BMAD 65 Posted March 26, 2017 Report Share Posted March 26, 2017 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? 1 Quote Link to post Share on other sites

1 Solution bonanova 85 Posted March 26, 2017 Solution Report Share Posted March 26, 2017 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. Quote Link to post Share on other sites

1 Donald Cartmill 3 Posted March 26, 2017 Report Share Posted March 26, 2017 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%. Quote Link to post Share on other sites

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## BMAD 65

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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