Which payoff do you want to go for?

This is a trick question, both payoffs have same expected value of profit, which is: $1.

Consider this:

We will count the number of ways HT can occur. We can have:

HT_ _ _ : 8 ways

_ HT _ _ _ : 8 ways

_ _ HT _ : 8 ways

_ _ _ HT : 8 ways

Total number of ways: 32

Similarly for HH, the total number of ways in which it can occur is 32.

What this means is that the sum of the payoffs of HT and HH over the 32 possible outcomes is $32 for both. The payoffs may be distributed in different ways for HT and HH, but their expected value is same.

Mathematically, the expected value is same. But there is one more thing left to do: analyze the probability distribution. We can easily see that payoff 2 might be desirable because it tends to give large profits as compared to payoff 1 for certain events. eg: HHHHH gives $4 in p2 and 0 in p1. HHHHT gives $3 in p2 and $1 in p1.

But the thing is that these events are not very common. Using method of reflection and analyzing the 16 possibilities , we see that there are many events for which p2 gives $0 payoff. There are 13 outcomes for which p2 gives $0 payoff, but only 6 where p1 gives $0 payoff.

So If I'm given this choice, I'd take payoff 1, as then I have more chance of leaving with at least a dollar in my pocket.

Since this is in quantitative finance, should I talk about risk and stuff? I do not know.

Let $X$ be an indicator variable which takes on value 1 for each $HT$ pair and 0 otherwise, and let $Y$ be an indicator variable which takes on value 1 for each $HH$ pair and 0 otherwise. There are $5 - 2 + 1 = 4$ indicator variables (both $X$ and $Y$) in 5 coin tosses. The expected value of the payoff is: $\text{E}\left [\sum_{k = 1}^{4} X_{k} \right ], ~ \text{E}\left [\sum_{k = 1}^{4} Y_{k} \right ].$We can apply Linearity of Expectation here, which holds even for dependent events (nice proof of this is given in Brilliant Wiki Page), to get: $\sum_{k = 1}^{4}\text{E}\left [ X_{k} \right ] , ~ \sum_{k = 1}^{4}\text{E}\left [ Y_{k} \right ].$Expected value of both $X_{k}$ and $Y_{k}$ is: $\text{E}\left [ X_{k} \right ] = \text{E}\left [ Y_{k} \right ] = 1 \cdot \dfrac{1}{4} + 0 \cdot \dfrac{3}{4} = \dfrac{1}{4}.$It follows that expected payoffs in both cases are the same and they equal:$\sum_{k = 1}^{4}\text{E}\left [ X_{k} \right ] = \sum_{k = 1}^{4}\text{E}\left [ Y_{k} \right ] = 4 \cdot \dfrac{1}{4} = 1.$

Bonus: What about the variance of the payoff? It can also influence our decision. Is it also the same?

It turns out they are not! Let us first consider payoff 1 and indicator variable $X$:$\begin{aligned} \text{Var}[X] &= \text{E}\left [\left(\sum_{k = 1}^{4}X_{k}\right)^{2}\right ] - \text{E}\left [\sum_{k = 1}^{4}X_{k}\right ]^{2} \\ &= \text{E}\left [\left(\sum_{k = 1}^{4}X_{k}^{2}\right) + \left(\sum_{k \neq j}X_{k}X_{j}\right)\right ] - \text{E}\left [\sum_{k = 1}^{4}X_{k}\right ]^{2} \\ &= \text{E}\left [\left(\sum_{k = 1}^{4}X_{k}^{2}\right)\right] + \text{E}\left[ \left(\sum_{k \neq j}X_{k}X_{j}\right)\right ] - \text{E}\left [\sum_{k = 1}^{4}X_{k}\right ]^{2} \\ &= 1 + \text{E}\left[ \left(\sum_{k \neq j}X_{k}X_{j}\right)\right ] - 1 \\ &= \text{E}\left[ \left(\sum_{k \neq j}X_{k}X_{j}\right)\right ].\end{aligned}$The same is true for payoff 2 and $Y$. Now, let us calculate $\text{E}\left[ \left(\sum_{k \neq j}X_{k}X_{j}\right)\right ]$. We make two groups of $X_{k}X_{j}$:

• $\left | k-j \right | = 1$ representing two consecutive indicator variables which share one common coin. Notice that their product is always 0 since it's impossible to have two $HT$s in 3 consecutive coins.

• $\left | k-j \right | > 1$ representing two non-consecutive indicator variables which share no common coin. Their product is non-zero only when they are both non-zero which happens with probability $\left(\frac{1}{4}\right)^{2} = \dfrac{1}{16}.$

There are total of $4 \cdot 3 = 12$ $\left(X_{k}X_{j}\right)$ pairs, and $2\cdot 3$ of them which are consecutive ie. where $\left | k-j \right | = 1$. Hence, we have: $\text{Var}[X] = \text{E}\left[ \left(\sum_{k \neq j}X_{k}X_{j}\right)\right ] = 0 \cdot 6 + \frac{1}{16} \cdot 6 = 0.375.$

In the same way, we calculate $\text{Var}[Y] = \text{E}\left[ \left(\sum_{k \neq j}Y_{k}Y_{j}\right)\right ] = \frac{1}{8} \cdot 6 + \frac{1}{16} \cdot 6 = 1.125.$

So, variance of the payoff 2 is 3 times greater than variance of the payoff 1. If my reasoning is correct, then payoff 1 is somewhat safer option and it tends to the expected value, while payoff 2 is more riskier but it offers better chances of earning more than 1$. What do you thinks about this approach which takes variance into account, @Calvin Lin ?

After calculating the expected value and variance of the payoffs by hand using indicator variables (similar to some other people who have posted here), I became curious as to what the distributions of the payoffs would look like and how different strategies (not necessarily of length 2) would work out. For those interested, I wrote up an R-Jupyter notebook on it here.

There are 2^5 outcoms of 5 coin tosses.

For payoff 1, there are this many ways to toss a HT. There are {HTHTT,HTHTH,HTTHT,HTHHT,THTHT, HHTHT,HTHHH,HHHTH,HHHHT, HTTTT,THTTT,TTHTT,TTTHT,HTTTH,HHTTT, HHTTH} Expected paayyoff 1 = (6/32$2) + (10/32$1)= $11/16=$22/32

For payoff 2, there are this many ways to toss a HH( or HHH,HHHH and HHHHHH). There are {HHHHH, THHHH,HHHHT, THHHT,HTHHH,HHHTH, HHHTT, TTHHH, HHTHH, HHTTT, THHTT, TTHHT, TTTHH, THTHH, THHTH] Expected payoff 2 =(1/32$4 )+ ( 2/32$3) + ( 5/32$2) + (7/32$1)=$3/4= $27/32

Payoff 2 is better, I guess LOL @Calvin Lin

Total possible outcomes of 32 of which there is one combination [TTTTT] which has no payoff.

There example given implies that HHH is counted as two consecutive HHs. This seems questionable as in standard English two consecutive HHs means HHHH. Is this really what the scenario intended?

Expected value is same for both payoffs

I remember reading about an interesting experiment done by a French mathematician regarding payoffs and expected utility from lotteries. Allais Paradox?

Let Ii be an indicator rv for an event of getting H at ith toss. Let X be a rv which is how many times HT occured. Similarly, let Y be the same but for HH. Then, X = I1I2 +... +I4I5 and Y = I1(1-I2) +...+I4(1-I5). We need to compare EX and EY. Given EIi = 0.5, we can apply linearity of expectation (even for multiplication since events are independent) and get 1 for both expectations.

I would go for Payoff 2 , because For payoff 1 ..... one can get maximum number of $1 is 2 which equals $2 i.e for this outcome HTTHT ( MAXIMUM POSSIBLE 'HT' consecutive is 2). For payoff 2 ..... one can get maximum number of $1 is 4 which equals $4 i.e for this outcome HHHHH ( MAXIMUM POSSIBLE 'HH' consecutive is 4).

I think this is a combinatorics problem?!

From a statistical standpoint, I would choose payoff option 1. As the number of H tossed increases, the probability of the consecutively tossing another H decreases. That being said, the HH combination can lead to a higher payoff because HTHTH only results in $2 while HHHHH results in $5. The issue lies in the decreased probability of continuing to toss H. This is a good example related to risk, much like in the decision involved in buying a AAA bond earning 5% vs a B bond earning 13%.