Saturday, December 29, 2012

Probabilities of Phase 10

We were playing a card game the other day and I couldn't resist making some back of the napkin calculations of probabilities.

The game was Phase 10 -- a stupid game that I can never win. The point of the game is to collect certain combinations of cards faster than your opponents.

There are 108 cards in a Phase 10 deck:

  •   96 numbered cards numbered 1-12, 2 of each in four different colors
  •   4 skip cards
  •   8 wild cards

To begin each hand, 10 cards are dealt to each player. After dealing and several turns have been played, in one round I have come to have these cards:

What are my chances of drawing what I need?

In the picture above, you can see I have a wide variety of cards, and two wilds to use. I am trying to get a run of 9 consecutive cards, and am almost there. I'm wondering, what is the probability of me drawing the card I need from the deck at random? I have to do this because my opponents are being stingy and purposefully leaving me cards that are not helpful, because they have been observing what I've picked up in the past few turns.

Well, at this point in the game there have been 30 cards dealt out, and an additional 12 cards played, which means there are 108-30-12 or 66 cards remaining. I am in search of 3, an 8, a 9, a 12, or a wild card, as any of them would allow me to complete the run and play my hand. I have not seen any 3's, 8's, 9's, 12's, or wilds played by my opponents, but I wouldn't expect to see them -- and especially not wilds, as those are seldomly discarded.

At worst, my opponents could be secretly collecting these cards -- and have all of them in their hands -- or could they?  There are 8 each of the 3's, 8's, 9's, and 12's, and 6 more wild cards out there, which means there are a total of 38 cards that I could win with that are unaccounted for. Even if my opponents were both incredibly lucky (not likely) and incredibly mean (possible...) that would still leave 18 winning cards available out of the 66 in the deck, for a minimum probability of 27%.

At best, my opponents might have none of these cards -- leaving all 38 as possible cards to draw from in the deck of 66 -- yielding a maximum probability of 58%.

Chances are likely that something in between is true -- perhaps my opponents have a few of these cards, but not all of them. I'll have to assume that my potential winning cards have been evenly dispersed amongst all of the other cards I don't have -- and even though I can't draw from my opponents hands, I should include the cards they have in my probability calculations. That means there are 66 + 20 or 86 cards to "draw from" and 38 possible winning cards, or a probability of 44%.

I figure I can last another three turns or so before one of my opponents goes out first. This isn't as much of an arbitrary guess as you might think -- a lot of times you can tell by how many cards a player still has how many turns you have left.  What then are my chances of winning? I should be guarenteed right? 44*3 is over 100%!  Unfortunately it's not that easy, but must be calculated by subtracting my chance of losing from 1. My chance of losing is my chances of not getting a card I want 3 times in a row. There is a 100-44 or 56 percent chance that the card I draw next will not help me.  Therefore, my chances of losing are (.56)(.56)(.56) or 17%.  This means my chances of winning are pretty good -- or around 83%.

Unfortunately, I did not win, but when you're playing your wife, winning is always best, and sometimes losing is a victory. Occasionally you need to lose a battle to win the war. Er... to avoid war? I mean, I love you honey!

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