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 A game of coins (Posted on 2010-11-09)
Siddharth and Alok have 10 coins and 9 coins respectively. They are playing a game in which they throw up all their coins simultaneously and observe the number of coins which land heads up. The person with more heads wins. If there are no trick coins, what is the probability that Siddharth wins?
Additionally, calculate the probabity that Siddharth loses?

 No Solution Yet Submitted by Vishal Gupta No Rating

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 solution | Comment 1 of 4

There are 2^10 ways Siddharth's coins can land and 2^9 ways of Alok's coins landing. The number of ways a given number, r, of coins being heads out of n coins is C(n,r), each with a 1/2^n probability of happening. So the probabilities of given numbers of heads for the two contestants is:

`        Siddharth    Alok0       1/1024      1/5121       5/512       9/512         2       45/1024     9/128         3       15/128      21/128   4       105/512     63/256    5       63/256      63/256     6       105/512     21/128       7       15/128      9/128          8       45/1024     9/512           9       5/512       1/512           10      1/1024                        `

A grid could be made showing all the products of the probabilities of any given score for Alok and any given score for Siddharth, and add up those probabilities that make for an Alok win or a Siddharth win respectively, but to ease the calculation, the following program does this for us:

4   dim Sidd(10),Alok(9)
5   N=9
10   Allways=2^N
20   for R=0 to N
30    Alok(R)=combi(N,R)//Allways
40   next
105   N=10
110   Allways=2^N
120   for R=0 to N
130    Sidd(R)=combi(N,R)//Allways
140   next
200   for S=1 to 10
210     for A=0 to S-1
220       Stot=Stot+Sidd(S)*Alok(A)
230     next
240   next
250   print Stot,Stot/1
300   for A=1 to 9
310     for S=0 to A-1
320       Atot=Atot+Sidd(S)*Alok(A)
330     next
340   next
350   print Atot,Atot/1
400   print 1-Atot-Stot,(1-Atot-Stot)/1

It finds that Siddharth has a 1/2 probability of winning and Alok has 84883/262144 ~= 0.323802947998046875 probability of winning, leaving a 46189/262144 ~= 0.176197052001953125 probability of a tie.

 Posted by Charlie on 2010-11-09 15:26:53

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