Deal a deck of cards (maths question)

[Tuska]

Dear brain of PH.

I'm doing some research for a speech i have to make and am trying to illustrate a point.
Can anyone answer the following:-

If you shuffle and deal a standard deck of cards into 4 hands of 13. What are the odds on one of the hands being dealt as all one suit?

Its over to you......

[hornetrider]

13/52 x 12/51 x 11/50 etc etc.

...practically nil

[speedychrissie]

hornetrider said:

13/52 x 12/51 x 11/50 etc etc.

...practically nil

if that theory is correct (and i think it is), then the chance is 1:158 Billion (roughly)

that number is the chance of one of the collections being all the same suit. If you want to specify (ie the chance that all the diamonds will group together) then it is roughly 1:635 Billion


my calculations may be wrong but i hope that helps.

[Tuska]

speedychrissie said:

hornetrider said:

13/52 x 12/51 x 11/50 etc etc.

...practically nil

if that theory is correct (and i think it is), then the chance is 1:158 Billion (roughly)

that number is the chance of one of the collections being all the same suit. If you want to specify (ie the chance that all the diamonds will group together) then it is roughly 1:635 Billion


my calculations may be wrong but i hope that helps.

Perfect, thanks! 635 billion was the number i was given elsewhere.
It seemed massive to me, but then i guess it is.......

[Russian Rocket]

another way of looking at it is the number of different ways the hands can be delt if factorial 52 = 52! = 52*51*50....
= 12.E65 to 1 against

[JP_Midget]

Tuska said:

speedychrissie said:

hornetrider said:

13/52 x 12/51 x 11/50 etc etc.

...practically nil

if that theory is correct (and i think it is), then the chance is 1:158 Billion (roughly)

that number is the chance of one of the collections being all the same suit. If you want to specify (ie the chance that all the diamonds will group together) then it is roughly 1:635 Billion


my calculations may be wrong but i hope that helps.

Perfect, thanks! 635 billion was the number i was given elsewhere.
It seemed massive to me, but then i guess it is.......

But surely the first card doesn't matter what it is, so the calculation of all cards in a hand being any complete suit would be...

1 x 12/51 x 11/50 etc. etc.?

If you wanted to specify the actual suit then you would revert to

13/52 x 12/51 x 11/50 etc etc.?

[hornetrider]

Actually thinking about it, the odds I gave were dealing out one suit directly.

What you've said is deal out all cards into four piles then what is the likelihood one of those piles is one suit, correct? i.e., the three other piles can be mixed. That is far more complicated and I can't even fathom it.

Edited by hornetrider on Monday 23 February 17:55

[speedychrissie]

JP_Midget said:

Tuska said:

speedychrissie said:

hornetrider said:

13/52 x 12/51 x 11/50 etc etc.

...practically nil

if that theory is correct (and i think it is), then the chance is 1:158 Billion (roughly)

that number is the chance of one of the collections being all the same suit. If you want to specify (ie the chance that all the diamonds will group together) then it is roughly 1:635 Billion


my calculations may be wrong but i hope that helps.

Perfect, thanks! 635 billion was the number i was given elsewhere.
It seemed massive to me, but then i guess it is.......

But surely the first card doesn't matter what it is, so the calculation of all cards in a hand being any complete suit would be...

1 x 12/51 x 11/50 etc. etc.?

If you wanted to specify the actual suit then you would revert to

13/52 x 12/51 x 11/50 etc etc.?

you are right, and that is why I have given two different odds in my post. one for any suit and one for a named suit.

[pokethepope]

You could do a tree diagram. Have you got a piece of paper the size of Brazil?

[cazzer]

Whatever.
Don't put money on it

[The Moose]

cazzer said:

Whatever.
Don't put money on it

Yeah, but imagine putting a quid on and getting a squazillion back!!

[cazzer]

The Moose said:

cazzer said:

Whatever.
Don't put money on it

Yeah, but imagine putting a quid on and getting a squazillion back!!

It could be youuuuuuu.
Or not.

[JP_Midget]

hornetrider said:

Actually thinking about it, the odds I gave were dealing out one suit directly.

What you've said is deal out all cards into four piles then what is the likelihood one of those piles is one suit, correct? i.e., the three other piles can be mixed. That is far more complicated and I can't even fathom it.

Edited by hornetrider on Monday 23 February 17:55

Oh, so possibly

1 x 39/51 x 38/50 x 37/49 x 12/48 x 36/47 x 35/46 x 34/45 ...

[The Moose]

ok, four "sets" delt - S1, S2, S3 and S4.

The chances of S1 being all one suit:

13/52 * 12/51 * 11/50 * 10/49 * 9/48 * 8/47 * 7/46 & 6/45 * 5/44 * 4/43 * 3/42 * 2/41 * 1/40 = 1.57477 x 10^-12 = 0.00000000000157477 = 1 / 635,014,000,000.

The chances of S2 being all one suit:

13/52 * 12/51 * 11/50 * 10/49 * 9/48 * 8/47 * 7/46 & 6/45 * 5/44 * 4/43 * 3/42 * 2/41 * 1/40 = 1.57477 x 10^-12 = 0.00000000000157477 = 1 / 635,014,000,000.

The chances of S3 being all one suit:

13/52 * 12/51 * 11/50 * 10/49 * 9/48 * 8/47 * 7/46 & 6/45 * 5/44 * 4/43 * 3/42 * 2/41 * 1/40 = 1.57477 x 10^-12 = 0.00000000000157477 = 1 / 635,014,000,000.

The chances of S4 being all one suit:

13/52 * 12/51 * 11/50 * 10/49 * 9/48 * 8/47 * 7/46 & 6/45 * 5/44 * 4/43 * 3/42 * 2/41 * 1/40 = 1.57477 x 10^-12 = 0.00000000000157477 = 1 / 635,014,000,000.

ok the chances of S1 or S2 or S3 or S4 being all one suit:
0.00000000000157477 + 0.00000000000157477 + 0.00000000000157477 + 0.00000000000157477 = 0.00000000000629908 = 1 \ 158,753,000,000.

So 1 in 158 billion, 753 million times.

So a quid bet would return you roughly 158 and 3/4 billion quid!!

[RacingPete]

Surely it depends on how many times you shuffle the deck too... as IIRC after about 4 shuffles the pack starts to sort itself out a bit more thus affecting the randomness of the deal!

But that just starts to get stupid... so ^^^ What he said ^^^

[richyb]

MX5 ?

[leeb]

you could cheat the odds, and just use a brand new deck each time, then the chances are fairly high if you are clever with the dealing!

[mmm-five]

Surely the chances of one of the piles being of the same suit would not be going down in increments of one, as each pile is only added to every 4th deal.

Wouldn't it then be 12/51 x 11/47 x 10/43 x 9/39 x 8/35 x etc.