Coin triplets

To answer these questions we need to calculate, for each pair of triplets, the probability that one triplet appears before the other. Given that each triplet is equally likely, it may initially seem that each is equally likely to appear first. For an example of why this is not so, consider the triplets HHH and THH. The only way for HHH to appear before THH is if the first three tosses come up heads. Any other result will allow THH to block HHH. Therefore, the probability that HHH appears before THH is 1/8.

We may calculate the probabilities for each pair in a similar manner. Consider, for example, HTT versus HHT. The probability HTT appears first is the mean of that probability over the four possibilities for the first two coin tosses. Let, for example, p(HT) be the probability HTT appears first following HT.

Suppose the first two throws are HH. Then the third throw can be either H or T. If it’s H, then we are back in the same position: the preceding two throws are HH. But if it’s T, then HHT has won! So the probability of HTT winning in this case is 0.

Putting the two possibilities for the third throw together, as a weighted mean, the probability that HTT wins following HH is: p(HH) = ½×p(HH) + ½×0 = p(HH)/2.

Now suppose the first two throws are HT. If the third throw is H, then neither player has won, and the probability HTT will ultimately win is (by definition) p(TH). (The last two throws were TH.) On the other hand, if the third throw is T, then HTT has won!

So this time the weighted mean for the probability that HTT wins, following HT is: p(HT) = ½×p(TH) + ½×1 = p(TH)/2 + 1/2.

Continuing in this way, we obtain the results below:

(1) p(HH) = p(HH)/2
(2) p(HT) = p(TH)/2 + 1/2
(3) p(TH) = p(HH)/2 + p(HT)/2
(4) p(TT) = p(TH)/2 + p(TT)/2

(1) implies p(HH) = 0. (Intuitively, HTT can avoid losing only by hoping for an infinite string of heads!)
(3) implies p(TH) = p(HT)/2
(2) implies p(HT) = p(HT)/4 + 1/2 implies p(HT) = 2/3
(3) implies p(TH) = 1/3
(4) implies p(TT) = p(TH) implies p(TT) = 1/3

The mean of these four results gives us: probability of HTT appearing before HHT = 1/3.

Here is the full table. Notice the surprising non-transitivity. Example: HHT beats HTT beats TTH beats THH beats HHT.
Coin triplet probabilities 2\1 HHH HHT HTH HTT THH THT TTH TTT
HHH 1/2 2/5 2/5 1/8 5/12 3/10 1/2
HHT 1/2 2/3 2/3 1/4 5/8 1/2 7/10
HTH 3/5 1/3 1/2 1/2 1/2 3/8 7/12
HTT 3/5 1/3 1/2 1/2 1/2 3/4 7/8
THH 7/8 3/4 1/2 1/2 1/2 1/3 3/5
THT 7/12 3/8 1/2 1/2 1/2 1/3 3/5
TTH 7/10 1/2 5/8 1/4 2/3 2/3 1/2
TTT 1/2 3/10 5/12 1/8 2/5 2/5 1/2
a. What is the optimal strategy for each player? With best play, who is most likely to win?

The table shows that, for any triplet chosen by player 1, player 2 can always select a triplet that is more likely to appear first. In particular, the best response to each play by player 1, is:
HHH: THH wins with probability 7/8
HHT: THH wins with probability 3/4
HTH: HHT wins with probability 2/3
HTT: HHT wins with probability 2/3

The color coding illustrates the following strategy. Player 2 wants the final two coins of his triplet to be the first two in player 1′s, because then he blocks half the cases where player 1 could win on the next round, by winning first. Similarly, player 2 wants the first two coins in his triplet not to be the final two coins in player 1′s. This intuitive strategy is indeed supported by the probabilities listed in the above table.

The optimal strategy for player 1 is to choose triplet HTH or HTT, or their mirror images, THT or THH. This limits player 2′s probability of winning to 2/3.
b. Suppose the triplets were chosen in secret? What then would be the optimal strategy?

From the table above, an optimal strategy for both players is to choose at random, with probability 1/2 for each, between HTT, and its mirror image, THH. The choice must be random so that the other player cannot discern and exploit any pattern of switching between HTT and THH.
(Note: These statements, while true, require rigorous proof; to be added.)
c. What would be the optimal strategy against a randomly selected triplet?

The expected return of each triplet against a randomly chosen triplet can be calculated from the above table.

We must decide what to do if our play matches the randomly selected triplet. We may call this void and play again, or we may split the (notional) winnings. The decision does not affect our choice of best play, but it does slightly alter the expected return from each play.

For example, the expected return for HHH, if we choose to void matching triplets, is:
(1/2 + 2/5 + 2/5 + 1/8 + 5/12 + 3/10 + 1/2) / 7 = 317/840 is approximately equal to 0.377.
On the other hand, if we choose to split matching triplets, the expected return is:
(1/2 + 1/2 + 2/5 + 2/5 + 1/8 + 5/12 + 3/10 + 1/2) / 8 = 377/960 is approximately equal to 0.393.

The best play against a random triplet is HTT or THH. The table below shows the expected return and percentage expected profit from each play.
Summary of expected return and profit Play Expected return
(void) % expected profit
(void) Expected return
(split) % expected profit
(split)
HHH 317/840 −12.3 377/960 −10.7
HHT 469/840 5.8 529/960 5.1
HTH 407/840 −1.5 467/960 −1.4
HTT 487/840 8.0 547/960 7.0
Further reading(Adobe) Portable Document Format Optimal Penney Ante Strategy via Correlation Polynomial Identities

I’ve got another take on this question.

Given that, in this Q2 game, there is no first or second in choosing in secret, James’ table need reinterpreting.

Firstly, P1 and P2 must have an equal chance of winning, as there is no difference in their position. Thus each player could reasonably assume that the other player will use the same strategy. If the other player is recognised as having a different and better strategy, then the first player ought to use that strategy too.

Each player randomly choosing one of the central block of 4×4 in James’ table (HTH, HTT, THT, THH) has, provided the other player chooses the same, a 50% chance of winning in every case.

Likewise, each player randomly choosing one of the central block of 2×2 in James’ table (HTT, THH) also has, provided the other player chooses the same, a 50% chance of winning in every case.

Thus, one of these two strategies would look to be the most self-consistent.

However, if the other player chooses another strategy, things change. If, by unknown means, the other player learned of one’s choice before making his decision, it would still be best to have chosen one of the 4×4 block (because that is the best choice for P1 according to Question 1). If the other player chooses randomly, the 2×2 block becomes the best choice. If the other player chooses the 2×2 block, then a choice of the 2×2 block or the 4×4 block does not reduce your chance of losing. Likewise if the other player chooses the 4×4 block, then a choice of the 4×4 block or the 2×2 block does not reduce your chances of winning.

Thus, on balance, and with a bit of unhappiness, I reckon both players should choose from the 2×2 block for Q2: that is randomly one of: HTT or THH.

It would not surprise me if James has an answer that covers the 4×4 block for both players: but I currently think my solution covers more cases of cheating / inspired play.

Best regards

1. What is the optimal strategy for each player? With best play, who is most likely to win?
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As James points out, Player 1 (P1 with choice ABC) chosing HHH allows P2 (with choice XYZ) to choose THH, which blocks P1 from winning anywhere except for the first 3 coin flips. Likewise, P1′s TTT allows blocking by P2′s HTT.

As Lord T has already pointed out, this quiz centres around the subject of Game Theory. One of the main analyses in Game Theory leads to the Minimax Algorithm. This is where each player assumes that his opponent is perfectly informed and will make the best move possible: thus he makes his moves to minimise the maximum advantage to his opponent, assessed after his opponents next (responding) move. [Aside: From this, such games as noughts and crosses can be shown to have a guaranteed draw, starting off with the first player always choosing a corner position.]

Anyway, on this question, P2 (as he makes his move second) cannot be at a disadvantage to P1. Hence, if there is any player having an advantage (with best play), it must be P2.

Next, and assuming we are past the special cases around the first few throws, any 3-coin sequence is as likely as any other 3-coin sequence. It is only the overlap of sequences that can provide any advantage to P2.

On that advantage, P2 can gain by setting his YZ to be the same as P1′s AB. As far as I can see, P2 should therefore chose his XYZ to be ?AB, where ? is randomly chosen to be H or T, or forced to be one or the other so as not to conflict with P1′s choice. By doing this, P2 has a 50% chance of winning on the coin flip just before P1 wins.

However, also as far as I can see, P2 loses his 50% chance of winning (excluding the first few flips) with a random choice, because he is no longer making a random choice. If he chooses ?AB, he cannot win earlier than 1 flip before P1 also has a 50% chance of winning. It’s round about here that I start to wonder if, like noughts and crosses, neither side has the advantage given their best choices (but I later reverse that view, perhaps).

P2 also gains additionally by setting his Z to be the same as P1′s A. However, both of these can only happen in the same gain if P1 sets his A=B. Thus, P1 should always have different A and B.

This paragraph contains additional provisional arguments that I subsequently reconsider as doubtful, but let’s give it a go anyway. So: given that P1 chooses TH? or HT?, can we determine which one is it better for him to choose; let’s consider the case TH?. Whichever P1 chooses, he as a 50% chance of losing to P2 on the preceding flip (assuming P2 chooses the above best-looking strategy). If P2 does not win on the preceding flip, P1 choosing THH has a 50% chance of winning on that flip; if P1 does not win, he has 0% chance of winning on the next flip and a 25% chance of winning on the following flip (as the non-winning sequence must have been THT). If P2 does not win on the preceding flip, P1 choosing THT has a 50% chance of winning on that flip; if P1 does not win, he has 0% chance of winning on the next flip and a 0% chance of winning on the following flip (as the non-winning sequence must have been THH). Thus it is better for P1 to choose THH than to choose THT. Likewise, it is better for P1 to choose HTT than to choose HTH.

Now onto P2, given that he chooses ?TH to P1′s TH?, which should P2 choose between HTH and TTH. But … it is around here that I find conflicts with James’ table. So I’m going to trust the table and flip arguments. Remember that, so far, I find P1′s HTT to be preferable to HTH (and THH to THT).

James’ table (as I interpret it) gives the probability of P2 winning given the row choice given that P1 has made the column choice.

Thus, using the Minimax Algorithm and believing James’ table, let’s choose the maximum value in each column (as P2′s probability of winning given P1′s column choice). These give: 7/8, 3/4, 2/3, 2/3, 2/3, 2/3, 3.4, 7/8. Thus, P1 should make a choice that gives the minimum of the column maxima (which is 1/3) for P1′s choices of HTH, HTT, THH and THT.

So, from my more detailed analysis above, and stopping before I get doubtful, we have in answer to Question 1:

The optimal strategy for P2 is, as stated above, to chose ?AB, in response to P1′s ABC.

The optimal strategy is for P1 to chose ABC such that A and B are different.

From the above, P2 has a definite advantage in that his go is made after P1, and there is advantage to be had. Therefore P2 is most likely to win.

And from my application of the minimax algorithm, and trusting James’ table, we get a consistent but more precise solution. With more detail being what P2 should choose for his X in XYZ: namely:

P1 chooses HTH; P2 chooses HHT and wins with probability 2/3
P1 chooses HTT; P2 chooses HHT and wins with probability 2/3
P1 chooses THH; P2 chooses TTH and wins with probability 2/3
P1 chooses THT; P2 chooses TTH and wins with probability 2/3

P2 is most likely to win.

I’d love to carry on and work out why (apart from James’ table) P2 does not have 2 equal choices of response (as per my earlier detailed case). But I’ve run out of time; maybe someone else would like to have a go at that.

2. Suppose the triplets were chosen in secret? What then would be the optimal strategy?
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Time is now severely limited, so it’s just back to the Minimax Algorithm and James’ table. Remember that Minimax works on the basis of assuming a perfectly informed and rational opponent, but (as with stone, paper, scissors) P2 does not know P1′s choice; thus psychology can easily have an impact, especially in repeated goes at the game or with P1 and P2 having prior knowledge of the mindset of their opponents.

And, of course, clashing choices – and thus draws – must be allowed. This complicates things as the definition of optimum becomes a plurality: winning (same choice counts as losing), not losing (same choice counts as winning), or defining a points system for win, draw or lose (and I remember when (some) UK league football had 0,1,2 and now has 0,1,3 – so don’t tell me this is totally irrelevant), or just leaving out the clashes.

One goodish possibility is for P1 to assume that P2 will choose randomly. Thus P1 should choose the column in the table with the lowest average score probability of losing. I calculate the column average probabilities (ignoring clashes) as 0.62, 0.44, 0.51, 0.42, 0.42, 0.51, 0.44, 0.62. So P1′s best strategy is to choose HTT or THH.

On the same basis, if P2 assumes P1 chooses randomly, the row averages are: 0.377, 0.558, 0.485, 0.580, 0.580, 0.485, 0.558, 0.337. Thus P2 should chose, randomly between the best of these: HTT and THH.

Another possibility would be for P1 to assume P2 will choose HTT or THH randomly (for the average probability maximisation), and then choose to minimise P2′s benefit from that (randomly assuming between P1′s HTT or THH). This gives equal validity to the same choices for P1 as in Question 1.

I am struggling, on this Question 2, to see any one of these possible good strategies as preferable over any other, unless one has some statistics relating to overall human nature. Perhaps James does have a solution that does better, for rational cause.

3. What would be the optimal strategy against a randomly selected triplet?
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The answer to this is one of the possibilities considered for question 2): P2 should chose, randomly between the best row average: HTT or THH.

[Note: this is all so complicated and long that I've not checked all my work. I hope there are no mistakes, but ask readers to beware. Also, if there are, to allow for mitigation of length/effort.]

Best regards

However, I have a talk on MRI scanners to attend first. One of the few times the local IET branch puts on something I’m both interested in and available for.

Best regards