harey 8 Posted December 23, 2013 Report Share Posted December 23, 2013 Let's toss a fair coin and note the events. When H H H appears, I pay three bucks When T H H appears, you pay one buck. And then we start over. Who is willing to play? 1 Quote Link to post Share on other sites
0 Solution ThunderCloud 5 Posted December 24, 2013 Solution Report Share Posted December 24, 2013 The only way I win is if the first three flips are HHH. Once a T appears in the sequence, it becomes impossible for me to win, since at that point THH is guaranteed to occur prior to HHH. So overall, I have a 1/8 chance of winning $3 and a 7/8 chance of losing $1; not worth the gamble. Quote Link to post Share on other sites
0 witzar 18 Posted December 24, 2013 Report Share Posted December 24, 2013 If H appears on first three tosses I win, otherwise I lose. My EV in this game equals (1/8)*3$ - (7/8)*1$ = -0.5$. It's negative, so I'll pass. Quote Link to post Share on other sites
0 bonanova 85 Posted December 24, 2013 Report Share Posted December 24, 2013 Extra credit: for what p(H) is this a fair game? Quote Link to post Share on other sites
0 Nins_Leprechaun 0 Posted December 25, 2013 Report Share Posted December 25, 2013 when p(H)~=.63 this game is fair. There is a 1/4 chance I will win and a 3/4 chance I will loose. After an average 4 games I have won 3 dollars and lost 3 dollars netting me 0. Quote Link to post Share on other sites
0 harey 8 Posted December 25, 2013 Author Report Share Posted December 25, 2013 Now, I have a problem. witzar was the first to answer and his answer is correct. ThunderCloud answered 1 minute later and his explication is clearer. Can I attribute two Best Answers? Anyway, congratulations to both that they did not fell into the trap. Quote Link to post Share on other sites
0 ThunderCloud 5 Posted December 25, 2013 Report Share Posted December 25, 2013 Extra credit: for what p(H) is this a fair game? To be perfectly fair, I should win the game 25% of the time; that way on average I win as much as I lose. Since I win if and only if the first three flips are H, (p(H))^3 must be 1/4, so p(H) = 4^(-1/3). Quote Link to post Share on other sites
0 bonanova 85 Posted December 26, 2013 Report Share Posted December 26, 2013 Now, I have a problem. witzar was the first to answer and his answer is correct. ThunderCloud answered 1 minute later and his explication is clearer. Can I attribute two Best Answers? Anyway, congratulations to both that they did not fell into the trap. This happens from time to time. Only one can be marked. It's your choice. Best is not always first. Sometimes "showing your work" makes for an instructive as well as a correct answer. Quote Link to post Share on other sites
0 TimeSpaceLightForce 12 Posted December 26, 2013 Report Share Posted December 26, 2013 Interesting game.. toss regulation of high-low limits of toss,toss type, or surface nature shall be considered. For the player, it is his skill factor that is to be considered. Quote Link to post Share on other sites
0 superprismatic 11 Posted December 27, 2013 Report Share Posted December 27, 2013 This type of game is widely known as "Penney's Game". The Penney is Walter Penny who made quite a few puzzles in the 60's and 70's. I posted nearly 50 of his puzzles (from unpublished papers of his) a few years ago on this forum. You can look at some of them if you search for Penney on this site. They are quite clever puzzles! Quote Link to post Share on other sites
Question
harey 8
Let's toss a fair coin and note the events.
When H H H appears, I pay three bucks
When T H H appears, you pay one buck.
And then we start over.
Who is willing to play?
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