# Poker Odds [Solved!]

**dmf** 05 Feb 2020, 14:12

### My question

Hi

I have found odds for 5 card poker hands made from 2 player cards, five community cards (7 total cards), and two wild cards added for a 54 card deck.

My question is how to calculate the effect of adding one blank card to the deck that does not help make a hand, so that when this card appears in a player hand or in the community cards, it forces the participants to make their hands using the remaining six cards

Thank you in advance for your help.

DMF

### Relevant page

7 Card Poker Probabilities

### What I've done so far

I have not yet tried to solve this.

X

Hi
I have found odds for 5 card poker hands made from 2 player cards, five community cards (7 total cards), and two wild cards added for a 54 card deck.
My question is how to calculate the effect of adding one blank card to the deck that does not help make a hand, so that when this card appears in a player hand or in the community cards, it forces the participants to make their hands using the remaining six cards
Thank you in advance for your help.
DMF

Relevant page
<a href="http://www.durangobill.com/Poker_Probabilities_7_Cards.html">7 Card Poker Probabilities</a>
What I've done so far
I have not yet tried to solve this.

## Re: Poker Odds

**Murray** 07 Feb 2020, 17:14

@dmf: I suspect calculating the final probabilities would be quite involved, as the page you linked to indicates.

But as a general rule, we would need to make use of "choose `r` objects from `n` objects" as explained in the page Combinations.

Your pack now has 55 cards and you would be choosing 7 cards for your hand. You would have the following number of possible poker hands:

`C7^55 = (55!)/((7!)(48!)) = 202,927,725`

The probability of drawing a particular card in a hand of `m` cards with a deck of size `n` is `m/n` (Reference)

So the probability of choosing that blank card would be `7/55`

Now, maybe Durangobill would be able to tie all this into the already complex probability calculations for 7-card hands!

X

@dmf: I suspect calculating the final probabilities would be quite involved, as the page you linked to indicates.
But as a general rule, we would need to make use of "choose `r` objects from `n` objects" as explained in the page <a href="https://www.intmath.com/counting-probability/4-combinations.php">Combinations</a>.
Your pack now has 55 cards and you would be choosing 7 cards for your hand. You would have the following number of possible poker hands:
`C7^55 = (55!)/((7!)(48!)) = 202,927,725`
The probability of drawing a particular card in a hand of `m` cards with a deck of size `n` is `m/n` (<a href="https://math.stackexchange.com/questions/146187/probability-of-picking-a-specific-card-from-a-deck">Reference</a>)
So the probability of choosing that blank card would be `7/55`
Now, maybe Durangobill would be able to tie all this into the already complex probability calculations for 7-card hands!

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