# Coin flipping probability game ; 7 flips vs 8 flips

Your friend flips a coin 7 times and you flip a coin 8 times; the person who got the most tails wins. If you get an equal amount, your friend wins.

There is a 50% chance of you winning the game and a 50% chance of your friend winning.

How can I prove this? The way I see it, you get one more flip than your friend so you have a 50% chance of winning if there is a 50% chance of getting a tails.

I even wrote a little script to confirm this suspicion:

from random import choice

coin = ['H', 'T']

def flipCoin(count, side):
num = 0
for i in range(0, count):
if choice(coin) == side:
num += 1
return num

games = 0
wins = 0
plays = 88888

for i in range(0, plays):
you = flipCoin(8, 'T')
friend = flipCoin(7, 'T')
games += 1
if you > friend:
wins += 1

print('Games: ' + str(games) + ' Wins: ' + str(wins))
probability = wins/games * 100.0
print('Probability: ' + str(probability) + ' from ' + str(plays) + ' games.')


and as expected,

Games: 88888 Wins: 44603
Probability: 50.17887678876789 from 88888 games.


But how can I prove this?

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The "How can I prove this?" part at the beginning still doesn't make sense, since there is no "this" to begin with. Please don't add on a separate "EDIT" at the end; instead actually edit the post so that it makes sense on first reading. –  Prateek Jan 31 '14 at 4:41
This might be too late to say so since an answer got accepted in a rush (roughly 20 minutes after the question was posted, roughly 5 minutes after said answer was posted, did the OP really have the time to read it and understand it and digest it?), but anyway: the goal of the exercise is NOT to make one compute the 15-terms explicit sum and see that, as if by miracle, it evaluates to 0.5. What happens with the same question for (2013,2014) instead of (7,8)? Answers based on swapping all the results or on stopping the game before the last draw (both present on the page), each have more value. –  Did Jan 31 '14 at 7:37
Note that you have a problem with a trivial finite number of equally distributed outcomes, 32768 of them to be exact. In such a case you can iterate over each one of them and count the outcomes to calculate the exact probabilities. Unlike a Monte Carlo simulation it is a valid proof for the specific conditions. Though there is no formal way to extend it to a class of problems, and it doesn't provide much understanding, so a traditional proof is usually preferable. –  eBusiness Jan 31 '14 at 11:45
I just joined the exchange and wanted to add to the comment from eBusiness above, but my reputation isn't high enough yet. Interestingly enough, the method proposed by eBusiness is actually another way of looking at the rigorous proof of the accepted answer. The double-summation of the proof effectively iterates over all 32768 possible games and sums the resulting probabilities. Both very cool ways to look at it. –  user125204 Jan 31 '14 at 13:54
@Jason by the way, you don't need range(0, end). Python automatically starts the range generator at 0; you can just do range(end). –  rookie Jan 31 '14 at 14:47

The probability distribution of the number of tails flipped by you is binomial with parameters $n = 8$, and $p$, where we will take $p$ to be the probability of obtaining tails in a single flip. Then the random number of tails you flipped $Y$ has the probability mass function $$\Pr[Y = k] = \binom{8}{k} p^k (1-p)^{8-k}.$$ Similarly, the number of tails $F$ flipped by your friend is $$\Pr[F = k] = \binom{7}{k} p^k (1-p)^{7-k},$$ assuming that the coin you flip and the coin your friend flips have the same probability of tails.

Now, suppose we are interested in calculating $$\Pr[Y > F],$$ the probability that you get strictly more tails than your friend (and therefore, you win). An exact calculation would then require the evaluation of the sum \begin{align*} \Pr[Y > F] &= \sum_{k=0}^7 \sum_{j=k+1}^8 \Pr[Y = j \cap F = k] \\ &= \sum_{k=0}^7 \sum_{j=k+1}^8 \binom{8}{j} p^j (1-p)^{8-j} \binom{7}{k} p^k (1-p)^{7-k}, \end{align*} since your outcome $Y$ is independent of his outcome $F$. For such a small number of trials, this is not hard to compute: \begin{align*} \Pr[Y > F] &= p^{15}+7 p^{14} q+77 p^{13} q^2+203 p^{12} q^3+903 p^{11} q^4+1281 p^{10} q^5 \\ &+3115 p^9 q^6+2605 p^8 q^7+3830 p^7 q^8+1890 p^6 q^9+1722 p^5 q^{10} \\ &+462 p^4 q^{11}+252 p^3 q^{12}+28 p^2 q^{13}+8 p q^{14}, \end{align*} where $q = 1-p$. For $p = 1/2$--a fair coin--this is exactly $1/2$.

That seems surprising! But there is an intuitive interpretation. Think of your final toss as a tiebreaker, in the event that both of you got the same number of tails after 7 trials each. If you win the final toss with a tail, you win because your tail count is now strictly higher. If not, your tail count is still the same as his, and under the rules, he wins. But the chances of either outcome for the tiebreaker is the same.

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Can't the intuitive interpretation be used as the proof? –  Pacerier Jan 31 '14 at 3:37
Not to denigrate this excellent answer, but this is overkill on the level of using the General Number Field Sieve to find the prime factors of $8$. –  Newb Jan 31 '14 at 7:46
The lesson I tried to imply through my answer is that when we don't have an intuitive sense of what the solution should be, we can go through the problem-solving process in a more or less mechanical way--set up the model and solve it. If the answer is surprisingly simple, then that suggests there may be a simpler argument, one that does not require a flexible, general approach. It is not overkill for the sake of overkill. –  heropup Jan 31 '14 at 7:58
"Can't the intuitive interpretation be used as the proof? / It can be used as the basis for a formal proof, yes, but a formal proof requires some notation." FWIW, I disagree with this statement. (I seem to remember that Halmos wrote some rather well-known and quite eloquent paragraphs about (perfectly formal) proofs without (what you call) "notation".) –  Did Jan 31 '14 at 8:12
I would also be perfectly willing to accept the intuitive explanation as a proof. In fact, it's a strictly better proof than the one given, because it immediately generalizes to the case where your friend flips $n$ times and you flip $n+1$ times for $n\geq 0$. –  Chris Taylor Jan 31 '14 at 9:16

Well, let there be two players $A$ and $B$. Let them flip $7$ coins each. Whoever gets more tails wins, ties are discounted. It's obvious that both players have an equal probability of winning $p=0.5$.

Now let's extend this. As both players have equal probability of winning the first seven tosses, I think we can discard them and view the 8th toss as a tiebreaker. So let's give player $A$ the 8th toss: if he gets a tail, he wins, otherwise, he loses. So with $p = 0.5$, he will either win or lose this 8th toss. Putting it like this, we can see that the 8th toss for player $A$ is equivalent to giving both players another toss and discarding ties, so both players have winning probabilities of $0.5$.

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This is a highly underrated answer, both intuitive and perfectly rigorous. Very good. –  Did Jan 31 '14 at 8:13
This is a super easy way to think about it, and requires no calculations. I love that OP thinks you were just reading the question back to him. –  Kris Harper Jan 31 '14 at 14:17
This works, because $B$ wins even if she is just one tail ahead of $A$, since the best $A$ can do on his 8th toss, is getting equal, which still makes $A$ lose. That was the bit I needed to realize to make this solution click. –  SQB Jan 31 '14 at 14:28
Much better than the math-heavy answer, demonstrating reasoning and logic skills instead of just picking the right formula and plugging numbers in - which, if you don't have a good intuition, will be tough to check that you got the right answer. –  Claudiu Jan 31 '14 at 16:03
There is an important addition to make this answer work : there is only one situation after the first 7+7 tosses in which the last toss changes the situation, and it is when the 2 players are at a tie. Any situation where a player is winning after the 14 first tosses cannot be changed by the result of the last toss. –  njzk2 Jan 31 '14 at 18:54

Another way to think of it: By swapping all heads and tails, the winner is always reversed. For example, 2T 5H would beat 2T 6H, but 2H 5T would lose to 2H 6T. Or, 7H would beat 8H, but 7T would lose to 8T.

This gives a unique one-to-one mapping between the set where you win and the set where your friend wins. The sets are of equal size, so they must each have 50% probability.

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This is a highly underrated answer: the intuitive approach is clear and flawless. Very good. –  Newb Jan 31 '14 at 7:49
This argument is misleading. All you have established by your bijection is that the number of outcomes in which player 1 wins is the same as the number of outcomes in which player 2 wins. However, since you're describing your outcomes as the number of heads and tails, the outcomes are not equiprobable (3T4H is much more likely than 7T0H) so you cannot conclude that the players have equal probability of winning without further work. The key missing fact is that your bijection preserves probability (e.g., it maps 7T0H to 0T7H). This is true only because the coins are (assumed to be) fair. –  David Richerby Jan 31 '14 at 14:02
It's obvious that the bijection preserves probability, though, and of course all the other answers are wrong if there are unfair coins. –  hunter Jan 31 '14 at 19:35
@hunter It's obvious but it should be stated. "The sets are of equal size, so they must each have 50% probability" is not, in general true: equal probability does not follow from equal size. It's not "equal size therefore equal probability" but "equal size AND -- in this case, because of special properties that do not hold, in general -- equal probability". Yes, the other answers fail on biased coins but most have an explicit step that fails; this one has an implicit step that fails. –  David Richerby Feb 1 '14 at 1:13
@DavidRicherby: The bijection I'm imagining when I read this answer is not between things such as 3T4H and 4T3H, but between things such as THHTTHH and HTTHHTT. Those things are all equiprobable, so comparing sizes them makes sense. –  Henning Makholm Feb 1 '14 at 5:03

Consider the two cases:

1. The first seven tosses do not end in a tie. This decides the game because your eighth toss is irrelevant (your friend wins ties).

2. The first seven tosses end in a tie. Your eighth toss decides the game.

In each case, you have a $50\%$ chance of winning.

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+1. I think this is the cleanest of the several equivalent answers posted so far. –  Ilmari Karonen Jan 31 '14 at 11:50
This is a good answer. –  user1853181 Jan 31 '14 at 14:53
This was the best layman answer by far. –  Dss Feb 5 '14 at 15:45

Proof without any formulas or numbers: We just create a different game that will produce the exact same winner, and where it is obvious that you both have equal chances to win.

You and your friend each toss eight times. If one has more tails than the other, that one is the winner. If the number of tails are equal (four each), you remove your friends last toss; if you have more tails you are the winner, if the number of tails is equal he is the winner.

If a "tail" is removed, you are the winner. If "head" is removed, your friend is the winner. You had equal chances to be the winner after eight tosses, and if there was a draw after eight tosses, you have equal chances to end up the total winner as well.

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The probabilities have a binomial distribution, so we have $$\mathrm{Pr}[\text{friend gets } i \text{ tails}]=\binom{7}{i} 0.5^7$$ and $$\mathrm{Pr}[\text{I get } j \text{ tails}]=\binom{8}{j} 0.5^8.$$

Since these events are independent, we have \begin{align*} \mathrm{Pr}[\text{friend gets } i \text{ tails and I get } j \text{ tails}] &= \mathrm{Pr}[\text{friend gets } i \text{ tails}] \times \mathrm{Pr}[\text{I get } j \text{ tails}] \\ & = \binom{7}{i} \binom{8}{j} 0.5^{15}. \end{align*}

The possible $(i,j)$-values where I win are mutually exclusive, hence we find $$\mathrm{Pr}[\text{I get more heads than my friend}]=\sum_{i=0}^7 \sum_{j=i+1}^8 \binom{7}{i} \binom{8}{j} 0.5^{15}=0.5$$

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(I know the question already has an answer, but I enjoy the simplicity of this argument.)

Assume you flip simultaneously and label your throw by W if you get a tail when your friend gets a head. So to win you need to have at least 4W, but beware, if you get exactly 4W then they had to occur during the first 7 flips.

This gives a total number of winning combinations:

$${7\choose 4} + {8\choose 5} + {8\choose 6} + {8\choose 7} + {8\choose 8}$$

Now divide by the total number of combinations: $2^8$.

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Note (for clarity): also label your last throw a $W$ if you get a tail (your friend does not throw at all in this case). –  6005 Jan 31 '14 at 20:38

we can simplify the process by reducing the number of flips progressively 8,7..7,6..1,0 until we reach the extreme position of 1 toss versus no tosses. This by the rules is a 50/50 chance of a win for either player.

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Nice method, but harder to make rigorous than the one viewing the last coin as tie breaker. –  user21820 Nov 30 '14 at 6:04

Use a Markov chain.

For any outcome (n tails) after 7 flips with probability p(n), the eighth flip generates n tails with probability 0.5 p(n) and n+1 tails with probability 0.5 p(n).

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Here's an explicit proof that doesn't change the rules of the game:

A tosses 8 times, B tosses 7 times. If they get the same number of tails, B wins.

Consider the situation after each player has tossed 7 times. There's a probability $p$ that they've now got the same number of tails, i.e., that the score is tied. (One could calculate $p$, but it turns out that there's no need to.)

a) Case 1: The score is tied. In that case, A's 8th and final toss has a 50% chance of handing each of them the win.

b) Case 2: The score is not tied at this point. By symmetry, there's a 50% chance that A is ahead. Now note that A's 8th and final toss cannot change the outcome: If A was behind after seven rolls, the best A can achieve with another roll is a tie-- which still hands the game to B. Hence in case 2, each player has a 50% chance of winning the game.

Since in each case A has a 50% percent probability of winning, the over-all probability of a win is also 50%. (The over-all probability of a win is the weighted average of the two cases, which is just $0.5\cdot p + 0.5 \cdot (1-p)$, i.e. 0.5 regardless of $p$.)

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This reproduces an answer already posted. –  Did Feb 2 '14 at 1:52
@Did, which one do you mean? If I'm not adding anything new, I'll delete my answer. Many answers consider the point after 7+7-tosses, but I didn't spot any that are complete in both dividing into cases properly, and in discussing what the final toss can and cannot do. –  alexis Feb 2 '14 at 13:54
At least those by Newb and JimmyHe. –  Did Feb 2 '14 at 14:20
I'm surprised you think they're the same. "ties are discounted" changes the rules too much for my comfort, what can I say. JimmyHe's answer is very different: At a higher level, and more elegant, I must say-- but only if the core assertions are demonstrated, and I'm among those who don't find them self-evident. –  alexis Feb 2 '14 at 23:17

Another intermediate answer between the laymans and the mathematicians :-)

$P_7$ probability after $7$ flips
$P_8$ probability after $8$ flips
$me$ number of my tails
$you$ number of my friend's tails

$$P_8(win\_me) = P_7(me>you)+\frac{1}{2}^{\,*}P_7(me=you) = \frac{1}{2}P_7(me<>you)+\frac{1}{2}P_7(me=you) = \frac{1}{2}(1 - P_7(me=you))+\frac{1}{2}P_7(me=you) = \frac{1}{2}$$

($^*$) due to the extra flip

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Here's my attempt at a layman's explanation:

His I win a tie! advantage equals your I get one extra toss! advantage.

These advantages cancel each other out, and they cancel each other out no matter how many tosses there are.

After n tosses, you have three possibilities: he's ahead, you're ahead, or it's a tie

1) He's ahead. You've got an extra toss, but the best you could possibly do is a tie, and he wins ties. You've lost, he's won.

2) You're ahead. Game over, you won!

3) It's a tie. You get one more 50% chance

The odds of 1 happening is the same as 2 happening, because you've both had the same number of tosses.

And if 3 happens, there's only one extra toss, which you have a 50% chance of winning and a 50% chance of losing. The game goes on longer, but the odds are still half-and-half.

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I flip n+1 coins; my friend flips n. I only win if I get strictly more tails than my friend.

First, consider the number of tails I flip in my first n tosses as compared to the number of tails my friend flips in his n tosses. Let the event {flip the same number of tails in our first n tosses} have probability p. We do not need to calculate it explicitly.

By symmetry, the event {I flip strictly fewer tails in my first n tosses than my friend} has the same probability as the event {I flip strictly more tails in my first n tosses} and that probability is (1-p)/2.

Now I win if either a) we flip the same number of tails in the first n tosses, then I flip a tail on my (n+1)st toss, or b) I flip strictly more tails than my friend in the first n tosses. Any other possibility results in a loss for me.

Therefore P(I win) = p * (1/2) + (1-p)/2 = 1/2.

Drawing a diagram of the outcomes may help; the trick is to forget about computing the probability of a tie in the first 7 (or n) tosses; it drops out.

If you care, the probability of getting the same number of tails in n tosses between two players is (2n choose n)/(2^(2n)). It's a nice combinatorical argument.

Finally I apologize for the lack of good formatting.

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This reproduces an answer already posted. –  Did Jan 31 '14 at 9:29

consider an alternative game:

1. both toss 7 times
2. if one player has more tails he wins, otherwise the winner is chosen by an additional toss.

in this game it is obvious that there is a 50-50 chance for any of the players to win. Also if you think of your 8th toss as the additional toss in the first game, it is clear that the games are equivalent.

EDIT: I see that robjohn had the same Idea before me..

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Player A flips a coin 8 times, player B flips a coin 7 times.

"Either A throws more heads than B, or A throws more tails than B, but (since A has only one extra coin) not both. By symmetry, these two mutually exclusive possibilities occur with equal probability.

Therefore the probability that A obtains more heads than B is ½."

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Usually, an answer to a question shouldn't be solely a web link. Can you at least provide a summary of the solution? –  M Turgeon Jan 31 '14 at 20:39
You are right, that looks better now. –  Papagon Feb 1 '14 at 11:48

Look at it as 8 rounds, that you can either win or lose.

For the first 7 rounds:

Nobody wins if: You both flip the same.

On the 8th round:
You win if: you flip tails

The winner is whoever wins the most rounds.

The expected number of wins after the first 7 rounds is certainly the same as your opponent. And the expected number of wins on the 8th round is the same as your opponent. Therefore as you're both expected to win the same number of rounds, you have equal chance of winning.

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Here's a general proof that's similar to Jimmy He's, but uses a different principle to produce a bijection between wins and losses. It's less succinct and elegant than some other proofs here, but I think it reveals some interesting details about the way the extra toss affects the results. (Note also that this treats Heads as winning, rather than tails!)

First, player A tosses the coin n times; then player B tosses it n + 1 times. That makes a total of 2n + 1 tosses. Instead of treating these as separate turns, let's treat them as a partitioned sequence. For example:

\begin{align*} \mathrm{HTTHHTH} \mathrm{HHTHHTTH}\ -> \mathrm{HTTHHTH}\ \ |\ \ \mathrm{HHTHHTTH} \end{align*}

We can then determine the winner by asking whether there are strictly more heads after the partition than before. (In the above case, we have 4 and 5, and player B wins.) Now, what happens when we reverse the sequence? Crucially, there is an asymmetry: players swap all values except for the first one after the partition. I'll call this the asymmetrical value. Here are the relevant cases that cover all possibilities evenly.

1. The number of heads after the partition is greater than the number of heads before the partition by 2 or more.
2. There is one more head after the partition than before, and the asymmetrical value is a $\mathrm{T}$.
3. There is one more head after the partition than before, and the asymmetrical value is an $\mathrm{H}$.
4. There are an equal number of heads before and after the partition, and the asymmetrical value is a $\mathrm{T}$.
5. There are an equal number of heads before and after the partition, and the asymmetrical value is an $\mathrm{H}$.
6. The number of heads after the partition is strictly less than the number of heads before.

In cases 1, 2, 5, and 6, rotation always turns winners into losers, thus creating a bijection in those cases. In case 1, the asymmetrical value doesn't matter because player B's lead is so big; rotation always passes a winning count to player A. Likewise, in case 6, the asymmetrical value doesn't matter because player A's lead is so big.

In case 2, the asymmetrical value doesn't matter because it is a $\mathrm{T}$, so it doesn't affect the counts; player B's lead is transferred to player A. In case 5, the asymmetrical value does matter; it stays with player B upon rotation, and as a result, a loss for B (equal heads on both sides) turns into a win for B (two more heads after the partition than before).

Finally, we have cases 3 and 4. In these cases, we can't create a bijection directly because of the asymmetrical value. In case 3, player B has the lead, and maintains it after reversal because the asymmetrical value (an $\mathrm{H}$ in this case) doesn't change hands. In case 4, the counts are equal, making player A the winner; the asymmetrical value is a $\mathrm{T}$ so it doesn't affect the count upon reversal, and A still wins.

Here, instead of creating a bijection between reversals, we create a bijection between the two cases. In both cases, the counts on either side of the asymmetrical value are equal, so every case 3 sequence can be converted into a case 4 sequence by changing the asymmetrical value, and vice versa. When the asymmetrical value is a $\mathrm{T}$, A wins; when it's an $\mathrm{H}$, B wins.

That last move is similar to the "tiebreaker" move that other proofs have offered, but I think this approach shows how that tiebreaker works in a more detailed way.

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This is not a new answer but a mention of some generalizations and a defense of heropup's computational answer against the criticism it has received in the comments. I realize I'm bucking the tide by pointing out reasons to prefer the computational answer to the intuitive one.

The intuitive explanation relies on a numerical coincidence and does not generalize in a number of natural directions. To flesh out the intuitive approach, let $W$ be the event that I win, and let $T$ be the event that you and I are tied after seven flips. Then $$\Pr[W]=\Pr[T]\cdot\frac{1}{2}+(1-\Pr[T])\cdot\frac{1}{2}=\frac{1}{2}.$$ The numerical coincidence is the equality of the two factors of $\frac{1}{2}.$ The $\frac{1}{2}$ in the first term is the tails probability on the $8^\text{th}$ flip, while the $\frac{1}{2}$ in the second term is the conditional probability that I win the first seven flips given that they aren't tied, which, by symmetry, equals the conditional probability that you win the first seven flips. There is no logical connection between these two factors of $\frac{1}{2},$ but it is their accidental equality that makes the intuitive trick work. Specifically, the equality of the two factors results in the cancelation of $\Pr[T],$ which is unpleasant to compute.

The accidental nature of this trick is a problem: what if we were using coins with tails probability $p\ne\frac{1}{2}?$ In that case we have $$\Pr[W]=\Pr[T]\cdot p+(1-\Pr[T])\cdot\frac{1}{2},$$ and there is no cancelation of $\Pr[T].$ Or what if we use fair coins, but I have a two flip advantage? Let $A_{\ge1}$ be the event that I am ahead by one or more tails after seven flips, let $A_0$ be the event that we are tied after seven flips, and let $A_{-1}$ be the event that you are ahead by exactly one tail after seven flips. Then $$\Pr[W]=\Pr[A_{\ge1}]+\Pr[A_0]\cdot\frac{3}{4}+\Pr[A_{-1}]\cdot\frac{1}{4},$$ which admits to no major simplification. Admittedly, the original formulation does generalize nicely from $7$ and $8$ flips to $n$ and $n+1$ flips: the answer is always $\frac{1}{2}$ for any $n\ge0.$ The answer in the version where I have a two flip advantage, however, does depend on $n.$ It varies from $\frac{3}{4}$ when $n=0$ to $\frac{1}{2}$ as $n\to\infty.$ (An advantage of two, or any fixed number, of flips is negligible when $n$ is large.)

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Based on the answers so far you can get the following induction. If 2 players A and B play a game with these rules: most number of tails wins. Each player tosses the coin N times. We have P(Awin-Ntosses)=P(Bwin-Ntosses) = 50%. You tried to demonstrate that if player B gets an extra coin toss the odds of him winning will remain 50%. that would mean that P(Awin-Ntosses)=P(Bwin-N+1 tosses) If that is indeed true then using induction you can obtain P(Awin-Ntosses)=P(Bwin-N+x tosses). Induction is valid for any N or x. So therefore we can state that the following statement is true: Two players A and B play a game of toss. A tosses the coin 0 times and B tosses the coin 5 million times. The winner is the player that tosses most tails. Both players have the same 50% chance to win the game.

I strongly advise you to evaluate this statement and if you still consider it to be true, then it means you would be willing to play such a toss game for a small stake.

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Your premise, P(Awin-Ntosses)=P(Bwin-Ntosses) = 50%, is not the same as that in the question and the other answers. The other answers to not attempt to prove that P(Awin-Ntosses)=P(Bwin-Ntosses) = 50% => P(Awin-Ntosses)=P(Bwin-Mtosses) = 50% for all natural numbers N, M. –  Emil Lundberg Jan 31 '14 at 13:49
Your friend flips a coin 7 times and you flip a coin 8 times; the person who got the most tails wins. If you get an equal amount, your friend wins.There is a 50% chance of you winning the game and a 50% chance of your friend winning. How can I prove this? –  user3049400 Jan 31 '14 at 14:15
Yes. This statement is P(Awin-(N+1)tosses)=P(Bwin-Ntosses) = 50%, where B wins ties. It is not the same as what you are criticizing in your answer. –  Emil Lundberg Jan 31 '14 at 16:30
I'm only criticizing the many answers that state since P(aWin)= P(bWin)= 50% and since one has extra toss gives also the 50% chance to win the game. Truth is the chance remaining 50% is strongly dependant on the winning conditions (b wins if tied) and the number of tosses and the number of extra tosses. Another sweet example can be found in the risk game. Attacker throws 3 6-faced dice. Defender throws 2 dice. You take highes 2 dice from attacker and compare in pairs with the highest two for the defender. If equal or lower, attacker loses one army, otherwise attacker wins. –  user3049400 Feb 2 '14 at 8:52

Here's the long way:

Different combos for 8 heads: 2^8 = 256

• For 8 Tails: 1/256
• For 7 Tails: 8/256
• For 6 Tails: 7+6+5+4+3+2+1 → 28/256
• For 5 Tails: 6*2+5*4+4*6 → 56/256
• For 4 Tails: 5*2+4*4+3*3 + 4*2+3*4 + 3*2+2*2 + 2*2 + 1 → 1/70
• ..Symmetry for rest

Different combos for 7 heads: 2^7 = 128

• For 7 Tails: 1/128
• For 6 Tails: 7/128
• For 5 Tails: 6+5+4+3+2+1 = 21/128
• For 4 Tails: 5*2+4*4+3*3 = 35/128
• ..Symmetry for rest

Now, you can figure out the odds.

• Case 0 (1/128): Friend has 0 Tails → You win when you don't get 0 tails (1-1/256)→255/256
• Case 1 (7/128): You win (255-8)/256→247/256 times
• Case 2 (21/128): You win (247-28)/256→219/256 times
• Case 3 (35/128): You win (219-56)/256→163/256 times
• Case 4 (35/128): You win (163-70)/256→93/256 times
• Case 5 (21/128): You win (93-56)/256→37/256 times
• Case 6 (7/128): You win (37-28)/256→9/256 times
• Case 7 (1/128): You win (9-8)/256→1/256 times

If you add the probabilities of wins, you get:

((1/128)(255/256)) + ((7/128)(247/256)) + ((21/128)(219/256)) + ((35/128)(163/256)) + ((35/128)(93/256)) + ((21/128)(37/256)) + ((7/128)(9/256)) + ((1/128)(1/256))

Which equals .5, so your assumption is correct :)

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