Sequence/series coin flipping problem?

May 2021 3 41 Report

Sequence/series coin flipping problem?

In this particular coin-flipping contest, participants flip a fair coin until a head is flipped. The participant to first flip a head is the winner. This means, for example, if the participant who flips first gets a head on the first try, that participant wins without the second participant even having the opportunity to flip. If however, the first participant flips a tail, then the second participant may flip,getting either a head or tail. If you think about it, this game could go on forever.

1. Tom flips first. If you are given a mandate to win, who would you rather be, Tom or Ray? In particular,what is Tom’s exact chance of winning? And Ray’s too?

2. Is it possible to convert this unfair game to a fair game by converting the fair coin into an unfair coin. If so, what is the exact probability of getting a heads with the unfair coin? And the exact probability of getting tails? Or, if it is not possible, why not?

Science & Mathematics / Mathematics

Nice problem!

I will assume that both players are flipping the same coin. Let's generalize by letting p represent the probability of heads on a given flip of this coin. Clearly, for the game to eventually have a winner, we will need 0 < p <= 1 (if p = 0 there is never a winner).

Now let x represent the probability that the first player (Tom) wins, and condition on the result of the first player's (Tom's) first flip.

If this flip is heads, which occurs with probability p, then the first player wins automatically.

If this flip is tails, which occurs with probability 1-p, then the first player wins with probability 1-x because in that case the other player (Ray) will then get first opportunity to get heads.

Therefore, x = p(1) + (1 - p)(1 - x); x = 1 - (1 - p)(x); (2 - p)(x) = 1; x = 1/(2 - p).

So the first player (Tom) wins with probability 1/(2 - p). Then the second player (Ray) wins with probability (1 - p)/(2 - p).

Note that for 0 < p <= 1, 1/(2 - p) is always greater than 1/2, so the first player (Tom) will always have the advantage no matter how fair or unfair the coin is (assuming both players are flipping the same coin, and there is eventually a winner).

Specifically, if the coin is fair, the first player (Tom) wins with probability 1/(2 - (1/2)) = 2/3. Then the second player (Ray) wins with probability 1/3.

It makes perfect sense that Tom would have the advantage (assuming both players flip the same coin and there is eventually a winner). Think of it this way: let both players take turns. If they both get tails, they both take turns again. If exactly one of the two players gets heads, that player wins. However, if both players get heads, then Tom always wins that tie.

Have a blessed, wonderful day!

Mr. Smartypants

I'm not sure what you mean 'a mandate to win'.

Anyway, lets take just one round. Tom flips first and he has a 50/50 chance of winning right there. Then, only if Tom gets tails, Ray flips. So he had a 50/50 chance of even getting to play, and THEN a 50/50 chance of winning. So there's a 1/2 chance Tom will win, but a 1/4 chance Ray will win and a 1/4 chance NOBODY will win. So if I wanted to win I'd rather be Tom.

Now, an unfair coin. This is a theoretical construct. Let's say instead of flipping a coin you had a computer program that would pick heads or tails and it could be set to favor one or the other. For purposes of the discussion, let's also say Tom and Ray are both using the same lopsided odds.

Tom still has the advantage. Whatever the odds are of getting 'heads', Tom's win depends on only one test while Ray's win depends on two. So let's say it's 20% heads. Tom has a 20% chance of winning on the first coin flip, 20% overall. Ray now has an 80% chance of getting to play but then a 20% chance of winning. So his overall odds of winning are 0.8 * 0.2 or 0.16, much worse than Tom's.

OTOH Let's say the program is set to give 80% heads. Now Tom has 80% chance overall of winning. Ray has 20% chance of getting to play at all, and then 20% chance of winning. So his overall odds are 20% * 80%, in fact the same odds he had when the 'coin' was biased in the other direction!