# Probability of 5 fair coin flips having strictly more heads than 4 fair coin flips

This question was previously asked in Probability of $5$ fair coin flips having strictly more heads than $4$ fair coin flips

I know that the answer is 0.5, but I'm not following OP's and the accepted answer's logic.

Using the notation that André Nicolas used, there is a probability, $$p$$, that player 1 or player 2 wins if each player has 4 tosses (this can be generalized to N tosses), so the probability of tying is thus $$q=1-2p$$. In case anyone's curious, for 4 tosses, $$p=0.36328125$$, and the chance of tying is $$q=0.2734375$$

Consider the situation where the players are tied after each player has tossed the coin 4 times. Say player 2 is the one who gets a 5th toss. Allowing a 5th toss will reduce the chance of a tie, and it will increase player 2's chance to win. This is intuitive and obvious.

I do not understand how the answer from the old post used the formula

$$p + \frac{1}{2}(1 - 2p) = \frac{1}{2}$$

Could someone explain to me what this is doing?

This to me this is saying "with the addition of the 5th toss, we are reducing the chance of tying by half $$\frac{1-2p}{2}$$ and increasing player 2's probability of winning by the same." But the chance of tying does not become cut in half with the addition of the 5th toss. The chance of tying is $$0.25390625$$, about $$93\%$$ that of $$q$$.

The post explained it.   $$1-2p$$ is the probability that both have the same amount of heads when they have tossed their coins the same count of times.   But $$A$$ is allowed to toss one more time than $$B$$.   The $$\tfrac 12$$ factors in the probability that this additional toss gives $$A$$ one more head, allowing a win.

Allow $$A$$ and $$B$$ to flip their coins $$n$$ times. Either one will be have more heads, or they will tie. Then $$A$$ is allowed to toss one more coin.

Let the probability that $$A$$ is has more heads be $$p$$, an unknown value. If $$A$$ does have more heads, this will still be so after one more toss (because $$B$$ cannot gain another head).

• The probability that $$A$$ gets ahead when both toss $$n$$ coins, and wins after one more toss is: $$p$$

By symmetry whatever value is $$p$$, it is also the probability that $$B$$ is has more heads.   If this does happen, then $$A$$ cannot win with one more toss (because, at best, a tie will occur).

• The probability that $$A$$ falls behind when both toss $$n$$ coins, and wins after one more toss is: $$0$$.

The probability that they are tied when both toss $$n$$ coins is $$1-2p$$, that is the probability that neither is ahead by then. If they are tied, then after $$A$$ throws one more toss, $$A$$ will have more heads with probability of $$1/2$$.

• The probability that $$A$$ and $$B$$ are tied when both toss $$n$$ coins, and $$A$$ wins after one more toss is $$\tfrac 12(1-2p)$$

Thus the Total Probability that $$A$$ obtains more heads after tossing one more coin that $$B$$ is $$\tfrac 12$$

$$p+\tfrac 12(1-2p)~~=~~\tfrac 12$$

Remark: This symmetry only holds when the coins are unbiassed.

• I'm still confused, but let me clarify something first. Define game1 to be where both players A&B have $n$ flips. Define game2 to be where player A has 1 more flip than player B. Is your $p$ the probability that player A will win in game1 or game 2? I assume it is in game1 because the probability of them tying in game2 isn't $1-2p$. – anonuser01 Mar 27 '20 at 1:49
• $p$ is the probability that $A$ will win game 1 (where A and B have the same number of tosses). As this is also the probability that B will win that game, then $1-2p$ is the probability that they will tie in that game. – Graham Kemp Mar 27 '20 at 1:53
• Let $X$ count the heads among the first $n$ tosses by $A$, $Y$ count the heads among the first $n$ tosses by $B$, and $Z$ count the heads among that extra toss by $A$. These are pairwise independent random variables. Let $p=\mathsf P(X>Y)$. Then $\mathsf P(X<Y)=p$ and $\mathsf P(X=Y)=1-2p$.\begin{align}\mathsf P(X+Z>Y)&=\mathsf P(Z=0)~\mathsf P(X+Z>Y\mid Z=0)+\mathsf P(Z=1)~\mathsf P(X+Z>Y\mid Z=1)\\[1ex]&=\mathsf P(Z=0)~\mathsf P(X>Y)+\mathsf P(Z=1)~\mathsf P(X\geq Y)\\[1ex]&=(\mathsf P(Z=0)+\mathsf P(Z=1))~\mathsf P(X>Y)+\mathsf P(Z=1)~\mathsf P(X=Y)\\[1ex]&=p+\tfrac 12~(1-2p)\end{align} – Graham Kemp Mar 27 '20 at 3:15
• No. That is correct. $\mathsf P(X>Y\cap (Z=0\cup Z=1))=p$ – Graham Kemp Mar 27 '20 at 3:37
• No. It is $p$. Just that. – Graham Kemp Mar 27 '20 at 3:48