Silly question but prob of someone else having ace in fr game - Gambling and Probability

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Silly question but prob of someone else having ace in fr game

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01-27-2016 , 05:21 PM

dodah

grinder

Join Date: Mar 2008 Posts: 497

Can't seem to find a straight answer but if youre dealt an ace , whats the probability sumone else has an ace in a 9 handed table? Appreciate the help

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01-27-2016 , 05:33 PM

whosnext

Carpal \'Tunnel

Join Date: Mar 2009 Posts: 6,733

Do you want to work through the answer yourself?

How many aces are in the deck?

How many aces do you have?

How many aces are not in your hand?

How many cards are in your opponents' hole cards at a 9-handed table?

How many cards are in the remaining deck after your hole cards are dealt? [how many of these are aces and how many of these are non-aces]

How can you express the probability that there is at least one ace among your opponents' hole cards in terms of the probability that there are no aces among your opponents' hole cards?

So what is the probability that there are no aces among your opponents' hole cards?

Finally, then, what is the probability that there is at least one ace among your opponents' hole cards?

Last edited by whosnext; 01-27-2016 at 05:43 PM.

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01-27-2016 , 06:44 PM

dodah

grinder

Join Date: Mar 2008 Posts: 497

am with you all the way until " express the probability" lol

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01-27-2016 , 06:47 PM

whosnext

Carpal \'Tunnel

Join Date: Mar 2009 Posts: 6,733

Getting at least one of something is the "complement" of getting zero.

So the probability of getting at least one of something is 1 - Probability of getting zero

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01-28-2016 , 07:12 AM

David Lyons

old hand

Join Date: May 2003 Posts: 1,459

Keep it simple OP.

You have one card that's an ace, and another that's not. That leaves 50 unknown cards containing three aces and 47 non-aces.

Sixteen of those cards are in the hands of other players. What are the chances all sixteen of the cards are not aces?

First card: 47/50
Second card: 46/49
Third card: 45/48
...
Sixteenth card: 32/35

Multiply all those together and you get the chance of zero aces in all other hands.

Last edited by David Lyons; 01-28-2016 at 07:19 AM.

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01-28-2016 , 05:45 PM

Buzz

gramps

Join Date: Sep 2002 Posts: 18,254

Quote:

Originally Posted by David Lyons

First card: 47/50
Second card: 46/49
Third card: 45/48
...
Sixteenth card: 32/35

Multiply all those together and you get the chance of zero aces in all other hands.

All that quickly reduces to 34*33*32/50/49/48=0.305.
Then you subtract that from 1 to get the probability of at least one opponent having at least one ace. (It's about 70%).

Buzz

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01-29-2016 , 01:05 PM

whosnext

Carpal \'Tunnel

Join Date: Mar 2009 Posts: 6,733

In case OP or anyone else wants total closure on this problem, I will put my answer into a spoiler ...

Spoiler:

How many aces are in the deck? 4

How many aces do you have? 1

How many aces are not in your hand? 4-1=3

How many cards are in your opponents' hole cards at a 9-handed table? 2*8=16

How many cards are in the remaining deck after your hole cards are dealt (how many of these are aces and how many of these are non-aces)? 50 cards total, 3 aces and 47 non-aces

How can you express the probability that there is at least one ace among your opponents' hole cards in terms of the probability that there are no aces among your opponents' hole cards?

Since getting at least one of something is the "complement" of getting exactly zero of something, the prob of getting at least one ace is equal to one minus prob of getting exactly zero aces

So what is the probability that there are no aces among your opponents' hole cards?

Method 1 using combinatorics (the "Choose" function):

Prob of 0 aces out of 16 cards from a deck of 3 aces and 47 non-aces is the ratio of the number of ways to draw 16 cards from 47 non-aces to the number of ways to draw 16 cards from 50 cards:

= C(47,16) / C(50,16)

= 30.5306122%

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Method 2 using "direct" probabilities:

prob of 1st of 16 cards being a non-ace = 47/50

prob of 2nd of 16 cards being a non-ace (given 1st card was non-ace) = 46/49

prob of 3rd of 16 cards being a non-ace (given cards 1-2 were non-aces) = 45/48

prob of 4th of 16 cards being a non-ace (given cards 1-3 were non-aces) = 44/47

prob of 5th of 16 cards being a non-ace (given cards 1-4 were non-aces) = 43/46

prob of 6th of 16 cards being a non-ace (given cards 1-5 were non-aces) = 42/45

prob of 7th of 16 cards being a non-ace (given cards 1-6 were non-aces) = 41/44

prob of 8th of 16 cards being a non-ace (given cards 1-7 were non-aces) = 40/43

prob of 9th of 16 cards being a non-ace (given cards 1-8 were non-aces) = 39/42

prob of 10th of 16 cards being a non-ace (given cards 1-9 were non-aces) = 38/41

prob of 11th of 16 cards being a non-ace (given cards 1-10 were non-aces) = 37/40

prob of 12th of 16 cards being a non-ace (given cards 1-11 were non-aces) = 36/39

prob of 13th of 16 cards being a non-ace (given cards 1-12 were non-aces) = 35/38

prob of 14th of 16 cards being a non-ace (given cards 1-13 were non-aces) = 34/37

prob of 15th of 16 cards being a non-ace (given cards 1-14 were non-aces) = 33/36

prob of 16th of 16 cards being a non-ace (given cards 1-15 were non-aces) = 32/35

So the prob of getting all 16 cards to be non-aces is the product of these 16 terms. After a whole bunch of "cancellation" of numerators and denominators, this is seen to be:

= (34*33*32) / (50*49*48)

= 35,904 / 117,600

= 30.5306122%

Finally, then, what is the probability that there is at least one ace among your opponents' hole cards?

= 1 - prob found above

= 1 - 35,904 / 117,600

= 81,696 / 117,600

= 69.4693878%

Last edited by whosnext; 01-29-2016 at 01:17 PM.

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