# Probability of getting more number of heads

Mercedes and Edmond are playing a game.They toss 25 coins and count the number of heads and tails.If the number of heads is more,mercedes wins,while if the number of tails is more,edmond wins.What is the probability that mercedes wins,given all coins are unbiased ?

I faced this question today in an aptitude exam.

My approach was like this !

Mercedes will win if number of heads is greater.The possibilities are 13-12,14-11,.....24-1,25-0...for mercedes and edmond correspondingly.

Sample space will 25 because there can be 25 possible splits in heads and tails.

Is the probability 13/25 ?

Can somebody help ? I think i'm 90% wrong ?

Thanks.

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You said the coins are biased? ... Can your answer involve a parameter $p$? – snarski Jan 27 '14 at 14:41
Very sorry for the inconvenience.All are unbiased. – vaidy_mit Jan 27 '14 at 14:42

Note that there are not $25,$ but $26$ possible outcomes, ranging from $0$ heads to $25$ heads. Since the coins are all unbiased, and Mercedes will win in exactly $13$ of the possible outcomes, and because the outcomes' probabilities are symmetrically distributed between the two players, then....

To be more rigorous, let's suppose that $H$ represents the number of heads in a given round. Then (for integers $0\le n\le 25$) there are $\binom{25}n$ ways to choose the $n$ coins that will be heads, and each coin has a probability of $0.5$ of being heads (tails), so the probability that $H=n$ is $$\Bbb P(H=n)=\binom{25}n\cdot0.5^n\cdot0.5^{25-n}=\binom{25}n\cdot0.5^{25}.$$ Mercedes will win if $13\le n\le 25$ and Edmond will win if $0\le n\le 12$ (i.e.: $13\le 25-n\le 25$). Since $\binom{25}{25-n}=\binom{25}n,$ then the conclusion readily follows.

Added: The important thing to keep in mind, here, is the symmetric distribution of winning outcomes between the two players. It is not so simple as simply taking $13$ out of $26$ distinguishable outcomes to get the probability that Mercedes wins, even though that gives the correct answer (in this case).

I mention distinguishable outcomes because there are (for example) $25$ different ways to get exactly one coin landing heads, but there may not be any way to tell such outcomes apart. If each coin was distinguishable from the rest (e.g.: had a different date, denomination, color, shape), then our different outcomes would all be equally likely, and we would only have to figure out the total number of such outcomes, and the number of those which gave Mercedes the win--$2^{25}$ and $2^{24},$ respectively, as it turns out. Unfortunately, we don't know how (or if) we may distinguish between such outcomes, and even if we did, it would be a ludicrous task to try to actually list them all (though counting them is not terribly difficult).

We need also keep in mind that the outcomes in this situation are not of equal probability! For example, while there are $25$ different (but indistinguishable) ways to flip exactly $1$ head, there is only $1$ way to flip all tails, so it is more likely that $1$ head will be flipped than no heads at all. This is ultimately why we can't simply take $13/26$ to get the answer. If the game had been such that Mercedes would win if there were $0,1,2,3,4,5,6,20,21,22,23,24,$ or $25$ heads, then despite the fact that she would win in exactly $13$ of the $26$ possible distinguishable outcomes, the outcomes' probabilities are not symmetrically distributed among the players. In fact, Edmond has more than a 99% chance of winning in such a scenario!

For a simpler example, suppose they were just flipping $3$ identical coins. Note that we have $1$ way for no heads to be flipped, $3$ ways for exactly one head to be flipped, $3$ ways for exactly two heads to be flipped, and $1$ way for three heads to be flipped. (Why?) This gives us $8$ total ways to flip the coins, even though there are only $4$ distinguishable outcomes. In particular, $$\Bbb P(H=0)=\frac18,\Bbb P(H=1)=\frac38,\Bbb P(H=2)=\frac38,\Bbb P(H=3)=\frac18.$$ Now, let's consider two possible games.

Game 1. If there are more heads than tails, Mercedes wins; otherwise, Edmond wins:

$$\Bbb P(\text{Mercedes wins})=\Bbb P(H=2)+\Bbb P(H=3)=\frac38+\frac18=\frac12.$$

Game 2. If there are only heads or only tails, Mercedes wins; otherwise, Edmond wins:

$$\Bbb P(\text{Mercedes wins})=\Bbb P(H=0)+\Bbb P(H=3)=\frac18+\frac18=\frac14.$$

In both cases, Mercedes won in two of the four distinguishable outcomes, but in the second case, she was given the two least likely outcomes, so was at a disadvantage from the start.

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So,1/2(13/26) is the answer ? – vaidy_mit Jan 27 '14 at 14:45
That's correct! – Cameron Buie Jan 27 '14 at 14:49
Although $\frac12=\frac{13}{26}$ is true, $\frac{13}{26}$ is a rather misleading way to put the answer, since the $26$ possible splits are far from being equally likely. It is just a basic symmetry observation that tells that $\frac12$ is the only possible correct answer. – Marc van Leeuwen Jan 27 '14 at 14:58
@Marc: My thanks. I should have drawn more attention to the fact that the outcomes were not of uniform probability in the first place. – Cameron Buie Jan 27 '14 at 15:03
The comparison of 13/26 possible outcomes and having a $< 1%$ chance of winning is a great one for building understanding, +1. – snarski Jan 27 '14 at 15:15

Since there is no chance of a tie, the answer -- $P(\text{M wins}) = 1/2$ -- can be found by symmetry: both $M$ and $E$ have an equal chance of winning.

\begin{gather*} 1 = P(\text{M wins}) + P(\text{M doesn't win}) \\ = P(\text{M wins}) + P(\text{E wins}) \\ = P(\text{M wins}) + P(\text{M wins}) \\ = 2P(\text{M wins}). \end{gather*}

Solve for $P(\text{M wins}).$

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WoW ! Cool answer :D Thanks :D – vaidy_mit Jan 27 '14 at 14:48