# If I have to win 3 more coin flips than my friend to win, but he has to win only 2 more than me to win, am I more likely to win?

My friend and I flip a coin when we have lunch to see who buys. My friend did not win a coin flip the whole month. I came up with this plan to give him a better than 50/50 chance of winning. That is based on the fact he only has to gain and advantage 2 for me to buy lunch but i have to gain an advantage of 3 for him to buy lunch. My friend says he really has no advantage. We do not have the ability to figure this out mathematically since our math knowledge ended with basic Algebra and that was 50 years ago when we were in high school. Yikes! Now We flip a coin until either my friend has correctly picked 2 decisions more than I OR I have correctly picked 3 decisions more than him. Here is an example with my friend winning, i.e. I buy lunch he picked correctly 2 more times than me before I could pick 3 more times correctly than him.Here is an example of the game:

Note: He always flips the coin and picks the side he wants.We both agree it has no bearing on the outcome who flips the coin or picks the side.

1st flip: My friend is correct - Friend is +1 2nd flip: My friend is incorrect - We are now even 3rd flip: My friend is incorrect - I am +1 4th flip: My friend is incorrect - I am +2 5th flip: My friend is correct - I am +1 6th flip: My friend is correct - we are now even 7th flip: My friend is correct - Friend is +1 8th flip: My friend is incorrect - We are now even 9th flip: My friend is correct - Friend is +1 10th flip: My friend is correct - Friend is + 2 and wins the game. I buy lunch.

My friends son mentioned something about this being a random walk and good luck figuring that out.

Thank you for your time. Cecil

• It's pretty difficult to understand your description of the problem, yet even more difficult to understand what you're asking. But I guess you want to hear that your friend is wrong. This really does give him advantage. About the random walk, yes, this is a random walk, but you don't need to know much about it to compute the probability of winning/losing. – Tunococ Mar 14 '14 at 4:11
• I appreciate your time to look at my question. He can win lunch with a minimum of two coin flips. I must have a minimum of three coin flips to win lunch (he picks wrong three straight times) Most of the time it takes us about 5 - 7 flips to determine the winner. So far the most number of flips till we determined a winner was 16. What do you approximate my friends advantage is as a %? Any idea? Seems like he would win 2times for my one??? – Cecil Hanson Mar 14 '14 at 4:41

It sounds like your approach is that you flip a coin and your friend has to win $2$ more than you have won to win. You have to win $3$ more than your friend to win. If the coin flip is random, it doesn't matter who picks what wins, so make it you win heads and your friend wins tails (first big idea). Now you are in a classic gambler's ruin situation (second big idea). You win $\frac 2{2+3}$, your friend wins $\frac 35$