+299 Alice and Bob each have a coin. Suppose Alice flips hers 1000 times, and Bob flips his 999 times. What is the probability that the number of heads Alice flips will be greater than the number Bob flips?
+2864 According to MathStackExchange ---> Answered here
"Apparently the answer is 1/2 by an argument which looks like this, Since Alice tosses one more coin that Bob, it is impossible that they toss both the same number of heads and the same number of tails. So Alice tosses either more heads than Bob or more tails than Bob (but not both). Since the coins are fair, these events are equally likely by symmetry, so both events have probability 1/2."
EITHER BOB TOSSES MORE
or ALICE TOSSES MORE
so its 1/2
+2864 The solution in the link is not exactly like question you asked, so that's why I pulled out a quote that answers your question correctly.
Alice and Bob each have a coin. Suppose Alice flips hers 1000 times, and Bob flips his 999 times. What is the probability that the number of heads Alice flips will be greater than the number Bob flips?
When both Bob and Alice have flipped 999 times, say that each of them has flipped the same number of heads, and they're tied. Alice flips one more time. The probability that her coin will land heads (giving her more heads than Bob) is 1/2. So the answer to the problem is 1/2.
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