XIV APPENDIX THE FIRST. 



RULE. Divide 30 times x by 56 times the square of 

 a, and from the quotient subtract -1049. If the result 

 be the common logarithm of s, then ss 1 to 1 are the 

 odds in favour of the event. 



EXAMPLE. The number of stakes is 45, and the 

 play equal : what is the chance that one or other is 

 ruined before 1520 games have been played ? 



a = 45, oc = 1520, 56 x a x a = 1 13400 



45600 

 x x 30 = 45600, = -4021 



4021 -1049 = -2972 = logarithm of 1 -982. 

 Answer, -982 to 1, or 982 to 1000, 



PROBLEM. The stakes being equal, and also the play, 

 as before, what is the number of games in which it is n 

 to 1 that one party or the other will have been ruined ? 



RULE. To the common logarithm of one more than 

 n add *1049 : multiply the result by 56 times the square 

 of the number of stakes, and divide by 30, which gives 

 the number required, very nearly. 



EXAMPLE. Both parties have 50 stakes: in what num- 

 ber of games is it 10 to 1 that one or other will be ruined? 



a = 50, n= 10, n + 1 =11, log. 11 = 1-0414 

 1-0414 + -1049=1-1463, 56 x a x a = 140000 



160482 

 1-1463 x 140000= 160432: ^ = 5349 



Answer: in about 5349 games. 



To find the number of games in which it is an even 

 chance that one or other will be ruined, from three- 

 fourths of the square of the number of stakes, subtract 

 its hundredth part. Thus, if both parties have 40 

 stakes, then 40 x 40 being 1600, three- fourths of which 

 is 1200, from 1200 subtract 12, which gives 1188 for 

 the number of games (very nearly) in which it is an 

 even chance that one or other will be ruined. 



If a player with a stakes play with one of unlimited 

 means, the chances being the same for both, it is an 

 even chance that he is ruined in a number of games 



