webayarea
Joined: Nov 30, 2018
• Posts: 6
November 30th, 2018 at 9:28:21 PM permalink
Looking for help in figuring the odds for a simple card game.

With 10 cards available to select from, A-10,

What are odds/payout of the following card combinations

A) 3 cards selected.
- odds of 3 in a row (aka 3 card Straight)?
- ie 2,3,4 or 7,8,9
- is it 8 out of 140? Or 15:1?

B) 4 cards selected
- odds of 3 out of 4 ina row (aka 3 card Straight)?
- ie 2,3,7,4 or 7,8,2,9
- 5:1 range? need help here

C) 5 cards selected
- odds of 3 out of 5 in a row (aka 3 card straight)?
- ie 2,3,7,4,8 or 7,8,2,9,3
- 2:1 range? Need help here

Thanks!!!
DogHand
Joined: Sep 24, 2011
• Posts: 1311
Thanks for this post from:
December 2nd, 2018 at 11:57:14 AM permalink
Quote: webayarea

Looking for help in figuring the odds for a simple card game.

With 10 cards available to select from, A-10,

What are odds/payout of the following card combinations

A) 3 cards selected.
- odds of 3 in a row (aka 3 card Straight)?
- ie 2,3,4 or 7,8,9
- is it 8 out of 140? Or 15:1?

B) 4 cards selected
- odds of 3 out of 4 ina row (aka 3 card Straight)?
- ie 2,3,7,4 or 7,8,2,9
- 5:1 range? need help here

C) 5 cards selected
- odds of 3 out of 5 in a row (aka 3 card straight)?
- ie 2,3,7,4,8 or 7,8,2,9,3
- 2:1 range? Need help here

Thanks!!!

webayarea,

I'll do the first two, and leave the third to you.

(A) If we consider only combinations, 8 3-card straights are possible: A-low, 2-low, ..., 8-low. Each combination has 6 permutations: for example, the 3-low straight can be drawn in any of these orders:

345, 354, 435, 453, 534, and 543.

So, the total number of ways to draw the straight is 8*6, while the total permutations is 10*9*8. Thus, the probability of a straight for a 3-card draw is

P(3) = (8*6)/(10*9*8) = 1/15, or 15:1 against, as you said.

(B) Now for a 4-card draw, we still have only 8 3-card straight combinations. However, each of these can be accompanied by 7 "4th" cards, and the "4th" card can be drawn first, second, third, or fourth. So the numerator is now (8*6*7*4), while the denominator (total permutations) is (10*9*8*7), so for a 4-card draw the 3-card straight probability is

P(4) = (8*6*7*4)/(10*9*8*7) = 4/15, or 15:4 against.

Hope this helps!

Dog Hand
webayarea
Joined: Nov 30, 2018
• Posts: 6
December 2nd, 2018 at 1:51:20 PM permalink
Dog! Thanks so much! Have a great day!
gordonm888
Joined: Feb 18, 2015
• Posts: 4802
December 2nd, 2018 at 4:39:14 PM permalink
When you say:

"C) 5 cards selected
- odds of 3 out of 5 in a row (aka 3 card straight)?"

Does that mean exactly 3 in a row? or would a 4 or 5 card straight qualify?
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
webayarea
Joined: Nov 30, 2018
• Posts: 6
December 3rd, 2018 at 10:23:31 AM permalink
That would be leading to my next questions;

Odds of a 4/4 straight and 4/5 straight and then, biggest Longshot of 5/5

Therefore payouts on a single unit bet would be:

3/3 = 15:1
3/4 = 3.75:1
3/5= TBD (guessing 3/2? Maybe 5/4?)
LONGSHOTS
4/4= TBD
4/5 = TBD
5/5= TBD
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 5944
December 3rd, 2018 at 4:59:31 PM permalink
Quote: DogHand

(B) Now for a 4-card draw, we still have only 8 3-card straight combinations. However, each of these can be accompanied by 7 "4th" cards, and the "4th" card can be drawn first, second, third, or fourth. So the numerator is now (8*6*7*4), while the denominator (total permutations) is (10*9*8*7), so for a 4-card draw the 3-card straight probability is

P(4) = (8*6*7*4)/(10*9*8*7) = 4/15, or 15:4 against.

You forgot to take two things into account:
First, that the three cards in the 3-card straight can be in any order;
Second, that a 4-card straight contains two 3-card straights, each of which you are counting separately.

I prefer using combinations (where order does not matter) instead of permutations.
For the 4-card hand, there are (10)C(4) = 210 different hands.
There are 8 3-card straights; for each one, there are 7 different cards for the "fourth card", so that is 56 - but again, I am counting the four-card straights twice.
(For example, 2-3-4 and 5 as a fourth card is 2-3-4-5, but 3-4-5 and 2 as a fourth card, which is counted separately, is also 2-3-4-5.)
Subtract 7 (the number of 4-card straights) from the number of hands with 3-card straights; the result is 49.
P(4) = 49 / 210 = 7 / 30, or odds of 23-7 against.

Also note that, say, a probability of 4/15 is odds of (15-4) - 4, or 11-4, against, not 15-4 against.
Otherwise, the odds of a coin toss being heads would be 2-1 against.
7craps
Joined: Jan 23, 2010
• Posts: 1977
December 3rd, 2018 at 5:47:17 PM permalink
Quote: webayarea

Looking for help in figuring the odds for a simple card game.

With 10 cards available to select from, A-10,

one can make a list of all the combinations
and see which ones are the events you are after
https://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html

end up being the numerator being the ones that satisfy your event
the denominator would be either Combin(10,3) or Combin(10,4) or Combin(10,4) (120,210,252)

the math can get messy when one over-counts events
see if it helps out
`List has 120 entries.1,2,31,2,41,2,51,2,61,2,71,2,81,2,91,2,101,3,41,3,51,3,61,3,71,3,81,3,91,3,101,4,51,4,61,4,71,4,81,4,91,4,101,5,61,5,71,5,81,5,91,5,101,6,71,6,81,6,91,6,101,7,81,7,91,7,101,8,91,8,101,9,102,3,42,3,52,3,62,3,72,3,82,3,92,3,102,4,52,4,62,4,72,4,82,4,92,4,102,5,62,5,72,5,82,5,92,5,102,6,72,6,82,6,92,6,102,7,82,7,92,7,102,8,92,8,102,9,103,4,53,4,63,4,73,4,83,4,93,4,103,5,63,5,73,5,83,5,93,5,103,6,73,6,83,6,93,6,103,7,83,7,93,7,103,8,93,8,103,9,104,5,64,5,74,5,84,5,94,5,104,6,74,6,84,6,94,6,104,7,84,7,94,7,104,8,94,8,104,9,105,6,75,6,85,6,95,6,105,7,85,7,95,7,105,8,95,8,105,9,106,7,86,7,96,7,106,8,96,8,106,9,107,8,97,8,107,9,108,9,10List has 210 entries.1,2,3,41,2,3,51,2,3,61,2,3,71,2,3,81,2,3,91,2,3,101,2,4,51,2,4,61,2,4,71,2,4,81,2,4,91,2,4,101,2,5,61,2,5,71,2,5,81,2,5,91,2,5,101,2,6,71,2,6,81,2,6,91,2,6,101,2,7,81,2,7,91,2,7,101,2,8,91,2,8,101,2,9,101,3,4,51,3,4,61,3,4,71,3,4,81,3,4,91,3,4,101,3,5,61,3,5,71,3,5,81,3,5,91,3,5,101,3,6,71,3,6,81,3,6,91,3,6,101,3,7,81,3,7,91,3,7,101,3,8,91,3,8,101,3,9,101,4,5,61,4,5,71,4,5,81,4,5,91,4,5,101,4,6,71,4,6,81,4,6,91,4,6,101,4,7,81,4,7,91,4,7,101,4,8,91,4,8,101,4,9,101,5,6,71,5,6,81,5,6,91,5,6,101,5,7,81,5,7,91,5,7,101,5,8,91,5,8,101,5,9,101,6,7,81,6,7,91,6,7,101,6,8,91,6,8,101,6,9,101,7,8,91,7,8,101,7,9,101,8,9,102,3,4,52,3,4,62,3,4,72,3,4,82,3,4,92,3,4,102,3,5,62,3,5,72,3,5,82,3,5,92,3,5,102,3,6,72,3,6,82,3,6,92,3,6,102,3,7,82,3,7,92,3,7,102,3,8,92,3,8,102,3,9,102,4,5,62,4,5,72,4,5,82,4,5,92,4,5,102,4,6,72,4,6,82,4,6,92,4,6,102,4,7,82,4,7,92,4,7,102,4,8,92,4,8,102,4,9,102,5,6,72,5,6,82,5,6,92,5,6,102,5,7,82,5,7,92,5,7,102,5,8,92,5,8,102,5,9,102,6,7,82,6,7,92,6,7,102,6,8,92,6,8,102,6,9,102,7,8,92,7,8,102,7,9,102,8,9,103,4,5,63,4,5,73,4,5,83,4,5,93,4,5,103,4,6,73,4,6,83,4,6,93,4,6,103,4,7,83,4,7,93,4,7,103,4,8,93,4,8,103,4,9,103,5,6,73,5,6,83,5,6,93,5,6,103,5,7,83,5,7,93,5,7,103,5,8,93,5,8,103,5,9,103,6,7,83,6,7,93,6,7,103,6,8,93,6,8,103,6,9,103,7,8,93,7,8,103,7,9,103,8,9,104,5,6,74,5,6,84,5,6,94,5,6,104,5,7,84,5,7,94,5,7,104,5,8,94,5,8,104,5,9,104,6,7,84,6,7,94,6,7,104,6,8,94,6,8,104,6,9,104,7,8,94,7,8,104,7,9,104,8,9,105,6,7,85,6,7,95,6,7,105,6,8,95,6,8,105,6,9,105,7,8,95,7,8,105,7,9,105,8,9,106,7,8,96,7,8,106,7,9,106,8,9,107,8,9,10List has 252 entries.1,2,3,4,51,2,3,4,61,2,3,4,71,2,3,4,81,2,3,4,91,2,3,4,101,2,3,5,61,2,3,5,71,2,3,5,81,2,3,5,91,2,3,5,101,2,3,6,71,2,3,6,81,2,3,6,91,2,3,6,101,2,3,7,81,2,3,7,91,2,3,7,101,2,3,8,91,2,3,8,101,2,3,9,101,2,4,5,61,2,4,5,71,2,4,5,81,2,4,5,91,2,4,5,101,2,4,6,71,2,4,6,81,2,4,6,91,2,4,6,101,2,4,7,81,2,4,7,91,2,4,7,101,2,4,8,91,2,4,8,101,2,4,9,101,2,5,6,71,2,5,6,81,2,5,6,91,2,5,6,101,2,5,7,81,2,5,7,91,2,5,7,101,2,5,8,91,2,5,8,101,2,5,9,101,2,6,7,81,2,6,7,91,2,6,7,101,2,6,8,91,2,6,8,101,2,6,9,101,2,7,8,91,2,7,8,101,2,7,9,101,2,8,9,101,3,4,5,61,3,4,5,71,3,4,5,81,3,4,5,91,3,4,5,101,3,4,6,71,3,4,6,81,3,4,6,91,3,4,6,101,3,4,7,81,3,4,7,91,3,4,7,101,3,4,8,91,3,4,8,101,3,4,9,101,3,5,6,71,3,5,6,81,3,5,6,91,3,5,6,101,3,5,7,81,3,5,7,91,3,5,7,101,3,5,8,91,3,5,8,101,3,5,9,101,3,6,7,81,3,6,7,91,3,6,7,101,3,6,8,91,3,6,8,101,3,6,9,101,3,7,8,91,3,7,8,101,3,7,9,101,3,8,9,101,4,5,6,71,4,5,6,81,4,5,6,91,4,5,6,101,4,5,7,81,4,5,7,91,4,5,7,101,4,5,8,91,4,5,8,101,4,5,9,101,4,6,7,81,4,6,7,91,4,6,7,101,4,6,8,91,4,6,8,101,4,6,9,101,4,7,8,91,4,7,8,101,4,7,9,101,4,8,9,101,5,6,7,81,5,6,7,91,5,6,7,101,5,6,8,91,5,6,8,101,5,6,9,101,5,7,8,91,5,7,8,101,5,7,9,101,5,8,9,101,6,7,8,91,6,7,8,101,6,7,9,101,6,8,9,101,7,8,9,102,3,4,5,62,3,4,5,72,3,4,5,82,3,4,5,92,3,4,5,102,3,4,6,72,3,4,6,82,3,4,6,92,3,4,6,102,3,4,7,82,3,4,7,92,3,4,7,102,3,4,8,92,3,4,8,102,3,4,9,102,3,5,6,72,3,5,6,82,3,5,6,92,3,5,6,102,3,5,7,82,3,5,7,92,3,5,7,102,3,5,8,92,3,5,8,102,3,5,9,102,3,6,7,82,3,6,7,92,3,6,7,102,3,6,8,92,3,6,8,102,3,6,9,102,3,7,8,92,3,7,8,102,3,7,9,102,3,8,9,102,4,5,6,72,4,5,6,82,4,5,6,92,4,5,6,102,4,5,7,82,4,5,7,92,4,5,7,102,4,5,8,92,4,5,8,102,4,5,9,102,4,6,7,82,4,6,7,92,4,6,7,102,4,6,8,92,4,6,8,102,4,6,9,102,4,7,8,92,4,7,8,102,4,7,9,102,4,8,9,102,5,6,7,82,5,6,7,92,5,6,7,102,5,6,8,92,5,6,8,102,5,6,9,102,5,7,8,92,5,7,8,102,5,7,9,102,5,8,9,102,6,7,8,92,6,7,8,102,6,7,9,102,6,8,9,102,7,8,9,103,4,5,6,73,4,5,6,83,4,5,6,93,4,5,6,103,4,5,7,83,4,5,7,93,4,5,7,103,4,5,8,93,4,5,8,103,4,5,9,103,4,6,7,83,4,6,7,93,4,6,7,103,4,6,8,93,4,6,8,103,4,6,9,103,4,7,8,93,4,7,8,103,4,7,9,103,4,8,9,103,5,6,7,83,5,6,7,93,5,6,7,103,5,6,8,93,5,6,8,103,5,6,9,103,5,7,8,93,5,7,8,103,5,7,9,103,5,8,9,103,6,7,8,93,6,7,8,103,6,7,9,103,6,8,9,103,7,8,9,104,5,6,7,84,5,6,7,94,5,6,7,104,5,6,8,94,5,6,8,104,5,6,9,104,5,7,8,94,5,7,8,104,5,7,9,104,5,8,9,104,6,7,8,94,6,7,8,104,6,7,9,104,6,8,9,104,7,8,9,105,6,7,8,95,6,7,8,105,6,7,9,105,6,8,9,105,7,8,9,106,7,8,9,10`
winsome johnny (not Win some johnny)
7craps
Joined: Jan 23, 2010
• Posts: 1977
December 3rd, 2018 at 6:02:41 PM permalink
Quote: webayarea

That would be leading to my next questions;

Odds of a 4/4 straight and 4/5 straight and then, biggest Longshot of 5/5

Therefore payouts on a single unit bet would be:

3/3 = 15:1

looks to me you may be mixing probability and odds. they are not the same.
https://www.mathsisfun.com/definitions/odds.html
https://www.mathsisfun.com/definitions/probability.html

for 3/3
8/120 = probability of success = 1 in 15
so 14 no against 1 yes
14 to 1 against
14:1 would be a fair payoff

your 15:1 payoff would have an edge
unless that is what you are after

why all the straights?
winsome johnny (not Win some johnny)
webayarea
Joined: Nov 30, 2018
• Posts: 6
December 4th, 2018 at 11:17:01 AM permalink
Thanks All. Lots to digest and I appreciate the input. Please keep it coming!

For the 15:1 payout point, I agree and see the house edge being here and possibly setting at 15 FOR 1 to be player fair.

Im interested in the 'straights stats' as I've been thinking of a table game that involves straight combinations between 3-5 among a random 5 card draw from dealer. (Main thing is dealer only has A-10; no faces.)

I would have these lined up as likely single unit side bets and Longshot bets as a secondary part to a more primary focused game between dealer and player around the 5 cards in another capacity.
unJon
Joined: Jul 1, 2018
• Posts: 4343
December 4th, 2018 at 11:23:06 AM permalink
Quote: webayarea

Thanks All. Lots to digest and I appreciate the input. Please keep it coming!

For the 15:1 payout point, I agree and see the house edge being here and possibly setting at 15 FOR 1 to be player fair.

Im interested in the 'straights stats' as I've been thinking of a table game that involves straight combinations between 3-5 among a random 5 card draw from dealer. (Main thing is dealer only has A-10; no faces.)

I would have these lined up as likely single unit side bets and Longshot bets as a secondary part to a more primary focused game between dealer and player around the 5 cards in another capacity.

The calculations above assume the dealer only has 10 cards. If the dealer has multiple suits/decks on A-10 then the calculations above are not correct, and you need to specify the total number of A-10 decks.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
webayarea
Joined: Nov 30, 2018
• Posts: 6
December 4th, 2018 at 11:39:52 AM permalink
Correct. Dealer only has 10 cards or one A-10 deck. Thanks
7craps
Joined: Jan 23, 2010
• Posts: 1977
December 4th, 2018 at 11:54:12 AM permalink
Quote: webayarea

Correct. Dealer only has 10 cards or one A-10 deck. Thanks

have you tried to actually play this as a game and how to shuffle just 10 cards and have it a random shuffle?

a computer could do it well.
I tried shuffle by hand and it seemed not a random shuffle (very awkward)
then I tried using 20 cards (2 each ranks) and that seemed easier to shuffle

I say 10 cards works well for a computer game as will 20 cards
but the game has to be fun to play (some anticipation built into the playing of the game or player interaction or both)
"I would have these lined up as likely single unit side bets and Longshot bets as a secondary part to a more primary focused game between dealer and player around the 5 cards in another capacity."

20 cards might be a better table game offering
the math would still be doable
20 cards
2 each 1-10
20C3 = 1140 combinations (210 Unique)
20C4 = 4845 combinations (615 Unique)
20C5 = 15504 combinations (1452 Unique)
this would also offer pairs

the coefficients shows the number of unique combinations when there are more than 1 to choose from
example: 1452*x^5 means that there are 1452 unique combinations when drawing 5 cards
` gp > (1+x+x^2)^10%1 = x^20 + 10*x^19 + 55*x^18 + 210*x^17 + 615*x^16 + 1452*x^15 + 2850*x^14 + 4740*x^13 + 6765*x^12 + 8350*x^11 + 8953*x^10 + 8350*x^9 + 6765*x^8 + 4740*x^7 + 2850*x^6 + 1452*x^5 + 615*x^4 + 210*x^3 + 55*x^2 + 10*x + 1`

a long journey still begins with one small step
good luck to you
Last edited by: 7craps on Dec 4, 2018
winsome johnny (not Win some johnny)
gordonm888
Joined: Feb 18, 2015
• Posts: 4802
December 4th, 2018 at 2:58:57 PM permalink
Quote: 7craps

one can make a list of all the combinations
and see which ones are the events you are after
https://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html

You need to be careful, I think, about applying that link to the ranks in a deck of cards, because with cards the Aces can be either high or low when forming a straight.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
7craps
Joined: Jan 23, 2010
• Posts: 1977
December 5th, 2018 at 9:16:32 AM permalink
Quote: gordonm888

You need to be careful, I think,

certainly. even when counting
Quote: gordonm888

about applying that link to the ranks in a deck of cards, because with cards the Aces can be either high or low when forming a straight.

The OP has not mentioned that an A could also be high and 1,9,10 would also be counted as a 3 card straight. (that is an interesting one)

the list of combinations still would not change as one would still add the events as they would be mutually exclusive (can not happen at the same time)
`1,2,31,2,41,2,51,2,61,2,71,2,81,2,91,2,101,3,41,3,51,3,61,3,71,3,81,3,91,3,101,4,51,4,61,4,71,4,81,4,91,4,101,5,61,5,71,5,81,5,91,5,101,6,71,6,81,6,91,6,101,7,81,7,91,7,101,8,91,8,101,9,10`
1,2,3 would be counted as a straight
and A as high
1,9,10
only straights I see in that section

that could easily be an option for the side bets
using A as low and high
winsome johnny (not Win some johnny)
7craps
Joined: Jan 23, 2010
• Posts: 1977
December 5th, 2018 at 11:51:15 AM permalink
Quote: ThatDonGuy

P(4) = 49 / 210 = 7 / 30, or odds of 23-7 against.

I can not agree on this
unless I (and excel) counted wrong (or do not understand the 3 out of 4)
you lost me on the math part (not hard to do)

I get 42/210 (1/5) for a 3 card straight in 4 cards drawn (not counting the 7 4 card straights)
that is 1 in 5
so for a fair payoff: 4 to 1

the list
`List has 210 entries.1,2,3,4 <<<<1,2,3,5 <<<1,2,3,6 <<<1,2,3,7 <<<1,2,3,8 <<<1,2,3,9 <<<1,2,3,10 <<<1,2,4,51,2,4,61,2,4,71,2,4,81,2,4,91,2,4,101,2,5,61,2,5,71,2,5,81,2,5,91,2,5,101,2,6,71,2,6,81,2,6,91,2,6,101,2,7,81,2,7,91,2,7,101,2,8,91,2,8,101,2,9,101,3,4,5 <<<1,3,4,61,3,4,71,3,4,81,3,4,91,3,4,101,3,5,61,3,5,71,3,5,81,3,5,91,3,5,101,3,6,71,3,6,81,3,6,91,3,6,101,3,7,81,3,7,91,3,7,101,3,8,91,3,8,101,3,9,101,4,5,6 <<<1,4,5,71,4,5,81,4,5,91,4,5,101,4,6,71,4,6,81,4,6,91,4,6,101,4,7,81,4,7,91,4,7,101,4,8,91,4,8,101,4,9,101,5,6,7 <<<1,5,6,81,5,6,91,5,6,101,5,7,81,5,7,91,5,7,101,5,8,91,5,8,101,5,9,101,6,7,8 <<<1,6,7,91,6,7,101,6,8,91,6,8,101,6,9,101,7,8,9 <<<1,7,8,101,7,9,101,8,9,10 <<<2,3,4,5 <<<<2,3,4,6 <<<2,3,4,7 <<<2,3,4,8 <<<2,3,4,9 <<<2,3,4,10 <<<2,3,5,62,3,5,72,3,5,82,3,5,92,3,5,102,3,6,72,3,6,82,3,6,92,3,6,102,3,7,82,3,7,92,3,7,102,3,8,92,3,8,102,3,9,102,4,5,6 <<<2,4,5,72,4,5,82,4,5,92,4,5,102,4,6,72,4,6,82,4,6,92,4,6,102,4,7,82,4,7,92,4,7,102,4,8,92,4,8,102,4,9,102,5,6,7 <<<2,5,6,82,5,6,92,5,6,102,5,7,82,5,7,92,5,7,102,5,8,92,5,8,102,5,9,102,6,7,8 <<<2,6,7,92,6,7,102,6,8,92,6,8,102,6,9,102,7,8,9 <<<2,7,8,102,7,9,102,8,9,10 <<<3,4,5,6 <<<<3,4,5,7 <<<3,4,5,8 <<<3,4,5,9 <<<3,4,5,10 <<<3,4,6,73,4,6,83,4,6,93,4,6,103,4,7,83,4,7,93,4,7,103,4,8,93,4,8,103,4,9,103,5,6,7 <<<3,5,6,83,5,6,93,5,6,103,5,7,83,5,7,93,5,7,103,5,8,93,5,8,103,5,9,103,6,7,8 <<<3,6,7,93,6,7,103,6,8,93,6,8,103,6,9,103,7,8,9 <<<3,7,8,103,7,9,103,8,9,10 <<<4,5,6,7 <<<<4,5,6,8 <<<4,5,6,9 <<<4,5,6,10 <<<4,5,7,84,5,7,94,5,7,104,5,8,94,5,8,104,5,9,104,6,7,8 <<<4,6,7,94,6,7,104,6,8,94,6,8,104,6,9,104,7,8,9 <<<4,7,8,104,7,9,104,8,9,10 <<<5,6,7,8 <<<<5,6,7,9 <<<5,6,7,10 <<<5,6,8,95,6,8,105,6,9,105,7,8,9 <<<5,7,8,105,7,9,105,8,9,10 <<<6,7,8,9 <<<<6,7,8,10 <<<6,7,9,106,8,9,10 <<<7,8,9,10 <<<<`
Last edited by: 7craps on Dec 5, 2018
winsome johnny (not Win some johnny)
7craps
Joined: Jan 23, 2010
• Posts: 1977
December 5th, 2018 at 12:44:35 PM permalink
Quote: webayarea

That would be leading to my next questions;

Odds of a 4/4 straight and 4/5 straight and then, biggest Longshot of 5/5

Therefore payouts on a single unit bet would be:

3/3 = 15:1
3/4 = 3.75:1
3/5= TBD (guessing 3/2? Maybe 5/4?)
LONGSHOTS
4/4= TBD
4/5 = TBD
5/5= TBD

here is what I got. the math is not hard, but lots of room for errors
I hope my code was done correctly
(I did not double check the 5 card draw as I lost interest quickly in the math)

3/3 = (8/120 = 1/15 / fair: 14 to 1)
3/4 = (42/210 = 1/5 / fair: 4 to 1)
3/5 = (89/252 / fair: 252/89 to 1)
LONGSHOTS
4/4 = (7/210 = 1/30 / fair: 29 to 1)
4/5 = (30/252 = 5/42 / fair: 42/5 to 1)
5/5 = (6/252 = 1/42 / fair: 41 to 1)

the list
`List has 252 entries.1,2,3,4,5 <<<<<1,2,3,4,6 <<<<1,2,3,4,7 <<<<1,2,3,4,8 <<<<1,2,3,4,9 <<<<1,2,3,4,10 <<<<1,2,3,5,6 <<<1,2,3,5,7 <<<1,2,3,5,8 <<<1,2,3,5,9 <<<1,2,3,5,10 <<<1,2,3,6,7 <<<1,2,3,6,8 <<<1,2,3,6,9 <<<1,2,3,6,10 <<<1,2,3,7,8 <<<1,2,3,7,9 <<<1,2,3,7,10 <<<1,2,3,8,9 <<<1,2,3,8,10 <<<1,2,3,9,10 <<<1,2,4,5,6 <<<1,2,4,5,71,2,4,5,81,2,4,5,91,2,4,5,101,2,4,6,71,2,4,6,81,2,4,6,91,2,4,6,101,2,4,7,81,2,4,7,91,2,4,7,101,2,4,8,91,2,4,8,101,2,4,9,101,2,5,6,7 <<<1,2,5,6,81,2,5,6,91,2,5,6,101,2,5,7,81,2,5,7,91,2,5,7,101,2,5,8,91,2,5,8,101,2,5,9,101,2,6,7,8 <<<1,2,6,7,91,2,6,7,101,2,6,8,91,2,6,8,101,2,6,9,101,2,7,8,9 <<<1,2,7,8,101,2,7,9,101,2,8,9,10 <<<1,3,4,5,6 <<<<1,3,4,5,7 <<<1,3,4,5,8 <<<1,3,4,5,9 <<<1,3,4,5,10 <<<1,3,4,6,71,3,4,6,81,3,4,6,91,3,4,6,101,3,4,7,81,3,4,7,91,3,4,7,101,3,4,8,91,3,4,8,101,3,4,9,101,3,5,6,7 <<<1,3,5,6,81,3,5,6,91,3,5,6,101,3,5,7,81,3,5,7,91,3,5,7,101,3,5,8,91,3,5,8,101,3,5,9,101,3,6,7,8 <<<1,3,6,7,91,3,6,7,101,3,6,8,91,3,6,8,101,3,6,9,101,3,7,8,9 <<<1,3,7,8,101,3,7,9,101,3,8,9,10 <<<1,4,5,6,7 <<<<1,4,5,6,8 <<<1,4,5,6,9 <<<1,4,5,6,10 <<<1,4,5,7,81,4,5,7,91,4,5,7,101,4,5,8,91,4,5,8,101,4,5,9,101,4,6,7,8 <<<1,4,6,7,91,4,6,7,101,4,6,8,91,4,6,8,101,4,6,9,101,4,7,8,9 <<<1,4,7,8,101,4,7,9,101,4,8,9,10 <<<1,5,6,7,8 <<<<1,5,6,7,9 <<<1,5,6,7,10 <<<1,5,6,8,91,5,6,8,101,5,6,9,101,5,7,8,9 <<<1,5,7,8,101,5,7,9,101,5,8,9,10 <<<1,6,7,8,9 <<<<1,6,7,8,10 <<<1,6,7,9,101,6,8,9,10 <<<1,7,8,9,10 <<<<2,3,4,5,6 <<<<<2,3,4,5,7 <<<<2,3,4,5,8 <<<<2,3,4,5,9 <<<<2,3,4,5,10 <<<<2,3,4,6,7 <<<2,3,4,6,8 <<<2,3,4,6,9 <<<2,3,4,6,10 <<<2,3,4,7,8 <<<2,3,4,7,9 <<<2,3,4,7,10 <<<2,3,4,8,9 <<<2,3,4,8,10 <<<2,3,4,9,10 <<<2,3,5,6,7 <<<2,3,5,6,82,3,5,6,92,3,5,6,102,3,5,7,82,3,5,7,92,3,5,7,102,3,5,8,92,3,5,8,102,3,5,9,102,3,6,7,8 <<<2,3,6,7,92,3,6,7,102,3,6,8,92,3,6,8,102,3,6,9,102,3,7,8,9 <<<2,3,7,8,102,3,7,9,102,3,8,9,10 <<<2,4,5,6,7 <<<<2,4,5,6,8 <<<2,4,5,6,9 <<<2,4,5,6,10 <<<2,4,5,7,82,4,5,7,92,4,5,7,102,4,5,8,92,4,5,8,102,4,5,9,102,4,6,7,8 <<<2,4,6,7,92,4,6,7,102,4,6,8,92,4,6,8,102,4,6,9,102,4,7,8,9 <<<2,4,7,8,102,4,7,9,102,4,8,9,10 <<<2,5,6,7,8 <<<<2,5,6,7,9 <<<2,5,6,7,10 <<<2,5,6,8,92,5,6,8,102,5,6,9,102,5,7,8,92,5,7,8,102,5,7,9,102,5,8,9,10 <<<2,6,7,8,9 <<<<2,6,7,8,10 <<<2,6,7,9,102,6,8,9,10 <<<2,7,8,9,10 <<<<3,4,5,6,7 <<<<<3,4,5,6,8 <<<<3,4,5,6,9 <<<<3,4,5,6,10 <<<<3,4,5,7,8 <<<3,4,5,7,9 <<<3,4,5,7,10 <<<3,4,5,8,9 <<<3,4,5,8,10 <<<3,4,5,9,10 <<<3,4,6,7,8 <<<3,4,6,7,93,4,6,7,103,4,6,8,93,4,6,8,103,4,6,9,103,4,7,8,9 <<<3,4,7,8,103,4,7,9,103,4,8,9,10 <<<3,5,6,7,8 <<<<3,5,6,7,9 <<<3,5,6,7,10 <<<3,5,6,8,93,5,6,8,103,5,6,9,103,5,7,8,9 <<<3,5,7,8,103,5,7,9,103,5,8,9,10 <<<3,6,7,8,9 <<<<3,6,7,8,10 <<<3,6,7,9,103,6,8,9,10 <<<3,7,8,9,10 <<<<4,5,6,7,8 <<<<<4,5,6,7,9 <<<<4,5,6,7,10 <<<<4,5,6,8,9 <<<4,5,6,8,10 <<<4,5,6,9,10 <<<4,5,7,8,9 <<<4,5,7,8,104,5,7,9,104,5,8,9,10 <<<4,6,7,8,9 <<<<4,6,7,8,10 <<<4,6,7,9,104,6,8,9,10 <<<4,7,8,9,10 <<<<5,6,7,8,9 <<<<<5,6,7,8,10 <<<<5,6,7,9,10 <<<5,6,8,9,10 <<<5,7,8,9,10 <<<<6,7,8,9,10 <<<<<`

hope this gives you more homework
and helps some
winsome johnny (not Win some johnny)
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 5944
December 5th, 2018 at 2:24:41 PM permalink
Quote: 7craps

I can not agree on this
unless I (and excel) counted wrong (or do not understand the 3 out of 4)
you lost me on the math part (not hard to do)

I get 42/210 (1/5) for a 3 card straight in 4 cards drawn (not counting the 7 4 card straights)

We may be interpreting the problem two different ways. I did count the 4-card straights as "containing a 3-card straight."
webayarea
Joined: Nov 30, 2018
• Posts: 6
June 21st, 2019 at 1:56:17 PM permalink
Reopening this thread to the helpful and brilliant minds of the Forum...

My concept game has now established a Straight Bet of 3 cards - out of 5 cards selected from a 10 card deck (A-10)

3/5 = (89/252 / fair: 252/89 to 1)

For a payout POV there are two ways:

3 to 1 payout (Player Edge) and 5 to 2 (House Edge)

The SECOND portion of the game is based on the SUM TOTAL of the 5 selected cards being OVER or UNDER 27.5

A single Unit bet on Under or Over will pay even money (1:1) as the combination of 5 random selected cards from A-10 will be between 15 and 40

For example:
A + 2 + 3 + 4 + 5 = 15 (UNDER)
5 + 9 + A + 2 + 4 = 21 (UNDER)
9 + 8 + 3 + 6 + 2 = 28 (OVER)
6 + 7 + 8 + 9 + 10 = 40 (OVER)

QUESTION FOR GROUP:

Need Breakdown of likelihood that if a single, random card is chosen (A - 10) what are odds that if an additional random 4 cards are added to it, that the TOTAL will be over or under 27.5?

For Example:
5 Cards from (A - 10) are randomly selected
I bet one unit on the SUM Total of those 5 cards being OVER 27.5
I've been dealt a 10. The next 4 cards (A - 9) will need to be at least 18 for a my wager to WIN.
10 + 4 + 8 + 9 + 2 = 33 (WIN)
10 + 4 + 3 + 2 + A = 20 (LOSE)

Inversely, If I bet UNDER 27.5 as the total and am dealt an Ace to start, the % of the TOTAL being under 27.5 would equal the chances that a 10 leads to OVER 27.5

(2 = 9 ; 3 = 8 ; 4 = 7 etc)