permutations/combinations question

[magickingdom]

if anyone one can settle an argument....

20 cards in a deck no 1-20 whats the chance of pulling a 7 if you have 12 attempts

[AFS]

12/20 = 0.6 = 60%

I think, provided the cards are replaced each time.

[Death Valley]

Are you replacing cards after each go? ie if you pull a 2 on go one do you put it back in the deck

[magickingdom]

yes you replace cards so there is a 5% chance of pulling a 7 everytime.
afs if your right and one had 20 goes instead of 12 it would be 20/20 = 100%. that would not be right

[Death Valley]

Tis a good question, easy at first sight. I believe it envolves a long series of combinations & permutations ie on go one the ch ances you dont pull out a 7 are 19/20 * go 2 you dont pull out a 7 19/20 * go 3 lets say you do pull out a 7 1/20 ie 361/8000. Then add up all these very combinations all 12. I cant be arsed meself. Come to think of it there is probably an easier way, been out of the game a long time now!.

1         2                  3                                4                                   up to 12
1/20 +19/20*1/20+ 19/20*19/20*1/20 + 19/20*19/20?*19/20*1/20+.........

[The Watcher Pat]

12/20 = 0.6 = 60%

I think, provided the cards are replaced each time.

That's too simplistic

You have a 1/20  chance the first pick the first time, 1/19 the second pick and so on..

I think if you add them all up 1/20 + 1/19 + 1/18 etc you will end up with about 88/100 therefor a 88% chance...

Thats if my thinking is right..

Would be interested to know if thats correct or not.

[Rossie11]

Pat I reckon the method you have is correct if you are not replacing the drawn card.

[Rossie11]

19/20 *19/20 twelve times works out at 54% approx.
Thats 54% of a chance that you wont pull a 7 in 12 goes.
46% that you will.

20 picks would give you 35% it wont happen thus still only a 65% chance it will.
Would that be right??

[dec]

1 - (19/20)^12 = 0.4596

ie the 46% that Rossie mentioned

[harlechman]

Tis a good question, easy at first sight. I believe it envolves a long series of combinations & permutations ie on go one the ch ances you dont pull out a 7 are 19/20 * go 2 you dont pull out a 7 19/20 * go 3 lets say you do pull out a 7 1/20 ie 361/8000. Then add up all these very combinations all 12. I cant be arsed meself. Come to think of it there is probably an easier way, been out of the game a long time now!.

1         2                  3                                4                                   up to 12
1/20 +19/20*1/20+ 19/20*19/20*1/20 + 19/20*19/20?*19/20*1/20+.........

This is correct. Works out at 46%.

[Niall Quinn]

if anyone one can settle an argument....

20 cards in a deck no 1-20 whats the chance of pulling a 7 if you have 12 attempts

You won't pull a 7 in 12 trys - normally comes out ~ 15th iteration.
8, 4, and 6 are shoe ins.

[magickingdom]

19/20 *19/20 twelve times works out at 54% approx.
Thats 54% of a chance that you wont pull a 7 in 12 goes.
46% that you will.

20 picks would give you 35% it wont happen thus still only a 65% chance it will.
Would that be right??

46% is my answer too...

[Gnevin]

if anyone one can settle an argument....

20 cards in a deck no 1-20 whats the chance of pulling a 7 if you have 12 attempts

You've a 1 in 20 chance every time or 5%
If you add the card back and you do it 12 times  you've a 60% chance,
You do it 20 time you've a 100% chance.

[Bogball XV]

if anyone one can settle an argument....

20 cards in a deck no 1-20 whats the chance of pulling a 7 if you have 12 attempts

You've a 1 in 20 chance every time or 5%
If you add the card back and you do it 12 times  you've a 60% chance,
You do it 20 time you've a 100% chance.

correct, probability is 1, that does not mean it will happen as each draw is a new draw and the chances are still 1 in 20 for each draw.

[thebigfella]

1-(1 - 1/20)^12 which I believe is 46% chance of it occurring if you only draw 12 cards.

As for 20,

1-(1 - 1/20)^20 = 64% chance it occurring.