538 riddler question about coin flipping game
I work with a few math PhDs who insist the answer to this question is 100%. Anyone think differently?
I flip a coin. If it’s heads, I’ve won the game. If it’s tails, then I have to flip again, now needing to get two heads in a row to win. If, on my second toss, I get another tails instead of a heads, then I now need three heads in a row to win. If, instead, I get a heads on my second toss (having flipped a tails on the first toss) then I still need to get a second heads to have two heads in a row and win, but if my next toss is a tails (having thus tossed tails-heads-tails), I now need to flip three heads in a row to win, and so on. The more tails you’ve tossed, the more heads in a row you’ll need to win this game.
I may flip a potentially infinite number of times, always needing to flip a series of N heads in a row to win, where N is T + 1 and T is the number of cumulative tails tossed. I win when I flip the required number of heads in a row.
What are my chances of winning this game? (A computer program could calculate the probability to any degree of precision, but is there a more elegant mathematical expression for the probability of winning?)
3 Replies
That’s a really fascinating probability setup — the deeper I think about it, the more it feels like a self-adjusting Markov chain where each tail increases the “winning streak threshold.” I hadn’t seen it explained with that level of recursive logic before, so thanks for sharing the reference!
While exploring these types of probabilistic coin flip models, I actually came across a few simulation tools that helped visualize how fast the required streak grows. One that worked well for me is
— it lets you flip virtually infinite coins and even track streaks of heads/tails in real time. It’s a nice way to experiment with patterns like this one hands-on.
It’s incredible how quickly the probability converges to near zero after a few tails — makes you appreciate how these “simple” games hide deep math behind them. Great post! 👏
Game selection is key 👍
Not offering an elegant solution but felt obliged to stop the "p=1" nonsense. You don't win 100% of the times.
PairTheBoard
