Expected value of 3 cards dealt

froggy21 · 2009-11-16 02:38:49

If each card on a regular deck has points that corresponds to their number (like 2 of hearts is 2 points, 7 of clubs is 7 points), the Jack, Queen, King each being 10 points...what's the expected value of your opponent's hand if you deal them 3 cards? I know the empirical expected value...but I'd like to know -how- to get the theoretical expected value, please : )

Help please D: I've been stewing over this question for days now. The only way I can think of doing this is by doing a tree diagram to get each probability but that'll have like 1000 end branches -headdesk-

mathsyperson · 2009-11-16 03:30:45

You're safe to ignore conditional probability, because it'll cancel itself out.

eg. In the unlikely case that you draw an ace first, you're less likely to draw another ace subsequently.
But in the likely case that you don't draw an ace first, you're slightly more likely to draw one in the future. So all in all, the expected probabilities are the same.

That means that you just need to work out the expected value of one card, then multiply by 3.

simplyjasper · 2009-11-16 06:03:48

Since this is not Monty Hall....
I don't see why u need to have 1000 end branches...

Considering that every number of different suits yield the same points
First row should only include 1 - 9 pts with probability of 1/13 each and (10, J, Q, K) = 10pts with a probability of 4/13

bobbym · 2009-11-16 10:10:39

Hi;

The average value of each card is 6.53846 or 85 / 13. So the expected value of the sum of 3 cards is 255 / 13 or 19.6153846.

Computer simulation for 10 million trials = 19.6153 very close.

Last edited by bobbym (2009-11-17 20:53:26)

froggy21 · 2009-11-16 14:50:01

Thank guys! I wasn't sure if ignoring the conditional probability was ok but since ya'll confirmed that the difference is relatively negligible, I think it should be fine. Thank you very much!

bobbym · 2009-11-16 15:08:57

Hi froggy;

I now think that there is no difference. The 3 cards dealt were also random so questions of replacement do not apply. We could have just as easily picked the 34th, 2nd and bottom card.

To simplify the model supposing you had 2 cards in your deck. A 1 and a 2. What is the average value of the the top card? It is clearly 1.5, the average value of all the cards in the deck.

froggy21 · 2009-11-17 19:56:47

Hey bobbym,

Sorry, I understood your initial explanation because it coincided with mathsyperson's explanation, but I don't understand what you meant about it having no difference at all. Cause the probabilities will change won't it not? Counted, it's a very small and negligible difference. Like, if you draw a 1 for the first card, then the probability for 1 in the probability distribution of the deck for the second card will change from 1/13 to 3/52. Or did you mean that it's no difference -because- the change in probabilities is tiny enough to ignore?

Thank you though for your help : )

____________________________

(To simplyjasper,

But won't it turn out to be about 1000 end branches because each number on the 1st row would have a 10 branches to them, which also applies to each number on the 2nd row?)

bobbym · 2009-11-17 20:22:45

Hi froggy21;

You can't speak of replacements when you don't know what was taken out.

To simplify the model supposing you had 2 cards in your deck. A 1 and a 2. What is the average value of the the top card? It is clearly 1.5, the average value of all the cards in the deck.

I think the above quote explains it . Their is no difference (not even negligible). Look at my 2 card deck explanation (Do a small tree). Since the cards chosen are random it exactly balances out. Think of a 3 card deck with a one, two and three in it. What is the average value of the sum of the top 2 cards. Clearly it is 4. The average value of each card times 2. Do a small tree and you will see what I mean.

Also I ran off by computer all 22100 possible 3 card combinations from a 52 card deck. Added them all up using your point system and divided by 3 and got 255 / 13 (same as the average method). This is, I believe the exact answer.

Last edited by bobbym (2009-11-17 22:07:23)

gckc123 · 2009-11-17 20:41:29

You can imagine a simpler scenario.
There are 10 balls in an urn. 5 are red and 5 are white.

You get 1 point when you get a red and get 0 when you get a white.
You now draw two balls from the urn. What is the expected points that you will get?

Cheers

bobbym · 2009-11-17 22:12:51

Hi froggy21;

Look at my post #8 and gkc123's post #9. The average value of a ball is 1/2 so the expected point total is 1 for 2 balls. Tree it and you will see.

Hi gkc123;
Haven't bumped into you yet, so I haven't had a chance to say hello. Welcome to the forum!

froggy21 · 2009-11-18 08:55:27

Oooh, I get it now. The tree diagram really helped.

Thank you very much, everyone for all the help! : )

gckc123 · 2009-11-18 11:23:40

Hi Bobbym.

I am new to this forum.
Getting bored with my exams and i found this when i surf the web aimlessly

Math Is Fun Forum

#1 2009-11-16 02:38:49

Expected value of 3 cards dealt

#2 2009-11-16 03:30:45

Re: Expected value of 3 cards dealt

#3 2009-11-16 06:03:48

Re: Expected value of 3 cards dealt

#4 2009-11-16 10:10:39

Re: Expected value of 3 cards dealt

#5 2009-11-16 14:50:01

Re: Expected value of 3 cards dealt

#6 2009-11-16 15:08:57

Re: Expected value of 3 cards dealt

#7 2009-11-17 19:56:47

Re: Expected value of 3 cards dealt

#8 2009-11-17 20:22:45

Re: Expected value of 3 cards dealt

#9 2009-11-17 20:41:29

Re: Expected value of 3 cards dealt

#10 2009-11-17 22:12:51

Re: Expected value of 3 cards dealt

#11 2009-11-18 08:55:27

Re: Expected value of 3 cards dealt

#12 2009-11-18 11:23:40

Re: Expected value of 3 cards dealt

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