GAMING. 



-f 1. Now since A could give B 2 out 

 of 3, A might undertake to win three 

 rows running ; and, consequently, the 

 chances in this case will be as z3 to 

 3 z a -f 3 + 1. Hence, z3 = 3 z 1 + 

 3 z + 1 ; or, 2 z3 = z3 -f- 3 z 2 -f 3 z 

 -+- 1. And, therefore, z ^/ 2 = z + 1; 



and, consequently, z = - -. The 



chances therefore, are 



and 1, 



respectively. 



Again, suppose I have two wagers de- 

 pending, in the first of which I have 3 to 

 2 the best of the lay, and in the second, 

 7 to 4, what is the probability I win both 

 wagers ? 



1. The probability of winning the first 

 is ^, that is, the number of chances I 

 have to win divided by the number of all 

 the chances : the probability of winning 

 the second is T T_ . therefore', multiplying 

 these two fractions together, the product 

 will be II , which is the probability of 

 winning both wagers. Now, this fraction 

 being subtracted from 1, the remainder is 

 .| which is the probability I do not win 

 both wagers: therefore the odds against 

 me are 34 to 21. 



2. If I would know what the probability 

 is of winning the first, and losing the se- 

 cond, I argue thus : the probability of 

 winning the first is |, the probability of 

 losing the second is_4_: therefore, mul- 

 tiplying I by ^-, the product .| will be 

 the probability of my winning the first, 

 and losing the second ; which being sub- 

 tracted from 1, there will remain ^1, 

 which is the probability I do not win 

 the first, and at the same time lose the 

 second. 



3. If I would know what the probability 

 is of winning the second, and at the same 

 time losing the first, I say thus : the pro- 

 bability of winning the second is JL ; 

 the probability of losing the first is | ; 

 therefore, multiplying these two fractions 

 together, the product ^ is the proba- 

 bility I win the second, and also lose the 

 first. 



4. If I would know what the proba- 

 bility is of losing both wagers, I say, 

 the probability of losing the first is *L 

 and the probability of losing the second 



* ; therefore, the probability of los- 

 ing them both is ^ ; which being sub- 

 tracted from 1, there remains 47 . there- 



fore, the odds of losing both wagers is 47 

 to 8. 



This way of reasoning is applicable to 

 the happening or failing of any events 

 that may fall under consideration. Thus, 

 if I would know what the probability is 

 of missing an ace four times together with 

 a die, this I consider as the failing of 

 four different events. Now the proba- 

 bility of missing the first is *, the se- 

 cond is also *, the third |, and the 

 fourth |; therefore the probability of 

 missing it four times together is 1 x | X 

 | X f = j-$ s ? which being subtract- 

 ed from 1, there will remain -S-JUg- for 

 the probability of throwing it once or of- 

 tener in four times; therefore the odds 

 of throwing an ace in four times, is 671 to 

 625. 



But if the flinging of an ace was under- 

 taken in three times, the probability 

 of missing it three times would be * X 

 I X | = if | : which being subtract- 

 ed from 1, there will remain $* for 

 the probability of throwing it once or 

 oftener in three times; therefore the 

 odds against throwing it in three times 

 are 125 to 91. Again, suppose we would 

 know the probability of throwing an ace 

 once in four times, and no more : since 

 the probability of throwing it the first 

 time is i, and of missing it the other 

 three times is % X % X , it follows that 

 the probability of throwing it the first 

 time, and missing it the other three suc- 

 cessive times, is i-xXX = T ^; 

 but because it is possible to hit it every 

 throw as well as the first, it follows, that 

 the probability of throwing it once in 

 four throws, and missing the other three, 

 . 4 X 125 500 , . , . . 



-1296 = 1296 ;whlch bein S Sub - 

 tracted from 1, there will remain -Z-96 

 for the probability of throwing it once, 

 and no more, in four times: therefore, 

 if one undertake to throw an ace once, 

 and no more, in four times, he has 500 

 to 796 the worst of the lay, or 5 to 8 very 

 near. 



Suppose two events are such, that one 

 of them has twice as many chances to 

 come up as the other, what is the proba- 

 bility that the event, which has the great- 

 er number of chances to come up, does 

 not happen twice before the other hap- 

 pens once, which is the case of flinging 

 7 with two dice before 4 once? Since 

 the number of chances are as 2 to 1, the 



