## probability question: incorrectly guessing card values

For the discussion of math. Duh.

Moderators: gmalivuk, Moderators General, Prelates

### probability question: incorrectly guessing card values

I'm trying to figure out the probability of guessing wrong every value of every card in a standard shuffled deck, when guessing the value of the card that is least likely to show up next in the deck. But I'm having trouble getting my head around having only 13 possible values and 52 cards. Any help on how to deal with this would be amazing.
keenin

Posts: 6
Joined: Wed Aug 11, 2010 12:04 am UTC

### Re: probability question

0, because when there is only one card left, there is only one value to guess, so you have to get it right, therefore it is impossible to get it wrong.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.
tomtom2357

Posts: 429
Joined: Tue Jul 27, 2010 8:48 am UTC

### Re: probability question

... I do not think that was the point of the puzzle. I think you can call values that are certain to not show up, so if you get all 4 aces straight away, you can call ace all the way (the chance of an ace showing up is 0, whereas all others are >0, so 0 is smallest), and get a clear.
As for strategy, the optimal one is calling the one that has been shown most so far, and if there are several with an equal amount left in the deck, pick one at random. This minimalizes the chance of you being right.. Just currently unable to figure out the odds.

t1mm01994

Posts: 293
Joined: Mon Feb 15, 2010 7:16 pm UTC
Location: San Francisco.. Wait up, I'll tell you some tales!

### Re: probability question: incorrectly guessing card values

Hint:
Spoiler:
Before drawing the first card, you have 0 values which occur 0 times in the stack, 0 values which occur 1 time, ..., 13 values which occur 4 times. This can be written as (0,0,0,0,13).
You guess a random value for the first card. With 1/13 probability, the card has this value, and you lose. Otherwise, some other value has only 3 cards left, which gives you (0,0,0,1,12).
Now one value is less frequent than the others, therefore you will guess this value.
How do these numbers develop? Can you calculate the probability, based on this development?

This might need some computer assistance. And store your calculated intermediate values to re-use them.

I get ~27% as result.
mfb

Posts: 803
Joined: Thu Jan 08, 2009 7:48 pm UTC