# Card question

1. May 14, 2006

### blumfeld0

This question is driving me crazy
It is a bayesian probability question
what is the probability of guessing the color of the 51st card correctly IF you have already drawn 50 cards and know exactly what colors each one cameout to be

thank you

khurram

2. May 15, 2006

### siddharth

What are your ideas or thoughts on this problem? You need to show your work to get help.

3. May 15, 2006

### arunbg

Just look at what colours the 51st and 52nd cards can turn out to be.
If they are of same colour you can guess the colour of 51st card correctly
(since you should have drawn all cards of the other colour earlier).
If they are of diff. colours you can easily find the probability that your guess is right. Multiply to get the probability.

4. May 15, 2006

### Curious3141

As Arun has stated, you have to consider what colors the next to last and last cards ought to be. They can be similar (both blacks or both reds) or dissimilar (red, black or black, red). Find the probability of each case. Caution : the prob. of similar is not equal to dissimilar.

5. May 15, 2006

### blumfeld0

I dont get it
the probability of him guessing the 51st card corrrectly is then 25/50 *26/52?
what is the probability of getting both blacks or red?
what is the probability of getting red, black and black red?
thanks
blumfeld

6. May 15, 2006

### Curious3141

As a general result, for a set of N items (N even), with half of one sort and half of the other, I get the probability of guessing the last two right after being shown the first (N-2) as being :

$$p = \frac{3N-4}{4(N-1)}$$

7. May 15, 2006

### Curious3141

I'm afraid the answer isn't correct.

Take it in steps. I found it easiest to see the problem from a combinatorics perspective. What are the total number of ways you can arrange N/2 cards of one type and N/2 cards of the other to give a deck of N cards?

How many of those arrangements will have the same type of card for the last two?

How many will have one of each type for the last two?

Hence work out the probabilities of each scenario materialising.

Now for each of those scenarios, what is the probability of guessing the (N-1)th card correctly?

8. May 15, 2006

### balakrishnan_v

you have 2 cards left.
Since you already know the colors of all other cards,you know the colors that the last 2 cards can take
each of the card can take one of the 4 possible colors and you know the color pair
C1 C1
C1 C2
.
.
.16 pairs

for C1,C1 C2C2 C3C3 C4C4 you will judge it perfectly(4 out of 16)
for any other color pairs(12 out of 16),there is a 1/2 probability of getting it right

So the net probability is 1/4*1+3/4*1/2=5/8

9. May 15, 2006

### Curious3141

I think you're mixing up 'suits' with 'colors'. A standard deck of cards has four suits (diamonds, hearts, spades and clubs) but only two colors (red and black).

10. May 15, 2006

### balakrishnan_v

I did not know that.I just looked up the net and somewhere it was written cards have 4 colors.
In any case for a 2 color suite,I guess the answer would be 3/4

11. May 15, 2006

### Curious3141

OK, here's how I did it. I prefer to derive a more general result here.

First of all, it's immediately obvious that in a deck of N cards (N is even) with N/2 of one color and N/2 of the other color, if all N/2 cards of one color have been drawn by the time the (N-2)th card has been drawn, there is no guesswork entailed in predicting the last two cards (prob. of correct guess = 1). Let's call this Scenario 1.

If, in contrast (Scenario 2), exactly (N/2 - 1) of one type and (N/2 - 1) of the other had been drawn by that time, the prob. of a correct guess is only half. So the overall probability has to be somewhere between half and one.

I found it difficult to predict the respective probabilities of Scenario 1 and 2 without using a combinatorics approach. With that approach, it becomes easy.

We're given a deck of N cards, well shuffled. Array them out in order.

The total number (N) of possible distinct arrangements of these N cards is $$N = \frac{N!}{({(\frac{N}{2})!)}^2}$$

In the case where the last two cards are dissimilar (Scenario 2), (N-2)/2 cards of each sort have to be arranged to fill the first (N-2) places .

The number of ways that this can happen is $$n_2 = 2\frac{(N-2)!}{({(\frac{N-2}{2})!)}^2}$$ (The multiplication by two is to take into account both the (red, black) and (black, red) possibilities).

In Scenario 1, (N/2) cards of one color and (N-4)/2 cards of the other color have to be arranged.

The number of ways for Sc. 1 is $$n_1 = 2\frac{(N-2)!}{(\frac{N}{2})!(\frac{N-4}{2})!}$$

Again, the expression is multiplied by two to consider the cases where one color predominates, and vice versa.

The respective probabilities for Sc 1 and 2 are :

$$p_1 = \frac{n_1}{N} = \frac{N-2}{2(N-1)}$$
$$p_2 = \frac{n_2}{N} = \frac{N}{2(N-1)}$$

(Alternatively, you could simply take $$p_2 = 1 - p_1$$ but this way you can check your work).

If Scenario 1 materialises, the probability of guessing the last two cards right is one (no guesswork). If Scenario 2 materialises, the probability of correctly guessing the last two right is 1/2.

So the overall probability of a correct guess is $$p = (1)p_1 + \frac{1}{2}p_2 = \frac{3N-4}{4(N-1)}$$

For a deck of 52 cards, that gives $$p = \frac{38}{51}$$.

12. May 15, 2006

### Curious3141

This would only hold if the chances of getting two similar and dissimilar cards in the last two places were equal, which isn't the case. This is why I specifically cautioned the OP against assuming they were equal.

The answer is close to 3/4 though. This is because 52 is a fairly large number. The limit of p as N tends to infinity is 3/4.

13. May 16, 2006

### arunbg

Nice job there Curious.

To find the probability of the card colours turning out a certain way, an easier approach would be to reverse the deck and deal two cards from the top. What is the probability that the two cards are of the same colour? Of diff. colours?
I haven't tried this ( too lazy ) but you ought to get the same answer shouldn't you ?

Last edited: May 16, 2006
14. May 16, 2006

### Curious3141

Yes, you are completely correct. The situation is completely reversible. Reversing the draw, the first card choice is immaterial. The prob of getting a similar card with only 25 of that sort left in the deck is 25/51. Prob of dissim is therefore 26/51, and everything works out the same, with much less fuss.

That's a much nicer job than mine. :grumpy: :rofl:

15. May 16, 2006

### arunbg

Ok I tried it and it worked beautifully.
Here's what I did.

To find the required probability, rather than dealing cards till the 51st and 52nd, we can deal the cards in the opposite sense to give the same probability as I mentioned in my earlier post.

Simply find the probability of getting two cards of the same colour and different colours separately(in two draws) , multiply with the probability that you get your guess is right and you have the answer. Mathematically, this comes out as
P = P(getting red and black)P(guessing right) +
P(getting both black or both red)P(guessing right)
= 26/51*1/2 + 25/51*1 = 38/51

which is the same answer Curious obtained. Pretty simple isn't it ?
Of course Curious' expression for the general probability is better when the cards for which you have to calculate the probability are more towards the middle.

Edit: Curious didn't see your post

Last edited: May 16, 2006
16. May 16, 2006

### HallsofIvy

Staff Emeritus
Of course, this problem had nothing to do with "Bayesian probability". I'm wondering why blumfeld0 would talk about "Bayesian probability" when in a more recent thread he has claimed not to know how calculate the probability of drawing a red ball from an urn containg 4 red and 4 white balls!

17. May 16, 2006

### Curious3141

"Bayesian" has sort of a nice esoteric ring to it. It's the same reason why I've been seeing threads entitled "Relativity Theory" that discuss nothing more abstruse than F = ma. :rofl:

18. May 16, 2006

### blumfeld0

Thanks for all the great replies
it was in the same chapter as bayesian probability so i figured it was bayesian probability