All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Winning the Competition (Posted on 2009-01-25)
There are 3(A,B,C) participants participating in the competition. The first question is asked to A, next one to B and the next to C and this repeats. There are 3 ways in which an answer is scored, wrong guess: 1 point, partial answer: 2 points and correct answer: 3 points. The person who scores 4 points or above first wins the competition. The probability of scoring 1, 2 or 3 points for a question is same and also the same for every participant. Find the probability that A wins the competition if a question with a wrong guess or partial answer doesn't pass on to the others.

 See The Solution Submitted by Praneeth Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 solution Comment 3 of 3 |

Any given participant, if allowed to continue, would achieve four or more points by taking at least 2 turns and at most 4.  A's advantage is that he goes first and those who haven't achieved 4 when one person gets 4 don't get "last licks"--there are no ties.

Going back to the individual participant achieving the goal in 2 to 4 rounds:

`It can be achieved in 2 rounds by getting:1 and 3               prob 1/92 and either 2 or 3   prob 2/93 and anything        prob 1/3                      --------                      prob 2/3                      `

At the other end of the luck (or skill) scale is taking 4 turns to get to the goal. It requires getting 1 point in each of the first three rounds and anything in the fourth.

This has probability 1/27.

That leaves the probability of achieving the goal on the third round as 1 - 2/3 - 1/27 = 8/27.

Back to having three participants:

How can A win?

He can achieve the goal on round 2. This doesn't give any of the others a chance, so it has probability 2/3, the same as if he were playing alone.

He can achieve the goal in round 3 after neither of his opponents has succeeded in round 2. The probability that both opponents will have failed in round 2 is (1/3)^2 = 1/9. A's probability of 8/27 of hitting his goal on round 3 already includes the consideration of his having failed by round 2, so it's the only other number we need. The probability, in the overall scheme, that A will win against his competition at round 3 is (1/9)*(8/27) = 8/243.

The remaining possibility for A's win is that each of the three competitors takes the full four rounds to reach the goal. That has probability (1/27)^3 = 1/19683.

That makes A's overall probability of winning the competition 2/3 + 8/243 + 1/19683 = 13771/19683, a number that has a very long period in decimal representation:

0.69963928262968043489305492048976273941980389168317837727988619620992734847330
18340700096530000508052634252908601331097901742620535487476502565665802977188436
72204440380023370421175633795661230503480160544632423919118020626936950668089214
04257481075039374079154600416603160087385053091500279428948839099730732103845958
44129451811207641111619163745364019712442209012853731646598587613676776914088299
54783315551491134481532286744906772341614591271655743535030229131738048061779200
32515368592186150485190265711527714271198496164202611390540059950210841843214957
06955240562922318752222730274856475130823553320123964842757709698724787888025199
41065894426662602245592643397856017883452725702382766854646141340242849159172890
31143626479703297261596301376822638825382309607275313722501651171061321952954326
06818066351674033429863333841385967586241934664431235075953868820809835898999136
31052177005537773713356703754508967128994563836813493877965757252451353960270284
00142254737590814408372707412487933749936493420718386424833612762282172433064065
43717929177462785144540974444952497078697353045775542346187064979931920947010110
24742163288116648884824467814865620078240105674947924604989076868363562465071381
39511253365848701925519483818523599044861047604531829497535944723873393283544175
17654829040288573896255652085556063608189808464156886653457298176091043032058121
22135853274399227759995935578925976731189351216786059035716100187979474673576182
49250622364476959813036630594929634710155972158715642940608647055834984504394655
28628765940151399685007366763196667174719300919575267997764568409287202154143169
23233246964385510338871107046690037087842300462327897170146827211299090585784687
29360361733475588070924147741706040745821267083269826754051719758166946095615505
76639739877051262510796118477874307778285830412030686379108875679520398313265254
28034344358075496621449982218157801148198953411573439008281257938322410201696895
79840471472844586699182035258852817151856932378194380937865162830869278057206726
61687750850988162373621907229588985418889396941523141797490219986790631509424376
36539145455469186607732561093329268912259310064522684550119392369049433521312808
006909515825839556978102931463...

where the ellipsis indicates repetition starting again with the 699639 of the beginning.

Here are the results of 3 million trials (one million at a time, three times):

`A wins  B wins  C wins700155  225907   73938699798  225898   74304699766  225901   74333`

This agrees with the calculated probability for A, and also shows that B has about a 22.6% probability of winning and C has about a 7.4% probability.

Simulation program:

DEFDBL A-Z
RANDOMIZE TIMER
PRINT
FOR trial = 1 TO 1000000
REDIM pts(3)
DO
FOR i = 1 TO 3
pt = INT(3 * RND(1) + 1)
pts(i) = pts(i) + pt
IF pts(i) >= 4 THEN winCt(i) = winCt(i) + 1: EXIT DO
NEXT
LOOP
NEXT
FOR i = 1 TO 3
PRINT winCt(i);
NEXT
PRINT

 Posted by Charlie on 2009-01-25 19:47:58
Please log in:
 Login: Password: Remember me: Sign up! | Forgot password

 Search: Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information