Three backgammon players of equal game skills compete for a prize.
The prize will be awarded to the winner of two games in a row.
A and B, following a drawing, play the first game, then the winner will
face C.
Next game, if needed, will be by C and the player who lost the 1st
game and so on.
Determine a priori chances of winning for each of the 3 players, assuming
Lady Luck treats them without discrimination.
Rem: There are no draws in backgammon.
By the symmetry of the situation, A and B have equal likelihood of winning, with C having the remaining probability. So if we find the probability that C will win, we can split the remaining probability in two, for A and B.
In a given set of three games, there is a 1/2 probability that the winner of the first game (A or B) will also win the second game and therefore the prize. In the remaining 1/2 probability (that C wins the second game) there is the further, conditional, probability of 1/2 that C would also win the next one, for a probability of 1/4 of C winning on this round. In the 1/4 remaining probability of no winner thus far, there is a 1/2 conditional probability that the winner of the third game (who must be A or B, as C would have had to have won the second game and lost the third to proceed this far) will be the winner of the fourth game, which is the first game of the next round. So that's an additional 1/8 probability that it's not C, leaving 1/8 undecided.
The between-round difference (if the match continues) between the winner of the last game in the preceding round and the winner of the first game of the next round does not affect the calculations in the subsequent round, so the cycle continues from here.
So in any given round, C has a 1/4 probability of winning, and 1/8 that the match will continue to another round (counting the round to begin actually on the second game, as that's what determines the first of our win types).
In situations like this, the ultimate probabilities are proportional to the probabilities within one round. In this case, the within-round probabilities are 5/16, 5/16 and 4/16, or a ratio of 5:5:4. Therefore overall, A and B each have a probability of 5/14 of getting the prize and C has a probability of 4/14. These are approximately .3571428571428572 and .2857142857142857, respectively.
A simulation verifies these probabilities:
DECLARE SUB play3 (round!)
DIM SHARED s$, ct(3), tct
DO
play3 1
LOOP
SUB play3 (round)
STATIC pl$
SELECT CASE round MOD 3
CASE 1
pl$ = "ab"
CASE 2
pl$ = RIGHT$(s$, 1) + "c"
CASE 0
IF LEFT$(pl$, 1) = "a" THEN
pl$ = "b" + "c"
ELSE
pl$ = "a" + "c"
END IF
END SELECT
r = INT(RND(1) * 2 + 1)
w$ = MID$(pl$, r, 1)
IF round = 1 THEN prev$ = "": ELSE prev$ = RIGHT$(s$, 1)
s$ = s$ + w$
IF prev$ = w$ THEN
ct(INSTR("abc", w$)) = ct(INSTR("abc", w$)) + 1
tct = tct + 1
PRINT s$; TAB(15); ct(1); ct(2); ct(3), ct(1) / tct; ct(2) / tct; ct(3) / tct
s$ = ""
ELSE
play3 round + 1
END IF
END SUB
The last few result lines, when the program was interrupted, were:
this round's cumulative
winners a wins b wins c wins p(a) p(b) p(c)
------------ ------ ------ ------ ------------------------------
aa 969973 970264 776622 .35702 .3571271 .2858529
bb 969973 970265 776622 .3570199 .3571273 .2858528
bcc 969973 970265 776623 .3570197 .3571272 .2858531
acc 969973 970265 776624 .3570196 .3571271 .2858533
bcaa 969974 970265 776624 .3570198 .357127 .2858532
aa 969975 970265 776624 .3570201 .3571268 .2858531
aa 969976 970265 776624 .3570203 .3571267 .285853
aa 969977 970265 776624 .3570206 .3571266 .2858529
aa 969978 970265 776624 .3570208 .3571264 .2858528
acc 969978 970265 776625 .3570206 .3571263 .2858531
bb 969978 970266 776625 .3570205 .3571265 .2858529
bb 969978 970267 776625 .3570204 .3571268 .2858528
acbb 969978 970268 776625 .3570203 .357127 .2858527
bb 969978 970269 776625 .3570201 .3571272 .2858526
bcabcabcaa 969979 970269 776625 .3570204 .3571271 .2858525
aa 969980 970269 776625 .3570206 .357127 .2858524
bcc 969980 970269 776626 .3570205 .3571268 .2858527
acc 969980 970269 776627 .3570203 .3571267 .2858529
aa 969981 970269 776627 .3570206 .3571266 .2858528
bcc 969981 970269 776628 .3570204 .3571264 .2858531
aa 969982 970269 776628 .3570207 .3571263 .285853
bcabb 969982 970270 776628 .3570206 .3571266 .2858529
bcc 969982 970270 776629 .3570204 .3571264 .2858531
bb 969982 970271 776629 .3570203 .3571267 .2858531
bb 969982 970272 776629 .3570202 .3571269 .2858529
bb 969982 970273 776629 .35702 .3571271 .2858528
acbaa 969983 970273 776629 .3570203 .357127 .2858527
bb 969983 970274 776629 .3570201 .3571272 .2858526
acc 969983 970274 776630 .35702 .3571271 .2858529
acbb 969983 970275 776630 .3570199 .3571273 .2858528
bb 969983 970276 776630 .3570198 .3571276 .2858527
aa 969984 970276 776630 .35702 .3571275 .2858526
bcc 969984 970276 776631 .3570198 .3571273 .2858528
bb 969984 970277 776631 .3570197 .3571275 .2858527
acbb 969984 970278 776631 .3570196 .3571278 .2858526
bb 969984 970279 776631 .3570195 .357128 .2858525
bb 969984 970280 776631 .3570193 .3571283 .2858524
bcc 969984 970280 776632 .3570192 .3571281 .2858527
aa 969985 970280 776632 .3570194 .357128 .2858526
aa 969986 970280 776632 .3570197 .3571279 .2858525
aa 969987 970280 776632 .3570199 .3571277 .2858524
bcaa 969988 970280 776632 .3570201 .3571276 .2858523
bcabcaa 969989 970280 776632 .3570204 .3571275 .2858522
acbacc 969989 970280 776633 .3570202 .3571273 .2858524
|
|
Posted by Charlie
on 2010-05-06 13:46:30 |
solution
