Boring probability lesson, while I wait to go home and check todays post.* Lets imagine the odds for each entry are 1 in 5 (as is commonly suggested). On your first entry the calculation is simple. 1 in 5 = 0.2 or a 20% chance. On your second entry you don't add your "20%"s together to give 40% (because that would guarantee a place after 5 entries which is wrong). And you don't multiply the 0.2 values (because that would lower the probability which is wrong). We multiply probabilities when we want the probability of x **and** y happening. 0.2 * 0.2 gives the probability of getting a place in year 1 *and* in year 2 (4%). Instead you can think about it like this. What are the odds of **not** getting a place in year 1 **and** **not** getting a place in year 2? 4/5 * 4/5 = 16/25 = 64% . So the odds of getting a place at least once in either of the two years of entry are the opposite of that, 36% (because the probability of getting a place plus not getting a place has to add up to 100%). So after 1 entry your chance of getting a place is: 1-(4/5)=0.2=20% After 2 entries: 1-(4/5*4/5)=36% After 3 entries: 1-(4/5*4/5*4/5)=49% 4 entries: 59% 5 entries: 67% 6 entries: 74% 7 entries: 79% 8 entries: 83% 9 entries: 87% 10 entries: 89% So (again, assuming it is about 1 in 5 each time) after 10 entries, the chance that you were successful at least once is 89%. After 12 entries you had a 93% chance of being successful at least once, or a 7% chance of STILL not getting a place, about 1 in 15. But of course you're never guaranteed a place even after 100 entries. And, perhaps confusingly, even after 9 fails (or 99 fails), the chance on the 10th time (or the 100th time) is still 1 in 5 or 20% (like Sussex Runner NLR says). The unlikely thing has already happened by then (as Keith says). * although I bequeathed so I'll probably have to wait a bit longer... |