Horse racing question probability Been thinking about this for a while.


*

*Horse Campaign length: 10 starts

*Horse Runs this campaign: 5

*Horse will is guaranteed to win 1 in 10 this campaign


Question: what is the Probability of winning at the sixth start if it hasn't won in the first five runs?
20%? 
Thanks for looking at this.
I would love to how to calculate this.
 A: it depends, do you mean that the horse is guaranteed to win AT LEAST one race, or that the horse is guaranteed to win ONE and ONLY ONE race. 
If it is one and only one race then the answer is simply 1/(number of remanning races) = 1/5 = 0.2 or 20% if you prefer. If you mean that the horse is guaranteed to win at least one then the problem becomes a bit more complicated and you will need to draw a probability tree
if you want to know about one or more wins more information is required as you need the probability that the horse will win each round.
A: It depends what you mean by  "Horse will is guaranteed to win 1 in 10 this campaign". 
For the sake of argument, let's imagine that you're watching a summary of the campaign after the event, but you have been told by a reliable friend that that particular horse only won once in its 10 starts. 
Then, while you're watching the documentary, if you say just before the horse's 6th race that this will be the one it wins - having seen it not win the first five - then there's a 20% chance that you're right, because there are 5 alternatives left for a known event to occur.
But this is stretching the concepts a little. 
A: I assume, that the horse will win only one race. And the probability of the (garanteed) win and a start at a race are equally like for every start/race.
The probability, that the horse has had no starts ($s_0$) is 
$\frac{{5 \choose 0}\cdot {5 \choose 5}}{{10 \choose 5}}$ and the probabilty that then the horse will win at the 6th start is 1 divided by the number of remaining starts. 
$P(w_6 \cap s_0)=P(s_0)\cdot P(w_6|s_0)=\frac{{5 \choose 0}\cdot {5 \choose 5}}{{10 \choose 5}}\cdot \frac{1}{5}$
The probability, that the horse has had one start ($s_1$) is 
$\frac{{5 \choose 1}\cdot {5 \choose 4}}{{10 \choose 5}}$ and the probabilty that then the horse will win at the 6th start is 1 divided by the number of remaining starts. 
$P(w_6 \cap s_1)=P(s_1)\cdot P(w_6|s_1)=\frac{{5 \choose 1}\cdot {5 \choose 4}}{{10 \choose 5}}\cdot \frac{1}{4}$
And so on ...
The probability of winning at the 6th start is $P(w_6)=P(w_6 \cap s_0)+P(w_6 \cap s_1)+P(w_6 \cap s_2)+P(w_6 \cap s_3)+P(w_6 \cap s_4)$
