Random Statistical Question

**BramSToker** · 01-26-2015, 08:54 PM

Greetings forum posters and lurkers,

I just watched the movie "This is the end" and it got me thinking about a statistical nuance that may be shared in blackjack that somebody here would be able to explain. If you haven't seen the movie, I'm referring to when they draw the matches to decide who has to go and find water/food. They burn one match, flip it upside down, and let each person pick a match and whoever has the burned one loses, basically drawing straws.

My question is, is there an advantage to going last in this situation? To me, the person going first is the only one who has a 100% chance of there still being the burnt one, whereas each person after only has to go if the person before him chooses a regular match. If there are 6 matches, for example, the first person has a 1/6 chance of losing, which would make it so the other people don't even have to go. Each time somebody chooses they're facing a chance of getting it that the last person doesn't have to take on, so it makes sense that the last person has a good advantage.

Part 2 of the question is, would this relate in any way to a blackjack table and positioning on the bases? For example, if you are side counting aces or just in a high count, the first base player/second base player ect until it gets to you has a chance to get that card before you do, and thus it hurts you slightly. I am aware this has been examined before but I'm not entirely sure I understand why it is fallacious. Let's say there are 3 players acting ahead of you and the count is +6 so you get a large bet out, those players could (possibly likely to bc of the count) easily take one, two or even three of your tens before you even get a chance, and the true count could take a dip by the time you even get to your hand. Since these are dependent events, it seems like you'd have to consider the chances of each player getting that ten/ace before you, sort of like the person drawing the first match. Clearly the impact would be amplified by less decks and deeper into the shoe.

Thanks for any thoughts,
Bram

**stopgambling** · 01-26-2015, 09:46 PM

This has been explained before. You info is before any cards were dealt out . How would you know the order of the cards ? you just know there are 6 high card per deck in the remaining shoe. You do not know the order. Someone will come to explain soon , maybe Don or one of the other guys that know their stuff.

**Nikky_Flash** · 01-26-2015, 11:53 PM

there's other players at the table ?

**MidniteToker** · 01-27-2015, 05:19 AM

Before the first person picks, you know there is 1 out 6 burnt matches, therefore 16.67% chance to pick it.

When the second person picks, there are 5 matches left. 1 time out of 6 (16.67%), all 5 matches are unburnt, so no chance to pick it. 5 times out 6 (83.33%), we have 1/5 chance to pick the burnt match.
83.33% * 1/5 + 16.67% * 0 = 16.67% chance to get a burnt match.

When the third person picks, there are 4 matches left. 1 time out of 6 (16.67%), 5 matches were unburnt when the 2nd was picked. 5 times out of 6, there was 1 burnt and 4 unburnt, so 1/5 chance we now have 4 unburnt matches (1/5 * 5/6 = 16.67%) and 4/5 we have 1 burnt and 3 unburnt (4/5 * 5/6 = 66.67%). So: 16.67% + 16.67% = 33.3% chance there is no burnt match in the lot of 4, and 66.67% chance there is with 1/4 chance to pick it:
66.67% * 1/4 + 33.3% * 0 = 16.67% chance to pick a burnt match.

So:
1st person to pick: 16.67% chance
2nd person to pick: 16.67% chance
3rd person to pick: 16.67% chance

You can see where this is going. We have 1/6th chance of getting a burnt match regardless of when we pick. We cannot evaluate our odds otherwise without more information changing what we know of the situation.

**Three** · 01-27-2015, 05:47 AM

1) They all have the same chance.
First person 1/6 chance of drawing burnt.
Second person 1/6 chance times 0/5 chance of drawing a burnt plus 5/6 1/5 chance of drawing burnt
= (1/6)*0 + (5/6)*1/5 = 1/6

2) It is the same as the match sticks. On the one side they might draw high cards but they also might draw low cards. One helps your chances the other hurts your chances. They balance out over all. Every spot makes their bet with the same information. Their would only be an advantage if you made your bet with more information. The guy at 3rd base does have a playing advantage for decisions but it is slight. The count has less of a chance to change between when he makes his decision and when the dealer acts. The difference is the action is made with extra info. Everyone bets with the same info. 3rd base may have extra info when he plays his hand that affects the dealers chances when the dealer plays his hand and factors that into his decision while the other spots made decisions without the same info. First base might make a playing decision based on a TC of +2 and have the dealer play his hand at TC +4 (if the other spots take cards and draw a lot of low cards) but 3rd base will have just his play alter the TC which is already factored into the decision. It is about what info you have when you make your decision. All bet with the same info.

Go to the scratch off lottery. 1 jackpot winner is among all the printed tickets. Is there an advantage to waiting to get a ticket? If you can choose not to buy a ticket there is a big advantage to waiting. The chances of getting the jackpot go up every time someone scratches off a ticket that doesn't win. If the jackpot is won you just don't bet. If you have already bought a ticket but just haven't selected it yet there is no advantage in waiting or getting it early. As the odds that your ticket will be a winner goes up IF the jackpot hasn't been won, the odds that you will have a 0% chance of getting the jackpot ticket because it was already scratched off also goes up. In the one lottery case you can make your bet of purchase or not purchase based on extra info but if you already paid for your ticket you must play regardless of the extra info you have before you get your ticket. The difference is whether you could change your bet based on the extra information. A subscriber has already paid for his tickets but a walk up player can choose to change his bet to 0 if the ticket has been claimed. Every spot at the table makes their bets with the same information. It doesn't change your odds by playing one spot over another. Sometimes little cards come out sometimes big cards but in the long run it averages out to not have an affect.

Remember the dealer has the same chance as you do to get good cards when you make your bet just like every other spot. The dealer gets his cards last. Your advantage isn't that you have a better chance of getting good cards than the dealer when the count is high. It mostly comes from getting 3:2 when you get a BJ while the dealer just wins or pushes your bet, more frequent and better results from doubles and splits from index plays while betting more in high counts and being able to play your hand different as the count gets high.

**DSchles** · 01-27-2015, 12:29 PM

You received a lot of good answers, but some are a bit long. Here's a shorter one:

You are six players at the table, and the dealer is about to deal the hands from a TC = +6 situation. You're worried about being the last to receive your cards. You're afraid the dealer may "run out" of good ones by the time he gets to you. So, we deal differently: we throw all 14 cards (six players plus the dealer) face down on the table, and we ask everyone to dig in and find two for his hand. So, no early-hand, late-hand bias now, right?

Now, ask yourself how getting your cards this way, as opposed to standard dealing, could remotely change anything whatsoever insofar as the probabilities are concerned. :-)

Don

**BramSToker** · 01-27-2015, 01:08 PM

Ahh, very interesting. I think I was trying to parlay the idea of a high count implying 10/A's coming sooner rather than later with the idea that as the amount of cards increases it will tend to approach a more neutral count. But, I guess since we can't know what the cards are going to do or what order they are in, it is just a form of superstition. Still, if it's me, I'm not going first on the match draw, because I'd sooner give somebody else the chance to get unlucky first.

Thank you all for the responses, Tthree and MidniteToker for the excellent detailed analyses, as well as stopgambling and Don for the concise, practical answers. Also Nikky thanks for the bump.

Bram

**Three** · 01-27-2015, 02:19 PM

Your welcome.

RC tends to approaches 0 as the deck is depleted but TC tends to stay the same. As in the match stick example depletion of a portion of the whole is balanced by depletion of the whole. In other words if you draw 13 cards they expected to represent the TC which moves the RC toward 0 but has no affect on the TC. The True Count Theorem says the RC tends to approaches 0 while the TC tends to stay the same. Basically removing card(s) is expected to not change the TC on average. At the same time we know that if the TC is not expected to change then removing cards which makes the number of cards less in the denominator would be expected to reduce the magnitude of the RC to keep te expected TC the same.