# Choosing TT vs TH problem

Discussion in 'Quant Interviews' started by tuanl, 8/18/12.

1. ### tuanl Member

Suppose you roll a fair coin 5 times. You can either be paid the number of consecutive tails "TT" or the number of "TH"'s. Which one would you choose and why? For example, the sequence "TTHTT" would count 1 instance of "TH" and 2 instances of "TT."

Generalize your result for n flips. Is the result different now?

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2. ### IlyaKEightSix Active Member

TT. Think about it. If you choose TH, then at most, you can have 2 sequences in your 5 flips. But if you choose TT, consider the case of getting lucky with five tails in a row: TTTTT.

How many instances of TT are there? Not 2, but 4. The reason for that is when you roll TH, then the next combination can't possibly be another TH. However, when you have TT, then the second T can be the first T in the next sequence. This basically gives you twice the chance.

Now if it was TH AND HT vs. TT (or HH for that matter), we'd have another discussion.

3. ### tuanl Member

I think "TH" is the answer based on calculating the P(getting at least one TH out of 5 tosses)$$= 26/32=13/16$$>P(getting at least one TT out of 5 tosses)($$=19/32$$). Notice that expected payoff of TT=the expected payoff of TH=1. So if you're less risk averse, we can choose TT and hope to get lucky with TTTTT.

I haven't worked on the general case yet, but I'm pretty sure the same result still holds(we should still choose TH).

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5. ### pruse Active Member

It ain't about googling, it's about sharing and discussing; I'm sure that's why tuanl posts his problems here.

Use indicator variables: For each of the first $$n-1$$ tosses, out of $$n$$, let $$X_1=1$$ if that toss and the next are TH, 0 otherwise. Then the expected number of TH's is equal to $$E(X_1+\cdots+X_{n-1})=(n-1)E(X_1)=(n-1)P(X_1=1)=\frac{n-1}{4}$$. The same argument shows that the number of TT's is expected to be the same. So it's the variance that will decide.

Now for the variance. $$Var(X_1+\cdots+X_{n-1})=E((X_1+\cdots+X_{n-1})^2)-\left(\frac{n-1}{4}\right)^2$$. $$E((X_1+\cdots+X_{n-1})^2)=(n-1)E(X_1^2)+2\sum_{i<j}E(X_iX_j)$$. Clearly $$(n-1)E(X_1^2)$$ will be the same in both the TT and the TH cases. We need to see how the $$\sum E(X_iX_j)$$ part differs.

1) TH case. When $$j-1=1$$, $$X_iX_j$$ clearly must equal 0. In all other cases, disjoint pairs of tosses are involved and these are independent, so $$E(X_iX_j)=P(X_iX_j=1)=P(X_i=1,X_j=1)=P(X_i=1)P(X_j=1)=\frac{1}{4}\cdot\frac{1}{4}=\frac{1}{16}$$.

2) TT case. This is clearly larger as the disjoint pairs of tosses give the same $$E(X_iX_j)$$ as in the TT case, and the overlapping pairs give a positive $$E(X_iX_j)$$, whereas in the TH case these are 0.

So it's better to pick TH.

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6. ### IlyaKEightSix Active Member

How is P(TT) < P(TH)? P(T)=50%. P(H)=50%. P(T)*P(T)=25%. P(T)*P(H)=25%.

7. ### tuanl Member

I said P(getting AT LEAST 1 TT)<P(getting AT LEAST 1 TH). And no, the coin is definitely fair.

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8. ### IlyaKEightSix Active Member

Well, what about the probability of getting 2 TTs vs. 2 THs? With 2 TTs, you can have TTTHH HTTTH, HHTTT, TTHTT, HTTTT, and so forth...

With TH, you can only get two of them at MOST on five flips.

9. ### tuanl Member

Nice solution, but the computation of the variance really bothers me. I know it's a common technique to compare the variance of two events to make a decision when the expected values of these events are equal. But I wonder if my reasoning above also works in this case, since we clearly see that probability of not getting any money when choosing TT is greater than that of TH.

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10. ### tuanl Member

But HTTTT counts as 3 TTs. In fact, probability of getting exactly 2 THs=probability of getting exactly 2 TTs. But we have to calculate the total probability of all the possibilities that can happen with TH and TT. Basically, choosing TH always give you a better chance to get some money than choosing TT. Notice that this reasoning works, in my opinion, since the expected value of TT=expected value of TH. The proof for this, for general value of n, was presented above by pruse above.

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11. ### pruse Active Member

The probability of not getting any money doesn't tell the whole story, as it's possible for the cases that do make money to make disproportionately larger amounts.

Your expected winnings, as well as the volatility of those winnings, are what matter. Now, between two alternatives with the same expected value, the one with the smaller volatility is generally preferred since your confidence interval is smaller.

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12. ### IlyaKEightSix Active Member

Hmmm, I did a dirty computational solution in R and it seems no matter what happens, when you add up the sum over all the combinations, the sums of both THs and TTs come out identical, but the variance of the THs are lower.

Seems I'm mistaken.

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14. ### Kangxi New Member

Hey, my answer for 5 tosses is as follows:
1 TT: 3+2+2+3 = 10
2 TT: 1+(2+1+2) = 6
3 TT: 1+1 = 2
4 TT: 1
So the expected money received if choosing TT will be (10*1 + 6*2+2*3+4*1)/(2^5)=1

1 TH: (2^3-4) + (2^3-2)+(2^3-2)+(2^3-4) = 20
2 TH: 2+2+2 = 6
So the expected money received if choosing TH will be (20*1 + 6*2)/(2^5)=1

Therefore, no matter what I choose, I will be paid \$1 in expectation.

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15. ### tuanl Member

But choosing TH gives you smaller variance than TT, which means TH is less volatile than TT. So you should choose TH

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16. ### Kangxi New Member

Yes, I think so.

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