BrainDen.com - Brain Teasers

## Question

"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.

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• 0

Assuming there are 52 cards, four of which are aces, and that the cards are shuffled randomly and not in any order that we know about:

drawing one ace: 4/52

If he did draw an ace, he takes it away, now there are 51 cards- and 3 aces

drawing another ace: 3/51

(in other words, Alex is correct)

In order to get both, it is AND, and 'and' is multiplication: (4*3) / (51*52) or 12/2652 or 1/221

Assuming the deck isnt touched, is still random, has 52 cards, etc:

To keep drawing and draw two CONSECUTIVE aces, you WILL draw one ace. The chances to draw an ace on the next card is n/# where n is the number of aces left in the deck and # is the number of cards left in the deck.

So now we have to figure out the average probability.

Say Card 1 is an ace.

We have a 3/51 chance. If it is not an ace, we have 3/50 for the next draw, etc. This method is nice for looking and understand, but not for getting the answer.

We have to look at the overall positioning of the deck, saying its random, since nothing is moved or replaced between drawings. Now it is clear that the cards being randomly distributed is important.

The chance of one particular ace being a particular spot: 1/52

The chance of another partciular ace of being in a particular spot (above or below): 1/52

Thus one would think that there are 1/(52^2) possibilities. But they would be forgetting that there are FOUR aces and TWO spots- above and below the first card. First the above and below:

First card: 1/52

Second card: 2/51 (twice chance of first)

however now we factor in that there are FOUR and then it would be down to THREE if one is drawn:

First card: 4/52

Second card: 6/51 (3*2)

thus 24/2652 or 2/221

Remember the first bet was 1/221

This only twice as likely... but neither are very likely.

If he did go for the bet, to make it even and if he pays a penny (\$0.01) to do it:

2/221 = 1/110.5

He should win \$1.10 or \$1.11 if he wins, depending on how you round

am i close to right? lol ;p

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• 0
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.

What Matt proposed was not that Alex had to get aces

on the first two cuts, but only that two successive cuts

be aces as he continued thru the deck. The first 30 cards

could be non-aces, for example, followed by aces on the

32nd and 33rd cuts.

I think that's what you calculated. And now I must confess I haven't

done the calculation, which I will take care of, and then post my result. I think the calculation is made easier by considering the

probability that a particular ace [hearts, say] is followed

immediately by one of the other aces in a well-shuffled deck,

then multiply that by four. [not followed or preceded;

I think that leads to double counting.]

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• 0

However hearts may not be the first one... but i dont think you could multiply it by 4 exactly cuz of the other 3.... lol we'll see if our calculations match

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• 0

Alex wins. By a lot.

Here's why:

Label the aces as:

H-ace of hearts,

D-ace of diamonds,

C-ace of clubs.

There are 12 equally likely consecutive-ace [CA] possibilities:

SH SD SC

HS HD HC

DS DH DC

CS CH CD

where SH means ace of spades followed by ace of hearts. etc.

Given the [equal] probability of any of these sequences, say p[sH],

the probability for any two consecutive aces is

p[CA] = 12 x p[sH]

Denoting position in the deck by 1, 2, 3, ..., 51, 52,

p[sH] = p[1,2] + p[2,3] + ... + p[51,52]. where p[1,2] is prob of spade being card 1, heart being card 2, etc.

The probability of a card being in a specified place in the deck is 1/52.

The probability of a 2nd card being in a different specified place in the deck is 1/51.

So each term in the exp​ression for p[sH] is equal to [1/52] x [1/51].

There are 51 terms, so

p[sH] = 51 x [1/52] x [1/51] = 1/52.

and so

p[CA] = 12 x p[sH] = 12/52 = 3/13.

In the proposed bet, the odds favor Davey by 4 1/3 to 1.

If he gives 100 to 1 odds [a dollar for a penny], then

Alex can expect to win about 23 cents on his penny.

Once again, Alex has the better of the deal.

Shall we call him a Smart Alex?

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• 0 Alex wins. By a lot.

Here's why:

Label the aces as:

H-ace of hearts,

D-ace of diamonds,

C-ace of clubs.

There are 12 equally likely consecutive-ace [CA] possibilities:

SH SD SC

HS HD HC

DS DH DC

CS CH CD

where SH means ace of spades followed by ace of hearts. etc.

Given the [equal] probability of any of these sequences, say p[sH],

the probability for any two consecutive aces is

p[CA] = 12 x p[sH]

Denoting position in the deck by 1, 2, 3, ..., 51, 52,

p[sH] = p[1,2] + p[2,3] + ... + p[51,52]. where p[1,2] is prob of spade being card 1, heart being card 2, etc.

The probability of a card being in a specified place in the deck is 1/52.

The probability of a 2nd card being in a different specified place in the deck is 1/51.

So each term in the exp​ression for p[sH] is equal to [1/52] x [1/51].

There are 51 terms, so

p[sH] = 51 x [1/52] x [1/51] = 1/52.

and so

p[CA] = 12 x p[sH] = 12/52 = 3/13.

In the proposed bet, the odds favor Davey by 4 1/3 to 1.

If he gives 100 to 1 odds [a dollar for a penny], then

Alex can expect to win about 23 cents on his penny.

Once again, Alex has the better of the deal.

Shall we call him a Smart Alex?

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 also can't multiply by 12 because you can't have SH and SC, for instance. In other words, those probabilities aren't independently distributed. There is also a small issue in that Alex wouldn't get paid twice if, for example, SH and DC came up in the same deck.

Long story short, the answer is not 3/13. It is much smaller than 3/13. I've looked at this a little while, and I don't see any way to solve it without getting into some ridiculously long and complicated expressions and just brute forcing. It would probably be faster to do it experimentally.

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• 0

Hi Chuck,

You have some interesting objections. I'll try to explain.

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].

Obviously.

And he doesn't need for that to happen.

Alex gets paid if it occurs as [1,2] or as [2,3], ... or as [51,52].

That's 51 equally likely winning situations.

If 51 events have equal probabilities pi, 1=1, 51, the probability of [event 1 or event 2 or ... or event 51] happening is 51p1.

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.

If the aces occur in the [2,3] position, for example, the probability that they do not occur somewhere else is unity.

You also can't multiply by 12 because you can't have SH and SC, for instance.

Obviously.

Fortunately, nothing in the bet requires impossible situations to occur.

Alex gets paid if he has SH or SC.

There is also a small issue in that Alex wouldn't get paid twice if, for example, SH and DC came up in the same deck.

That's not really an issue.

The bet states that Alex gets paid if he cuts consecutive aces.

The odds of that happening - whether or not it happens again or a 3rd time - is 3/13.

According to the bet, he should never be paid double or triple.

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• 0 Hi Chuck,

You have some interesting objections. I'll try to explain.

Obviously.

And he doesn't need for that to happen.

Alex gets paid if it occurs as [1,2] or as [2,3], ... or as [51,52].

That's 51 equally likely winning situations.

If 51 events have equal probabilities pi, 1=1, 51, the probability of [event 1 or event 2 or ... or event 51] happening is 51p1.

If the aces occur in the [2,3] position, for example, the probability that they do not occur somewhere else is unity.

Obviously.

Fortunately, nothing in the bet requires impossible situations to occur.

Alex gets paid if he has SH or SC.

That's not really an issue.

The bet states that Alex gets paid if he cuts consecutive aces.

The odds of that happening - whether or not it happens again or a 3rd time - is 3/13.

According to the bet, he should never be paid double or triple.

...I hate probability! hahaha

I'll take calculus over it any day...

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• 0 First, I have two different approaches to this puzzles.

The independent way:

Since the deck is already set, the result of one card really in a way is indepedent of the next because everything is set. So that leaves the original result of 1/221.

The dependent way:

The second way I approached this is similar to bonanova way but might be simpler to understand:

In every deck, there is 51 pairs of cards (drawing one by one, ie pair of card [1,2] [2,3]

Then I wrote out all the possibilities with the order

2,2..2,3.....2,A

For each set there is 13 different pairs, every pair except for matching has 16 different possibilities. The matching pairs have only 12 different possibilites.

That means, 12+(16*12) = 204 different combinations for each card.

13 different cards, 204*13 = 2,652

Since there are only 12 different combinations of Aces, and 51 different trys to achieve it you have the following:

(12*51)/2652 resulting in Bonanova result of 3/13 or 1 in 4.33 odds.

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• 0 i remember you had like 80 posts yesterday.. i fink lol

anyhu i love card odds

well readin them more than anything

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• 0 If 51 events have equal probabilities pi, 1=1, 51, the probability of [event 1 or event 2 or ... or event 51] happening is 51p1.

Here's a counterexample I hope will help you see the flaw in this reasoning: If you roll a standard dice, the probability of getting an even number is 1/2. The probability of getting a number less than or equal to three is also 1/2. However, the probability of getting a number that is even or less than or equal to three is not the sum of those probabilities. 1/2 + 1/2 is 1, but there is a 1/6 chance the numb er you roll will be neither even nor less than or equal to three (5).

Since those events are not disjoint, you need to subtract the probability of their union - that is, Pr(roll is even or <= 3) = Pr(roll even) + Pr(roll <=3) - Pr(roll even AND <= 3).

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• 0 Alex wins. By a lot.

Here's why:

Funny, the "eyeballing it" math works out exactly the same.

52 cards pulled, 51 changes you get an ace and a shot at a 2nd ace. First time you get 3 chances at another ace, then 2, then 1.

But what about when you find the aces? Does it come up early ( */51) or late ( */1).

"Eyeball" method (best math I am able to fathom, to be honest), is to just try and ballpark it by adding the 2 ends (51 and 1), divide by 2, and roll with it and see what you get.

3/26 + 2/26 + 1+26 = 6/26.

Or... 1 out of 4.33.

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• 0
Here's a counterexample I hope will help you see the flaw in this reasoning: If you roll a standard dice, the probability of getting an even number is 1/2. The probability of getting a number less than or equal to three is also 1/2. However, the probability of getting a number that is even or less than or equal to three is not the sum of those probabilities. 1/2 + 1/2 is 1, but there is a 1/6 chance the numb er you roll will be neither even nor less than or equal to three (5).

Since those events are not disjoint, you need to subtract the probability of their union - that is, Pr(roll is even or <= 3) = Pr(roll even) + Pr(roll <=3) - Pr(roll even AND <= 3).

Thanks for clarifying what you're having difficulty with.

Your die analysis is correct, and in fact nicely illustrates the matter. You state that p[even] is 1/2. How do you arrive at that result?

Simply by noting that p = 1/6, p = 1/6, and p = 1/6; and these events are disjoint.

So, paraphrasing my previous comment, with the disjoint property stated for clarity,

If 3 [disjoint] events have equal probabilities pi, i=1, 3, the probability of [event 1 or event 2 or event 3] happening is 3p1.

The three events are rolling a 2, rolling a 4 and rolling a 6, and their individual probabilities are all 1/6.

p[even] = p[2 or 4 or 6] = p + p + p = 3p = 1/2.

When you combine p[even] with p[<=3] i.e. apples and oranges, it's a different matter, as you point out.

Hope that helps.

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• 0 Alright, let's see if your method works on another, simpler problem. Suppose that instead of having 4 aces out of 52 cars, you had 3 aces out of 5 cards. What would be the probability of getting two consecutive aces then?

Using your method, you would first find the number of permutations of those 3 aces, which is 6: SH, HS, SD, DS, HD, DH (for example). For each of those permutations, you would say that there are 4 pairs of cards which could be consecutive aces; the probability of the correct ace being in the first spot being 1/5, and the probability of the correct ace being in the second spot being 1/4. This would lead to a total probability of 6*4*(1/5)*(1/4), which comes out to 6/5. Obviously, that method doesn't work here, since probabilities must be, by definition, in the interval [0,1].

I'll demonstrate the easiest solution I can think of for the original problem (which is, of course, not that easy at all) on this simpler example (although there is an easier way to do it for this example). To simplify the notation, I'll call aces A and other cards N. I'll try to format this so you can follow what I'm saying - the sequence will be on the left, and the probability of that sequence given the sequence before it (in other words, the probability of the last card pulled being what's shown) is on the right. I'll truncate the tree when the outcome (success or failure) is determined.

A 3/5 AA 2/4

AN 2/4 ANA 2/3

ANAAN 1/2

ANANA 1/2

ANN 1/3

ANNAA 1

N 2/5 NA 3/4 NAN 1/3 NANAA 1

NAA 2/3

NN 1/4 NNAAA 1

Multiplying all these probabilities out gives 3/5(2/4 + 2/4(2/3*1/2 + 1/3)) + 2/5(3/4(1/3 + 2/3) + 1/4). This gives a total probability of 9/10. We can confirm this by taking the probability to be 1-Pr(failure), since the only way to fail is the ANANA sequence. This sequence has probability (3/5) (2/4) (2/3) (1/2) = 1/10.

As you can see, going through this calculation, or any similar method, for a 4 ace in 52 card problem would be exceedingly tedious. There is probably some software available that will solve the original problem easily, but I don't know of it. In any case, I'm very confident there is no feasible way to solve the original problem by hand. There are just too many conditional probabilities.

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• 0 Alright, let's see if your method works on another, simpler problem. Suppose that instead of having 4 aces out of 52 cars, you had 3 aces out of 5 cards. What would be the probability of getting two consecutive aces then?

Using your method, you would first find the number of permutations of those 3 aces, which is 6: SH, HS, SD, DS, HD, DH (for example). For each of those permutations, you would say that there are 4 pairs of cards which could be consecutive aces; the probability of the correct ace being in the first spot being 1/5, and the probability of the correct ace being in the second spot being 1/4. This would lead to a total probability of 6*4*(1/5)*(1/4), which comes out to 6/5. Obviously, that method doesn't work here, since probabilities must be, by definition, in the interval [0,1].

I'll demonstrate the easiest solution I can think of for the original problem (which is, of course, not that easy at all) on this simpler example (although there is an easier way to do it for this example). To simplify the notation, I'll call aces A and other cards N. I'll try to format this so you can follow what I'm saying - the sequence will be on the left, and the probability of that sequence given the sequence before it (in other words, the probability of the last card pulled being what's shown) is on the right. I'll truncate the tree when the outcome (success or failure) is determined.

A 3/5 AA 2/4

AN 2/4 ANA 2/3

ANAAN 1/2

ANANA 1/2

ANN 1/3

ANNAA 1

N 2/5 NA 3/4 NAN 1/3 NANAA 1

NAA 2/3

NN 1/4 NNAAA 1

Multiplying all these probabilities out gives 3/5(2/4 + 2/4(2/3*1/2 + 1/3)) + 2/5(3/4(1/3 + 2/3) + 1/4). This gives a total probability of 9/10. We can confirm this by taking the probability to be 1-Pr(failure), since the only way to fail is the ANANA sequence. This sequence has probability (3/5) (2/4) (2/3) (1/2) = 1/10.

As you can see, going through this calculation, or any similar method, for a 4 ace in 52 card problem would be exceedingly tedious. There is probably some software available that will solve the original problem easily, but I don't know of it. In any case, I'm very confident there is no feasible way to solve the original problem by hand. There are just too many conditional probabilities.

If you up that to 10 cards, 2 aces. His formula would result in the following:

1/10*1/9*9(different pairs)*2=

2/10 or .2

On the other hand, listing out every combination results in 45 unique order without order mattering (like your method) and you get 45 different ways, with 9 being valid. 9/45 = .2

While bononova method may not work with the lower extremities, it does well at estimating large numbers faster, Will it be precise? No. Will it be close enough to place a bet on it, yea especially with those odds.

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• 0

Hi Chuck,

I'll meet you halfway on this one.

Multiplying by 51 is entirely correct in that it gives the probability of Spade ace followed by Heart ace anywhere in the deck.

And it works to multiply this result by 2 to have those two suits in either order.

When we limit things to those two suits, there is no possibility of double counting - there are no other simultaneously occurring events of relevance.

But when it's opened up to the other two suits, multiplying by 12 does allow multiple counting.

These are the multiple-counting cases:

 .....AA.....AA..... [two occurrences]

 ....AAA.....A...... [two occurrences]

 ......AAAA......... [three occurrences]

I've tinkered with enumerating these cases to get a closed form solution.

While I'm doing that I computed 10 million shuffles, which provides a fair degree of confidence in the results.

Summing the CAOs gives 2308409 for a CAO probability of .23084..., which agrees with 12/52 = .23077...

Summing the decks that had at least one CAO gives 2174267 for a probability of .2174...

So the number of decks that have at least one CAO is about 6% fewer than the number of CAOs.

That is, ignoring the multiple occurrences over-counts by about 6%.

Thanks for a lively discussion, and apologies for not seeing the flaw in my thinking until now.

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• 0 Out of curiosity, what software did you use to do that simulation, and is it free?

Or, alternatively, how many Malaysian children do you have slaving away for 16 hours a day in your probability factory for pennies an hour shuffling decks of cards, and what are your monthly expenses for operating and maintaining this facility?

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• 0
Out of curiosity, what software did you use to do that simulation, and is it free?

Or, alternatively, how many Malaysian children do you have slaving away for 16 hours a day in your probability factory for pennies an hour shuffling decks of cards, and what are your monthly expenses for operating and maintaining this facility?

I can't respond to that ... INS may monitor these boards. APL.

You prob never heard of it.

Very handy, not free.

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• 0

To be honest here, I started reading this thread and then thought:

Can we reduce all this into:

"In a well shuffled deck, what is the probability to get 2 consecutive aces somewhere inside?"

This way if we draw them one by one, we will get 2 consecutive aces at sometime...

Edited by roolstar

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• 0
To be honest here, I started reading this thread and then thought:

Can we reduce all this into:

"In a well shuffled deck, what is the probability to get 2 consecutive aces somewhere inside?"

This way if we draw them one by one, we will get 2 consecutive aces at sometime...

coz that's very easy to calculate: Let's chose a pair (Ace of Hearts ; Ace of Spades).

A = Possibilities of shuffled decks = Permutations of 52 cards = 52!

B = Possibilities of shuffled decks with the 2 aces in a row:

You consider the pair of aces to be 1 card and you get a 51 cards deck

==>B = Permutations of the two aces in the pair * pemutations of a 51 cards deck = 2!*51!

Prob(2 consecutive aces of a specified kind in a shuffled deck) = B / A = 2*51!/52! = 1/26

Now since we have 6 possible pairs of aces (Spades + Hearts, Spades + Diamond.....)

P(2 consecutive aces) = 6 * B = 6 * 1/26 = 3/13

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• 0
coz that's very easy to calculate: Let's chose a pair (Ace of Hearts ; Ace of Spades).

A = Possibilities of shuffled decks = Permutations of 52 cards = 52!

B = Possibilities of shuffled decks with the 2 aces in a row:

You consider the pair of aces to be 1 card and you get a 51 cards deck

==>B = Permutations of the two aces in the pair * pemutations of a 51 cards deck = 2!*51!

Prob(2 consecutive aces of a specified kind in a shuffled deck) = B / A = 2*51!/52! = 1/26

Now since we have 6 possible pairs of aces (Spades + Hearts, Spades + Diamond.....)

P(2 consecutive aces) = 6 * B = 6 * 1/26 = 3/13

And voila Yah, but now CR is gonna get all over you for counting ace pairs instead of counting decks.

See post 17.

Courtesy of a brilliant colleague, the proper calculation goes something like this:

Imagine a well shuffled deck in this way: ......A...[k]...A...[l]...A...[m]...A...[j]... where the dots are non Ace cards.

i, j, k, l and m are the numbers of non-Ace cards around the four Aces.

Consider the case of zero Ace pairs: at least one card separating the Aces -- k, l and m are nonzero.

How many ways can that happen?

N[zero ace pairs] = sum(i=0,45) [ sum(j=0,45-i) [ sum(k=1,46-i-j) 47-i-j-k ]] since there are 47-i-j-k ways we can assign l and m.

Permitting ace pairs simply means allowing k, l and m to be zero:

N[permit ace pairs] = sum(i=0,48) [ sum(j=0,48-i) [ sum(k=0,48-i-j) 49-i-j-k ]] - now there are 49 ways to assign l and m.

p[at least one ace pair] = 1 - p[no ace pairs] = 1 - N[no ace pairs]/N[permit ace pairs]

p[at least one ace pair] = 1-(C(49,4)/C(52,4)) = 1201/5525 = .217... which is a little less than 3/13.

And .217... agrees with my 10-million simulated shuffle results.

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• 0 alex wins! w00t!~

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• 0 Ok if I am reading the question right- every time the deck is cut the cut card is removed from the deck. So lets say that you cut to the top card in the deck, then you cut to the 2nd card if both are Aces you would win the bet if not then you would continue cutting the top card of the deck untill you get 2 Aces in a row or you see 3 Aces and the next card isn't an Ace, at which you would pay the winner and shuffle the deck again and start over. Well the odds of getting 2 Aces in a row in a randomly shuffled deck is the same as being delt hole Aces in holdem', because being delt aces is pulling 2 aces that are spaced evenly apart, in your riddle the 2 aces would be spaced together. So the same odds would apply as in holdem'. While I don't know how the equasion works I know that the odds are 220 to 1. So if both players agree that they will keep playing until Alex wins for Alex not to lose money he would need at leat 2.21 to his .01.

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• 0 alex of course

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• 0 I don't know about you guys but when I cut a deck I don't pull the card from the deck. In that case the math is MUCH easier.

I suppose the question then becomes "If Alex had to pay a penny a cut would it be worth it to bet \$1?

The odds of getting the first ace is 4/52 (1/13), the odds of getting the second ace is still 1/13. This makes the overall odds 1/13^2 or 1/169... Alex would pay an average of \$1.69 per dollar won. Of course he could win after paying \$0.02 or never...

Of course this is assumed he has no magician-like skills with a deck of cards.

The Lottery is a tax on people that failed Statistics. ## Join the conversation

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