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## Question

Yesterday while playing poker with friends, I suddenly came up with this problem.

There are 4 players on the table (including the dealer). A deck is shuffled and cards are dealt. What's the probability that each of the 4 players has received an Ace (as hole card)?

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(4*48 * 3*47 * 2*46 * 1*45) / (48*47*0.5 * 46*45*0.5 * 44*43*0.5 * 42*41*0.5)

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There are exactly 2,652 two-card combinations. 52 * 51

Correction – it would be 2,652 combinations only if the ORDER of the two cards mattered. Divide that by 2, and that’s the number of combinations without consideration of the order of the 2 cards.

There are exactly 1,326 two-card combinations without regards to the order of the cards. (52 * 51) / 2

How many of those two-card combinations contain exactly 1 ace?

Ace of clubs plus any of 48 non-Ace cards.

Ace of diamonds plus any of 48 non-Ace cards.

Ace of hearts plus any of 48 non-Ace cards.

Ace of spades plus any of 48 non-Ace cards.

That’s 192 two-card combinations that include exactly one Ace.

Deal 8 cards to 4 players. It doesn’t matter how. Let’s take each player 1 at a time.

Player A has a 192 out of 1,326 chance that he has one of those two-card combinations.

Assume that he does. Now there are only 50 cards left in the deck, only 3 of which are Aces. There are only 1,225 two-card combinations left [ (50 * 49) / 2 ], and only 3 Aces times only 47 non-Ace cards to create the right hand. That’s 141 two-card combinations left that contain exactly 1 Ace.

Player B, therefore, has a 141 out of 1,225 chance that he has one of the remaining two-card combinations of a single Ace.

Assume that he does. Now there are only 48 cards left in the deck, only 2 of which are Aces. There are only 1,128 two-card combinations left, [ (48 * 47) / 2 ], and only 2 Aces times only 46 non-Ace cards to create the right hand. That’s 92 two-card combinations left that contain exactly 1 Ace.

Player C, therefore, has a 92 out of 1,128 chance that he has one of the remaining two-card combinations of a single Ace.

Assume that he does. Now there are only 46 cards left in the deck, only 1 of which is the remaining Ace. There are only 1,035 two-card combinations left, [ (46 * 45) / 2 ], and only 1 Ace times only 45 non-Ace cards to create the right hand. That’s 45 two-card combinations left that contain exactly 1 Ace.

Player D, therefore, has a 45 out of 1,035 chance that he has one of the remaining two-card combinations of a single Ace.

Multiply all the chances together since they all MUST happen to meet your criteria.

It has a 0.0059100563302243974512882075907286% chance of happening.

Put another way, it will happen only once per every 16,920.3125 deals.

The first formula ends up giving you a 1 out of 8,484.4375 chance. I'm just not sure where I'm going wrong.

I THOUGHT I was right until I read the first person to answer. If I were a betting man, I'd probably figure I was most likely to be wrong. At any rate, there MUST be an easier way to do this problem than the way my wee brain drew it up. Can someone enlighten me as to where (if at all) I went wrong, and what the easier way to think of this problem is?

Thanks all! I enjoy this forum even though it's just a bit over my head.

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Each person needs to get an ace and a non-ace.

Person 1: Chance of getting an ace = 4/52, Chance of getting a non-ace = 48/51 (the order of these doesn't matter, if he gets a non-ace first then the chances are 48/52 and 4/51, which when multiplied together give the same thing)

Person 2: Chance of getting an ace = 3/50 (three aces left, only 50 cards left in the deck), Chance of getting a non-ace = 47/49

Person 3: Chance of getting an ace = 2/48, Chance of getting a non-ace = 46/47

Person 4: Chance of getting an ace = 1/46, Chance of getting a non-ace = 45/45 = 1

Multiply all of these together:

(4/52)*(48/51)*(3/50)*(47/49)*(2/48)*(46/47)*(1/46)*1 = 3.69 * 10^-6 = 0.00000369 = 0.000369% chance

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about a .000069857 or .0069857% chance of each person getting an ace.

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Each person needs to get an ace. There are four aces in a deck of 52 cards.

Person 1: Chance of getting an ace = 4/52.

Person 2: Chance of getting an ace = 3/51 (3 aces remaining after 1 is dealt as a hole card. With 1 card already removed from the selection there remains 51 other cards).

Person 3: Chance of getting an ace = 2/50.

Person 4: Chance of getting an ace = 1/49.

The probability of the event occurring is the product:

(4/52)*(3/51)*(2/50)*(1/49) = 24/6497400 = 1/270725 ~= 0.000369379%.

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Each person needs to get an ace and a non-ace.

Person 1: Chance of getting an ace = 4/52, Chance of getting a non-ace = 48/51 (the order of these doesn't matter, if he gets a non-ace first then the chances are 48/52 and 4/51, which when multiplied together give the same thing)

Person 2: Chance of getting an ace = 3/50 (three aces left, only 50 cards left in the deck), Chance of getting a non-ace = 47/49

Person 3: Chance of getting an ace = 2/48, Chance of getting a non-ace = 46/47

Person 4: Chance of getting an ace = 1/46, Chance of getting a non-ace = 45/45 = 1

Multiply all of these together:

(4/52)*(48/51)*(3/50)*(47/49)*(2/48)*(46/47)*(1/46)*1 = 3.69 * 10^-6 = 0.00000369 = 0.000369% chance

Each person needs to get an ace. There are four aces in a deck of 52 cards.

Person 1: Chance of getting an ace = 4/52.

Person 2: Chance of getting an ace = 3/51 (3 aces remaining after 1 is dealt as a hole card. With 1 card already removed from the selection there remains 51 other cards).

Person 3: Chance of getting an ace = 2/50.

Person 4: Chance of getting an ace = 1/49.

The probability of the event occurring is the product:

(4/52)*(3/51)*(2/50)*(1/49) = 24/6497400 = 1/270725 ~= 0.000369379%.

Each player gets to cards and they are not given their cards two at a time.

Edited by rlgandy
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It doesn't matter in what order the cards are dealt. It just matters WHAT cards are dealt.

But don't you have to take into account the fact that each person gets two cards? You have to remove the possibility that, say, the first person gets dealt two cards. For each person to have an ace, NOBODY can be dealt two cards. So we have to deal with the probability of getting two aces.

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Each person needs to get an ace. There are four aces in a deck of 52 cards.

Person 1: Chance of getting an ace = 4/52.

Person 2: Chance of getting an ace = 3/51 (3 aces remaining after 1 is dealt as a hole card. With 1 card already removed from the selection there remains 51 other cards).

Person 3: Chance of getting an ace = 2/50.

Person 4: Chance of getting an ace = 1/49.

The probability of the event occurring is the product:

(4/52)*(3/51)*(2/50)*(1/49) = 24/6497400 = 1/270725 ~= 0.000369379%.

Don't have time to do the Math but I'll say

4/52. That is the probability that his first card will be an ace. If he gets any other card, you still need to consider his chances of getting an ace with the second card. This would be dependent on how many aces are left (I'm sure it could be simplified).

I think the easiest way to calculate this is to model all of the ways to deal them out (ignoring suits)

1. A,A,A,A,x,x,x,x

2. A,x,A,A,x,A,x,x

3. A,A,x,A,x,x,A,x

etc.

then calculate each prob and add 'em up.

Again, that second round of cards changes the probability. A good way to look at it is think, If all of the cards were dealt out, would the probability still be .000369? Obviously it would be much higher.

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The poker variant being played was not mentioned. Was the hint that each player has a hole card mean the poker game was twenty-one? Could the poker game not be five card stud? Or, perhaps another poker variant. In most variants, and in the two aforementioned (twenty-one and five card stud) the first cards dealt would be the hole cards. This would mean for the event in question that the first four cards dealt would be aces.

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I assume by the title the game is Texas Hold'em (two hole cards). Though you are very correct in that the variant matters.

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I got the same answer as pdqkemp but did not read all the way through his post.

You can look at the problem as 8 cards out of 52 so the winning combinations would be:

Ax Ax Ax Ax

Ax Ax Ax xA

Ax Ax xA Ax

Ax xA Ax Ax

Ax Ax xA xA

Ax xA Ax xA

Ax xA xA Ax

Ax xA xA xA

Ax Ax Ax Ax

Ax Ax Ax xA

Ax Ax xA Ax

Ax xA Ax Ax

Ax Ax xA xA

Ax xA Ax xA

Ax xA xA Ax

Ax xA xA xA

In other words there are 16 combinations of each player having a single ace irregardless of suit.

Now there are also four different aces so there are 4 x 3 x 2 x 1 or 24 permutations there.

And the other hole card of each hand yields 48 x 47 x 46 x 45 permutations there.

The total number of permutations of 8 out of 52 cards is 52 x 51 x 50 x 49 x 48 x 47 x 46 x 45.

so the odds would then be (maybe):

(16 x 24 x 48 x 47 x 46 x 45)/(52 x 51 x 50 x 49 x 48 x 47 x 46 x 45) or

(16 x 24)/(52 x 51 x 50 x 49 x 48) = .0000591

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I got the same answer as pdqkemp but did not read all the way through his post.

You can look at the problem as 8 cards out of 52 so the winning combinations would be:

Ax Ax Ax Ax

Ax Ax Ax xA

Ax Ax xA Ax

Ax xA Ax Ax

Ax Ax xA xA

Ax xA Ax xA

Ax xA xA Ax

Ax xA xA xA

Ax Ax Ax Ax

Ax Ax Ax xA

Ax Ax xA Ax

Ax xA Ax Ax

Ax Ax xA xA

Ax xA Ax xA

Ax xA xA Ax

Ax xA xA xA

In other words there are 16 combinations of each player having a single ace irregardless of suit.

Now there are also four different aces so there are 4 x 3 x 2 x 1 or 24 permutations there.

And the other hole card of each hand yields 48 x 47 x 46 x 45 permutations there.

The total number of permutations of 8 out of 52 cards is 52 x 51 x 50 x 49 x 48 x 47 x 46 x 45.

so the odds would then be (maybe):

(16 x 24 x 48 x 47 x 46 x 45)/(52 x 51 x 50 x 49 x 48 x 47 x 46 x 45) or

(16 x 24)/(52 x 51 x 50 x 49 x 48) = .0000591

Excel and I got

0.000059101

I agree with PG on the different ways to get an ace for every player. But I calculated the prob of each way and added them up.

The prob of each turns out to be 0.000003694

For example the first way A A A A x x x x is 4/52*3/51*2/50*1/49 *48/48 47/47 *46/46 *45/45 = .000003694

The second way A A A x x x x A is 4/52 * 3/51 * 2/50 * 48/49 * 47/48 * 46/47 *45/46 * 1/45 = .000003694

.000003694*16 = .000059101

Edited by maurice
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My previous answer was wrong (too fast), so let me correct myself and explain solution.

To obtain the answer we need to find:

A = total number of ways it is possible do deal four players two hole cards each (out of 52-card deck)

B = as above but with additional condition: each player gets an ace.

Then B/A will be our probability.

Both A and B are easy and straightforward to calculate (though I made strange mistake calculating A in my first answer).

Let's call our four players: SB, BB, CO, BTN.

To find A we have to choose 2 cards out of 52 for SB, then 2 cards out of remaining 50 for BB, and so on:

A = Comb(52,2) * Comb(50,2) * Comb(48,2) * Comb(46,2)

A = 52*51*0.5 * 50*49*0.5 * 48*47*0.5 * 46*45*0.5

To find B we have to choose suit of the ace for SB, BB, CO and BTN (4*3*2*1) and choose remaining hole card for each player (48*47*46*45):

B = 4*3*2*1 * 48*47*46*45

B/A = (4*3*2*1 * 48*47*46*45) / (52*51*0.5 * 50*49*0.5 * 48*47*0.5 * 46*45*0.5)

(You can paste above expression in google to evaluate it, if you need the number.)

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