# Two men for one gold coin

Two men find an old gold coin and want to have a coin toss with it to decide who gets it. The only problem is the coin is heavier on one side so it comes up heads more than tails. What is a fair way for the men to toss the coin and decide who gets the coin?

• take a modern coin and use it to decide :) – Novarg Mar 2 '15 at 17:33
• too many possible answers in my opinion. you should add restrictions such as "minimize the number of tosses", and "no skill-based solution". – user3453281 Mar 2 '15 at 19:25
• Do they know which side is favored? If not, it seems like just flipping it once would be "fair." – wchargin Mar 3 '15 at 4:15
• @WChargin They don't know which side is favored (they can easily discovered it by flipping the coin multiple times), but flipping it once would not be "fair" because there isn't 100% chance that the men who choose the heavier side wins. ;) – Pierre-Arthur Ferraro Mar 3 '15 at 9:43
• @Pierre-ArthurFerraro Right, but if they don't know how the coin is favored, then no one can force an advantage on the other. – wchargin Mar 3 '15 at 15:10

They could toss it twice

• HT = first player wins; stop
• TH = second player wins; stop
• HH or TT = ignore these two tosses; repeat the procedure
• Just to be clear, for this to work they need to discard the previous results if they repeat the procedure. In other words, if they get HH then TH, the second player wins even though HT shows up first in HHTH. – Rob Watts Mar 2 '15 at 18:24
• What if the probability distribution is very asymmetric, say 1/99. Then the average number of needed tosses with this method can be very high. – fibonatic Mar 2 '15 at 21:04
• @RobWatts Yes, I'm pretty much sure that's what Gamow meant. – JLRishe Mar 3 '15 at 6:41
• @JLRishe I agree, but I felt like people who see the answer might not catch that. It's important enough that I wanted to call attention to it. – Rob Watts Mar 3 '15 at 8:00
• Can you show mathematics behind this? I want to know if the probabilities of them winning are equal theoretically. – Bhaskar Mar 3 '15 at 10:27

The answer is to simply flip the coin once, but make sure you catch it without letting it bounce.

The question makes the assertion that a weighted coin will be biased, but that only happens if the coin is allowed to bounce and/or spin (on a surface). For a flipped/caught coin, there is no significant bias.

The law of conservation of angular momentum tells us that once the coin is in the air, it spins at a nearly constant rate (slowing down very slightly due to air resistance). At any rate of spin, it spends half the time with heads facing up and half the time with heads facing down, so when it lands, the two sides are equally likely (with minor corrections due to the nonzero thickness of the edge of the coin).

[This] explains why weighting the coin has no effect here (unless, of course, the coin is so light that it floats like a feather): a lopsided coin spins around an axis that passes through its center of gravity, and although the axis does not go through the geometrical center of the coin, there is no difference in the way the biased and symmetric coins spin about their axes.

• The question says it comes up more heads than tails. Even though it doesn't say how it happened but it happens. I understand the question as if a coin comes up more heads than tails, how can you do you fairly, regardless of why it comes up heads or tails – Huangism Mar 2 '15 at 21:28
• @Huangism Sure, that's one interpretation. I think my real world answer is also valid. It's fair, and simple since it minimizes flip count. – Set Big O Mar 2 '15 at 21:53

Each person tosses the coin 3 times, whoever gets more heads wins, if tie then do it again. If ties keep occurring then toss it 5 times or more to decide

OR

each person tosses it once whoever gets tail wins. If tie or no one gets tail then each person tosses it again until someone wins.

Each man "tosses" the coin. Whoever "tossed" it the farthest (or closest to a fixed point) gets to keep it (assuming they don't lose it in the process). This will eliminate any bias introduced by the coin.

NOTE: When I initially posted my answer it was tagged as a 'riddle'
so I was playing off the semantics of 'coin toss' to literally
mean 'to toss a coin'. If that absolutely bothers you that is
fine, but if you are going to down vote at least be courteous

• Might makes right? – Paul Rowe Mar 2 '15 at 19:12
• @PaulRowe I believe that the question asks for "a fair way for the men to toss the coin and decide who gets the coin" and if the two men were to agree that whoever tossed it further got to keep it, then that would make it a fair way to do so. Would it not? – Warlord 099 Mar 2 '15 at 19:38
• Certainly, if both participants agree that such a method is fair. Another scenario would award the coin based on the accuracy of the "toss". – Paul Rowe Mar 2 '15 at 20:46

They play rock paper scissors with the coin, where heads is rock, tails is paper and somebody ran away with the scissors.

A game consists of each of them tossing the coin. The first one to get tails when the other gets heads wins.

oooh we could use the same rules as NHL shoot outs! in hockey, the goaltender has a statistical advantage over the shooting player, but since each team takes turn shooting agaisnt the opponent's goalie, the overall result of the shoot out is unaffected by the statistical advantage of the goalie.

Each team names three shooters. If the game remains tied after the three shooters are done, teams continue shooting in "sudden death" mode. The game cannot end until each team has taken the same number of shots.

so, let's say that Heads = save and Tails = goal. the players simply take turn flipping the coin as if they were playing hockey shoot outs.

• Neat analogy. Do you mean that they keep tossing it for a best of three till one of them wins, switching sides revert time? – user230452 Mar 20 '16 at 11:31