**From The Sunday Times, 28th April 1974** [link]

A party of six racegoers arrived on the course with precisely £6 each with which to speculate. There were six races on the card and six runners in each race, so they decided to hold sweepstakes among themselves on each of the six races, the stake being £1 per race per person.

The runners in each race were numbered 1, 2, 3, 4, 5 and 6, and each of the racegoers drew one of these numbers out of a hat. Each player’s number remained the same throughout the six races. There were thus, in effect, six separate sweepstakes, the holder of the winning number drawing £6 on each race. To add a little interest to the proceedings it was arranged that the winner on any one of the races would be permitted to buy (at cost price of £1) one additional chance in one or more of the races subsequent to his win, from one of the other players. Only a player who had not yet had a win could sell his chance.

At the conclusion of the events it transpired that three players had made a net profit; the holder of No. 1 who won £4, the holder of No. 5 who won £1, and the holder of No. 4. The holder of No. 2 lost £3, and the holder of No. 6 lost £6.

Three winning chances were sold, but none of these by the holders of Nos 1, 4 and 5. The holder of No. 1 did not have any transaction with the holders of Nos 2 and 3. There were no dead-heats and no number appeared in the winner’s frame in consecutive races.

What, in order, were the numbers of the six winning horses?

This puzzle was included in the book **The Sunday Times Book of Brain-Teasers: Book 2** (1981, edited by Victor Bryant and Ronald Postill). The puzzle text above is taken from the book.

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## Jim Randell 8:38 am

on24 November 2022 Permalink |If the total times are A, B, C, D, E, F. Then we derive:

From which we get 2 chains (shortest to longest time):

However we are told that A was in 3rd place, but we see there are at least 3 competitors (B, C, E) who finished in a shorter time, which would imply the best A could have placed was 4th.

So, assuming positions are ranked according to time, either the puzzle is flawed, or the goal of the race is to achieve the longest time, not the shortest.

In this latter case the finishing positions (longest to shortest time) are:

Solution:Dave won second prize.LikeLike