# A little math fun....

Eazzy
Red Chipper Posts:

**1,004**✭✭✭✭
You have a deck of cards ...you turn over cards till an Ace comes.....

whats the odds that the next card will be another Ace....

whats the odds that the next card will be another Ace....

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## Comments

85✭✭1,145✭✭✭✭34✭✭P(A) on the first card is .08 (4/52), P(A) on the second card given !A on the first card is (3/51*48/52), each successive probability of first Ace is 4/remaining * sum(P(A)) on any earlier card.

P(second A) is 3/remaining cards * P(first A | card number)

Total prob = sum(P(A, 1_52)) * P(second A | P(A, 1_52))

1,145✭✭✭✭2,964-1,004✭✭✭✭no the question is before you deal any cards......whats the odds that the next card after the first ace is dealt will be another ace....

2,964-34✭✭It's a finite series (the first A has to come in 49 cards) so a spreadsheet solves.

I'm on the edge of my seat to see if my answer was right @Eazzy

2,964-Right. Ironically it's easier to do it analytically if the series is infinite, but that would require an unusual deck of cards.

Still wondering if there's some sort of argument from symmetry to do this quicker tho.

1,004✭✭✭✭I you shuffle a deck of cards.....

what the odds that the bottom card is an ace....

Same deck now deal out the cards till you turn over an ace.....

whats the odds the bottom card is an ace....

There is no difference between the bottom card or the card after an ace once an ace is turned over (in terms of this question).

its still 13 to 1.....

2,964-2,964-Suppose you restrict yourself to a 4-card deck:

We can again ask, if we deal out the cards, what is the probability that an A will immediately follow the first A? The trap is to assume we are now choosing one A from 3 cards giving a probability of 1/3, whereas the correct probability is actually 1/2 - the equivalent of 1/13 in the full-deck example.

You can also ask the question, what is the probability that the card dealt after the is the ? Is the answer now 1/3? 1/2? Or something else?

1,004✭✭✭✭If you look for a particular Ace that changes the problem because the AH could be dealt out before you get to the As

Another way to look at the solution is to realize...that the cards dealt before the the first ace on average, will offset the fact that the first Ace takes an ace out of the deck...

In the 4 card example

if the first card was and ace its 2 to 1 the next card is an ace

If its dealt 2nd then it its even odds...

and if the ace is dealt 3rd its 100 %.....

On average thats 0% of the time....

2,964-4,947✭✭✭✭✭Assume we have turned over X cards until we found the ace. The probability of the next card being an ace is 3/(52-X).

While you seem to be looking for some sort of average number, it will rarely be the case that the odds of turning over an ace after you've found the first one will be 1/13. The only time that will be true is if you just happened to find the first ace on the 13th card.

4,947✭✭✭✭✭1 out of 13 and 13 to 1 are not the same thing.

4,947✭✭✭✭✭2,964-No, it's always 1/13. Or if you prefer, the average is 1/13 and a probability is in some sense an average.

4,947✭✭✭✭✭Well obviously this isn't true. "You turn over cards until an ace comes. whats the odds that the next card will be another Ace"

The first card in the deck is an ace. The probably the next card is an ace is 3/51.

The 23rd card in the deck is the first ace. The probability the next card is an ace is 3/29.

Probably just poorly worded first attempt. Reading through the thread I think he is actually asking what is the chance the first 2 aces are back to back.

4,239✭✭✭✭✭2,964-The point is, a probability for this general situation cannot then say "let's suppose the first card is an A." It's like an EV equation. We don't claim the probability of hitting a flush draw is 1 when we hit it and 0 when we miss. Similarly, in this general problem, on any given realization you can claim the probability is not 1/13 once the first A has been revealed - it then follows the formula you gave.

If you do this with my 4-card simplification and simply write down every permutation, you can see this explicitly simply by averaging all the 0s and 1s.

2,964-4,947✭✭✭✭✭2,964-For a related problem: What's the probability that the is dealt immediately after:

1) Another Ace

2) The

2,964-Compared to what I'd be doing otherwise, this is definitely fun.

4,947✭✭✭✭✭Yes, but I meant that still has to jibe with a real math answer that adds up all the probabilities. I'm sure there's an equation.

2,964-Sure just using brute force would get there.

119✭✭71✭✭I must be missing something. Lets say we turn over cards until only four cards are left the odds of turning over all aces is now 100%.

Edit: O.K. looks like the OP wants to know the odds before we start to turn over cards? Would it be 1/17 (16 to 1 odds)? We know of one Ace removed but there are 51 more cards in the deck and 3 of them are aces. Its the odds of being dealt a pocket pair of Aces except we are guaranteed the first Ace.

71✭✭