I think the probability is 12/47 or .255319. (25.5319%) but I'm not a math expert.

Quote:CharmedQuarkYou are trying to get one card (As) in one draw out of 12 cards with 47 cards in total. You are holding 5 cards in hand whose values are known and not part of the original 52 so they should not count in the calculation. The makeup of the tableau is not taken into consideration..

I think the probability is 12/47 or .255319. (25.5319%) but I'm not a math expert.

I think it depends on how you phrase the question. To me, a tableau, either face-up or -down, is dealt at the same time as the player hands. So, if you want a specific card on the tableau, you're not caring/knowing whether it's in your hand or not. So 12/52.

If, OTOH, you're looking at your cards, and the tableau is face down, or dealt after player hands, and you want a specific card that's not in your hand to show up, then 12/47 would be correct.

I’m posting on this thread because my question seems related.

In my NBA mobile phone game (that I spend way too much time on), I can earn picks of “prizes” on a board of 22 “cards.” One of the picks is a “top” prize, and resets the board. So I know the odds of getting the top prize on a fresh board is 1 in 22, and those occurrences seem about right.

But what are the odds that the top prize is chosen on the 22nd pick of the board? This seems to happen much more frequently than I think it should, so I feel like the picks aren’t truly random.

Quote:smoothgrhHi!

I’m posting on this thread because my question seems related.

In my NBA mobile phone game (that I spend way too much time on), I can earn picks of “prizes” on a board of 22 “cards.” One of the picks is a “top” prize, and resets the board. So I know the odds of getting the top prize on a fresh board is 1 in 22, and those occurrences seem about right.

But what are the odds that the top prize is chosen on the 22nd pick of the board? This seems to happen much more frequently than I think it should, so I feel like the picks aren’t truly random.

Combinatorics is a neat way to do this. How many combinations of 22 cards are there and how likely is it to pick 21 of them without picking the other one?

https://web2.0calc.com/

nCr(21,21)*nCr(1,0)/nCr(22,21) = 0.0454545454545455

Same thing. 1 in 22 starting from a fresh board to pick all of them before picking that one.

Or, you could:

(21/22)*(20/21)*(19/20)*(18/19)*(17/18)*(16/17)*(15/16)*(14/15)*(13/14)*(12/13)*(11/12)*(10/11)*(9/10)*(8/9)*(7/8)*(6/7)*(5/6)*(4/5)*(3/4)*(2/3)*(1/2) = 0.0454545454545455

Of course, this is assuming that the cards stay gone until you pick the one that resets the board, which is how I took what you said.

Thanks much for doing the math and describing it!

(And yes, the picked cards stay gone.)

So 4.5%? That’s a lot higher than I expected! So maybe it’s random after all.

Quote:smoothgrhHi Mission,

Thanks much for doing the math and describing it!

(And yes, the picked cards stay gone.)

So 4.5%? That’s a lot higher than I expected! So maybe it’s random after all.

The easy way to think about it is the chance of picking it first is the same as the chance of picking it last is 1/22.

Or run it this way: Pick one card and decide that’s the card you will turn over last. Obviously it’s 1/22 that the card you picked first (to check last) is 1/22.

Quote:unJonThe easy way to think about it is the chance of picking it first is the same as the chance of picking it last is 1/22.

GAH! How could I be so daft! Of course it’s the same odds of picking it first as picking it last. I knew the formula would be (21/22)*(20/21)*(19/20)and-so-on…but for some reason I thought the occurrence would be far less than 1/22.

It’s the same odds as if you were trying to avoid that card!

Sorry for causing eye rolls.

Quote:smoothgrhGAH! How could I be so daft! Of course it’s the same odds of picking it first as picking it last. I knew the formula would be (21/22)*(20/21)*(19/20)and-so-on…but for some reason I thought the occurrence would be far less than 1/22.

It’s the same odds as if you were trying to avoid that card!

Sorry for causing eye rolls.

Don't feel bad! You should see some of the stuff that I miss when trying to answer posts before I go back and realize what mistake I made! When doing that Sette y Mezzo, for example, I forgot to keep running probabilities for one of the results and was banging my head off of my desk (not literally) trying to figure out why it wasn't coming out to a greater probability. As it turns out, I forgot to add back in LITERALLY the most likely thing that could occur in that particular situation.

Oh, and then the Craps post I recently answered...I couldn't figure out why the House Edge he was getting kept coming out the same in the poster's equations. But then, I realized since he was doing it based on one unit bet he just kept getting the EXPECTED $$$ LOSS, not the house edge...it's just with one unit bet (without odds) the Expected $$$ Loss and House Edge were the same result.

My numbers were coming out 100% in a way that made sense. That's why I usually like to use the actual bet amounts of the games, like $5 for a Pass Line bet at Craps, then I know what is the Expected $$$ Loss and what is the house edge without confusing the two.