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# Geometric series probability watch

1. Can anyone point me in the right direction please?

''I have football games, all of which are independent from one another. I can win (0.5), draw (0.2) or lose (0.3).

What is the probability of obtaining one win, followed by at least four draws.''

I'm not entirely sure where to start.
2. (Original post by Bugzy)
Can anyone point me in the right direction please?

''I have football games, all of which are independent from one another. I can win (0.5), draw (0.2) or lose (0.3).

What is the probability of obtaining one win, followed by at least four draws.''

I'm not entirely sure where to start.
Is it not just (0.5)x(0.2)^4?

(Probability of a win) x (Probability of 4 draws)

?
3. I wonder if that would work actually.

Then since as the value is 'at least four draws,' we would need to play about with a geometric series after, since r will converge to some point.
4. any other thoughts?
5. I don't think you need to sum the series; you have at least 4 draws if and only if none of the first 4 games following the initial win are be either a win or a loss.
6. (Original post by DFranklin)
I don't think you need to sum the series; you have at least 4 draws if and only if none of the first 4 games following the initial win are be either a win or a loss.
The clue given in the question was an application of a/(1-r) hence why I'm curious as if they want me to sum the series? R is less than 1 (0.2) so can we not assess the point to which the probability eventually converges to and then manipulate/take that and use it?

If I'm honest - I don't know exactly what the question wants me to do. I've worked with 'at least' questions for binomial etc. but never with this kind of event.
7. You can do it either way. To be honest, I'd probably sum the series myself, even though it's arguably less 'elegant'.
8. (Original post by DFranklin)
You can do it either way. To be honest, I'd probably sum the series myself, even though it's arguably less 'elegant'.
Sorry to keep coming back to this - if I were to use a geometric method, how would I go about it?

When I use a/(1-r), is my a going to be the probability of my first draw or of the win that precedes the draws? I'm guessing I sum the series for all the draws, using r as 0.2 and then multiply that by the probability of winning the first game that sets it off?

Or multiply the a/(1-r) result by the probability of one win & four draws - since this satisfies the criteria of 'at least.'

Scallym is right. Calculating P(win, draw, draw, draw, draw) covers all the cases where you have at least 4 draws (basically, whatever happens after that sequence, you know you have at least 4 draws).

I was thinking "calculate P(exactly 4 draws) + P(exactly 5 draws) + ...", but that is going to be much much more work.

Apologies again.

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