What are the odds of backing .........

three winners in a row @ 4/1 each winner?

[Mark]

Not going to get trapped into what the odds/probablility/chances are, but the return is 125.

Excellent young man.

Now you really have to think beyond what is written in this question. (Not trying to be a smart a..se)

We have established that the return / odds of backing three 4/1 winners in a row is equal to 125/1. (125/1 is the return, 124/1 the odds if you ,like ..... not quibbling over 1 point).

My Q is. What are the chances of backing a 125/1 winner compared with the chance of backing three 4/1 winners in a row?

Please, I would love comments.

[KennyVictor]

The chances of betting on three 4 to 1 chances in a row and having them win, without doing any maths and just going on experience, would have to be about 20% on each or (coincidentally) 1 in 125. I haven't bet on a lot of 125 to 1 horses but I would feel lucky if one won in a couple of hundred bets. I'd say you'd be better betting on the 4 to 1 chances.

[AssumeTheCrown]

According to my data a 4/1 chance has an average win strike rate of around 17%(no other factors taken into account). This would mean that the probability of backing 3 consecutive winners would be .17x.17x.17 which is about 0.5% or 1 in 200(assuming independence between results).

My data suggests that 125/1 chances win with a probability of .0025 which equates to 1 in 400(very small sample however).

So in answering your question i would say that the former is more likely to happen than the latter remembering the assumption has been made that 4/1 chances win less than 1 in 5 and 125/1 chances win less than 1 in 126 times.