Math Is Fun Forum

Fruityloop

Member

Registered: 2009-05-18

Posts: 143

Trouble understanding probabilities with bridge hand.

A bridge hand has 13 cards.
Can someone explain these equations?
The probability of holding at least 1 ace...

The probability of holding just 1 ace...

The probability of holding at least 2 aces...

Bob

Administrator

Registered: 2010-06-20

Posts: 10,626

Re: Trouble understanding probabilities with bridge hand.

hi Fruityloop

This has me puzzled too but I can offer this as a temporary way forward.

P(at least 1 ace) = 1 - P(no aces)

So I would find P like this:

Want card 1 to be a non-ace = 48/52
And the next                        = 47/51
And the next                        = 46/50

........ and so on until

13th card                             = 36/40

By further cancelling of factors

I'm thinking there must be another way to calculate the probability where you add two probs and subtract a third.  But I haven't found it yet.

Where did it come from? Maybe there's a clue there.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Bob

Administrator

Registered: 2010-06-20

Posts: 10,626

Re: Trouble understanding probabilities with bridge hand.

A little progress:

Probabliity of 4 aces in 13 cards is {choose 1st = 4} then {second = 3} then {third = 2} then no choice for the last. Then choose 10 non aces. Finally, the aces may occur in any spaces within the thirteen slots so 13 C 4

Continuing to cancel factors:

So where do the other two terms come from?

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

sologuitar

Member

Registered: 2022-09-19

Posts: 467

Re: Trouble understanding probabilities with bridge hand.

A little progress:

Probabliity of 4 aces in 13 cards is {choose 1st = 4} then {second = 3} then {third = 2} then no choice for the last. Then choose 10 non aces. Finally, the aces may occur in any spaces within the thirteen slots so 13 C 4

Continuing to cancel factors:

So where do the other two terms come from?

Bob

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