Robb's 3000th Post

Originally Posted by MadMojoMonkey

Same question with the 3-suited deck (39 cards).

After your 3-year-old f**ks up your only deck of cards two minutes before your home game starts, you can play 39-card poker by switching the flushes < straights, and no quads obv. Here's the breakdown.

We have 39c5 = 575757 hands - interesting number - who knew?

• Exactly 1 Pair = 13c1 * 3c2 * 12c3 * (3c1)^3 / 39c5 = 231660 / 575757 ~ 40.24% (standard deck is 42.26%)
• Exactly 2 Pair = 13c2 * (3c2)^2 * 11c1 * 3c1 / 39c5 = 23166 / 429000 ~ 4.02% (standard deck is 4.75%)
• Exactly Trips = 13c1 * 3c3 * 12c2 * (3c1)^2 / 39c5 = 7722 / 575757 ~ 1.34% (standard deck is 2.11%)
• Straight (no straight flushes) = (10 * (3c1)^5 - 30 ) / 39c5 = 2400 / 575757 ~ .41% (standard deck is .39%)
• Flush (no straight flushes) = (3c1 * 13c5 - 30) / 39c5 = 3831 / 575757 ~ .67% (standard deck is .20%)
• Full House = 13c1 * 3c2 * 12c1 * 3c3 / 39c5 = 15600 / 575757 ~ .08% (standard deck is .14%)
• Straight Flush = 30 / 575757 ~ .01% (standard deck is .0015%)

Don't ground the 3-year-old or send him to bed without dinner. Easy game.

Edit: fixed flush calculation MadMoJo pointed out below.

Originally Posted by Robb

• Flush (no straight flushes) = (3c1 * 13c5 - 30) / 39c5 = 5118 / 575757 ~ .89% (standard deck is .20%)

Found a typo in the 3-suit frequencies: the formula is correct, but the result should be 0.67%

Also: Is the numerator for the straight flush eq's based on:
10c1 * [n]c1
where [n] is the number of suits, or are there missing terms (all of which are 1, so moot)?

Originally Posted by MadMojoMonkey

Found a typo in the 3-suit frequencies: the formula is correct, but the result should be 0.67%

Also: Is the numerator for the straight flush eq's based on:
10c1 * [n]c1
where [n] is the number of suits, or are there missing terms (all of which are 1, so moot)?

You are exactly right. Another way to do it is simply to count them since they're so few. There are 10 possible straights, 5-high through A-high, and there is one of each variety in each suit, so 10 * [n] is the correct number.

Either way, you're right.