How to calculate Place v Win odds advantage

Hi, this is my first post here, and I am not an experienced gambler, so please be gentle with me. :)

I'm a motorsport fanatic and only bet on NASCAR and Formula 1.

I am trying to work out how to calculate if place odds are better value than win odds.


To use last Sunday's NASCAR race as an example, I wanted to bet on Juan Pablo Montoya and had these options;

Win @ 21/1
Top 3 @ 6.2/1
Top 4 @ 5.5/1 (must be combined with an each way win bet)
Top 5 @ 3.1/1


There are 43 starters in the field, of which 10 have a realistic chance of winning on merit and another 20 are good enough to pick up a win by being in the right place at the right time. So a total of 30 chances, if that is relevant here.

Any help with creating an Excel formula to work this out would be greatly appreciated.

Sorry to reply to myself, but maybe if I pose the question a bit more simply it could help.


Does a bet to win paying $22.00 make better or worse statistical sense than a Top 3 bet paying $7.20 in a field of 43?

you are probably running into a group of people unfamiliar with Nascar betting.

It doesnt seem much similar to horse racing, not that I would be up to date with that either.

Quote: odiousgambit

you are probably running into a group of people unfamiliar with Nascar betting.

It doesnt seem much similar to horse racing, not that I would be up to date with that either.

Thanks for the reply, and I think you might be right.

I did think it would be a fairly common question though, but have searched around quite a bit and can't get a lead on how to work it out.

Surely horse racing punters think about it. Unless the answer is just simply that the odds of filling 1 of 3 places is 1/3 of winning only? It can't be that simple can it?

I'd think to accurately answer this question you would need to have some sort of probabilities of winning for each driver included - as well a probabilities of each driver finishing in the top 3. In the most simple format, you would say each of the 30 drivers has an equal chance of winning.

Sorry, I know that doesn't help much....

Quote: ruascott

I'd think to accurately answer this question you would need to have some sort of probabilities of winning for each driver included - as well a probabilities of each driver finishing in the top 3. In the most simple format, you would say each of the 30 drivers has an equal chance of winning.

Sorry, I know that doesn't help much....

I do have a probability figure for each driver, to win.

Knowing how to re-calculate that to produce a top 3 probability would give me the exact answer I'm seeking. :)

Not sure it will help, but here is my ratings sheet for the top 20 in last weeks race.

WR = My rating. TO = True odds. AO = Actual Odds on offer. OR = Over Round. PROB = Win probability. STAKE = Kelly Bet or Lay amount recommended (lays in red).

In paramutuel horse racing, each bet, win, place, show, trifecta etc has its own betting pool. Thus the probability of, say to win, depends on the composition of the win pool. There is no connection, I believe, between the odds in one pool and the odds in another. It is possible for a horse to pay 100 to 1 to win but only pay 1 to 1 or less to show. I don't know if your nascar bookie operates pools this way.

Quote: matilda

In paramutuel horse racing, each bet, win, place, show, trifecta etc has its own betting pool. Thus the probability of, say to win, depends on the composition of the win pool. There is no connection, I believe, between the odds in one pool and the odds in another. It is possible for a horse to pay 100 to 1 to win but only pay 1 to 1 or less to show. I don't know if your nascar bookie operates pools this way.

Nah, I usually use Betfair.

Thanks for the replies. When I work out the answer I'll come back and post it.

Persistence pays off! I have the solution. (thanks to Ganchrow on the SBR Forum)

Which is the better value bet on a runner rated as a $17.00 chance to win, in a field of 30?

Win @ $21.00
Top 3 @ $6.20
Top 4 @ $5.50
Top 5 @ $4.10


First calculate the win probability from rated true odds ($17.00);

1/17 = 5.88235% chance of this selection winning the race. And conversely means that the other 29 runners each have a (1-5.88235%)/29 ≈ 3.24544% average probability of winning the race.

Then calculate the selection's probability of placing 2nd, conditioned on it not placing 1st;

5.88235% / (1 - 3.24544%) = 6.0797%

Then calculate the absolute probability of finishing 2nd, by multiplying by the probability of not finishing 1st.

6.0797% * (1 - 5.88235%) = 5.72204%

or to put it another way

2nd = 5.88235% / (1-3.24544%) * (1-5.88235%) = 5.72204%

Then calculate the probability of 3rd, 4th and 5th place finishes in the same manner, but multiply by (1 - sum of probability of higher finishes);

3rd = 5.88235% / (1-2*3.24544%) * (1-11.60439%) = 5.56068%
4th = 5.88235% / (1-3*3.24544%) * (1-17.16507%) = 5.39823%
5th = 5.88235% / (1-4*3.24544%) * (1-22.56330%) = 5.23465%

Then sum 1st+2nd+3rd etc to get the probability of a top 3, Top 4 and Top 5 finish, and multiply by the available odds to calculate the edge on each bet;

Win = 5.88235% * 21 - 1 = 23.5294%
Top 3 = 17.16507% * 6.2 - 1 = 6.42343%
Top 4 = 22.56330% * 5.5 - 1 = 24.09815%
Top 5 = 27.79795% * 4.1 - 1 = 13.97160%


So to answer the original question, A Top 4 bet was best value, followed by Win, Top 5 then Top 3.



Another interesting calculation I ran across doing this might help people making Head to Head bets in multi runner markets like this...

Using our selection above rated at $17.00 (5.88235%) as 'A' versus 'B' at $19.50 (5.12821%)

First, to get the probability that each runner will finish ahead of ANY different selected runner calculate;

Win Probability / (Win Probability + Any Other Runner probability)

For A, Any Other Runner prob = (1-5.88235%)/29 = 3.24544%
For B, Any Other Runner prob = (1-5.12821%)/29 = 3.27144%

Therefore...

Aprob = 5.88235% / (5.88235% + 3.24544%) = 64.44444%
Bprob = 5.12821% / (5.12821% + 3.27144%) = 61.05266%

And then we can calculate the Probability of 'A' finishing higher than 'B' like this;

(Aprob*(1-Bprob))/(Aprob*(1-Bprob)+(1-Aprob)*Bprob)

(.6444444*(1-.6105266))/(.6444444*(1-.6105266)+(1-.6444444)*.6105266) = 53.62315%


Meaning the Fair Odds for A winning the matchup are $1.86 (-116), and for B winning are $2.16 (+116)


Hope that helps someone else sometime. :)



That's not really the full answer I wanted though.

Next step in the project is to try to calculate for the different chances each other runner has of winning, rather than assuming the average applies for each of them.

This is beyond my regular Excel spreadsheet though. I need to write a program to step through each combination... which is really going to test my learning skills. :\

Any comments about if that effort is justified, or if the accuracy level using the above method should be "accurate enough" would be welcome.

As would any tips for a novice trying to create a program like this for the first time. (or comments on the validity of the calcs)