Card game Probability: Having a second war after a first war

Question

I was wondering about the probability of getting two wars. The way the game works is, both players get $26$ cards, and flip one card over, whoever has the largest card value, takes both cards and puts them underneath the deck.

Now the tricky part comes when the two cards flipped over have an equal value, this is called a "war". At a war, the players each flip another card on the top of their deck over, and whoever has the highest value card gets all the cards on the table. My question is, what is the chance that these two cards also have equal value? So, the probability of having two cards out of $52$, be equal, then the next two cards be equal.

I keep arriving at this number: $\frac{3}{49} + \frac{1}{49} = \frac{4}{49} = 0.0816$.

Thanks for the help!

Answer 1

I think you can ignore everything except for the two flips that result in war. Given that there was a war on the first flip, the deck has lost two cards of equal rank.

Then $50$ cards remain. There are $12$ complete ranks of $4$ cards each. Each of those ranks has $\binom{4}{2}=6$ possible pairs. The $13$th rank is the one that was drawn for the first war, and there is only $1$ remaining pair of that rank.

Overall there are $(12\cdot6)+1=73$ pairs left in the deck of $50$ cards.

So the probability of a second war is $\frac{73}{\binom{50}{2}}=\frac{73}{1225}\approx5.96\%$

Answer 2

This answer is in response to the OP's request, in comments below DreiCleaner's answer, to explain what is wrong in their (mis)calculation ${3\over49}+{1\over49}$.

The OP is quite right that if the third card flipped is matches the rank of the cards in the first war, then there is only a $1/49$ chance the fourth card will match the third. But this (the third card matching the first to) only happens with probability $2/50$ -- that is, the third card must be one of the $2$ cards of the same rank from the first war, out of $50$ cards available. Similarly, if the third card is different from the rank of the first war, which happens with probability $48/50$, then the fourth card will match it with probability $3/49$. So a correct calculation of the probability of a second war is

$${48\over50}\cdot{3\over49}+{2\over50}\cdot{1\over49}={24\cdot3+1\over25\cdot49}={73\over1225}$$

as in DreiCleaner's answer.