Money Billiards and Arithmetic

Interesting correspondence has appeared during the latter end of June in The Evening Standard concerning an elusive question in billiards arithmetic of the sort that invariably sets tongues a-wagging or pens a-writing. The correspondence, which is headed: "A Billiard Problem," was as set out below:—

At a certain billiard saloon, the price of billiards is 8d. per 100. The marker was one day playing a customer 100 up, and had a half-crown bet with him on the game, which the marker won. The customer then said he would play another 100, double or clears on the lot (table money included). The marker again won. A third 100 was played on the same conditions, and the marker won again. How much did the loser have to pay?

I have put the question to at least fifty persons, and not one has given the correct answer.—J. A. R.

• At the end of the first game the loser owed 3s. 2d.; at the end of the second game 6s. 4d., and 8d. for the game—7s. together; at the end of the third 14s., and 8d. the game 14s. 8d. together.—W. H. Bellamy.
• I make it the loser has to pay 15s. 2d.—W. Halford.
• According to my calculation the sum he should pay at the end of the third game is 16s. 8d.—E. L. S.
• I should say the player would owe £1 4s. 2d.—C. Rowe.
• I agree with none of the answers given, but make it 12s. —that is, 10s. on the bets and 2s. table money.—W. Archibald Boyes.
• I make it 16s. 8d. After the first game he owes 3s. 2d., in the second he has to play for this plus table money, so has to pay 3s. 10d. and 8d. for the second table, making him 7s. 8d. to the bad. In the third game he must play for 8s. 4d., he loses this and table money, making him 16s. 8d. down.—J. L. Scott.
• I take it the first game would be 2s. 6d. plus 8d.—3s. 2d.; second game, 5s. plus 1s. 4d.—6s. 4d.; third game, 10s. plus 2s. 8d.—total, 12s. 8d.—Fred Thompson.
• The answer is surely 14s.—i.e., 10s. for bets and 4s. the equivalent of the 2s. cost of table on the basis of double or quits. The player stands in the last game to pay 10s. or nothing for bets and 4s. or nothing for the table—loses and pays 14s. Is there any doubt? It seems simple.—A. H. H.
• The amount is 15s. 4d.; your correspondents can think it out, as so far they are all wrong.—W. V. W.
• I think the problem works out as follows—namely, first game, 8d. plus 2s. 6d.—3s. 2d.; second game, 8d. plus 2s. 6d. -3s. 2d.; third game, 6s. 4d. plus 8d. In other words, the marker won 13s. 4d.—J. 1. P.
• As the 8d. for each game was the liability of the customer, and went to the saloon and not to the marker, I calculate that the customer lost £1 7s. 6d. as follows:—First game, 2s. 6d., plus 8d.—3s. 2d.; second game, 5s. plus 1s. 4d. plus 1s. 4d.—7s. 8d.; third game, 10s. plus 2s. 8d. plus 2s. 8d. plus is. 4d.—16s. 6d.; total, £1 7s. 6d.—H. N. R.
• The billiard problem seems simple. The gentleman player loses 2s. 6d. on the first game; then, playing "double or quits," and again losing, his liability becomes 5s. Once more repeating the process, with like result, his final liability becomes. 10s., plus three times 8d. for table money (2s.); total, 12s. Of course, the bets and table money are separate propositions.—Wm. Potter.
• From reading the problem as it stands, I should say the customer would have to pay altogether £1 8s. 0d.—that is, 3s. 2d. for the first game, plus 6s. 4d. for the second (double, table money included), plus 19s. for the third game (double on the lot, table money included).—"Regular Reader."

At this point the propounder of the original question intervened with what he termed the correct answer (namely, 14s. 8d.), but this only served to re-start the discussion.

Thus J. L. Scott again writes:—

It is distinctly stated in the question that the customer played the second game "double or quits on the lot (table money included)." Surely this is meant to imply that had he won the game he would have been quits on the two games, table money included. To arrive at this he must pay double or quits the table money as well as double or quits his losses on the first game. The same thing applies to the third game, bringing the total to 16s. 8d.

J. A. R.'s solution simply allows for his losing 3s 2d. again on the second game plus 8d. table money, therefore had he won the second game he would have been 8d. to the bad on the meeting instead of "Quits on the lot, table money included" as I read the question.—J. L. Scott.