Okay, so I came across an interesting probability question that has left me stumped to come to a solid conclusion.

Assume that you are playing a card game with two opponents, $A$ and $B$. Assume that the probability of winning against $A$ is greater than winning against $B$. Probabilities of winning against $A$ or $B$ are independent. You will play these opponents one at a time in alternating order. You receive a payout only if you are able to win two games in a row out of a three game series.

Given the choice, would you choose to pick $A$ (i.e. Would play $ABA$) or $B$ first (Would play $B A B$)

My instinct tells me that you would be better of choosing $A$ first, but since any two wins in a row would involve beating $A$ and $B$, I am not convinced that it matters.

I am curious if there is a more mathematical explanation than my instinct.


Let $p$ be the probability of winning a round against $A$, and $q$ that against $B$, for some $p>q$.

The probabilility of winning two games in a row out of three, if you start playing $A$ is: $$P_A~{=\mathsf P(wwl\cup www\cup lww\mid ABA)\\= pq+(1-p)qp\\= (2-p)pq}$$

The probabilility of winning two games in a row out of three, if you start playing $B$ is: $$P_B~{=\mathsf P(wwl\cup www\cup lww\mid BAB)\\= qp+(1-q)pq\\= (2-q)pq}$$

Since $p>q$ , therefore $P_A<P_B$

Intuitively, you chances of winning two rounds in a row is improved if the outside rounds are the easiest to win; since you must certainly win the middle round if you wish to win the game.   If you can win that, then you have two oportnities win against the other player; so make those the easiest.

Or in simpler terms: to win the game you must win against one player twice and the other player once, and the best way to manage this is to choose the easiest target as the one to play against twice.

  • $\begingroup$ The intuitive answer is that you have to beat B to win two in a row, so two chances of that are better than one. This computation clearly demonstrates it. I think people are confused by the better chance of winning two games if you play ABA but that is not what the question asks. +1 $\endgroup$ – Ross Millikan Jun 2 '17 at 3:36

We can compute the probabilities directly.

There are three configurations that have two wins in a row: $WWW,$ $WWL$ or $LWW.$ Let $p_A$ be the probability of beating $A$ and $p_B$ for $B$ so that $p_A>p_B.$

For $ABA$ the probabilities are $$P(WWW) = p_Ap_Bp_A \\P(WWL) = p_Ap_B(1-p_A) = P(LWW).$$ Adding the three up gives a total probability of $ p_Ap_B(2-p_A).$

For $BAB$ you similarly get $p_Ap_B(2-p_B).$

So it looks like you are better off with $BAB!$

What's going on here? Since your ability to get two in a row depends pretty crucially on getting the middle one, it's actually best to have the easiest in the middle and taking your chances with $B$ on the outside.


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