Why is coin tossing combination without repetition?

Hello all,

As per the title, for a traditional statistics question like:

Find the probability of landing 7 heads in 10 throws of biased coin, P(head) = 0.2

No problems with this being a combination (as opposed to a permutation), but I cannot convince myself why it's the WITHOUT repetition sub category of combinations. Thus allowing one to use the Binomial Distribution via nCr. As opposed to using the n+r-1Cr distribution for WITH repetition.

What exactly does "without repetition" physically mean in this real life example?

Reply 1

Astronomical

14

Original post by Choochoo_baloo

Hello all,

As per the title, for a traditional statistics question like:

Find the probability of landing 7 heads in 10 throws of biased coin, P(head) = 0.2

No problems with this being a combination (as opposed to a permutation), but I cannot convince myself why it's the WITHOUT repetition sub category of combinations. Thus allowing one to use the Binomial Distribution via nCr. As opposed to using the n+r-1Cr distribution for WITH repetition.

What exactly does "without repetition" physically mean in this real life example?

Well you're trying to calculate P(7 heads out of 10 throws) = P(7 heads and 3 tails) = (number of distinct ways of getting 7 heads and 3 tails)*P(7 heads)*P(3 tails)

i.e.

P = \frac{10!}{7!3!}(0.2)^7(0.8)^3

which is of course the binomial formula. That is, without repetition (as far as I can tell) just means you want distinct ways of achieving the correct number of heads and tails.

Reply 2

rayquaza17

17

Original post by Astronomical

That is, without repetition (as far as I can tell) just means you want distinct ways of achieving the correct number of heads and tails.

This is why.

Without repetition means you class HTH the same as HHT (i.e. two heads and a tail).

Reply 3

Choochoo_baloo

OP

6

Thanks both.

I'm afraid I remain unconvinced.

rayquaza17 isn't that the point that HHT is NOT the same was HTH i.e. heads on throws 1 & 2 vs heads on throws 1 & 3. You would therefore say that 3C2 = 1 (which is obviously wrong)? Ie in my questions, 10C7 = 120 includes (for example)

HHHHHHHTTT & HHHHHHTHTT being discrete outcomes.

Hopefully I conveyed my query alright. So because we are factoring in all 120 ways of arranging 7H & 3T in the calculation, the question remains: what does the without repetition refer to?

Thanks in advance.

Reply 4

rayquaza17

17

Original post by Choochoo_baloo

Thanks both.

I'm afraid I remain unconvinced.

rayquaza17 isn't that the point that HHT is NOT the same was HTH i.e. heads on throws 1 & 2 vs heads on throws 1 & 3. You would therefore say that 3C2 = 1 (which is obviously wrong)? Ie in my questions, 10C7 = 120 includes (for example)

HHHHHHHTTT & HHHHHHTHTT being discrete outcomes.

Hopefully I conveyed my query alright. So because we are factoring in all 120 ways of arranging 7H & 3T in the calculation, the question remains: what does the without repetition refer to?

Thanks in advance.

I think I may have explained it incorrectly.

This link may help you understand: http://math.stackexchange.com/questions/158755/how-to-know-if-its-permutation-or-combination

Reply 5

franciscolopez

I think the intuition is that you don't have one coin and toss it 10 times but 10 coins previously tossed. You work with the group of 10 items, so every time you grab one of them to add it to the combination you take it out of its original group. You can't take twice the same coin. Therefore no repetition. Does this make sense?