# ? for math inclined Texas Hold 'em Players **URL:** \ **Category:** The Game Room\ **Created:** [September 22, 2010, 2:45am UTC](https://boards.straightdope.com/t/for-math-inclined-texas-hold-em-players/554606 "2010-09-22T02:45:48Z")\ **Posts on this page:** 13\ **Page:** 1
**Author:** ![etv78](https://avatars.discourse-cdn.com/v4/letter/e/d78d45/32.png) [@etv78](https://boards.straightdope.com/u/etv78)\ **Post date:** [September 22, 2010, 2:45am UTC](https://boards.straightdope.com/t/for-math-inclined-texas-hold-em-players/554606/1 "2010-09-22T02:45:48Z")
What are the odds of a “rainbow board” after the turn? Rainbow=all 4 suits represented. ---
**Author:** ![Gray\_Ghost](https://avatars.discourse-cdn.com/v4/letter/g/c0e974/32.png) [@Gray\_Ghost](https://boards.straightdope.com/u/Gray_Ghost)\ **Post date:** [September 22, 2010, 4:10am UTC](https://boards.straightdope.com/t/for-math-inclined-texas-hold-em-players/554606/2 "2010-09-22T04:10:35Z")
I think the answer is this: (39/51) \* (26/50) \* (13/49) = 0.105498 10.5% of the time. Start with any suit (A) for the first card on the board. For the second card, the card can be any suit of B, C, or D to satisfy your prop. Let’s pick on B. There are 51 cards left in the deck and 39 of suits B-D, hence 39/51. For the third card, the suit must then be C or D—use C, there are 50 cards left, and 26 satisfy the prop. Fourth card is similar, suit D is all that’s left, there are 49 cards to choose from and 13 satisfy the prop. All probabilities are multiplied together since all are independent events, and all must come true for the rainbow board to occur. ---
**Author:** ![digs](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/digs/32/14889_2.png) [@digs](https://boards.straightdope.com/u/digs)\ **Post date:** [September 22, 2010, 4:22am UTC](https://boards.straightdope.com/t/for-math-inclined-texas-hold-em-players/554606/3 "2010-09-22T04:22:20Z")
> [@etv78](#): > > What are the odds of a “rainbow board” after the turn? Rainbow=all 4 suits represented. The chances of ALL of _MY_ cards being of different suits, and having absolutely no relationship to each other: 99.44%. ---
**Author:** ![Amblydoper](https://avatars.discourse-cdn.com/v4/letter/a/fbc32d/32.png) [@Amblydoper](https://boards.straightdope.com/u/Amblydoper)\ **Post date:** [September 23, 2010, 6:27am UTC](https://boards.straightdope.com/t/for-math-inclined-texas-hold-em-players/554606/4 "2010-09-23T06:27:53Z")
> [@Gray\_Ghost](#): > > I think the answer is this: > > (39/51) \* (26/50) \* (13/49) = 0.105498 10.5% of the time. Is this accurate? Do you need to account for the cards already in play, or the burn cards? Or does their unknown status make them irrelevant? ---
**Author:** ![Chipacabra](https://avatars.discourse-cdn.com/v4/letter/c/e19b73/32.png) [@Chipacabra](https://boards.straightdope.com/u/Chipacabra)\ **Post date:** [September 23, 2010, 6:44am UTC](https://boards.straightdope.com/t/for-math-inclined-texas-hold-em-players/554606/5 "2010-09-23T06:44:03Z")
Their unknown status makes them irrelevant. When the deck is shuffled, an arbitrary 4 cards end up in the spots that will become the flop and the turn, the other cards have no effect on that. 10.5% of all hands of hold 'em will (if played out to the turn) have a rainbow board. Note that once you’ve looked at your hole cards and seen if they are the same suit or not, you now know a little more about the rest of the deck and can come up with more precise odds that this particular hand will be a rainbow board. The math for that one is a little fussy, I don’t want to work it out right now. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [September 23, 2010, 7:01am UTC](https://boards.straightdope.com/t/for-math-inclined-texas-hold-em-players/554606/6 "2010-09-23T07:01:31Z")
Because each set of four cards is equally as likely to come up as any other, the probability for a rainbow board is the number of ways to draw cards of four different suits divided by the number of ways to draw four cards altogether; that is, 13^4/(52 choose 4) = what **Gray Ghost** said. Since **Gray Ghost** already said it, why am I repeating it? Because this presentation also makes calculating the probability after seeing your hole cards just as easy: If they’re the same suit: The number of ways to draw four more cards of different suits is 13^3 \* 11 [13 possibilities for every other suit, 11 possibilities for the suit you have] = 24167 If they’re different suits: The number of ways to draw four more cards of different suits is 13^2 \* 12^2 [13 possibilities for the suits you don’t have, 12 possibilities for the suits you do have] = 24336 Either way: The number of ways to draw four more cards at all is (50 choose 4) = 230300 Thus, the probability of a rainbow after the turn given that your two hole cards are of the same suit is 24167/230300 = approximately 10.49%, and the probability of a rainbow after the turn given that your two hole cards are of different suits is 24336/230300 = approximately 10.57%. So, not a huge difference either way, but there you go. ---
**Author:** ![septimus](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/septimus/32/410_2.png) [@septimus](https://boards.straightdope.com/u/septimus)\ **Post date:** [September 23, 2010, 7:15am UTC](https://boards.straightdope.com/t/for-math-inclined-texas-hold-em-players/554606/7 "2010-09-23T07:15:03Z")
> [@Chipacabra](#): > > Their unknown status makes them irrelevant. When the deck is shuffled, an arbitrary 4 cards end up in the spots that will become the flop and the turn, … I hate to be branded a nitpicker, but strictly speaking it’s not _completely_ irrelevant. The fact that there’s a 4th card at all may make the 3-card flop slightly less likely to be all the same suit. With same-suit flop, a player might have a flush or bluff, might have raised strongly, and might have won the pot with no “turn card” ever coming. If the point is unclear, here’s a much more extreme example of a slightly similar phenomenon: The chance any craps roll is seven is 1-in-6, right? The chance a shooter’s 1st roll is seven is 1-in-6; the chance his 11th roll is seven is 1-in-6, etc. Now, what’s the chance his _last_ roll is seven? 😃 Hint:Proud of their income tax-free state, Nevadans have rewritten a famous adage: “The only things certain are Death and Seven Out!” ---
**Author:** ![Chipacabra](https://avatars.discourse-cdn.com/v4/letter/c/e19b73/32.png) [@Chipacabra](https://boards.straightdope.com/u/Chipacabra)\ **Post date:** [September 23, 2010, 7:18am UTC](https://boards.straightdope.com/t/for-math-inclined-texas-hold-em-players/554606/8 "2010-09-23T07:18:38Z")
For the sake of this problem, it’s best to just consider the straight odds, and ignore player behavior, which can’t be quantified easily. Even if the players all fold before the turn, that doesn’t change what the card WOULD have been if they had stayed in. ---
**Author:** ![septimus](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/septimus/32/410_2.png) [@septimus](https://boards.straightdope.com/u/septimus)\ **Post date:** [September 23, 2010, 7:41am UTC](https://boards.straightdope.com/t/for-math-inclined-texas-hold-em-players/554606/9 "2010-09-23T07:41:36Z")
Mine was intended as an interesting aside, and certainly not to impugn the correct calculations that preceded my post, but if we’re going to play defensive … > [@Chipacabra](#): > > For the sake of this problem, it’s best to … ignore player behavior, which can’t be quantified easily. “Because the lights better there”? > [@Chipacabra](#): > > Even if the players all fold before the turn, … Does that happen often? 😃 ---
**Author:** ![notfrommensa](https://avatars.discourse-cdn.com/v4/letter/n/f14d63/32.png) [@notfrommensa](https://boards.straightdope.com/u/notfrommensa)\ **Post date:** [September 23, 2010, 7:40pm UTC](https://boards.straightdope.com/t/for-math-inclined-texas-hold-em-players/554606/10 "2010-09-23T19:40:39Z")
[quote=“septimus, post:7, topic:554606”] > [@Chipacabra](#): > > The chance any craps roll is seven is 1-in-6, right? The chance a shooter’s 1st roll is seven is 1-in-6; the chance his 11th roll is seven is 1-in-6, etc. Now, what’s the chance his _last_ roll is seven? 😃 > > Hint:Proud of their income tax-free state, Nevadans have rewritten a famous adage: > “The only things certain are Death and Seven Out!” What is the chance that the shooters next to last roll is a seven ? :p:D ---
**Author:** ![Hari\_Seldon](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/hari_seldon/32/5173_2.png) [@Hari\_Seldon](https://boards.straightdope.com/u/Hari_Seldon)\ **Post date:** [September 23, 2010, 11:31pm UTC](https://boards.straightdope.com/t/for-math-inclined-texas-hold-em-players/554606/11 "2010-09-23T23:31:56Z")
I make it C(4;1,1,1,1)/4^4 = 24/256, around 9.38%. C(4;1,1,1,1) = 4!/1!1!1!1! is the number of ways of choosing one from each suit, while 4^4 is the number of choices (ignoring rank of cards). ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [September 23, 2010, 11:49pm UTC](https://boards.straightdope.com/t/for-math-inclined-texas-hold-em-players/554606/12 "2010-09-23T23:49:01Z")
I’m afraid you’re mistaken, **Hari**. Your analysis ignores that, for example, Heart-Heart-Heart-Heart is less likely than Heart-Heart-Heart-Diamond (for after having already gotten three hearts, there’s less hearts left in the deck than diamonds). Applying the same analysis to a deck with only two suits and 1 rank and drawing two cards would tell us that 2!/2^2 = 1/2 is the probability of a rainbow draw. But of course, that isn’t true; in this scenario, the rainbow draw is guaranteed, since you draw all the cards (all two of them) in the entire deck! ---
**Author:** ![Hari\_Seldon](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/hari_seldon/32/5173_2.png) [@Hari\_Seldon](https://boards.straightdope.com/u/Hari_Seldon)\ **Post date:** [September 24, 2010, 12:53am UTC](https://boards.straightdope.com/t/for-math-inclined-texas-hold-em-players/554606/13 "2010-09-24T00:53:42Z")
> [@Indistinguishable](#): > > I’m afraid you’re mistaken, **Hari**. Your analysis ignores that, for example, Heart-Heart-Heart-Heart is less likely than Heart-Heart-Heart-Diamond (for after having already gotten three hearts, there’s less hearts left in the deck than diamonds). > > Applying the same analysis to a deck with only two suits and 1 rank and drawing two cards would tell us that 2!/2^2 = 1/2 is the probability of a rainbow draw. But of course, that isn’t true; in this scenario, the rainbow draw is guaranteed, since you draw all the cards (all two of them) in the entire deck! You’re right. Sorry. The computation would have correct from an infinite deck.