# Probability of A staying ahead of B

Alice and Bob toss a coin 11 times. Heads are a point for Alice, tails are a point for Bob. After 11 tosses, Alice wins 7-4.

What is the probability that Alice was never behind in the score (ties allowed)?

This is not homework and I have an answer, but I would like an unbiased second opinion.

There are $\binom{11}{7}=\binom{11}{4}=330$ possible sequences.

There are $29$ sequences where Alice is not behind Bob:

• $1$ sequence after the $1$st toss ($1:0$)
• $2$ sequences after the $2$nd toss ($2:0,1:1$)
• $2$ sequences after the $3$rd toss ($3:0,2:1$)
• $3$ sequences after the $4$th toss ($4:0,3:1,2:2$)
• $3$ sequences after the $5$th toss ($5:0,4:1,3:2$)
• $4$ sequences after the $6$th toss ($6:0,5:1,4:2,3:3$)
• $4$ sequences after the $7$th toss ($7:0,6:1,5:2,4:3$)
• $4$ sequences after the $8$th toss ($7:1,6:2,5:3,4:4$)
• $3$ sequences after the $9$th toss ($7:2,6:3,5:4$)
• $2$ sequences after the $10$th toss ($7:3,6:4$)
• $1$ sequence after the $11$th toss ($7:4$)

Hence the probability that Alice is never behind Bob is $\frac{29}{330}$.