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BMAD

More heads?

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BMAD    62

Player A has one more coin than player B.  Both players throw all of their coins simultaneously and observe the number that come up heads.  Assuming all the coins are fair, what is the probability that A obtains more heads than B?

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2 answers to this question

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bonanova    76
Spoiler

Let B have n coins and let A have n+1 coins.

Keep in mind that coin tosses are independent events. Thus, tossing them simultaneously is not material to the outcome.

Suppose A keeps one of his coins in his hand while all the other 2n coins are tossed simultaneously. By symmetry, the expected number of heads tossed by B will equal the expected number of heads tossed by A. Now A tosses his remaining (n+1)st coin. It will be heads with probability of 1/2.

The probability that A obtains more heads than B (obtains) is 1/2.

 

 

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50% chance of more heads by (A) ;           " X " equal  number of coins by (A) and (B)  are 50/50 ,  Therefore the results of  a single coin toss is the answer ,which is  50%.                  

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