Probability and Rolling a die

[PhilipLeitch]

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?


Hint - The answer is not 5/36.

Philip
___________________
Nobody can hear you scream in Euclidean space.

[TomGutman]

I get 275/1296.
__________________
� � � � Tom Gutman

[PhilipOakley]

I say Sue is irrelevant 😉 [She never throws a relevant die]

Bob roll die until he gets a 6 (infinite sum of probabilities). Leads to - what is the particular probability it is on the second throw.

Philip Oakley

[AlvaroDíaz]

On 3/25/2009 4:37:54 AM, philipoakley wrote:
>I say Sue is irrelevant 😉
>[She never throws a relevant
>die]

>Bob roll die until he gets a 6
>(infinite sum of
>probabilities). Leads to -
>what is the particular
>probability it is on the
>second throw.

Sue not roll a 6 (p = 5/6)and Bob not roll a 6 (p = 5/6) and Sue not roll a 6 (p = 5/6) and then Bob rolls a 6 (p = 1/6) (this is the 2nd, isn't?).

So P = Prod(p) = 5/6*5/6*5/6*1/6 = 125/1296


Alvaro

[PhilipOakley]

On 3/25/2009 5:33:44 AM, adiaz wrote:
>On 3/25/2009 4:37:54 AM, philipoakley
>wrote:
>I say Sue is irrelevant
>;-)
>[She never throws a
>relevant
>die]

>Bob roll die until he
>gets a 6
>(infinite sum
>of
>probabilities). Leads to -
>what is
>the particular
>probability it is on
>the
>second throw.

Sue not roll a 6 (p
>= 5/6)and Bob not roll a 6 (p = 5/6) and
>Sue not roll a 6 (p = 5/6) and then Bob
>rolls a 6 (p = 1/6) (this is the 2nd,
>isn't?).

So P = Prod(p) =
>5/6*5/6*5/6*1/6 = 125/1296

Alvaro

Sue definately did not roll a 6 on first turn, so probability = 1.0.

Sue definately did not roll a 6 on any other turn (remember Bob won), so again probability = 1.0.

However, we must be certain we are taking the same ratio ...

Philip Oakley

[PhilipLeitch]

Another hint - Tom's got the right answer.

Tom - don't give it away - but see if anyone else can come up with the right reasoning.

Philip
___________________
Nobody can hear you scream in Euclidean space.

[AlvaroDíaz]

On 3/25/2009 8:33:55 AM, pleitch wrote:
>Another hint - Tom's got the
>right answer.
>
>Tom - don't give it away - but
>see if anyone else can come up
>with the right reasoning.
>
>Philip

Yes, Tom have the right answer. I email you one (supposed wright) reasoning to waiting others posts.

Alvaro.

[PhilipOakley]

On 3/25/2009 4:37:54 AM, philipoakley wrote:
>I say Sue is irrelevant 😉
>[She never throws a relevant
>die]
>
>Bob roll die until he gets a 6
>(infinite sum of
>probabilities). Leads to -
>what is the particular
>probability it is on the
>second throw.
>
>Philip Oakley

I misread the question. I had thought that it said that bob won....

Philip Oakley

[TomGutman]

>>I misread the question. I had thought that it said that bob won....<<

It did. Which provides the information that Sue did not roll a six on her two rolls.
__________________
� � � � Tom Gutman

[PhilipOakley]

On 3/25/2009 6:03:17 PM, Tom_Gutman wrote:
>>>I misread the question. I had thought that it said that bob won....<<
>
>It did. Which provides the
>information that Sue did not
>roll a six on her two rolls.
>__________________
>� � � � Tom Gutman

OK, so I go back to my first point. Sue is irrelevant. Because Bob rolls a dice until he gets a 6 (the first 6). What is the chance that it is on his second roll.

That is my reading of the question - which would result in a probability of (5/6)*(1/6) which PhilipL said it wasn't...

Philip Oakley

[PhilipLeitch]

No - that dosn't reason out...
That is:
1) it is given he won
2) the likelysood of sue winning is zero.
3) the chance of him winning in the second round is the chance of him not getting a 6 on his first throw bug getting a 6 on his second throw.

So Sue never had a chance of winning? No. She did have a chance of winning. But she didn't win.

This is one of those "game is over" events (see the Boy Girl Paradox - previous post in the Puzzels & Games area.

That is to say:
1) They played a game.
2) Bob won.
3) What was the chance that Bob won in the 2nd round?

The probability that XY will occur is NOT the same as "The probability of XY given X, occured?"

Therefore - Sue's rolls can't be discounted, they occured. Therefore - she DID have a probability of winning. Therefore, it is moe likely that Bob won on his 2nd throw than his "random" probability of rolling a 6 on his second throw because Sue didn't win but could have.

To crack it you just have to work out what the probability of Bob winning in any particular round is.

Philip
___________________
Nobody can hear you scream in Euclidean space.

[LouP]

This is a conditional probability = Pr(Bob on second toss given that Bob is the winner). this is not the same as the prob. that bob wins on the second toss.

My answer is 275/1296.

Lou

[AlvaroDíaz]

Oh, an anser post. This version is without series.

Alvaro.

[ptc-1368288]

This is too raw an example. The problem is who is going to be the winner on the long run. Then the solution is 500 years old with Fermat. The "enjeu" was how to split the gain when one of the players goes sucking roots. On an infinite game, neither Sue or Bob wins and the their last throw each are equally probable � ... and the game starts all over again ... and the infinite answer � does not need more calculation than knowing in advance.

jmG

[TomGutman]

>>The problem is who is going to be the winner on the long run.<<

That is not the question that was asked. Nor is ½ the answer to this question. In the long run the persion who throws first will win with a probability of 6/11.
__________________
� � � � Tom Gutman

[PhilipLeitch]

It's been a little while since I've had time to come back.

I like Lou's approach. Alvaro's is right too, and shows the probabilities in a nice piece of logic.

Here's the solution I came up with which includes a monte carlo simulation to confirm the results.

As I said earlier, I didn't get the right result earlier (I got the same conclusion as Philip Oakley), which is why I did the simulation, to see what the data was actually doing.

Philip
___________________
Nobody can hear you scream in Euclidean space.