# Ask The Wizard #364

You travel one week back in time in an attempt to win the lottery. It’s worth \$10 million, and each ticket costs a dollar. Note that if you win, your ticket purchase is not refunded. All of this sounds pretty great.

The problem is, you’re not alone. There are 10 other time travelers who also know the winning numbers. You know for a fact that each of them will buy exactly one lottery ticket. Now, according to the lottery’s rules, the prize is evenly split among all the winning tickets (i.e., not evenly among winning people).

How many tickets should you buy to maximize your profits?

Gialmere

Remember the quotient rule:

[f(x)/g(x)] d/dx = (f'(x)*g(x) - f(x)*g'(x))/(g(x))^2

9,990

Let n = number of tickets you purchase.

Let w(n) = Net win if you buy n tickets = 10,000,000 × (n/(n+10)) - n.

To find the point where one extra ticket does not change the net win, we take the derivative of w(n). Here use the quotient rule (see hint), where the numerator is n and the denominator is (n+10).

w'(n) = 10,000,000 * (1*(n+10) - n*1)/(n+10)^2 - 1 = 0

10,000,000 * 10/(n+10)^2 - 1 = 0

10,000,000 * 10/(n+10)^2 = 1

100,000,000 = (n+10)^2

10,000 = n+10

n = 9,990

The question is asked and discussed in my forum at Wizard of Vegas.

I see somebody on your forum is claiming to have witnessed 60 consecutive winning hands in a row in blackjack. How many hands would one need to play, on average, to witness that? Also, how does it compare with the event mentioned in Ask the Wizard column 363 of somebody allegedly witnessing 18 totals of 11 in a row in craps?

anonymous

To answer your question, we must first assume some rules for the blackjack game. I will use what I call the "liberal strip rules," which are as follows:

• Six decks
• Dealer stands on soft 17
• Double after split allowed
• Re-splitting aces allowed
• Surrender allowed

Under these rules, assuming perfect basic strategy, here are the probabilities of a net win/loss:

• Net win: 42.43%
• Net push: 8.48%
• Net loss: 49.09%

I questioned the witness to this event and he said the 60 hands does not include pushes. In other words, 60 resolved hands. I am also going to assume that if the player splits it still counts as one hand only and there was a net win among all the hands the player split to.

As explained in the question on the 18 consecutive totals of 11, the formula for the expected waiting time is:

Expected waiting time = [(1/p)^(n+1) - 1] / [(1/p) - 1] - 1, where:
n = number of consecutive wins
p = probability of win

In this case, the probability of a net win per hand, given a hand resolved is 42.43%/ (43.43% + 49.09%) = 46.36%. To be more specific, 0.46359564.

Using the formula above, the expected waiting time to observe this event is (1/0.46359564)^61 - 1]/[(1/0.46359564) - 1] - 1 = 200,941,772,393,648,000,000 hands.

By comparison, the waiting time to observe 18 totals of 11 in a row in craps is 41,660,902,667,961,000,000,000 rolls, which is 207 times more. So, the 18 yo's still holds the record as the tallest tale told at the Wizard of Vegas forum, as far as I know.

So, I had \$300 in Comps and I invited some Friends to use my Comps up at a Restaurant. We all had a good time until the bill came to about \$386 with tax and tip. My Friends purposely didn't bring any money and just assumed I would pay the tax and tip for everyone. I was upset and told them I thought everyone should split the tax and tip since Comps do not cover tax and tip.

They argued they felt I should pay all of the tax and tip since I invited them and these were my Comps. I argued common courtesy would be to pay a fair share of the tax and tip. My Friends were disgusted and thought I was being unreasonable and refused to chip in for tax and tip. I paid the roughly \$86 in tax and tip as it was clear none of my Friends would pay the tax and tip. And the Waiter was getting anxious. Instead of thanking me for using my Comps with them, and paying the tax and tip myself, they are all upset with me for even asking them to chip in for tax and tip. Was I in the wrong here?

Riverjordan

I haven't had an etiquette question in a while, so thank you.

You make your friends sound pretty cheap for not chipping in towards the tax and tip. By the way, the patron is not charged tax on a comped meal in Nevada, so this must have taken place in another state. However, you should have made clear your expectations at the time of the invitation. So, I see fault both ways.

This at least makes for a teachable moment. I have been a part of comped meals in Vegas at least 100 times, sometimes when I provide the comp and sometimes when someone else does. The etiquette I follow, which is usually agreed to, is the one with the comp explains any restrictions on the maximum comp value at the time of the invitation. Then, everyone who was invited pays any portion beyond the maximum comp and the tip.

This question was asked and discussed in my forum at Wizard of Vegas.