[bonanova]

It had been a long time since Alex had dropped by Morty's.

Too long, as they say. So when he happened by last night,

he had some explaining to do.

Well I spent a couple days looking over the puzzles the blokes

there at BrainDen have been discussing. This particular one

was about flipping a coin three times. There are only eight ways

it will turn out, and none of them is more likely than another.

Still, they didn't seem able to grasp what would happen if they

raced each other one on one. Head-to-head, you might say,

he added with a wink.

Well it isn't all that hard to figure out, really. And so now I've got

a challenge: you pick an outcome - any one you like - say three

heads [HHH]. Then I'll take second choice - one of the others,

say two heads followed by a tail [HHT].

Then one of us flips a coin three times. If neither pattern shows,

he flips again. And again. Until the last three flips match one of

the choices. Kind of like a horse race, to see which comes first.

Whichever pattern shows first is the winner.

Since you get first pick, you really should offer me some odds.

But since I'm a nice kind of fellow, I'm offering to play you even up.

Jaime looked skeptical at first, and Ian wasn't saying anything.

Finally Davey spoke up. I'll play; but only if you offer me the odds:

your 8 quid against my 5.

Alex thought for a moment. You drive a hard bargain. But hey,

it's only a game, and we're friends, aren't we? You're on!

Has Davey, finally, gotten the better of Alex?

[preflop]

Well i'm not sure that I have arrived at an answer but here my initial thoughts

The person picking second does have the advantage statistically speaking. The second chooser should always chose the same as the first player except switching the last flip. So if Davey chose HTH for example, Alex would then want to chose HTT. This is because Davey has a 1 in 8 chance of being right for any 3 flips. While Alex is really only guessing on the outcome of the 1 coin (the last and different one) and so has a 1 in 2 chance.

So on the surface If I am Davey I would want 4 to 1 odds (or better) from Alex (note I haven't really thought through this last part, but i'm guessing it is correct).

[HoustonHokie]

Well i'm not sure that I have arrived at an answer but here my initial thoughts

The person picking second does have the advantage statistically speaking. The second chooser should always chose the same as the first player except switching the last flip. So if Davey chose HTH for example, Alex would then want to chose HTT. This is because Davey has a 1 in 8 chance of being right for any 3 flips. While Alex is really only guessing on the outcome of the 1 coin (the last and different one) and so has a 1 in 2 chance.

So on the surface If I am Davey I would want 4 to 1 odds (or better) from Alex (note I haven't really thought through this last part, but i'm guessing it is correct).

My first thoughts were the similar, but then I thought about it again, and I don't know.

Comparing Davey's 1/8 chance on any 3 flips with Alex's 1/2 on the last one is comparing apples and oranges, I think. They both have the same probability of the first 2 flips matching their chosen combinations, and then they both have a 1/2 chance on the last coin. I'm beginning to think the odds are even, no matter who chooses first.

[Guest]

It had been a long time since Alex had dropped by Morty's.

Too long, as they say. So when he happened by last night,

he had some explaining to do.

Well I spent a couple days looking over the puzzles the blokes

there at BrainDen have been discussing. This particular one

was about flipping a coin three times. There are only eight ways

it will turn out, and none of them is more likely than another.

Still, they didn't seem able to grasp what would happen if they

raced each other one on one. Head-to-head, you might say,

he added with a wink.

Well it isn't all that hard to figure out, really. And so now I've got

a challenge: you pick an outcome - any one you like - say three

heads [HHH]. Then I'll take second choice - one of the others,

say two heads followed by a tail [HHT].

Then one of us flips a coin three times. If neither pattern shows,

he flips again. And again. Until the last three flips match one of

the choices. Kind of like a horse race, to see which comes first.

Whichever pattern shows first is the winner.

Since you get first pick, you really should offer me some odds.

But since I'm a nice kind of fellow, I'm offering to play you even up.

Jaime looked skeptical at first, and Ian wasn't saying anything.

Finally Davey spoke up. I'll play; but only if you offer me the odds:

your 8 quid against my 5.

Alex thought for a moment. You drive a hard bargain. But hey,

it's only a game, and we're friends, aren't we? You're on!

Has Davey, finally, gotten the better of Alex?

I'm not sure it matters, but after the first 'triple flip', the coin is flipped once each round and only the 3 most recent results are considered, correct? As opposed to 3 flips no match, three more flips no match, etc.

I must be missing something, because I too think that the odds are even if he picks the first 2 the same as Davey and then the last one different. Any time they get to the last flip and the first two have lined up for them, the final flip is 50-50.

I also think the patterns are all equivalent - if the weren't then Davey would be free to choose the best one anyway.

So I'll say that yes, Davey got the better of Alex this time.

[preflop]

My first thoughts were the similar, but then I thought about it again, and I don't know.

Comparing Davey's 1/8 chance on any 3 flips with Alex's 1/2 on the last one is comparing apples and oranges, I think. They both have the same probability of the first 2 flips matching their chosen combinations, and then they both have a 1/2 chance on the last coin. I'm beginning to think the odds are even, no matter who chooses first.

HH - I kind of agree with you, but let me further rationalize my why i think my first thought is correct.

The problem here is in perspective you take when viewing the odds from each player. Yes the 1/8 viewpoint is different than the 1/2 and it can be seen as apples and oranges, but (and someone please correct me if I am wrong here) in a statistical analysis they are comparable.

The way I see it is the two players are making entirely different bets. Alex by selecting the same first two flips as Davey has reduced his bet to just one coin flip. He does not care what the first two are, only that they match Davey's and as the flips happen it doesn't matter to Alex what they are until we hit the point that Davey's first two come in order. Then it becomes a 1 coin flip bet and so Alex has a 50/50 shot. Davey has a 50/50 shot at that point but he is still in the middle of his 1/8 bet, so his overall bet is still that chance. I think it is more of a matter of *when* each player made his bet.

These statistics problems always seem a little illogical from a certain point of view, at least to me they do.

[EventHorizon]

I built a state diagram, where the initial state is chosen randomly.

The first thing to notice is that the only way to get to HHH or HHT (if you don't start there) is from THH, so if the opposing player chooses either HHH or HHT, I'd choose THH. Similarly if TTT or TTH, I'd choose HTT. Both of these situations have 3/4 chance of winning (if you don't start there, you need to go through my state).

I haven't analyzed the rest yet, but I'll post again soon with further probability analysis. Though from what I'm thinking now, you'd definitely want to choose one of the two states that preceeds the one chosen by the opponent.

Another good one from Bonanova.

Edited February 27, 2009 by EventHorizon

[Guest]

Since you get first pick, you really should offer me some odds.

But since I'm a nice kind of fellow, I'm offering to play you even up.

Jaime looked skeptical at first, and Ian wasn't saying anything.

Finally Davey spoke up. I'll play; but only if you offer me the odds:

your 8 quid against my 5.

Has Davey, finally, gotten the better of Alex?

I don't have time right now to calculate who the 8:5 odds favor, but I am reasonably sure I would rather be the player that chooses last. I think that whatever the first player bets on, the other player can choose something that has a comparative advantage.

[Prof. Templeton]

I don't have time right now to calculate who the 8:5 odds favor, but I am reasonably sure I would rather be the player that chooses last. I think that whatever the first player bets on, the other player can choose something that has a comparative advantage.

can always make a choice based on player one's pick to give at least a 2/3 win probability. So 8 to 5 is not good enough. Whatever player one picks as their first two outcomes, player two should pick as their last two. If Davey goes with HHH Alex would pick THH as EventHorizon said and he would win 87.5% of the time.

[EventHorizon]

Another good one from Bonanova.

I built a state diagram, where the initial state is chosen randomly.

The first thing to notice is that the only way to get to HHH or HHT (if you don't start there) is from THH, so if the opposing player chooses either HHH or HHT, I'd choose THH. Similarly if TTT or TTH, I'd choose HTT. Both of these situations have 3/4 chance of winning (if you don't start there, you need to go through my state).

I haven't analyzed the rest yet, but I'll post again soon with further probability analysis. Though from what I'm thinking now, you'd definitely want to choose one of the two states that preceeds the one chosen by the opponent.

That didn't take long...

If HHH or TTT are chosen, the second player has a 7/8 chance of winning if he chooses well.

If HHT or TTH are chosen, the second player has a 3/4 chance of winning if he chooses well.

If THH, HTT, HTH, or THT are chosen the second player has a 2/3 chance of winning if he chooses well.

Here's a quick analysis for if player 1 chooses THH and player 2 chooses TTH.

If it starts at THH player 1 wins.

If it starts at HTT, TTT, or TTH, player 1 cannot win.

If it starts at THT, HHT, or HHH...theres a 50% chance to get to win, plus a 50% chance of a 50% chance of losing or returning.

So A = .5 + .5(.5*0+.5*A) = .5 + .25A => .75A = .5 => A = 2/3. Thus there's a 2/3 chance of player 1 losing.

If it starts at HTH, there's a 50% chance of going to a state with 2/3 chance of winning, and a 50% chance of losing. So 1/3 chance that player 1 will lose.

Summing these up (1+3*0+3*(1/3)+(2/3))/8 = (1+0+1+(2/3))/8 = (8/3)/8 = 1/3. So player 1 will win one third of the time.

The others are similar.

Alex risks 8 which Davey will win at most 1/3 of the time, and Davey risks 5 which he will lose at least 2/3 of the time.

8/3 - 5*(2/3) = 8/3 - 10/3 = -2/3. Davey shouldn't have played...

Edited February 27, 2009 by EventHorizon

[EventHorizon]

can always make a choice based on player one's pick to give at least a 2/3 win probability. So 8 to 5 is not good enough. Whatever player one picks as their first two outcomes, player two should pick as their last two. If Davey goes with HHH Alex would pick THH as EventHorizon said and he would win 87.5% of the time.

I lumped HHH and HHT in the same group, but it is even worse for HHH (and similarly TTT) as Prof. Templeton points out. 1/8 of a chance of winning for player 1.

[Guest]

can always make a choice based on player one's pick to give at least a 2/3 win probability. So 8 to 5 is not good enough. Whatever player one picks as their first two outcomes, player two should pick as their last two. If Davey goes with HHH Alex would pick THH as EventHorizon said and he would win 87.5% of the time.

OK, but let's assume that Davey isn't dumb enough to choose TTT or HHH.

For anything else, there are two possible strings that could precede Davey's. So, this strategy has a 50% chance of winning that way, and the rest of the time, Alex has a 50% chance of winning on the next toss. It's not obvious to me exactly how the remaining 25% of the cases are distributed.

If we include the chance each player has to win with the first three tosses, all I can say right now is that Davey wins at least 1/2 of the time, and Alex at least 5/16 of the time, which leaves 3/16 of the time for which I don't have the energy to spare at the moment. I don't see where your 2/3 is coming from. Can you explain that, please? -- D.

[Prof. Templeton]

OK, but let's assume that Davey isn't dumb enough to choose TTT or HHH.

For anything else, there are two possible strings that could precede Davey's. So, this strategy has a 50% chance of winning that way, and the rest of the time, Alex has a 50% chance of winning on the next toss. It's not obvious to me exactly how the remaining 25% of the cases are distributed.

If we include the chance each player has to win with the first three tosses, all I can say right now is that Davey wins at least 1/2 of the time, and Alex at least 5/16 of the time, which leaves 3/16 of the time for which I don't have the energy to spare at the moment. I don't see where your 2/3 is coming from. Can you explain that, please? -- D.

Davey picks HTH so Alex would pick HHT.

H (Alex Wins)
/
H H (Davey Wins)
/ \ /
1 T
\ \
T T (start over)
(start over)

P(to win with initial T) = p

P(to win with HH) = 1

P(to win with HTT) = p

P(T) = 1/2

P(HH) = 1/4

P(HTT) = 1/8

p= p x 1/2 + 1 x 1/4 + p x 1/8

p = (2 + 5 x p)/8

p = 2/3

If T is the first outcome we have to go back to the begining, likewise if HTT is the outcome after three flips we have to go back to the begining since neither of these cases match for both players.

[Guest]

can always make a choice based on player one's pick to give at least a 2/3 win probability. So 8 to 5 is not good enough. Whatever player one picks as their first two outcomes, player two should pick as their last two. If Davey goes with HHH Alex would pick THH as EventHorizon said and he would win 87.5% of the time.

Come to think of it, if that's going to be your strategy, I'll wager $5 on TTH. Do you want to put your $8 on HTT or TTT? ;P

[Prof. Templeton]

Come to think of it, if that's going to be your strategy, I'll wager $5 on TTH. Do you want to put your $8 on HTT or TTT? ;P

I'd pick HTT and win 75% of the time.

[EventHorizon]

OK, but let's assume that Davey isn't dumb enough to choose TTT or HHH.

For anything else, there are two possible strings that could precede Davey's. So, this strategy has a 50% chance of winning that way, and the rest of the time, Alex has a 50% chance of winning on the next toss. It's not obvious to me exactly how the remaining 25% of the cases are distributed.

If we include the chance each player has to win with the first three tosses, all I can say right now is that Davey wins at least 1/2 of the time, and Alex at least 5/16 of the time, which leaves 3/16 of the time for which I don't have the energy to spare at the moment. I don't see where your 2/3 is coming from. Can you explain that, please? -- D.

I had some explanation in my post, but I guess it was insufficient or switched between player 1 and player 2 too much. Here's the same thing with better/more explanation.

Here's a quick analysis for if player 1 chooses THH and player 2 chooses TTH.

If it starts at THH player 1 wins.

If it starts at HTT, TTT, or TTH, player 1 cannot win.

If it starts at THT, HHT, or HHH...theres a 50% chance for player 1 to lose (or, equivalently, to move to a place in the state diagram where player 2 will surely win).

If the other path was taken, then there's a 50% chance of winning for player 1, and a 50% chance of returning to a state similar to where we started.

Let A be the chance that player 1 will win if the first three flips are one of the three situations listed above.

So A = 0*.5 + .5(.5*1+.5*A) = .25 + .25A => .75A = .25 => A = 1/3. Thus there's a 1/3 chance of player 1 winning.

If it starts at HTH, there's a 50% chance of going to a state with 1/3 chance of winning, and a 50% chance of winning. So .5*(1/3) + 1*.5 = (1/6) + (3/6) = 2/3 chance that player 1 will win.

Summing these up (1+3*0+3*(1/3)+(2/3))/8 = (1+0+1+(2/3))/8 = (8/3)/8 = 1/3. So player 1 will win one third of the time, and player 2 will win 2/3 of the time.

Hope that helps.

Edited February 27, 2009 by EventHorizon

[Guest]

I'd pick HTT and win 75% of the time.

Really? Because I win if the first three tosses are TTT or TTH, which happens exactly 25% of the time. So, you are saying you win every other case. Hmmm. Interesting.

[Guest]

Really? Because I win if the first three tosses are TTT or TTH, which happens exactly 25% of the time. So, you are saying you win every other case. Hmmm. Interesting.

Oh, yeah. You do, don't you. I don't want to play anymore.

[EventHorizon]

Really? Because I win if the first three tosses are TTT or TTH, which happens exactly 25% of the time. So, you are saying you win every other case. Hmmm. Interesting.

Notice that if there is an H in either the first or second position, you need to get two T's in a row and then something else.

Prof. Templeton simply needs to wait for the two T's in a row.

[Guest]

Notice that if there is an H in either the first or second position, you need to get two T's in a row and then something else.

Prof. Templeton simply needs to wait for the two T's in a row.

Yeah, I get it now. Thanks guys. -- D