Probability problem for card players

[Patrick H. Crowe]

Take an ordinary deck of 52 cards (leave jokers out). As you turn up each card call the numbers, ace, duce, three and so on. What are the chances that the number you call will match the card you turn over. For example what are the chance that the seventh card will be a 7. In general what are the chances you can call all 52 cards and never hit one that matches the number you called?

How does one make the calculation? I do not have the answer. I know from repeated attempts that it is almost certain you'll match a number to a card before you get through the deck.

Patrick H. Crowe

 

[Mitch]

The probability of getting one card correct is 1/52 or 1.9%. The probability of getting two in a row is 1/52 X 1/52 or 0.00037%. So for the whole deck, you're looking at (1/52)exp52,or 1.9 e48 %. That is .000000000000000000000000000000000000000000000000019% with 47 zeros. This is what you learn in genetics class in college. Very useful information!

 

[chadrpalmer]

hmmmm, i dont think that answer is correct, as the problem does not mention suit, your first propability should be one in 1 in 13...yes?

 

[Parker]

yeah it would be 1 in 13 for the first card which makes it like a 7% or 8% chance of guessing it right

 

[MEP001]

yeah it would be 1 in 13 for the first card which makes it like a 7% or 8% chance of guessing it right

It would be 1 in 13 with each card if you guessed the same number each time - if your guess is the next successive card each time, the chances of getting it right drop significantly.

As far as chances, no one can accurately calculate what the odds are for a single pass through a deck since there's a randomness involved. One could only do it over and over and average the results to estimate the probable chances.

Of course if you start with a new, unshuffled deck, the odds are exactly 50/50 for the entire deck.

 

[Patrick H. Crowe]

It seems to me that at least at the very start your answer that the probability is 1 in 52 is obviously dead wrong. Isn't it 4 in 52? There are four aces, right?No offense to genetics but this is a math problem. The thing I relish about the problem is that it seems simple but is counter intuitieve, as your answer clearly proves.

 

[Patrick H. Crowe]

Suit is omitted by design. It has no bearing on the problem.

 

[Patrick H. Crowe]

MEPOO1:

I see poker on TV and each time the audience is given the probability of a player winning. I suspect, but do not know, that these are calculations, i.e. they are not based on having dealt these exact hands over and over again. The permutations are tooooooo many.

It seems to me somone who knows far more probability theiry than I do (let alone gentics) could solve this problem and explain the solution.

 

[MEP001]

The probability of a win in poker is based on the cards remaining and the known cards that have been revealed, therefore it's a simple process of elimination and odds of what card might come up. If you're simply naming a card in succession then turning one up to see if it's a match, prediction for the next card won't even be the same throughout the deck because at some point the card you predict will not even be in the deck. You could predict the chance of a card being a match or not by the same process used in poker, but predicting from the start whether or not there will be a match is not possible. Only a probability derived from an average can be used.

 

[Patrick H. Crowe]

Dear MEP:

Many thanks for your interest. It seems to me you recognize the complexity of the calculations. Obviously as each card is shown the probablity of a future hit changes. If the first four cards are seven then there's zero probability that when seven is called it will be a hit.

This has to be accounted for in the poker probabilities, i.e. what cards are shown impacts the outcome. In my days in graduate school there were no computers. Thus some solutions which were suggested got rejected on the grounds that the calculations were so long and tedious they were not practical.

Are they now?

 

[Mitch]

OK. That is why I did not major in math; I could never read the problem correctly. The probability of going through the whole deck and not matching any numbers without regards to suit is 1.56%. The highest probability occurs with four matches. Can anyone tell me why? These probabilities do not consider what cards have been dealt, only chance of numbers. I studied probabilites in statistics, but they only made sense to me when applied to genetics.

 

[rph9168]

I don't think their is an analogy with poker in regards to the odds of what card will show up next. The odds in poker are not based on one card but rather the numbers of "outs" or possible winning cards that are potentially left in the deck. This is seldom not just one card.

 

[Danny]

I agree with Mep, with each card that is turned the probability changes for the next card available. However in a "perfect" scenario the cards would be turned over in order (Ace, 2,3,4...King) giving you set probability rules, which no longer contains variables.

Meaning first turn for an ace would be:
1:13 (4 out of 52) getting the ace
Second turn for the 2 would be:
1:12.75 (4 out of 51) getting the 2
third turn for the 3 would be:
1:12.50 (4 out of 50) getting the 3
fourteenth turn would be for the ace again
1:13.667 (3 out of 41) getting the ace
fifteenth turn would be for the 2
1:13.333 (3 out of 40) getting the 2

and so on...

This can be done 52 times taking the percentage and multiplying the first percentage (1:13 is 7.69%) and multiplying it by the second percentage (1:12.75 is 7.84%) gives the chances for getting the first 2 in the sequence at .603% chance. Then (1:12.5 is 8%) times the .603 equals .04824% chances of getting the 3 on the third turn. If this is followed through all 52 steps (since I do not know the formula to do it for me, lol) it will give the percentage on a "perfect scenario" It will also give the drastic drop in percentage as each card is drawn in sequence to the chance that the next card will be correct.

 

[TheDoc]

I think it is great that some of you are smart enough to sit down and figure the calculation.
I would prefer to sit down with all of you and play poker. While you're figuring out the odds, I'll whip your butts and take your money!

Patrick, are you a wash operator?

 

[Patrick H. Crowe]

I was noit a strong probability student by any means. I found it extremely challenging, though I passed (barely). What I was known for was by no means my ability to solve complex probability problems but for being able to show why proposed solutions would never work.

As in poker what cards are showing obviously impacts the probability. If you need a seven to fill an inside straight and four sevens are alredy facr up then the probability of a seven is zero. Dah???

Thus I submit that an solution which involves the multiplication of individual probabilities (a common technique) must be wrong: Proof: If the first four cards are sevens the the chances of a hit when you call seven are zero and the product of all probabilities is then zero. Thus "products" can't be involved in a simple calculation, can they?

It seems to me that a sum of various probabilities, some of which could be zero, may be in the solution but it is a trick problem, right?

Patrick H. Crowe

 

[MEP001]

I don't see it as a "trick problem." The request was for a manner in which one could determine the chance of whether or not there would be a match. One sure-fire method would be to take every possible combination of a shuffled deck of cards and work out for each combination whether or not there would be a match, with a 1 or a 0 indicating match or not, then average the results.

 

[Patrick H. Crowe]

MEP001:

I meant to type is was a tricky (complex & challenging) problem, not a "trick" one. I apologize for the typo.

Patrick H. Crowe

 

[Sequoia]

It's a calculation that must be performed as progression occurs as it is the progression itself that can vary the odds.

btw, you did not specify what happens after you get to number 13. If you don't restart the count back to one (ace), then none of the numbers above 13 would match, obviously.

To make the calculation, the first step is to analyze the odds on the first card turned over. These odds are 4/52, which is ... umm ... different from 1/4. Read on. If the first card turned over is a deuce, then the odds on the next card are 3/51 (instead of 4/51 had a deuce not turned up.) Since the cards being turned up affect the odds in this way, the odds can only be accurately calculated if such progression is factored. You could perform a long calculation that would include the "average" odds considering expected results, but it would only be an ... average ... value.

You can go through each card in the deck, record the odds on each card prior to it being turned, and then use the right math equation to calculate the odds of various events .... for that shuffle only. Those events could include the odds of never hitting a particular card as the number was read, but again, it would be valid for that shuffle only.