Can you answer this mathematical poseur<?

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How can cards are expected to be turned over before the first snap with a standard deck of 52 cards?

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Type: Question • Score: 2 • Views: 4,900 • Replies: 55

[+3] - rg123 - 01/26/2009Looks like I may be able to successfuly post the program if I break it into more than one post... [code language='python'] # snapchance.py # Written and tested with Python 2.5.2 (not...
[+3] - rg123 - 01/26/2009Program continued. First line here is last line in previous post... [code] # now for the case where no existing groups are matched... if cards_remaining>current_numerator:...

@solipsister,

oops how many cards

1 Reply

@solipsister,

what is a 'snap' (in relation to a deck of cards)?

1 Reply

@aidan,

snap is two consecutive cards of the same value in 2 of the 4 suits

1 Reply

@solipsister,

I'll have to think about that one. I won't look back here until I get an answer.

1 Reply

@aidan,

Okay, I thought about it and I don't think you can answer that unless you know more about the deck of cards. Is it new - or has it been used to play a game and shuffled? Who shuffled - is the person who shuffled a thorough shuffler? All of that will change the odds.
Actually - I just did it and my first two cards were the eight of spades and eight of hearts.
But I use this deck to play golf- so that explains that.

1 Reply

@aidan,

aidan wrote:

Okay, I thought about it and I don't think you can answer that unless you know more about the deck of cards. Is it new - or has it been used to play a game and shuffled? Who shuffled - is the person who shuffled a thorough shuffler? All of that will change the odds.
Actually - I just did it and my first two cards were the eight of spades and eight of hearts.
But I use this deck to play golf- so that explains that.

I think the question assumed, or should assume, that the order of the cards is randomized.

1 Reply

@Brandon9000,

yes very random

1 Reply

@solipsister,

then I think the answer is two - tw0 cards of the same face value of two different suits could be expected to be found anywhere after (but possibly including) the first card.

but I think the wording of your question is confusing - do you mean- how many cards MUST be turned over before one can expect to find the first snap in a standard deck of 52 cards? That's how I read it.
Should I be reading it differently?

0 Replies

nope

1 Reply

@Bella Dea,

nope, I'm wrong - or nope I shouldn't be reading it differently?

I think the way it's worded it's totally subjective - how many card are expected to be turned over before finding the first snap in a deck of 52 cards would indicate different answers for different people who have different expectations.
I'm still gonna say two - because now that I did it this morning and found a snap in the first two cards of my deck of cards - that'll influence my expectation for the next time.

0 Replies

The answer is certainly statistical. Probably one would give the mean number of cards and the standard deviation, or a cofidence interval, or something like that.

1 Reply

@Brandon9000,

I just don't get that. It seems to me -at least as it's currently stated, that it's more of question of location than odds or probability or statistics.

How can you determine the statistical likelihood of an event when you're not given the factual components of the event- as in 'what is the probability or statistical likelihood of finding a snap within the first third (or whatever) of a standard deck of 52 cards?'
And if it's statistical - shouldn't it be something like, ' where would one be most likely to find the first snap in a standard deck of 52 cards?'
But I admit - I'm confused at what exactly is being asked. Maybe the question should be reworded- or maybe I'm just focusing too much on the wording and missing the point.
I still think my answer is right though - it CAN be expected to be found anywhere within the deck after one card has been turned over.
And it can change situationally.

1 Reply

@aidan,

aidan wrote:

I just don't get that. It seems to me -at least as it's currently stated, that it's more of question of location than odds or probability or statistics.

How can you determine the statistical likelihood of an event when you're not given the factual components of the event- as in 'what is the probability or statistical likelihood of finding a snap within the first third (or whatever) of a standard deck of 52 cards?'
And if it's statistical - shouldn't it be something like, ' where would one be most likely to find the first snap in a standard deck of 52 cards?'
But I admit - I'm confused at what exactly is being asked. Maybe the question should be reworded- or maybe I'm just focusing too much on the wording and missing the point.
I still think my answer is right though - it CAN be expected to be found anywhere within the deck after one card has been turned over.
And it can change situationally.

The question asked was:

"How can cards are expected to be turned over before the first snap with a standard deck of 52 cards?"

I assume he meant to ask, "how many." The answer is that the number of cards turned over before this condition is satisfied (a snap appearing) varies, and the manner in which it varies can be calculated with statistical analysis based on the composition of a standard deck of cards. This assumes that the deck is randomized. Events like this pertaining to cards, dice, and coins are a favorite subject of probability and statistics.

1 Reply

@Brandon9000,

I guess I'm stuck on the varies part, which I totally agree with. But I can't see how it wouldn't vary enough to make it almost statistically impossible to pinpoint a specific location for a snap to most likely or often occur within the deck.

But then again, I'm not a card shark. I didn't even know what a snap was. I'm sure there are people who have studied the odds- etc. I know that dice, cards, etc. are often used to illustrate probability - the wording is usually just more specific.

Do you know how to figure this out?

2 Replies

@aidan,

I suspect the wording should be, "on average, how many cards...."

0 Replies

This will require some more thought, but:

1. The chance of a snap on the first card is 0
2. The chance of a snap on the second card is 3/51.
3. The chance of a snap on the third card is 3/50. (We know that the first and second card don't match, so there are three of the second card remaining).

After that, it gets more complex....

1 Reply

@DrewDad,

On average, we're really getting there now.

3/51 for snap on card 2

(48/51)x(3/50) snap card 3 given no snap card 2

(48/51)x(47/50)x(3/49) snap card 4 less the probability that snap card 4 was not played as card 1

do continue

1 Reply

@aidan,

aidan wrote:

I guess I'm stuck on the varies part, which I totally agree with. But I can't see how it wouldn't vary enough to make it almost statistically impossible to pinpoint a specific location for a snap to most likely or often occur within the deck.

But then again, I'm not a card shark. I didn't even know what a snap was. I'm sure there are people who have studied the odds- etc. I know that dice, cards, etc. are often used to illustrate probability - the wording is usually just more specific.

Do you know how to figure this out?

I once did, but don't any longer. It's been a long time, since I remembered my probability and statistics in depth.

1 Reply

@Brandon9000,

Well, I just asked my friend to explain it to me over a cup of tea with a deck of cards and he did.
Brandon's right (in how he started it) but you'd have to do that for all fifty-two cards multiplying the possibility that it was a snap by the possibility that it wasn't. My friend suggested converting the fractional values to decimals and making a table to compare the values of each probability.

I asked him if he could make a prediction of where the snap would most likely occur - he said he couldn't right off the bat, but that he might do the whole problem and let me know.

He said that this was a very complex, high level problem and that if he'd given this to even the brightest of his a-level maths students - he can't think of one who'd even begin to know how to start it.

He gave me a really good lesson in probabilities - which is the conceptual part of this I was struggling with - even though the order might change, affecting the outcome (which is what I was focusing on) - the probability would stay the same.
Now I get what solipsister was asking.
I'm not gonna work it out though - I'm gonna let Rob do that and when he gives me the answer - I'll post it.

1 Reply

Can you answer this mathematical poseur<?

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