View Full Version : Tough nut (long)

12-19-2003, 07:25 AM
Hi all!

I have been thinking about this problem for a while, without reaching a solution. Maybe you all have some ideas?

Here in Sweden, we can gamble on something called v75. It's horseracing, and you should pick the winner in 7 races. You win if you can pick at least 5 winners (but 6 pays out more, and 7 the most). The total prizemoney is divided as follows:
7 races correct: 57%
6 races correct: 14%
5 races correct: 29%

The payout also depend on how many other rows share your number of correctly picked horses. This is unknown, but can be approximated since we know how how many other systems the horses are on.

You pick the horses on one cupon, and you can choose an arbitrary number of horses in each race (with a minimum of one). Each possible permutation of the horses you have selected is called a "row". For instance, if we choose horse 1 and 2 in all 7 races, we will end up with 2^7=128 rows. Each row costs 0.5 skr, so this system would cost about 64 skr (plus a small fee).

Let's assume we know the winning chance for all horses. Is this game "solveable"? How does one calculate the best possible EV? One obvious solution would be to calculate the probability of winning for each possible permutation, and compare that to the estimated payout. All these "rows" which had a positive EV would then be submitted on a separate cupon. A practical problem comes in here though. For each cupon submitted there is a small (yet significant) fee. This fee doesn't grow very much with each added row, but it is a few skr if you just submit one row on a cupon.

My guess is that since the payout is quite heavily biased to 5 winners, some "survival"-strategies would probably apply (similar to tournament poker).