Give it a toss

Probability Level 3

Suppose Mark and Geoff each have a fair coin, (one side being heads, the other tails). Mark tosses his coin 2017 2017 times and keeps a tally of the number of tosses that come up heads, while Geoff does the same except that he tosses his coin 2018 2018 times.

If P P is the probability that Geoff tosses more heads than Mark, then find ⌊ 1000 P ⌋ \lfloor 1000P \rfloor .


The answer is 500.

Brian Charlesworth by Brian Charlesworth , 55, Canada

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Let A A be the probability that both Mark and Geoff have tossed the same number of heads after 2017 2017 tosses. Then by symmetry, at this point the probability that Geoff has tossed more heads than Mark has is 1 − A 2 \dfrac{1 - A}{2} , and this status will not change after Geoff makes his final toss.

If they are tied after 2017 2017 tosses, then there is a 1 / 2 1/2 probability that Geoff will get a head on his final toss and end up with more heads. Thus the probability that Geoff ends up with more heads is

P = 1 − A 2 + 1 2 A = 1 2 P = \dfrac{1 - A}{2} + \dfrac{1}{2}A = \dfrac{1}{2} , and so ⌊ 1000 P ⌋ = 500 \lfloor 1000P \rfloor = \boxed{500} .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Haha... Great problem!

At first I thought it was going to involve some binomial coefficients of 2017 and 2018....

I suppose if we had both tossed 2018 coins it would have...

The only problem is, my finger now hurts from all that tossing! ;0)

Geoff Pilling - 4 years ago

Log in to reply

Haha Yeah, you and Mark were real troupers for doing all that tossing. :)

With the coin count differing by 1 we don't have to calculate A A , but in any other scenario we do. If both of you had n n coins then the probability of you tossing more heads would be 1 − A 2 \dfrac{1 - A}{2} where

A = ∑ k = 0 n ( ( n k ) 2 n ) 2 = 1 4 n × ( 2 n n ) A = \displaystyle\sum_{k=0}^{n} \left(\dfrac{\dbinom{n}{k}}{2^{n}}\right)^{2} = \dfrac{1}{4^{n}} \times \dbinom{2n}{n} ,

which for n = 2018 n = 2018 is a bit of a trick to calculate, (although I think there is a way to estimate it). I'll post this scenario sometime with each of you having, say, 20 coins, so as not to hurt your fingers too much. :)

Brian Charlesworth - 4 years ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...