how do you calculate the chances of this:

I believe 52 over 13 then 51 over 13 then 50 over 13 and so till 30 times.
I may be wrong.

65/13 x 64/13 x 63/13 (and so on for a total of 30times) -1 to 1

Yes.  It's a huge 30 leg acca.  The first leg being 3/1, then descending fractionally in price down to a short odds on final leg.  I worked it out to 240*10 to the power 43 but don't think this can be right. can it?

Anyway my mate did it.  If I worked it out right it may be the most unlikely statistical achievements in the history of the world. Or I may have just worked it out wrong.

It's not 65/13.  52/13 then downwards.

just over 700,000 / 1 that non of your 13 cards are matched in 30 turns

?

its

39/52 x 38/51 ..........     .....10/23

so

0.00000078332 chance....so over a 1.2 million to 1 shot


( if ive pressed the right buttons this time)

bigpop agrees with me

Feb 8, 2014 -- 12:30AM, donny osmond wrote:

its 39/52 x 38/51 ..........     .....10/23so0.00000078332 chance....so over a 1.2 million to 1 shot( if ive pressed the right buttons this time)

Agree.

the combinations of 30 from 39/combinations of 30 from 52 is another way to do it (obv much easier in excel)

Try my suggestion I believe im right (bigpop agrees) and work it out and please post the odds im crap on excel.

for the first draw there are 52 possible outcomes , 39 of which give no match

that becomes the odds for first draw

39/52

second draw there are 51 possible outcomes, 38 of which give no match

38/51

repeat 30 times

i disagree donny

52 over 13 which is 3/1      4.0 decimal
then 51 over 13              3.923 decimal
then 50 over 13              3.846 decimal
and so till 30 times 

multipy all decimal results eg    4.0 x 3.923 x 3.846 and so on

Why 13? There's 39 cards in the master deck that DON'T match the initial hand.

u have 13 cards and there are 52 cards in the master deck hence 52/over 13

you can simplify the question to using just the 4 aces, and you select 1 ace at random, then calculate odds for
3 draws without hitting your ace

odds are 3/4 x  2/3 x 1/2  = 1/4

your method gives

4/1 x 3/1 x2/1

ok lets do the aces.
lets say i had an ace and four aces where in front of me.
to pick my ace on go one would be 3/1
if i chose wrong and an ace was removed leaving 3 aces for pick two the odds would be 2/1
if i chose wrong again leaving two aces for pick three the odds would be evens.

If you never discarded an ace always leaving four aces the odds on each individual pick would be 3/1

Feb 9, 2014 -- 12:38AM, iprefertolay wrote:

u have 13 cards and there are 52 cards in the master deck hence 52/over 13

Except you want them to be different to your hand, not the same.

P(30 consecutive different cards)=P(1st card is different) x P (2nd card is different) x ... x P(30th card is different)

P(1st card is different) = 39 different cards / 52 available cards
P(nth card is different) = 39-(n-1) different cards / 52-(n-1) available cards

Therefore

P(30 consecutive different cards)=(39/52)*(38/51)*...*(10/23)

the combination method gives .000000783 - therefore donny is correct imo

Correct - donny is correct.

Who agrees with me?

Let me explain this in a different way.
Assume the 13 cards are all hearts.
How many different suits are in the masterdeck? answer 3
Therefore its 3/1 on the first card then reducing slightly on subsequent picks.
I maintain the answer is 52/13 then 51/13 and o on. Till 30 picks.

getting back to the aces example and you want to find the odds of 3 picks without the ace of hearts - your method will, as pointed out, be 1/4 x 1/3 x 1/1.  But you are missing the times that you DO pick the ace of hearts. 

If you have, for example, two independent goes at a 2-1 chance the odds are 5/4.  You have a 33.3% chance of hitting it first time and another 33.3% of the remaining 67.7%.

Accordingly the second term of your calc should I think be 51/13 x (1-13/52) and the third term be 50/13 x 1 - whatever the prob of the first two picks being the relevant card.

Dave the original question was what are the odds if you DONT match please address this not if you do.then you will see im right.

your system doesnt work for just the aces therefore it does not stand up at all

but continue to use it if you wish

but donny do u agree im correct on the original question plz dont blur the issue

ok your trolling bye bye then

no your system is wrong

if the maths works then it always works

Trolling or stupid. I give up.

If the OP wants a correct answer, which has been given at least four times, he'll post again.

donny u stated my system doesnt work for just the aces what about the original question?

no .....your system is wrong

thanks for the answers guys.  Pretty sure it's the 30 leg acca which starts at 3/1 and odds decrease slightly as cards are removed from the master deck.  I was getting the maths wrong, but I'll take the 1.2million/1 as an answer.

As I said - it happened to my mate.  It's a pub-quiz tie breaker bingo game (aim being to match all your 13).  My mate found it remarkable that he had no matches one night (seemed unusual) and the bloke running the quiz had never had it happen.  And now we know why; it's a million to one...

the odds of a match at stage 1 is 3/1  or 13/52




but then the odds of a match at stage 2 is 13/51, assuming you had no match at stage 1

but to get to this point you have a double at 39/52 x 13/51

donny thats what i said but u said i was wrong!!

Feb 10, 2014 -- 7:09PM, iprefertolay wrote:

donny thats what i said but u said i was wrong!!

What?

its not what you said !



the odds of a match at stage 2 is  ( note we dont bother with stage 2 if stage 1 gives a match)

39/52 x 13/51 = 0.191




and we know that a match at stage one is 0.25


so odds of no match after stage 2 is

1-(0.25+0.191)= 0.559



our other method gives no match after 2 events as

39/52 x 38/51 =

0.559

I agree with Donny Osmond ~(39!22!)/(52!9!) =7.8 [-07]