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[icon] My answer to the probability puzzle - Patti
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Subject:My answer to the probability puzzle
Time:02:33 pm


This reduces to a five-card problem-- four aces and the two of clubs.

Using our five-card deck, 20% of the time, the first card in the deck will be the two of clubs. In this case, we really care about the third card off the deck, because the second card will be the first ace. Obviously, the third card in the deck will never be the 2c when the first card is. 25% of this time, the third card will be the ace of spades. .20 * .25 = .05.

Running total: 5% ace of spades.

20% of the time, the first card in the deck will be the ace of spades. In that case, 25% of the time the next card will be the 2c, and the second card will never be the As when the first one is. Again, .2 * .25 = .05.

Running total: 5% ace of spades, 5% 2 of clubs.

The remaining 60% of the time, the first card in the deck will be one of the other aces. When that happens, the 2c will be the next card 25% of the time, and the As will be the next card 25%
of the time. .60 * .25 = .15 for the As, and .15 for the 2c.

Running total: 20% ace of spades, 20% 2 of clubs. That leaves 60% of the time for the next card to be neither of those two.

These numbers are for the five-card case. When you add the other 47 cards to the deck the absolute probabilities change, but the relative probabilities don't. In fact, the absolute probability of the next card being (any card) is 1/N, where N is the number of cards in the deck. In a 52-card deck, the next card after the first ace will be the ace of spades 1 time in 52, and it will be the two of clubs 1 time in 52.
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ronsrants
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Time:2009-08-20 09:58 pm (UTC)
So...was I right?

-R
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whipartist
Link:(Link)
Time:2009-08-20 10:02 pm (UTC)
When the first card is the 2c and the second card is the As, then the third card will (obviously) never be the As or the 2c. Please feel free to add a 0 to the .05 if you'd like.
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loser_variable
Link:(Link)
Time:2009-08-20 10:16 pm (UTC)
I'll take a shot. My assumptions: Flip until you get an ace, then is the AS or 2C more likely to be the NEXT card?

I believe you can reduce this to the 5 card deck and maintain correct RELATIVE probabilities although the absolutes would of course be different.

Five card deck (AS,AH,AD,AC,2C), FIRST card off is AS/2C/Other with 20/20/60 probability.

When first card is AS, next card is 2C 25% of the time.
Running wins AS = 0% 2c = 20%*25% = 5%

When the first card is the 2C, the next card will be a non-space ace 75% of the time, followed by AS 33% of the time.
Running wins 2C = 5%, AS = .2*.75*.33 = 4.95%

When the first card is a non-spade ace (60%), AS and 2C each follow with 25% probability.
Running totals 2C = 20%, AS = 19.95%.

2C weasels out the relative win which I suspect translates to the non-reduced problem. Maybe I'll try to write a sim. Assumptions are everything.

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timprov
Link:(Link)
Time:2009-08-20 10:19 pm (UTC)
You have rounding error. .75/3 = .25, not .2475.
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loser_variable
Link:(Link)
Time:2009-08-20 10:39 pm (UTC)
Good point, I didn't divide by 3, I multiplied by .33. Division is surely preferable and more accurate. You win.
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rmd
Link:(Link)
Time:2009-08-20 11:12 pm (UTC)
In a 52-card deck, the next card after the first ace will be the ace of spades 1 time in 52

I don't think so. It'll be 1 time in X, where X is the number of cards remaining after the first non-spade ace. It's only 1 in 52 if you're randomly picking a card, not removing cards from the set of cards that includes the As. (see also Monty Hall.)
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adb_jaeger
Link:(Link)
Time:2009-08-21 12:21 am (UTC)
No, it's 1/52, I think.

Here's the output of a sim I wrote:

Deck of 5 cards, 120 permutations
after first ace
next card is As: 24 times
next card is 2c: 24 times

Deck of 6 cards, 720 permutations
after first ace
next card is As: 120 times
next card is 2c: 120 times

Deck of 7 cards, 5040 permutations
after first ace
next card is As: 720 times
next card is 2c: 720 times

Deck of 8 cards, 40320 permutations
after first ace
next card is As: 5040 times
next card is 2c: 5040 times

Deck of 9 cards, 362880 permutations
after first ace
next card is As: 40320 times
next card is 2c: 40320 times

Deck of 10 cards, 3628800 permutations
after first ace
next card is As: 362880 times
next card is 2c: 362880 times


Note the pattern. For every deck size I tested, once you turn over an ace, the next card is the ace of spades with probability 1/n, where n is the deck size:

5: 24/120 = 1/5
6: 120/720 = 1/6
7: 720/5040 = 1/7
8: 5040/40320 = 1/8
9: 40320/362880 = 1/9
10: 362880/3628800 = 1/10

I assume it keeps growing in sequence so that for 52, it's 1/52, just like luckylefty said it would be, using the "bottom dealing" explanation.

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loser_variable
Link:(Link)
Time:2009-08-21 01:49 am (UTC)
SIMing is always the way. Theoretical calculations are nice, but a well designed Monte Carlo is generally what you want to bet on.
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rmd
Link:(Link)
Time:2009-08-21 02:59 am (UTC)
hm. i believe the math, but i haven't figured out where my conceptualization of the problem is failing.

i think i'm having the reverse of the problem many people have when encountering the monty hall problem.
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markgritter
Link:(Link)
Time:2009-08-21 12:54 am (UTC)
Nifty.

You could make the same puzzle about suits instead, with the same answer: Suppose you draw from the deck until a heart is shown. Is the next card more likely to be the Ah or the 2c?

The probability argument is good, but it doesn't show where intuition (that the As is less likely because we've seen one of the aces) goes wrong.
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thorfinn
Link:(Link)
Time:2009-08-21 01:40 am (UTC)
The other way to reframe it is:

1. randomness is introduced at the shuffle.
2. all the guff about "card after the first ace" translates to:
- Choose a random deck position from the second card to the 50th card.
3. The odds of *that* card being AS or 2C?

Exactly the same - 1/52.


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abostick59
Link:(Link)
Time:2009-08-21 01:54 am (UTC)
Suppose the first card is the 2c. Then Monty turns up (at his choice) either the fourth or the fifth card and it's a non-spade ace. Does that change your answer?
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[icon] My answer to the probability puzzle - Patti
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