If probability only increases with a miss, I think the final answer calculated in previous posts would be different...

I need to check that.

No..No.. wait

Oh now i get it...When i first thought of the problem, i had in mind that the probability only increases with a miss and not with a hit..

So, if probability only increases with a miss then the maximum one would be

(miss)(miss)(hit)(hit)

The problem's solved!!!

Thanks gAr, bobbym..

]]>It's (Miss)(Miss)(Miss)(Hit)(Hit)

Sorry but i didn't get the 4th case?..

]]>Okay, I came to know from your solution.

I was about to update.

Anyway, here's the probabilities I get:

]]>Yes, i just want to count only the cases where there are 2 consecutive hits.

]]>I guess that question is not clear.

There are many combinations:

Hit, Miss,Miss, Hit

Miss, Miss, Hit, Hit etc.

Do you want to count only the cases where there are 2 consecutive hits?

]]>Oh yes I think I was unknowingly answering the question "Death by which shots is most probable?"

Was I right on this one?

Thanks..

]]>I could verify the propability of death at supershot. I too did the same way.

But your answer seems to be answering a different question.

Is there something else you need?

For the answer in the first post..i just figured that to reach 50% probability, there is a more than 50% probability..like

1/8 is probability to hit, so 7/8 is probability to miss and also to increase chance to 2/8

so 7/8 * 6/8 * 5/8 would be probability to get 4/8.

I solved the following few equations for consecutive hits

1. two consecutive shots 1st and 2nd ->

2. two consecutive shots 2nd and 3rd ->

3.

4. the probability reaches a max here out of the six cases that i tested.

5.

6. the probability has already started falling...needless to put in the sixth case

7. Super Shot :

probability at Super Shot: 7/8 ----- (1)

to reach super shot probability

7/8 * 6/8 * 5/8 * 4/8 * 3/8 * 2/8 ----- (2)

Hence total probability = (1) * (2) = 0.0168

I would be grateful if any loopholes be pointed out..i think there are some..

Thanks in advance!

]]>Oh, okay.

Hope you find that bug!

Thanks for providing that.

I do not know what is wrong with it. I wrote the routine myself so it must have a bug in it that did not show up till now.

]]>I saw your answer, but couldn't verify it. How did you solve?

The matrices are cumbersome to do without a CAS. I hope your method to be simpler!

Hi bobbym,

This is my Q matrix:

and R:

The first 7 rows are for single shot.

What happened to your software?

]]>My software went kablooey so I could not compute the answers. What did your transition matrix look like?

]]>