would take you guys here.
You may thank me for that problem, but you might hate me for these. http://wiki.xkcd.com/irc/Puzzles

hivizman

Thanks for the link, Heggy.

I had a quick look - some of the puzzles are classics, like the Monty Hall puzzle (although I note a couple of new twists at the end), but others I don't recognise. Something for the dark evenings, I think (though the nights are getting lighter at the moment ))