[superprismatic]

You are offered a chance to win $100. All you have to do

is play 3 games and win any two consecutive games to get

the $100. The hitch is that you must play against two

players, A and B, alternately. That is, you must play

them in one of two orders, either ABA (A first, B second,

A third) or BAB (B first, A second, B third). Now, the

probability that you can beat A is known to be P_A and the

probability that you can beat B is known to be P_B.

Furthermore, P_A is less that P_B.

Which of the two orders, ABA or BAB, should you choose to

give yourself the best chance of winning the $100?

Note:

In the ABA order, you are playing the one you are most

likely to beat (B) only once, while you are playing the

one you are least likely to beat (A) twice.

In the BAB order, you are playing the one you are most

likely to beat (B) twice, while you are playing the

one you are least likely to beat (A) only once.

[Guest]

Doesn't matter. It's the gambler's fallacy to think otherwise.

[Guest]

Go with the ABA case.

I'm sure someone will come up with calculations about this , however my answer is entirely hunch-based.

In the BAB scenario , you've got a greater chance at winning two games , HOWEVER these two games are separated by another game which you're likely to lose.

If we for simplicity's sake consider that B = 100% and A = 50%, and given the fact you need two games in a row, it's MUCH safer to go with the ABA case since your streak-breaking game ( the middle one ) is a guaranteed ( or highy likely ) win.

[Molly Mae]

Go with the ABA case.

I'm sure someone will come up with calculations about this , however my answer is entirely hunch-based.

In the BAB scenario , you've got a greater chance at winning two games , HOWEVER these two games are separated by another game which you're likely to lose.

If we for simplicity's sake consider that B = 100% and A = 50%, and given the fact you need two games in a row, it's MUCH safer to go with the ABA case since your streak-breaking game ( the middle one ) is a guaranteed ( or highy likely ) win.

I agree with this statement after a booze-filled night of discussing this game.

[Guest]

ABA only ,

to maximise the chance of winning the two consecutive games. otherwise BAB if if u want to win any two games. U have to win the middle one game to ensure the two consecutive win and ABA is most feverate case for winning the middle game

Edited October 9, 2011 by superprismatic
put answer in spoiler. Please use spoilers.

[witzar]

You need one of the 3 following outcomes in consecutive games:

wwl, www, lww

where w=win and l=loss.

So playing in order ABA your probability of success is:

ab(1-a) + aba + (1-a)ba = ab(2-a)

where a and b are probabilities of winning in single game against A and B respectively.

Obviously your probability of success for order BAB is

ba(2-b)

Therefore ABA is better, since a<b.

[Guest]

I'll Go with ABA

Chance of losing to A for the first game will be highly expected ... i'll make this experience to know his moves, techniques and make him over confident that he can beat me anytime we meet. Next, with B which is expected to loss to me will gave me confidence to my next game and to practice my moves with some moves incorporation i've learned with A. the last game would be a 50-50 chance between me and A, he will be over confident and me knowing his best moves can counter his techniques and whoala maybe i can surprise him, to beat him.

[Molly Mae]

I'll Go with ABA

Chance of losing to A for the first game will be highly expected ... i'll make this experience to know his moves, techniques and make him over confident that he can beat me anytime we meet. Next, with B which is expected to loss to me will gave me confidence to my next game and to practice my moves with some moves incorporation i've learned with A. the last game would be a 50-50 chance between me and A, he will be over confident and me knowing his best moves can counter his techniques and whoala maybe i can surprise him, to beat him.

I would assume that when you play any player twice, the P of the second round is the same as the P of the first round.

Edited October 10, 2011 by Molly Mae

[superprismatic]

I would assume that when you play any player twice, the P of the second round is the same as the P of the first round.

Yes, you assume correctly.

[Guest]

Your hint was convincing, but misleading.

A B A

The probability of winning two consequtive games in ABA is 2*PA*PB - PA*PB*PA

This, in case of BAB is 2*PA*PB - PB*PA*PB

Since PA < PB, ABA has better probability of winning.

[Guest]

is there any set of probabilities for A and B in which BAB becomes perferable?

[Molly Mae]

is there any set of probabilities for A and B in which BAB becomes perferable?

I don't think so. The cases to be concerned about are 1) Where A approaches 1.0, 2) Where B approaches 0.00, and 3) Where B = 100 and A approaches 0. That gives a reasonably healthy range to test. In any case, ABA seems to be better.

Another interesting case is A and B both approaching .5, but ABA is still better. =/

Edited October 11, 2011 by Molly Mae

[Guest]

BAB since B has a better a better Pb. There is still a chance either way you'll loose to B so its better so play him twice than A twice. You are playing 3games not 2 so the whole AB BA blah doesn't matter. your chances of winning altogether are better with BAB. but if you break it down to just 2 games each then BAB turns to BA or AB and ABA turns to AB or BA to win and in that sense its the same chance you;ll win in any set of those two game sets.

[Guest]

BAB since B has a better a better Pb. There is still a chance either way you'll loose to B so its better so play him twice than A twice. You are playing 3games not 2 so the whole AB BA blah doesn't matter. your chances of winning altogether are better with BAB. but if you break it down to just 2 games each then BAB turns to BA or AB and ABA turns to AB or BA to win and in that sense its the same chance you;ll win in any set of those two game sets.

The key to this problem is really the deal with winning two in a row.

Again , using my overly simplified example where pA is 50% and pB is 100% , consider this:

In the BAB case, you've got one chance at failing. If you fail there, the two definite wins you've got are wasted since they're separated by that single fail.

In the ABA case, you've got two chances at failing. If you fail on the first try , you've got a definite win letting you continue onwards to your second try. If you fail on the second try , it does not matter since you've already claimed a win in the first pA game.

If we continue observing the ABA case and again go by the implications that pB is 100%, we can rule out the B since it does not play an important role in our chances of winning.

If we denote the consequences of each game with 1 for win and 0 for a loss, we get the following outcomes:

1 / 1 | 1 / 0 | 0 / 0 | 0 / 1

Okay now we put the pB back into these four outcomes, ofcourse replacing it with a 1. We get this:

1 1 1 | 1 1 0 | 0 1 0 | 0 1 1

Out of these four outcomes, the one we don't look forward to is the third one. We've got three winning situations out of four total.

This leaves us with a 3/4 winning ratio, compared to the 1/2 winning ratio of the BAB case.

Edited October 12, 2011 by Razorspined

[CaptainEd]

Look back at witzar's answer--Probability of winning two in a row in ABA > prob of winning in BAB as long as P_A < P_B..