Spoiler for I looked at this puzzle last night...

and I came up with the same probabilities as curr3nt.

I thought perhaps combining the two darts would still have a larger variance and that would change things. I worked up formulas for the binomial distribution, combined them, and ended up with another binomial distribution of the new probability .98% (which really should have been obvious from the Bernoulli trial).

There surely was something I was missing.

I thought perhaps combining the two darts would still have a larger variance and that would change things. I worked up formulas for the binomial distribution, combined them, and ended up with another binomial distribution of the new probability .98% (which really should have been obvious from the Bernoulli trial).

There surely was something I was missing.

Since then I've decided that....

Spoiler for Alex is a MACHINE!

I noticed the word

Let B(k;n,p) be the probability of getting k successes in a Binomial distribution with n trials and a probability p of success.

Assuming he always gets exactly 97 out of 100, he will still lose since the probability the duo get 98, 99, or 100 is (B(100;100,.98) + B(99;100,.98) + B(98;100,.98)) which is slightly over 67%.

But will Alex ever get more than 97 out of 100?

Assuming Alex's distribution is uniform (so he gets 97, 98, 99, and 100 darts out of 100 with equal probability)...

The probability the duo will win is:

.25 * ( B(100;100,.98) + B(99;100,.98) + B(98;100,.98) ) +

.25 * ( B(100;100,.98) + B(99;100,.98) ) +

.25 * ( B(100;100,.98) ) +

.25 * 0

= about .303144

The probability they tie is :

.25* ( B(100;100,.98) + B(99;100,.98) + B(98;100,.98) + B(97;100,.98) )

= about .214740

This leaves the probability Alex win at .482116.

But that only works if Alex's distribution is uniform between 97 and 100 when throwing 100 darts.

But what if they only throw 33 darts at a time? Alex must always get all 33 or he will not have reproduced the result of a percentage greater than or equal to 97%. In this case he will never lose.

So that's my guess. They always throw darts in groups of 33. Alex always gets 97% or more, so he cannot miss. Alex gets some more beer money.

**reproducibly**, and that changes things since if he never gets less than 97% we are no longer dealing with the binomial distribution for him.Let B(k;n,p) be the probability of getting k successes in a Binomial distribution with n trials and a probability p of success.

Assuming he always gets exactly 97 out of 100, he will still lose since the probability the duo get 98, 99, or 100 is (B(100;100,.98) + B(99;100,.98) + B(98;100,.98)) which is slightly over 67%.

But will Alex ever get more than 97 out of 100?

Assuming Alex's distribution is uniform (so he gets 97, 98, 99, and 100 darts out of 100 with equal probability)...

The probability the duo will win is:

.25 * ( B(100;100,.98) + B(99;100,.98) + B(98;100,.98) ) +

.25 * ( B(100;100,.98) + B(99;100,.98) ) +

.25 * ( B(100;100,.98) ) +

.25 * 0

= about .303144

The probability they tie is :

.25* ( B(100;100,.98) + B(99;100,.98) + B(98;100,.98) + B(97;100,.98) )

= about .214740

This leaves the probability Alex win at .482116.

But that only works if Alex's distribution is uniform between 97 and 100 when throwing 100 darts.

But what if they only throw 33 darts at a time? Alex must always get all 33 or he will not have reproduced the result of a percentage greater than or equal to 97%. In this case he will never lose.

So that's my guess. They always throw darts in groups of 33. Alex always gets 97% or more, so he cannot miss. Alex gets some more beer money.

Though perhaps I'm reading too much into it...