# Math Help - Euchre Questions

1. ## Euchre Questions

Okay so I have a few data management questions relating the game of Euchre.
the rules of euchre are: the Euchre deck is A,K,Q,J,10,9 of all suits, and 5 cards are dealt to each of the players.

What is the chance of not being dealt any jacks?
whats the chance of being dealt atleast 1 jack?
what is the probability of being dealt atleast 1 jack and 1 ace?
what is the probability of being dealt neither jacks nor aces?
what is the probabiltiy of being dealt a jack knowing you are already dealt an ace

thanks in advance and im more interested in the way the solutions are gotten, this is pre-exam study

2. Hello, Ari!

The Euchre deck is A ,K, Q, J, 10, 9 in four suits,
and 5 cards are dealt to each player.

There are 24 cards in a Euchre deck.

There are: . $\displaystyle {24\choose5} \:=\:\frac{24!}{5!\,19!} \:=\:42,\!504 \text{ possible hands.}$

(a) What is the probability of being dealt no Jacks?

There are 4 Jacks and 20 Others.

We want 5 Others: . $\displaystyle {20\choose5} \:=\:\frac{20!}{5!\,15!} \;=\;15.\!504\text{ ways.}$

Therefore: . $P(\sim\!\text{Jacks}) \;=\;\dfrac{15,\!504}{42,\!504} \;=\;\dfrac{646}{1771}$

(b) What is the probability of being dealt at least 1 Jack?

This is the exact opposite of (a).

Therefore: . $P(\text{at least 1 Jack}) \;=\;1 - \dfrac{646}{1771} \;=\;\dfrac{1125}{1771}$

(c) What is the probability of being dealt at least 1 Jack and 1 Ace?

This one requires some Olympic-level gymnastics.

We use this formula:

$P(\sim\!\text{Jacks} \:\vee \sim\!\text{Aces}) \;=\;P(\sim\!\text{Jacks}) + P(\sim\!\text{Aces}) - P(\sim\!\text{Jacks} \:\wedge \sim\!\text{Aces})$

From (a): . $P(\sim\!\text{Jacks}) \:=\:\dfrac{15,\!504}{42,\!504}$

. . .Then: . $P(\sim\!\text{Aces}) \:=\:\dfrac{15,\!504}{42,\!504}$

From (d): . $P(\sim\!\text{Jacks}\: \wedge \sim\!\text{Aces}) \;=\;\dfrac{4,\!368}{42,\!504}$

We have: . $P(\sim\!\text{Jacks}\: \vee \sim\!\text{Aces}) \;=\;\dfrac{15,\!504}{42,\!504} + \dfrac{15,\!504}{42,\!504} - \dfrac{4,\!368}{42,\!504} \;=\;\dfrac{35,\!376}{42,\!504}$

This is the exact opposite of "at least one Jack and at least one Ace".

Therefore:

. . $P(\text{at least one Jack} \wedge \text{at least one Ace}) \;=\;1 - \dfrac{35,\!376}{42,\!504} \;=\;\dfrac{7,\!128}{42,\!504} \;=\;\dfrac{27}{161}$

(d) What is the probability of being dealt neither Jacks nor Aces?

There are: 4 Jacks, 4 Aces, and 16 Others.

We want 5 Others: . $\displaystyle {16\choose5} \:=\:\frac{16!}{5!\,11!} \:=\:4,\!368\text{ ways.}$

Therefore: . $P(\text{no Jacks and no Aces}) \;=\;\dfrac{4,\!368}{42,\!504} \;=\;\dfrac{182}{1771}$

This is getting too long . . . *yawn*
I'll let someone else explain part (e)