# Can you solve this (fun) math problem?

This one is just for fun—I saw it years ago on a very cool exam. You have a fair coin, meaning it’s equally likely to come up heads or tails. How can you design a game that involves only flipping the coin, that you have a one out of three chance of winning?

This isn’t a trick question. It’s possible to come up with many different honest solutions.

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## 29 Answers

You have to flip one side twice in a row?

Okay, I kinda just wrote the first idea that sprang to mind. I have work in the morning. Goodnight, Fluther.

You would have to flip 2 coins and only win when you get two heads. (Can you use two coins?)

Easy. It’s a three-sided coin.

I think it’s just one. :(

I swear I’m going to bed.

The game:

Each person flips the coin until it comes up tails. Count the number of heads in a row as that person’s score. If there is a tied score you repeat the process with the tied players until there is a definite winner.

You play with two other people.

okay I think I got it, what if player 1 gets heads-heads player 2 gets heads-tails player 3 gets tails tails then you flip the coin twice to see who won

@applesaucemanny

Who gets tails-heads? If it is player two then player two has a 50% chance of winning and players one and three have a 25% chance of winning.

Elsewise, first flip tails means that player three wins 50% of the time and players one and two each win 25% of the time (dividing the times it comes up heads).

I’m sorry, *fun* math question? Does not compute.

The starting probability is 0.5. To make the chances of winning 0.33… we need to multiply 0.5 by 2/3. Any form of the game that does this would give a viable answer.

I like @phoenyx reply of 3 players. No variation on the coins will give an appropriate answer, as no multiple of 0.5 approaches 0.33… Therefore we must have an external factor such as the number of players to satisfy this question.

Another possible answer is to use binary. Since 3 = 11, maybe we can make some way of saying heads = 1 and tails = 0 to make a factor of 3.

@phoenyx has a game that works. A very clever solution,, because it completely avoids having to do math. It’s just clear that every player has the same chance of winning.

@applesaucemanny—yours doesn’t quite work. In fact, it turns out that there can be no solutions that involve some fixed number of flips (note that @phoenyx‘s game could last arbitraily long)

There are still more solutions. There is one so simple, in fact, that I gasped when I heard it.

@elumas—you can use as many coins as you want.

carve an arrow into each side, then spin the coin between 3 people. Whoever the arrow points at wins

swallow the coin and drink 1/3 of a bottle of pepto bismol…see what happens

3 players flip the coin. Every time one player flips a heads or tails you drink!

Single player game:

You flip the coin twice. If it is heads both times, you win. If it is heads-tails or tails-heads you loose. If it is tails-tails you flip again in pairs until you win or loose.

Nice @phoenyx—that’s the incredibly simple one that made me gasp when someone told it to me.

Here’s another solution that a student of mine came up with the other day: Two people take turns flipping, and the first one to flip a head wins. The second person has a one third chance of winning. It’s a little trickier to see why that’s true though…

@finkelitis That doesn’t work, its still 1/4. Player 1 has 0.5:0.5 chance, and Player 2 has 0.5 x (0.5:0.5). 0.5×0.5 = 0.25.

No—they keep flipping until someone gets a heads. It might take thirty flips, for example.

Right. An easier way to see the math if you don’t know infinite series is to compare the player’s chances. Player one has a 1/2 chance of winning right away, and player two has a quarter chance of winning on their first flip (as you calculated @FireMadeFlesh ), calculating further for the later throws (as @phoenyx did), you notice that the first player always has twice as good a chance of winning as the second player. Since the odds must add up to 1, 2/3 and 1/3 must be their chances of winning.

@phoenyx That only works *if* there are multiple throws. To approach 1/3, there must be many throws, but probability tells us that the second player will likely get it first or second throw, at which point you are still stuck at 1/4 + 1/16 = 5/16.

Ow. 1/3 of my brain hurts.

@FireMadeFlesh

“Likely” is 5/16? That is pretty close to 5/15 or 1/3

;)

@phoenyx So it all comes down to limits of accuracy then huh? Still, the chance of the probability of winning approaching 1/3 is very low. Its a novel idea, but I tend to avoid such discrepancies. For round 2, you have a 2% error in your measured probability, which I guess is acceptable since we work on 5% (type 1, 20% type 2) in statistics.

Probability was never my best topic in maths, I could always find some line of reasoning to make me uneasy about how to go about it. Give me some good challenging calculus instead.

but if you make it a three player game and only one player can win, then it doesn’t matter what the other rules are. you have a 1 in 3 chance of winning, and so do the other 2 players. 1 will win, the other 2 will lose….

Well @alive, not all three player games are fair. But if it was a fair game, you’re right. That was @phoenyx ‘s first game, above.

@FiremadeFlesh, I think you’re bumping up to the idea of how to deal with infinite sequences (which often does come up in calculus!). But if you approach it by trying to get the exact answer, rather than a good estimate, it may be easier, paradoxically. You could check out the infinite sums from @phoenyx ‘s earlier post. Like: What is

1/4 + 1/16+ 1/64 + ... ?

@finkelitis I understand the infinite sums, but there is no point to them, since player 2 will win and the game will end within 2 rounds and therefore it is not an infinite sum, but a sum to 2 sequences.

Okay, so you had me a while back, I was just playing the devil’s advocate to clear up this point I’m uneasy about. Thanks guys.

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