Pair of Aces

[bonanova]

"I bet you can't cut two aces out of a shuffled deck," said

Davey to Alex down at Morty's last night. Davey was still

sore that Alex had flipped 10 Tails on Monday.

"I'm sure I wouldn't take that bet," Alex replied, "I happen

to know that my chances are [4/52]*[3/51], and I'll tell ya

right now that my dear momma didn't raise no fools!"

Matt the mathematician overheard the two boys talkin' and

pushed Alex a little farther: "But what if you could continue

cuttin' the deck until the cards were all gone? Do you know

what yer odds would be to cut consecutive aces, then?"

And he sipped his brew while Alex thought about it.

After a moment, Alex replied, "Give me a dollar against my

penny and yer on!"

Was Alex correct about his chances on the first bet?

What odds should he have demand to make the second bet fair?

Edited for clarity.

[Guest]

Now, I know I'm new to this board and this isn't exactly the best way to make friends, but you guys are all wrong.

The wording of the problem doesn't really make sense, and that's the main problem. What you guys have solved is, what are the chances of their being two consecutive aces anywhere in the deck. But, that's not the problem. The question asked what are the changes of being able to cut to two consecutive aces in the deck. There are a couple ways of considering this, depending on how you interpret cut consecutive aces:

1. The kind of cheating easy way. You don't remove cards when you cut the deck, so, if cut consecutive aces means make cut to aces in two consecutive cuts, you just cut until you get an ace and then cut again, if the next card is an ace you win, if not repeat. Since no cards ever leave the deck, you are an infinite favorite and will eventually get a two ace combination.

2. Assume we are going to be removing the cards we cut to and we are cutting to two cards that are consecutive in the deck. In this situation, there are really two probabilities to consider. The probability that there exists anywhere in the shuffled deck a two ace combination and the probability that we will cut to it. If we assume we are removing cards after we cut, it matters what counts as cutting to consecutive aces. If we only count when we cut to the top ace, then we need to figure out both probabilities separately and combine them. To give you an example, say a 10 card deck looks like this (A=ace, X=any other card):

XXXAAXXAXX

If cutting to 4 wins, but cutting to 5 loses (and also removes the pair from the deck), then the probability is the combined probability of the combination existing and us cutting to the right part of the deck to get credit for it.

3. If we can go in either direction, then there is just the probability that the two ace combination appears somewhere and there really is no point in cutting the deck, because you would find it eventually. But, that's not really true, because you still have the contigent probabilities of what happens when you cut to specific cards.

The problem is that by cutting to a particular card, you are changing the make-up of the deck, and so the probability of drawing consecutive aces is continuously changing contigent on the cards you draw. To give you an example, say a 10 card deck looks like this (A=ace, X=any other card):

XXAXXXAXXA

There are no consecutive aces here, so it doesn't matter what my first cut is, I'm not cutting to a consecutive ace. But, if my cut sequence is 8,8,7. By cutting and removing the cards between the aces, I'm creating my own consecutive ace situation.

I don't want to get into the math, but essentially what you would need to calculate in this situation is the probability of there existing a 2 ace combination anywhere in the deck PLUS the probability that you would cut to every card between two of the aces in the deck before you cut to that ace when it is non consecutive.

That feels like too contigent of a probability set to calculate because you would have to calculate the probabilities of drawing out all the intermediate combinations of cards for all the possible decks. You can't just calculate all the gaps because of situations like this:

XXAXXAXXAX

Where there are 2 two gaps, but if you were to remove the middle ace first, you would contigently create a 4 gap which is still possible to get.

With a monte carlo type simulation of 5,000,000 runs, I'm showing that you have about a 36% chance (1,818,282) of drawing consecutive aces if you could draw until the deck was exhausted.

So 100 to 1 greatly favors Alex. He could even ask for 3 to 1 and still be making money.

[Guest]

...

2. Assume we are going to be removing the cards we cut to and we are cutting to two cards that are consecutive in the deck. In this situation, there are really two probabilities to consider. The probability that there exists anywhere in the shuffled deck a two ace combination and the probability that we will cut to it. If we assume we are removing cards after we cut, it matters what counts as cutting to consecutive aces. If we only count when we cut to the top ace, then we need to figure out both probabilities separately and combine them. To give you an example, say a 10 card deck looks like this (A=ace, X=any other card):

XXXAAXXAXX

If cutting to 4 wins, but cutting to 5 loses (and also removes the pair from the deck), then the probability is the combined probability of the combination existing and us cutting to the right part of the deck to get credit for it.

...

With a monte carlo type simulation of 5,000,000 runs, I'm showing that you have about a 36% chance (1,818,282) of drawing consecutive aces if you could draw until the deck was exhausted.

So 100 to 1 greatly favors Alex. He could even ask for 3 to 1 and still be making money.

And just for kicks, I ran the simulation where you can't look both directions to determine whether you've cut to two consecutive aces (this also handles the situation where cutting consecutive aces means cutting exactly between them, which might be the straightforward understanding of the wording) and that comes out to 19% (988,847/5,000,000).

That's assuming you only remove one card. I also ran the simulation where if you don't cut exactly between the two aces, you remove both the top and bottom card and that reduces the chances down to 11% (552,037/5,000,000).

So, there are really huge differences in the chances, depending on how you interpret the problem, but I think the last reading above is the most strict and it's not even close to 100 to 1, so if Alex makes that penny to dollar bet, he's getting way the best of it.

[Guest]

watching bona and chuck go back and forth on this one was awesome. well done gentlemen. good puzzle.

[Guest]

alex wins! w00t!~

The math is precision, but taking a sharp axe to the deck is also cutting..

[Guest]

This is a tough one that has way to many permutations to spend my time doing the calculations. I read the problem as cutting the deck, checking the card. Cutting the deck again, checking the card and so on until you pull 2 aces in a row.

This of course means FIRST you start with the odds of getting any ace. 4/52 .. 4/4 (once you are down to 4 cards left your odds are unity).

Now once you draw an ace, the odds of getting the 2nd ace is 3/x (number of cards left).

But if you do not draw the 2nd ace then your odds change for drawing the 1st ace to 3/x and the odds of drawing the 2nd ace go to 2/(x-1).

If you once again do not get the 2nd ace your odds go to 2/x (number of cards left) and the odds for the 2nd card are 1/(x-1).

If you once again do not get the 2nd ace, game over.

So to calculate the odds you have to calculate the odds on every draw for 4,3, and 2 aces being left in the deck and then of 3, 2, 1 aces being left in the deck for 2nd ace.

Of course you can leave out some of the possibilities such as calculating the odds of drawing an ace on the first cut with less than 4 aces. Or of drawing an ace with 1 card left and 4 aces. First draw must be 4/52. and you can't go greater than unity.

Lot of work.

[Guest]

im not entirely sure that there is as mucha chance of drawing two as consectutive, anywhere in the deck.

most likely chances are that he would have none consequetive ending up with his chance of getting 2 consecutive further than 1/221.

[Guest]

Not surprisingly, both Bonanova's and roolstar's solutions were correct.

In regard to Chuck's argument against Bonanova's method:

Bonanova, your analysis is wrong because it double counts the probabilities. You can't take the SH as [1,2] probability and multiply it by 51 because if it occurs in the [1,2] spot, it obviously can't occur as [2,3] or [44,45]. To do it this way, you would need to multiply the second term by the probability that those two aces did not occur in the first spot (1 - first term), and then multiply the third term by probability that the two aces did not come up in the first two terms (1 - second term), etc.

You are correct that a proper way to directly calculate the probability of a Single Hand (Spades followed by hearts) would be to

take the probability that it occurs on the [1,2] + ![1,2]*[2,3] + ![1,2]*![2,3]*[3,4] + etc...

You'll find that this way actually also calculates out to 1/51 each time:

It's been established that p([1,2]) = 1/51. So the calculation as described above will be:

[1,2] + ![1,2]*[2,3] + ![1,2]*![2,3]*[3,4] + etc = (1/51) + (50/51)(1/50) + (50/51)(49/50)(1/49) + ... = (1/51) + (1/51) + (1/51) + ...

It works because everything cancels out. Hope that clears up confusion (if anyone is still looking at this topic).

[bonanova]

Not surprisingly, both Bonanova's and roolstar's solutions were correct.

In regard to Chuck's argument against Bonanova's method:

You are correct that a proper way to directly calculate the probability of a Single Hand (Spades followed by hearts) would be to

take the probability that it occurs on the [1,2] + ![1,2]*[2,3] + ![1,2]*![2,3]*[3,4] + etc...

You'll find that this way actually also calculates out to 1/51 each time:

It's been established that p([1,2]) = 1/51. So the calculation as described above will be:

[1,2] + ![1,2]*[2,3] + ![1,2]*![2,3]*[3,4] + etc = (1/51) + (50/51)(1/50) + (50/51)(49/50)(1/49) + ... = (1/51) + (1/51) + (1/51) + ...

It works because everything cancels out. Hope that clears up confusion (if anyone is still looking at this topic).

Perhaps also not surprisingly I am ...

I recall looking through Chuck's differing approach, still satisfied mine was all right.

Thanks for the vote of confidence, but I'm open to anyone showing me the error of my ways.

[Guest]

Perhaps also not surprisingly I am ...

I recall looking through Chuck's differing approach, still satisfied mine was all right.

Thanks for the vote of confidence, but I'm open to anyone showing me the error of my ways.

Pardon me if I am not current on this but 3/13 (=.2307692...) over counts the possibility of getting three aces in a row or two aces in a row twice in the same deck.

Follow your same procedure but consider at each stage the probability of not getting two aces in a row. These can be cleanly anded to get your answer, .2061189 . . . = 1-((2651/2652)^51)^12.

By the way your organization of the problem helped immensely.

[EventHorizon]

Now that this one is solved....why not solve a related one I posted about a week ago...

Question 3