I Shopping List Game: Probability Question

[tomwilliam2]

I have a probability question which has cropped up while playing a game called "Shopping List" with my sons. The game is played like this: you pick a fixed shopping list of 10 items, and three other players do the same. You have all of the items on cards, face down on the floor in-between the players. They take it in turns to pick up a card. If it's on their shopping list, they keep it. If it's not, it gets shuffled back in among the others.
Now, the chances of a player winning are just 1/4, with random picks.
However, if only three people play, and one of them takes two shopping lists, do they have a greater or lesser chance of winning? Instinctively I would say if completing one shopping list is enough, they have a 1/2 chance of winning, as they effectively represent two players. But if they have to complete both lists to win...is it harder or easier? They get a larger pool of cards that are good for them, but they have double the items to find. I can't think how to work out the exact probability, but my son asked me and I'd like to work it out!

 

[jbriggs444]

The player with two carts and two lists picks twice per rotation? Does he pick once for one cart/list and a second time for the other cart/list? Is he allowed to pick for one cart and rather than discarding an unwanted item, place it in the other cart if the item matches the other shopping list?

 

[PeroK]

I have a probability question which has cropped up while playing a game called "Shopping List" with my sons. The game is played like this: you pick a fixed shopping list of 10 items, and three other players do the same. You have all of the items on cards, face down on the floor in-between the players. They take it in turns to pick up a card. If it's on their shopping list, they keep it. If it's not, it gets shuffled back in among the others.
Now, the chances of a player winning are just 1/4, with random picks.
However, if only three people play, and one of them takes two shopping lists, do they have a greater or lesser chance of winning? Instinctively I would say if completing one shopping list is enough, they have a 1/2 chance of winning, as they effectively represent two players. But if they have to complete both lists to win...is it harder or easier? They get a larger pool of cards that are good for them, but they have double the items to find. I can't think how to work out the exact probability, but my son asked me and I'd like to work it out!

If you have to find all the items on two lists to win, assuming you get two picks to everyone else's one, then you effectively have to finish first and second to win. Your chance of finishing first is $1/2$ and your chance of then finishing second is $1/3$. So, your chance of winning is $1/6$.

 

[PeroK]

The player with two carts and two lists picks twice per rotation? Does he pick once for one cart/list and a second time for the other cart/list? Is he allowed to pick for one cart and rather than discarding an unwanted item, place it in the other cart if the item matches the other shopping list?

Good questions. I assumed the former: playing two independent hands.

 

[mfb]

Some points of the game are unclear. In total there are 40 items, the shopping lists are all without overlap, and the cards drawn are from a different stack, containing these 40 items once each? The first player to complete the shopping list wins?
That sounds like a boring game as there is absolutely no decision you can make.

Assuming you only draw once even with two lists, or you draw separately for both lists, you have a 50% chance that one of the lists wins.
If you need all 20, that is equivalent to two players getting first and second. There are 6 ways to choose 2 players out pf 4, and only one let's the player win -> 1/6 chance.

If you can draw twice and use both cards for both lists, however, you have a huge advantage.

 

[jbriggs444]

Some points of the game are unclear. In total there are 40 items, the shopping lists are all without overlap, and the cards drawn are from a different stack, containing these 40 items once each? The first player to complete the shopping list wins?
That sounds like a boring game as there is absolutely no decision you can make.

A trip to Google shows that the game of this name by Orchard Toys involves 32 items and shopping lists with 8 items each, non-overlapping. The game mechanics and skills involved seem to be very much like "Bingo".

http://www.jenniferslittleworld.com/2011/12/shopping-list-game-from-orchard-toys.html

 

[PeroK]

We used to play something similar with a pack of cards, which started with all 52 cards face down. Each player would turn over two cards and if they had the same face value they kept the cards and had another go. Otherwise the two cards were turned face down again.

We called it "Pelmanism".

 

[tomwilliam2]

Thanks for the responses: Yes, indeed there are 32 cards not 40...I was trying to simplify for myself, but I guess it shouldn't be any different.
Actually, we've been playing with him only getting one turn (one pick) per round, and having to complete both shopping lists to win, so I'm guessing his chances of winning will be less than 1/6?
On some of the questions: yes, there's no overlap between the lists (I'm not sure if that's the case for Bingo) and the cards are all unique and in the same stack. We all pick from that one pile.
Oh, and yes, it is a boring game. But it's my youngest son who loves it (aged 4), and who insists on having the extra list for himself!

 

[mfb]

It doesn’t matter (to an extremely good approximation*) if you make two separate draws for the lists or one draw for both. In both cases each list has 1/4 chance to get aother card.

*the correlation between the lists will be different, but that effect is negligible. As an example, with two separate draws you can win after 10 rounds, with one draw you cannot - but you are very unlikely to win that fast anyway.

If you want a fair game, your son could pick one list as his list amd the other list is a separate player who cannot win. But making the game fair is probably not the highest priority here... so let him win with either list.