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## A probability puzzle

“Colin and Dave are playing a game. Colin has a probability of 0.2 of hitting the target with any given shot; Dave has a probability of 0.3. Whoever hits the target first, wins. Colin goes first; what is his probability of winning?”

Yesterday I listened to the latest edition of “Wrong, but useful” (@wrongbutuseful), and the above is Is the puzzle set by the cohost Dave “The king of stats” Gale (@reflectivemaths).

It was pretty late on in the evening, but I decided to have a quick attempt at the puzzle nonetheless. Here you can see my back of the envelope workings, complete with the word “frustum” on the top of the envelope as I had added an extra r into it in my recent post. You will also see that approaching midnight on a Saturday after a day at the NTEN-RESEARCHED-YORK conference is not the most idea time for solving maths puzzles! My thinking was fairly valid, I think, but my tired brain has made an absolute ton of mistakes! (If you haven’t spotted them, go and have a look before reading on!)

When I looked at it this morning the first thing that jumped out at me was “0.14×0.028 does NOT equal 0.0364. Then I thought, “ffs, 10/43 is very definitely not 0.43!” This was closely followed by: “and why have you used 0.14 as r, the 0.2 is the probability he hits, if he hits the game is over, Cav you ploker!”

I think this is right now. (Although do feel free to correct me if you spot another error!) The probability Colin wins on the first go is 0.2, on his second go is the product of him missing (0.8) Dave missing (0.7) then him hitting (0.2) so 0.8×0.7×0.2. As the sequence goes on you are multiplying by 0.7 and 0.8 each time, so it gives rise to a geometric sequence with first term (a) as 0.2 and common ratio (r) as 0.56.

The total probably of Colin winning is the sum of the probabilities of him winning each time. This is because the probability he wins his the probability he wins on his first go OR his second go, OR his third etc. The game goes on until someone wins, so is potentially infinite, thus we need to sum the series infinitely.

As the series has a common ratio of 0.56, which is less than one, we can sum the series to infinity using s = a/(1-r) which gives 0.2/(1-0.56) = 0.2/0.44 = 5/11. Thus the probability Colin wins is 5/11 or 0.45recurring.

Categories: A Level, Maths, Teaching
1. May 5, 2014 at 1:53 am

Haven’t checked myself yet, but if your idea is right, then Dave’s probability MUST be 6/11. His infinite sequence should start with (0.8)*(0.3) = 0.24 and his ratio should be the same (right?) which is 0.56 so 0.24/(1 – 0.56) = 0.24 / 0.44 = 24/44 = 6/11
Well done!

2. May 5, 2014 at 6:00 am

Let p be the probability that Colin wins. Colin has a 0.2 chance of winning on the first attempt, but he can still win with probability p if both players fail on their first attempts. The probability is 0.8 * 0.7 = 0.56 that both players fail on their first attempts. Therefore p = 0.2 + 0.56 p, so p = 0.2/0.44 = 5/11.

3. May 19, 2015 at 9:51 am

Just found this through Twitter. I had a go before reading yours and agree. Will give it to my Further Mathematicians in their last lesson tomorrow! Thanks 🙂

1. May 7, 2014 at 11:56 am