non-trivial extension to a ruin problem

[boytoy]

You roll two die: What is the probability of rolling a 10 and an 11 before rolling a 7?

 

[koupparis]

Rolling a 10: 3 ways, an 11: 2 ways and a 7: 6 ways, for a total of 11 outcomes. Hence need either a 10 or an 11 for the 1st successful roll (could be after many rolls), then the other number for the 2nd successful roll, all before we get a 7.

So P(10 and 11 before 7) = P(10 or 11 before 7)*P(other number before 7) = 5/11*(2/11+3/11) = 25/121

 

[AlexandreH]

I have to think about it. But, a remark, this is an "AND" not an "OR" problem. U need to roll a 10 AND roll an 11 (in any order) before rolling a 7...

 

[koupparis]

I think I've answered it as an AND problem. No?

 

[AlexandreH]

Actually You answered as an "Or" but even that is wrong.
You need to think about the distribution here...
For the "Or" problem, the solution:
the solution is 1-P(7 first roll)-P(7 in second roll and not 10 or 11 or 7 in first roll)-P(7 in 3 roll and not 10 or 11 or 7 in first 2 rolls)...
Need a way to sum these.
Prob of rolling all other numbers (i.e. not 7, not 10 ,not 11) is (36-6-5)/36=25/36

Solution = 1-6/36-6/36*(25/36)^1-6/36*(25/36)^2-6/36*(25/36)^3...
=1-6/36 * ((25/36)^0+(25/36)^1+(25/36)^2+(25/36)^3....)

the sum is the geometric series. the total = 1/(1-r) = 1/(1-25/36)=36/11
Solution = 1 -6/36 * 36/11 = 1 - 6/11 = 5/11
What do you think? ( I ran this quickly so I may have missed something...)
for the "And" problem, it is more complicated. You need to hit both (10 and 11), so when u use the formula 1 - P( ) you could have one of them but not both...
I will think about it later and get back to you.

 

[koupparis]

Sure you can use an infinite series to solve this. I think my method gets around this by only considering what is needed. Ie taking a smaller sample space in which I only need 10, 11 and 7.

Here's the solution with AND and using infinite series:

Count the numbers of ways to win: Clearly you can only win on 2nd roll and after, to win on 3rd roll you need to have 10 or 11 on either 1st or 2nd roll, to win on 4th roll you need to have 10 or 11 on one of the previous 3 rolls etc.

So P(win) = 5/36*(2/36+3/36)+2*(5/36)*((36-5-6)/36)*(2/36+3/36)+3*(5/36)*((36-5-6)/36)^2*(2/36+3/36)+ ...
= 5/36*5/36*(1+2a+3a^2+ ...), where a = 25/36
=25/36^2*(36^2/11^2) = 25/121

 

[AlexandreH]

Your solution is still wrong. for the AND problem, you need to have both (10 and 11) before the 7, not one of them.
for example, to win on second roll you need 10 then 11 or 11 then 10.
P (winning on second roll) = P (10)* P (11) * 2 = 2/36 * 3/36 * 2
for winning above second roll you need to use the Combination
for example, for third roll, 3!/(2! * 1!) * 25/36 *

to win on third roll, you need to have 10 in the first or second roll, 11 in the third roll and any other (no 7 no 11) in the first or second + you need have 11 in the first or second roll, 10 in the third roll and any other (no 7 no 10) in the first or second
so for having 10 in the first or second roll, 11 in the third roll and any other (no 7 no 11) in the first or second = 3/36 * 28/36 *2/36 + 25/36 * 3/36 *2/36 = 3/36 * 2/36 * 53/36 = 53 / (6*36)
and for having 11 in the first or second roll, 10 in the third roll and any other (no 7 no 10) in the first or second = 2/36 * 27/36 *3/36 + 25/36 * 2/36 *3/36 = 2/36 * 3/36 * 52/36 = 52/(6*36)
so, for winning on third roll, solution is 105/(6*36)
what do u get with ur calculations

got my point?
having 10,10,10,10,5,11,7 is a win, which u don't take into consideration.
but having 5,6,11,7 is not, because you need both 10 and 11 before the 7. So, back to calculations!!!
Plus, we need to use Combination. This problem is solved by a Geometric Distribution.
so, to solve the problem, we need a formula that takes all that into consideration.
Again, this is not a trivial problem!!!

 

[pruse]

Recursion is ideal for a problem like this.

First, let (p=P(10 \text{ before } 7)), (q=P(11 \text{ before }7)).

Then

(p=\frac{3}{36}\cdot 1+\frac{27}{36}\cdot p\Rightarrow p=\frac{1}{3}),

(q=\frac{2}{36}\cdot 1+\frac{28}{36}\cdot q\Rightarrow q=\frac{1}{4})

Now, denote (r = P(10 \text{ and } 11 \text{ before } 7)). We have

(P(10 \text{ and } 11 \text{ before } 7)=P(10 \text{ on first})P(11\text{ before } 7)+P(11\text{ on first})P(10 \text{ before }7)+P(\text{NO } 10 \text{ or } 11\text{ or } 7 \text{ on first}) P(10 \text{ and } 11\text{ before } 7))

which using the above numbers becomes

(r = \frac{3}{36}\cdot \frac{1}{4} + \frac{2}{36}\cdot \frac{1}{3}+\frac{25}{36}r\Rightarrow r = \frac{17}{132})

 

[koupparis]

Peter's answer seems correct. I think my answer solves the problem of exactly one 10 and one 11 before a 7

 

[AlexandreH]

I am afraid Peter's solution is wrong too. Again, guys, this is more complicated that this. just to keep it simple, there is a higher probability of hitting 7 than 10. I hope we agree on this as a start. so for Peter's solution, p (10 before 7) cannot be 3/4. I can detail this if you want.
same for 11 before 7. now we need BOTH (thus the "AND") before 7, so that probability is a small number.

This is my solution, it could be wrong, but I think I am working it the right track. we have to use the Geometric series here.


I think the easiest way to solve it is by finding the probability of losing, i.e. getting a 7 before getting 10 and 11.
P (losing) = P (first 7 with no 11s)+ P (first 7 with no 10s) – P(neither 10s nor 11s before first 7)
we need to subtract the third case because we are double counting it.

First, the case of getting 7 on first roll is 6/36
Then for all other rolls (infinite number):

P(first 7 with no 11s) = 28/36 * 6/36 + (28/36)^2 * 6/36 + (28/36)^3 * 6/36 …=6/36 * ( (28/36)^0 + (28/36)^1+(28/36)^2…..)-6/36
The reason we remove 6/36 because we don’t have the item (28/36)^0 initially but we need it for the series.
So, Then P(7 with no 11s) = 6/36 * (1/(1-28/36))-6/36 = 6/36 * 36/8 – 6/36 = 3/4 – 1/6 =(9 – 2)/12=7/12

Then P(first 7 with no 10s) = 27/36 * 6/36 + (27/36)^2 * 6/36 + (27/36)^3 * 6/36 …=6/36 ( (27/36)^0 + (27/36)^1+(27/36)^2…..)-6/36
The reason we remove 6/36 because we don’t have the item (27/36)^0 initially but we need it for the series.
So, Then P(7 with no 10s) = 6/36 * (1/(1-27/36))-6/36 = 6/36 * 36/9 – 6/36 = 2/3 – 1/6 =1/2

We then need to add the case of 7 with no 10s or 11s because we are counting this twice.
Then P(7 with no 10s and no 11s) = 25/36 * 6/36 + (25/36)^2 * 6/36 + (25/36)^3 * 6/36 …=6/36 ( (25/36)^0 + (25/36)^1+(25/36)^2…..)-6/36
The reason we remove 6/36 because we don’t have the item (25/36)^0 initially but we need it for the series.
So, Then P(7 with no 10s and no 11s) = 6/36 * (1/(1-25/36))-6/36 = 6/36 * 36/11 – 6/36 = 6/11-1/6=(36-11)/66=25/66

P (win)= 1 – 6/36 – 7/12 -1/2 + 25/66 = (36-6-21-18)/36+25/66 =
-1/4+25/66= (-33+50)/132 = 17/132
That's my solution.

 

[pruse]

ha. my solution is not wrong; check again. all that was wrong was the numbers in the first two equations. I've fixed them and now gotten the same result as you. albeit, in far fewer words. so the problem is definitely not as complex as you proclaim it to be.

 

[AlexandreH]

I never said my solution was the easiest, this is how I know to solve it. Nice job on urs, doing it in a shorter version, but your solution was wrong when I read it first. As I said before, you cannot have a probability of hitting a 10 before a 7 over 0.5. Yours was 0.75. It doesn't make sense. Now those numbers seem correct (1/3 and 1/4). I was not criticizing the methodology (which is obviously smart and short) but the values did not make sense and I was right, so you ended up fixing them.
Now, I will take a look at yours and try to understand it.