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# Case Study: Texas Hold'Em

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WRITTEN ASSIGNMENT 3
CASE STUDY: TEXAS HOLD'EM
STA-201: PRINCIPLES OF STATISTICS

a. The probability that you are dealt pocket aces is 1/221, or 0.00452 to three significant digits. If you studied either Section 4.5 and 4.6 or Section 4.8, verify that probability.
1st Card- 4 cards that are aces out of the 52 cards in the deck; so 4/52 = 0.0769 or slightly less than an 8% chance that the first card is an Ace.
2nd Card- Since first card dealt was an Ace, there are only 3 aces out of 51 card remaining; so 3/51 = 0.0588 or slightly less than a 6% chance that a second ace will be dealt.
Therefore to find the probability that these two events will happen, you will need to take the results of both and multiply them together. 0.0769 x 0.0588= 0.0452172 rounded to three significant digits is 0.00452 is the probability of getting dealt pocket aces. So the probability of 1 in 221 is correct.

b. Using the result from part (a), obtain the probability that you are dealt "pocket kings."
Same probability as part (a).
I find that it is unnecessary to repeat the exact same calculations since the result will be the same as part (a.). However, since the problem is quite vague, I decided that calculating the result of pocket kings after pocket aces are drawn, is more fitting.
1st Card Ace- 4 cards that are aces out of the 52 cards in the deck; so 4/52 = 0.0769 or slightly less than an 8% chance that the first card is an Ace.
2nd Card Ace- Since first card dealt was an Ace, there are only 3 aces out of 51 card remaining; so 3/51 = 0.0588 or slightly less than a 6% chance that a second ace will be dealt.
0.00452 is the probability of getting dealt pocket aces (or pocket kings or any pocket pair for that matter).
3rd Card King- 4 cards out of 50 cards left in the deck; 4/50= 0.08 or an 8% chance to draw a king.
4th Card King- Only 3 kings out of 49 cards; 3/49= 0.0612 or slightly more than a 6% chance to draw the second king after two aces are drawn.
Therefore: 0.0769 x 0.0588 x 0.08 x 0.0612 = 0.0000221 or a 1/45,249 chance of these happening. Ironically this happened to me when playing. I drew the pocket kings and another guy drew the pocket aces. After the flop, I went all in, since there was a king in the flop. Only two guys called and one had pocket Aces. Damn if he didn’t get an Ace on the river card. Cost me \$100. Quit playing Hold-em after that.

c. Using the result from part (a) and your analysis in part (b), find the probability that you are dealt a "pocket pair," that is, two cards of the same denomination.
2nd Card Match- Since first card dealt has no relevance: there are only 3 aces out of 51 cards remaining; so 3/51 = 0.0588 or slightly less than a 6% chance of drawing pocket pairs.

d. contains at least 1 card of your denomination. (Hint: Complementation Rule.)
Using the Complementation Rule: P(E)=1-P(not E)
1st Card- There are 2 cards that's your denomination out of 50.
2nd Card- There are 48 cards that's not the pair denomination out of 49.
3rd Card- There are 44 cards that would not make a pair out of 48.
Since there are three cards on the flop, that means there are three possible outcomes and need to multiply the three events by three.
1-((2/50)(48/49)(44/48))x3=1-((0.04)(0.978)(0.916)x3)=1-0.108(0.10750176 actual), means that there is an 89.2% chance of trips Not happening.

e. gives you "trips," that is, contains exactly 1 card of your denomination and 2 other unpaired cards.
1st Card- There are 2 cards that's your denomination out of 50.
2nd Card- There are 48 cards that's not the pair denomination out of 49.
3rd Card- There are 44 cards that would not make a pair out of 48.
Since there are three cards on the flop, that means there are three possible outcomes and need to multiply the three events by three.
((2/50)(48/49)(44/48))x3=((0.04)(0.978)(0.916))x3=0.108(0.10750176 actual), means that there is a 10.8% chance of trips happening.

f. gives you “quads,” that is, contains 2 cards of your denomination
1st Card- There are 2 cards that's your denomination out of 50.
2nd Card- There are 1 card left that is your denomination out of 49.
3rd Card- There are 48 cards that would not make a pair out of 48.
Since there are three cards on the flop, that means there are three possible outcomes and need to multiply the three events by three.
((2/50)(1/49)(48/48))x3=((0.04)(0.0204)(0))x3=0.00245(0.002448 actual), means that there is a 0.25% chance of trips happening.

g. gives you a “boat,” that is, contains 1 card of your denomination and 2 cards of another denomination.
1st Card- There are 2 cards that's your denomination out of 50.
2nd Card- There are 48 cards not in your denomination out of 49.
3rd Card- There are 3 cards that would make a pair out of 48.
Since there are three cards on the flop, that means there are three possible outcomes and need to multiply the three events by three.
((2/50)(48/49)(3/48))x3=((0.04)(0.978)(0.0625))x3=0.00734(0.007335 actual), means that there is a 0.73% chance of trips happening. (Hmmmnnn… I would have thought this would be lower than four of a kind since four of a kind is a higher winning hand.)

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